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Chemistry MCQs
1. The process in which a solid directly changes to vapours without melting is called __________.
(Evaporation, Condensation, Sublimation)
2. The oxidation number of P in PO3-4 is __________.
(3+, 5+, 3-)
3. The pH of 0.001 M HCl is __________.
(2, 4, 3)
4. K ( rate constant) is dependent on __________.
(temperature, concentration, volume)
5. The universal indicator in water shows the colour __________.
(red, green, blue)
6. The pH of blood is __________.
(7.3, 8.4, 5.6)
7. The oxidation potential of hydrogen electrode is __________.
(0.0 volt, +0.76volt, -0.36volt)
8. __________ quantum number describes the shape of a molecule.
(Pricipal, Azimuthal, Spin)
9. An orbital can have the maximum number of two electrons but with opposite spin, it is called __________.
(Pauli’s Exclusion Principle, Hund’s Rule, Aufbau Principle)
10. When a-particle is emitted from the nucleus of radioactive element, the mass number of the atom __________.
(Increases, Decreases, Does not change)
11. Dissociation of KclO3 is a __________ process.
(Reversible, Irreversible)
12. The e/m ratio of cathode rays is the __________ when Hydrogen is taken in the discharge tube.
(Lowest, Highest)
13. The negative ion tends to expand with the __________ of negative change on it.
(Decreases, Increases)
14. Ionic compounds have __________ melting points.
(Low, High)
15. The allotropic forms of an element are called __________.
(Polymorphs, Isomorphs)
16. Absolute Zero is equal to __________.
(273.16°C, -273.16°C)
17. The compounds having hydrogen bond generally have __________ boiling points.
(High, Low)
18. Surface tension __________ with the rise of temperature.
(Increases, Decreases)
19. Mercury forms __________ meniscus in a glass tube.
(concave, convex)
20. The reactions with the high value of energy of activation are __________.
(slow, fast)
21. 2.000 has/have __________ significant figure(s).
(1, 4)
22. E + PV is called __________.
(Entropy, Enthalpy)
23. The shorter the bond length in a molecule, the __________ will be bond energy.
(Lesser, Greater)
24. Positive rays are produced from __________.
(anode, Cathode, Ionization of gas in a discharge tube)
25. __________ of the following contains the fewer number of molecules.
(1 gm of hydrogen, 4 gm of oxygen, 2 gm of nitrogen)
26. the true statement about the average speed of the molecules of hydrogen, oxygen and nitrogen confined in a container is __________.
(Hydrogen is quicker, Oxygen is quicker, The molecules of all the gases have the same average speed)
27. The correct statement about the glass is __________.
(It is crystalline solid, Its atoms are arranged in an orderly fashion, It is a super cooled liquid)
28. When a substance that has absorbed energy emits it in the form of radiation the spectrum obtained is __________.
(Continuous Spectrum, Line Spectrum, Emission Spectrum)
29. __________ of the overlap forms strong bond.
(S-S, P-S, P-P)
30. __________ compound has a greater angle between a covalent bond.
(H2O, NH3, CO2)
31. When sodium chloride is mixed in water then __________.
(pH is changed, NaOH and HCl are formed, Sodium and chloride ions become hydrated)
32. The boiling point of a liquid __________ with an increase in pressure.
(Decreases, Increases, remains constant)
33. An Azimuthal Quantum Number describes the __________.
(size of an atom, shape of an orbital, spin of orbital)
34. The rate of the backward reaction is directly proportional to the product of the molar concentration of __________.
(Reactants, Products, None of them)
Chapter 1
Introduction To Fundamental Concepts
1. The formula, which gives the simple ratio of each kind of atoms present in the molecule of compound, is called __________.
(Molecular Formula, Empirical Formula, Structural Formula)
2. The formula, which expresses the actual number of each kind of atom present in the molecule of a compound, is called __________.
(Empirical Formula, Molecular Formula, Structural Formula)
3. Mole is a quantity, which has __________ particles of the substance.
(One billion, 6.02 x 1023, 1.013 x 105)
4. The simplest formula of a compound that contain 81.8% carbon and 18.2% hydrogen is __________.
(CH3, CH, C2H6)
5. The empirical Formula of a compound __________.
(is always the same as the molecular formula, Indicates the exact composition, Indicates the simplest ratio of the atoms)
6. Very small and very large quantities are expressed in terms of __________.
(significant figures, Exponential Notation, Logarithm)
7. Two moles of water contains __________ molecules.
(6.02 x 1023, 1.204 x 1024, 3.01 x 1023)
8. One mole of Cl- ions contains __________ ions.
(6.02 x 1023, 1.204 x 1024, 3.01 x 1023)
9. 220 gms of CO2 contains __________ moles of CO2.
(One, Five, Ten)
10. In rounding off __________ figure is dropped.
(First, Last, No)
11. Precision is linked with __________.
(Individual measurements, Actual results, Accepted Value)
12. Accuracy refers to how closely a measured value agrees with __________.
(Individual result, Actual result, Average value)
13. 6600 contains __________ significant figures.
(2, 3, 4)
14. 3.7 x 104 contains __________ significant figures.
(2, 3, 5)
15. 9.40 x 10-19 contains __________ significant figures.
(2, 3, 5)
16. The figure 39.45 will be rounded off to __________.
(39.4, 39.5, 39)
17. __________ means that the result obtained in different experiments are very close to the accepted values.
(Accuracy, Precision, Significant Figure)
18. The average weight of atoms of an element as compared to the weight of one atom of carbon taken as __________ is called the atomic weight.
(12, 13, 14)
19. 58.5 is __________ of NaCl.
(Atomic weight, Formula Weight, Molecular Weight)
20. 18.0 a.m.u is the __________ weight of water.
(Atomic, Formula, Molecular)
21. 28 gms of nitrogen will have __________ molecules.
(6.02 x 1023, 12.04 x 1023, 3.01 x 1023)
22. 22.4 dm3 of CO2 is __________ 22.4 dm3 of SO2.
(Heavier than, Lighter than, Equal to)
23. 100 gms of water is equal to __________ moles.
(5.56, 27.78, 6.25)
24. The reactions, which proceed in both the directions are called __________ reactions.
(Reversible, Irreversible, Neutrilization)
25. The reactions, which proceed in forward direction only are called __________ reactions.
(Reversible, Irreversible, Ionic)
26. Molecular weight is used for __________ substances.
(Ionic, Non ionic, Neutral)
27. Formula weight is used for __________ substances.
(Ionic, Non ionic, Neutral)
28. The modern system of measurement is called __________ system.
(SI, Metric, F.P.S)
29. The S.I unit of mass is __________.
(kilogram, gram, pound)
30. One mole of glucose contains __________ gms.
(100, 180, 342)
Chapter 2
The Three States of Matter
1. __________ was the first scientist who expressed a relation between pressure and the volume of a gas.
(Charles, Boyle, Avogadro)
2. If the pressure upon a gas confined in a vessel varies, the temperature remaining same, the volume will __________.
(Vary directly as the pressure, Vary inversely as the temperature, Vary inversely as the pressure)
3. The statement concerning the relation of temperature to the volume of a gas under fixed pressure was first synthesized by __________.
(Boyle, Charles, Avogadro)
4. Absolute Zero is __________.
(273°C, -273°C, -273°K)
5. Gases intermix to form __________.
(Homoge\= ous mixture, Heterogenous mixture, compound)
6. Water can exists in __________ physical states at a certain condition of temperature pressure.
(One, Two, three)
7. The temperature at which the volume of a gas theoretically becomes zero is called __________.
(Transition temperature, Critical Temperature, Absolute Zero)
8. Gases deviate from ideal behaviour at __________ pressure and __________ temperature.
(Low, High, Normal)
9. Very low temperature can by produced by the __________ of gases.
(Expansionn, Contraction, Compression)
10. Boiling point of a liquid __________ with increase in pressure.
(increases, decreases, remains same)
11. 273°K = __________
(100°C, 273°C, 0°C)
12. -273°C is equal to __________.
(0°K, 273°K, 100°K)
13. Evaporation takes place at __________.
(All temperatures, At constant temperature, at 100°C)
14. __________ is the temperature at which the vapour pressure of a liquid becomes equal to atmospheric pressure.
15. The freezing point of water in Fahrenheit scale is __________.
(0°F, 32°F, 212°F)
16. All gases change to solid before reaching to __________.
(-100°C, 0°C, -273°C)
17. Pressure of the gas is due __________ of the molecules on the wall of the vessel.
(Collisionns, Attraction, Repulsion)
18. Boiling point of water in absolute scale is __________.
(212°K, 100°K, 373°K)
19. Boyle’s Law relates __________.
(Pressure and volume, Temperature and volume, Pressure and temperature)
20. Charles Law deals with __________ relationship.
(temperature and volume, pressure and volume, temperature and pressure)
21. Effusion is the escape of gas through __________.
(A small pin hole, Semi permeable membrane, porous container)
22. The expression P = P1 + P2 + P3 represents __________ mathematically.
(Graham’s Law, Avogadro’s Law, Dalton’s law of partial Pressure)
23. According to __________ equal volumes of all gases at the same temperature and pressure contain equal number of molecules.
(Graham’s Law, Avogadro’s Law, Dalton’s Law)
24. The boiling point of pure water is __________.
(32°C, 100°F, 373°K)
25. The internal resistance of a liquid to flow is called __________.
(Surface tension, Capillary action, Viscosity)
26. The existence of different crystals forms of the same substance is called __________.
(Isomorphism, Polymorphism, Isotopes)
27. Rate of Evaporation __________ on increasing temperature.
(Increases, Decreases, Remains same)
28. The temperature at which more than one crystalline forms of a substance coexist is called the __________.
(Critical Temperature, Transition Temperature, Absolute Temperature)
29. The gases which strictly obey the gas laws are called __________.
(Ideal gases, Permanent gases, Absolute gases)
30. Lighter gas diffuse __________ than the heavier gases.
(More readily, Less readily, Very slowly)
Chapter 3
Structure of Atom
1. The charge on an electron is __________.
(-2.46 x 104 coulombs, -1.6 x 10-19 coulombs, 1.6 x 10-9coulombs)
2. The maximum number of electrons that can accommodated by a p-orbital is __________.
(2, 6, 10)
3. A proton is __________.
(a helium ion, a positively charged particle of mass 1.67 x 10-27 kg, a positively charged particle of mass 1/1837 that of Hydrogen atom)
4. Most penetrating radiation of a radioactive element is __________.
(a-rays, b-rays, g-rays)
5. The fundamental particles of an atom are __________.
(Electrons and protons, electrons and neutrons, Electrons, Protons, Neutrons)
6. The fundamental particles of an atoms are __________.
(the number of protons, The number of neutrons, The sum of protons and neutrons)
7. “No two electrons in the same atoms can have identical set of four quantum numbers.” This statement is known as __________.
(Pauli’s Exclusion Principle, Hund’s rule, Aufbau Rule)
8. __________ has the highest electronegativity value.
(Fluorine, Chlorine, Bromine)
9. Principle Quantum number describes __________.
(Shape of orbital , size of the orbital, Spin of electron in the orbital)
10. Canal rays are produced from __________.
(Anode, Cathode, Ionization of gas in the discharge tube)
11. Electromagnetic radiation produce from nuclear reactions are known as __________.
(a-rays, b-rays, g-rays)
12. Cathode rays consist of __________.
(Electorns, Protons, Positrons)
13. The properties of cathode rays __________ upon the nature of the gas inside the tube.
(depend, partially depend, do not depend)
14. Anode rays consists of __________ particles.
(Negative, Positive, Neutral)
15. Atomic mass of an element is equal to the sum of __________.
(electrons and protons, protons and neutrons, electrons and neutrons)
16. Neutrons were discovered by __________.
(Faraday, Dalton, Chadwick)
17. The value of Plank’s constant is __________.
(6.626 x 10-34, 6.023 x 1024, 1.667 x 10-28)
18. P-orbitals are __________ in shape.
(spherical, diagonal, dumb bell)
19. The removal of an electron from an atom in gaseous state is called __________.
(Ionization energy, Electron Affinity, Electronegativity)
20. The energy released when an electron is added to an atom in the gaseous state is called __________.
(Ionization Potential, electron Affinity, Electronegativity)
21. The power of an atom to attract a shared pair of electrons is called __________.
(Ionization Potential, Electron Affinity, Electronegativity)
22. Electronegativity of Fluorine is arbitrarily fixed as __________.
(2, 3, 4)
23. The energy difference between the shells go on __________ when moved away from the nucleus.
(Increasing, decreasing, equalizing)
24. __________ discovered that the nucleus of an atom is positively charged.
(William Crooke’s, Rutherford, Dalton)
25. Isotopes are atoms having same __________ but different __________.
(Atomic weight, Atomic number, Avogadro’s Number)
26. __________ consists of Helium Nuclei or Helium ion (He++).
27. The angular momentum of an electron revolving around the nucleus of atom is __________.
(nh/2p, n2h2/2p, nh3/3p)
28. The wavelengths of X-rays are mathematically related to the __________ of anticathode element.
(atomic weight, atomic number, Avogadro’s number)
29. Lyman Series of spectral lines appear in the __________ portion of spectrum.
(Ultraviolet, Infra red, Visible)
30. According to __________ electrons are always filled in order of increasing energy.
(Pauli’s Exclusion Principle, Uncertainty Principle, Aufbau Principle)
Chapter 4
Chemical Bonding
1. The energy required to break a chemical bond to form neutral atoms is called __________.
(Ionization Potential, Electron Affinity, Bond Energy)
2. The chemical bond present in H-Cl is __________.
(Non Polar, Polar Covalent, Electrovalent)
3. A polar covalent bond is formed between two atoms when the difference between their E.N values is __________.
(Equal to 1.7, less than 1.7, More than 1.7)
4. The most polar covalent bond out of the following is __________.
(H-Cl, H-F, H-I)
5. __________ bond is one in which an electron has been completely transferred from one atom to another.
(Ionic, Covalent, co-ordinate)
6. __________ bond is one in which an electron pair is shared equally between the two atoms.
(Ionic, Covalent, Co-ordinate)
7. Bond angle in the molecule of CH4 is of __________.
(120°, 109.5°, 180°)
8. A molecule of CO2 has __________ structure.
9. The sigma bond is __________ than pi bond.
(Weaker, Stronger, Unstable)
10. The sp3 orbitals are __________ in shape.
(Tetrahedral, Trigonal, Diagonal)
11. The shape of CH4 molecule is __________.
(Tetrahedral, Trigonal, Diagonal)
12. The bond in Cl2 is __________.
(Non polar, Polar, Electrovalent)
13. Water is __________ molecule.
(None polar, Polar, Electrovalent)
14. Covalent bonds in which electron pair are shared equally between the two atoms is called __________ covalent bond.
(Non polar, Polar, Co-ordinate)
15. Each carbon atom in CH4 is __________ hybridized.
(Sp3, Sp2, Sp)
16. Each carbon atom in C2H4 is __________ hybridized.
(Sp3, Sp2, Sp)
17. Each carbon atom in C2H2 is __________ hybridized.
(Sp3, Sp2, Sp)
18. Oxygen atom in H2O has __________ unshared electron pair.
(One, two , three)
19. Nitrogen atom in NH3 has __________ unshared electron pair.
(One, two, three)
20. The cloud of charge that surrounds two or more nuclei is called __________ orbital.
(Atomic, Molecular, Hybrid)
21. A substance, which is highly attracted by a magnetic field, is called __________.
(Electromagnetic, Paramagnetic, Diamagnetic)
22. HF exists in liquid due to __________.
(Vander Waal Forces, Hydrogen bond, covalent Bond)
23. Best hydrogen bonding is found in __________
(HF, HCl, HI)
24. Shape of CCl4 molecule is __________.
(tetrahedral, Trigonal, Diagonal)
25. __________ bond is formed due to linear overlap.
(Sigma bond, Pi bond, Hydrogen bond)
26. __________ is defined as the quantity of energy required to break one mole of covalent in gaseous state.
(Bond energy, Ionization energy, Energy of Activation)
27. Repulsive force between electron pair in a molecule is maximum when it has an angle of __________.
(120°, 109.5°, 180°)
28. Repulsive force between electron pair in a molecule is maximum when it has an angle of __________.
(120°, 109.5°, 180°)
29. The sum of total number of electrons pairs (bonding and lone pairs) is called __________.
(Atomic Number, Avogadro’s Number, Steric Number)
30. Shape of __________ molecule is tetrahedral.
(BaCl2, BF3, NH3)
Chapter 5
Energetics of Chemical Reaction
1. The quantity of heat evolved or absorbed during a chemical reaction is called __________.
(Heat or Reaction, Heat of Formation, Heat of Combination)
2. An endothermic reaction is one, which occurs __________.
(With evolution of heat, With absorption of Heat, In forward Direction)
3. An exothermic reaction is one during which __________.
(Heat is liberated, Heat is absorbed, no change of heat occurs)
4. The equation C + O2 ® CO2 DH = -408KJ represents __________ reaction.
(Endothermic, Exothermic, Reversible)
5. The equation N2 + O2 ® 2NO DH = 180KJ represents __________ reaction.
(Endothermic, Exothermic, Irreversible)
6. Thermo-chemistry deals with __________.
(Thermal Chemistry, Mechanical Energy, Potential Energy)
7. Enthalpy is __________.
(Heat content, Internal energy, Potential Energy)
8. Hess’s Law is also known as __________.
(Law of conservation of Mass, Law of conservation of Energy, Law of Mass Action)
9. Any thing under examination in the Laboratory is called __________.
(Reactant, System, Electrolyte)
10. The environment in which the system is studied in the laboratory is called __________.
(Conditions, Surroundings, State)
11. When the bonds being broken are more than those being formed in a chemical reaction, then DH will be __________.
(Positive, Negative, Zero)
12. When the bond being formed are more than those being broken in a chemical reaction, then the DH will be __________.
(Positive, Negative, Zero)
13. The enthalpy change when a reaction is completed in single step will be __________ as compared to that when it is completed in more than one steps.
(Equal to, Partially different from, Entirely different from)
14. The enthalpy of a system is represent by __________.
(H, DH, DE)
15. The factor E + PV is known as __________.
(Heat content, Change in Enthalpy, Work done)
16. Heat of formation is represented by __________.
(Df, DHf, Hf)
17. The heat absorbed by the system at constant __________ is completely utilize to increase the internal energy of the system.
(Volume, Pressure, Temperature)
18. Heat change at constant __________ from initial to final state is simply equal to the change in enthalpy.
(Volume, Pressure, Temperature)
19. A system, which exchange both energy and energy with the surrounding, is __________ system.
(Open, Closed, Isolated)
20. A system, which only exchange energy with the surrounding but not the matter, is __________ system.
(Open, Closed, Isolated)
21. A system, which neither exchanges energy nor matter with the surroundings is __________ system.
(Open, Closed, Isolated)
22. __________ property of a system is independent of the amount of material concerned.
(Intensive, Extensive, Physical)
23. __________ property of a system depends upon the amount of substance present in the system.
(Intensive, Extensive, Physical)
24. DE = q – w represents __________.
(First Law of Thermodynamics, Hess’s Law, Enthalpy Change)
25. __________ is defined as the change in enthalpy when one gram mole of a compound is produced from its elements.
(Heat of Reaction, heat of Formation, Heat of Neutrilization)
Chapter 6
Chemical Equilibrium
1. At equilibrium the rate of forward reaction and the rate of reverse reaction are __________.
(Equal, Changing, Different)
2. Such reactions, which proceed to forward direction only and are completed after sometime are called __________ reaction.
(Irreversible, Reversible, Molecular)
3. Such reactions, which proceed to both the direction and are never completed, are called __________ reaction.
(Irreversible, Reversible, Molecular)
4. The rate of chemical reaction is directly proportional to the product of the molar concentration of __________.
(Reactants, Products, Both reactants and products)
5. “If a system in equilibrium is subjected to a stress, the equilibrium shifts in a direction to minimize or undo the effect of this stress. This principle is known as __________.
(Le-Chatelier’s Principle, Gay Lussac’s Principle, Avogadro’s Principle)
6. A very large value of Kc indicates that reactants are __________.
(very stable, unstable, moderately stable)
7. A very low value of Kc indicates that reactants are __________.
(very stable, very unstable, moderately stable)
8. The equilibrium in which reactants are products are in single phase is called __________.
(Homogenous Equilibrium, Heterogenous Equilibrium, Dynamic Equilibrium)
9. The equilibrium in which reactants and products are in more than one phases are called __________.
(Homogenious Equilibrium, Heterogenious Equilibrium, Dynamic Equilibrium)
10. Chemical Equilibrium is __________ equilibrium.
(Dunamic, Static, Heterogeneous)
11. In exothermic reaction, lowering of temperature will shift the equilibrium to __________.
(right, left, equally on both the direction)
12. In endothermic reaction, lowering of temperature will shift the equilibrium to __________.
(right, left, equally on both the direction)
13. A catalyst __________ the energy of activation.
(increases, decreases, has no effect on)
14. At equilibrium point __________.
(forward reaction is increased, backward reaction is increased, forward and backward reactions become equal)
15. NH3 is prepared by the reaction N2 + 3H2 Û 2NH3 DH = -21.9 Kcal. The maximum yield of NH3 is obtained __________.
(At low temperature and high pressure, at high temperature and low pressure, at high temperature and high pressure)
16. When a high pressure is applied to the following reversible process: N2 + O2 Û 2NO The equilibrium will __________
(shift to the forward direction, shift to the backward direction, not change)
17. The value of Kc __________ upon the initial concentration of the reaction.
(depends, partially depends, does not depend)
18. While writing the Kc expression, the concentration of __________ are taken in the numerator.
19. Solubility product constant is denoted by __________.
(Kc, Ksp, Kr)
20. “The degree of ionization of an electrolyte is suppressed by the addition of another electrolyte containing a common ion.” This phenomenon is called __________.
(Solubility Product, Common Ion Effect, Le-Chatelier’s Principle)
Chapter 7
Solutions and Electrolytes
1. Molarity is the number of moles of a solute dissolved per __________.
(dm3 of a solution, dm3 of solvent, Kg of solvent)
2. Molality is defined as the number of moles of solute dissolved per __________.
(dm3 of solution, kg of solvent, kg of solute)
3. The solubility of a solute __________ with the increase of temperature.
(increases, decreases, does not alter)
4. The loss of electron during a chemical reaction is known as __________.
(Oxidation, Reduction, Neutralization)
5. The gain of electron during a chemical reaction is known as __________.
(Oxidation, Reduction, Neutralization)
6. The ions, which are attracted towards the anode, are known as __________.
(Anins, Cations, Positron.
7. The pH of a neutral solution is __________.
(1.7, 7, 14)
8. A current of one ampere flowing for one minute is equal to __________.
(One coulomb, 60 coulomb, one Faraday)
9. A substance, which does not allow electricity to pass through, is known as __________.
(Insulator, Conductor, Electrolyte)
10. Such substances, which allow electricity to pass through them and are chemically decomposed, are called __________.
(Electrolytes, Insulators, Metallic conductors)
11. __________ is an example of strong acid.
(Acetic Acid, Carbonic Acid, Hydrochloric Acid)
12. __________ is an example of weak acid.
(Hydrochloric Acid, Acetic Acid, Sulphuric Acid)
13. When NH4Cl is hydrolyzed, the solution will be __________.
(Acidic, Basic, Neutral)
14. When Na2CO3 is hydrolyzed, the solution will be __________.
(Acidic, Basic, Neutral)
15. When blue hydrated copper sulphate is heated __________.
(It changes into white, it turns black, it remains blue)
16. Sulphur has the highest oxidation number in __________.
(SO2, H2SO4, H2SO3)
17. The reaction between an acid and a base to form a salt and water is called __________.
(Hydration, Hydrolysis, Neutralization)
18. __________ is opposite of Neutralization.
(Hydration, Hydrolysis, Ionization)
19. The substance having pH value 7 is __________.
(Basic, Acidic, Neutral)
20. An aqueous solution whose pH is zero is __________.
(Alkaline, Neutral, Strongly Acidic)
21. Solubility product of slightly soluble salt is denoted by __________.
(Kc, Kp, Ksp)
22. The increase of oxidation number is known as __________.
(Oxidation, Reduction, Hydrolysis)
23. The decrease of Oxidation number is known as __________.
(Oxidation, Reduction, Electrolysis)
24. One molar solution of glucose contains __________ gms of glucose per dm3 of solution.
* 180, 100, 342)
25. The number of moles of solute present per dm3 of solution is called __________.
(Molality, Molarity, Normality)
26. ‘M’ is the symbol used for representing __________.
(Molality, Molarity, Normality)
27. 1 mole of H2SO4 is equal to __________.
(98gms, 49gms, 180gms)
28. Buffer solution tends to __________ pH.
(Change, Increase, maintain)
29. The logarithm of reciprocal of hydroxide ion is represented as __________.
(pH, pOH, POH)
30. In __________ water molecules surround solute particles.
(Hydration, Hydrolysis, Neutralization)
Chapter 8
Introduction to Chemical Kinetics
1. The rate of chemical reaction __________ with increase in concentration of the reactants.
(Increases, Decreases, Does not alter)
2. Ionic reactions of inorganic compounds are __________.
(very slow, moderately slow, very fast)
3. The rate of __________ reactions can be determined.
(Very Slow, Moderately Slow, Very fast)
4. The sum of exponents of the concentrations of reactants is called __________.
(Order of reaction, Molecularity, Equilibrium Constant)
5. The rate of reaction generally __________ in the presence of a suitable catalyst.
(Increases, Decreases, remains constant)
6. The rate of a reaction __________ upon the temperature.
(depends, slightly depends, does not depends)
7. The minimum energy required to bring about a chemical reaction is called __________.
(Bond energy, Ionization energy, Energy of Activation)
8. Oxidation of SO2 in the presence of V2O5 in Sulphuric Acid industry is an example of __________.
(Homogenous catalyst, Heterogeneous catalyst, Negative catalyst)
9. Hydrolyses of ester in the presence of acid is an example of __________.
(Homogenous catalyst, Heterogeneous catalyst, Negative catalyst)
10. Concentration of the reactants __________ with the passage of time during a chemical reaction.
(Increases, Decreases, Does not alter)
11. Concentration of the products __________ with the passage of time during a chemical reaction.
(Increases, Decreases, Does not alter)
12. The rate constant __________ with temperature for a single reaction.
(Varies, Slightly Varies, Does not vary)
13. The rate of reaction at a particular time is called __________.
(Average Rate of reaction, Absolute rate of reaction, Instantaneous rate of reaction)
14. The specific rate constant K has __________ value for all concentrations of the reactant.
(Fixed, Variable, negligible value)
15. By increasing the surface area the rate of reaction can be __________.
(Increased, Decreased, Doubled)
16. MnO2 when heated with KClO3 __________.
(Gives up its own oxygen, Produces ozone O3, Acts as catalyst)
17. Reactions with high energy of activation proceed with __________.
(High speed, Moderately slow speed, slow speed)
18. The minimum amount of energy required to bring about a chemical reaction is called __________.
(Energy of ionization, Energy of Activation, Energy of Collision)
19. An inhibitor is a catalyst which __________ rate of reaction.
(Increases, Decreases, Does not alter)
20. __________ is the change of the concentration of reactant divided by the time.
(Rate of reaction, Velocity Constant, Molecularity)
Numericals
- Chemistry XI Karachi Board Numericals
Five Year Papers
1. Simplify according to the rule of significant figure .
2. The atomic mass of Zn is 65.4 a.m.u. Calculate (i) the number of moles and also the number of atoms in 10.9 gm of Zn. (ii) The mass of 1.204 x 1024 atoms of Zn in gm.
3. Adipic acid is used in the manufacture of Nylon. The acid contains 49.3%C, 6.9%H and 43.6%O by mass. The molecular mass of the acid is 146 a.m.u. Find the molecular formula of the Adipic Acid.
4. Calculate the value of R (Gas constant) with the help of Gas Equation when (i) the pressure is in atmosphere and the volume in dm3 or litre. (ii) the pressure is in Nm-2 and the volume is in cubic metre.
5. 400cm3 of helium gas effuse from a porous container in 20 seconds. How long will SO2 gas take to effuse from the same container? (Atomic Weight = S = 32, He = 4).
6. A system absorbs 200J of heat from the surroundings and does 120 J of work on the surroundings by expansions. Find the internal energy change of the system.
7. 1.2 gm of acetic acid (CH3COOH) is dissolved in water to make 200cm3 of the solution. Find the concentration of the solution in Molarity.
8. The solubility of calcium oxalate (CaC2O4) is 0.0016 g/dm3 at 25°C. Find the solubility product of calcium oxalate: CaC2O4 ® Ca2+ + C2O42-
9. Calculate H+ ion concentration of a solution whose pH = 5.6.
10. The rate constant (k) for the decomposition of nitrogen dioxide 2NO2(g) ® 2NO(g) + O2(g) is 1.8 x 103- dm3mole1-sec1-. Write down the rate expression and (i) find the initial rate when the initial concentration of NO2 is 0.75 M. (ii) Find the rate constant (k) when the initial concentration of NO2 is doubled.
11. Calculate the volume of nitrogen gas produced by heating 800 gm of ammonia at 21°C and 823 torr pressure. 2NH3 ® N2 + 3H2 (Atomic Weight = N = 14, H = 1)
12. In collection of 24 x 1025 molecules of C2H5OH. What is the number of moles. ( Atomic weight = C = 12, O = 16, H = 1)
13. Simplify using exponential notation: 43100 + 3900 + 2100.
14. A given compound contains 75. 2% carbon, 10.75% hydrogen and 14.05% oxygen. Calculate the empirical formula of the compound. (Atomic weight: C = 12, O = 16, H = 1)
15. Calculate the wave number of spectral line of hydrogen gas when an electron jumps from n = 4 to n= 2. (RH = 109678 cm-1)
16. 13.2 gm of gas occupies a volume of 0.918 dm3 at 25°C and 8 atm pressure. Calculate the molecular mass of the gas.
17. Calculate the heat of formation of benzene at 25°C when the heat of formation of CO2 and water and heat of combustion of benzene are given:
(i)
6C + 3H2 ® C6H6
DHf = ?
(ii)
C + O2 ® CO2
DH = -286KJ/mole
(iii)
H2 + ½O2 ® H2O
DH = -286KJ/mole
(iv)
C6H6 + 7.5O2 ® 6CO2 + 3H2O
DH = -3267 KJ/mole
18. The rate constant for the decomposition of nitrogen dioxide is 1.8 x 10-8 dm3 mole-1s-1. What is the initial rate when the initial concentration of NO2 is 0.50M? 2NO2 ® 2NO + O2.
19. Should AgCl precipitate from a solution prepared by mixing 400cm3 of 0.1M NaCl and 600cm3 of 0.03 M of solution of AgNO3? (Ksp for AgCl = 1.6 x 10-10 mole/dm3)
20. A sample of chlorine gas at S.T.P has a volume of 800cm3 calculate The number of moles of chlorine, the mass of the sample and the number of chlorine molecules in the sample.
21. How many atoms of carbon are present in 10 gm of coke?
22. The volume of the oxygen gas, collected over water at 24°C and 762mm pressure, is 128 ml. Calculate the mass in gm of oxygen gas obtained. The pressure of water vapour at 24°C is 22 mm.
23. Calculate the radius of orbit n = 3 for a Hydrogen atom in Armstrong unit. (h = 6.625 x 10-27 erg-sec, p = 3.14, m = 9.11 x 10-28gm, e = 4.8 x 10-10 esu)
24. For the reaction H2 + I2 ® 2HI. Kc is 49. Calculate the concentration of HI at equilibrium when initially one mole of H2 is mixed with one mole of I2 in one litre flask.
25. Determine the mass of HCl required to prepare 400 ml of 0.85M HCl solution.
26. Calculate pH value of 0.004M NaOH solution.
27. Kc for the reaction is 0.0194 and the calculated ratio of the concentration of the reactants and the product is 0.0116. Predict the direction of the reaction.
28. For the decomposition of ethyl chlorocarbonate ClCOOC2H5 ® CO2 + Cl.C2H5. Find the value of rate constant when initial concentration of Ethyl Chlorocarbonate is 0.25 M and the initial rate of the reaction is 3.25 x 10-4 mole/dm3/sec.
29. 1.0 gm of a sample of an organic substance was burnt in excess of oxygen yield 3.03 gm of CO2 and 1.55 gm of H2O. If the molecular mass of the compound is 58. Find the molecular formula.
30. Calculate the volume of the oxygen at S.T.P that may be obtained by complete decomposition of 51.3 gm of KClO3 on heating in presence of MnO2 as a catalyst. 2KClO3 ® 2KCl + 3O2. (Atomic mass of K = 39, Cl = 35.5, O = 16, Mn = 55)
31. Calculate the wave number of the Line in Lyman Series when an electron jumps from orbit 3 to orbit 1.
32. Calculate the heat of formation of ethane (C2H6) at 25°C from the following data:
(i)
2C + 3H2 ® C2H6
DHf = ?
(ii)
C + O2 ® CO2
DH = -394KJ/mole
(iii)
H2 + ½O2 ® H2O
DH = -286KJ/mole
(iv)
C2H6 + ½O2 ® 2CO2 + 3H2O
DH = -1560.632KJ/mole
33. At the equilibrium a 12 litre flask contains 0.21 mole of PCl5, 0.32 mole of Cl2 at 250°C. Find the value of Kc for the reaction. PCl5 Û PCl3 + Cl2.
34. A given compound contains C = 60%, H = 13.0% and O = 27%. Calculate its Empirical Formula.
35. How many grams of chlorine are required to prepare 7.75 dm3 of chloro benzene? The equation of the reaction is C6H6 + Cl2 ® C6H5Cl + HCl. (Atomic Number of C = 12, H = 1 and Cl = 35.5)
36. A mixture of helium and hydrogen is confined in a 12 dm3 flask at 30°C. If 0.2 mole of the helium is present, find out the partial pressure of each gas whereas the pressure of the mixture of gases is 2atm.
37. Calculate the radius by hydrogen atom by applying Bohr’s Theory. (h = 6.625 x 10-27 erg-sec, p = 3.14, m = 9.11 x 10-28gm, e = 4.8 x 10-10 esu)
38. Calculate the heat of formation of C2H2 from carbon and hydrogen from the following data:
(i)
2C + H2 ® C2H2
DHf = ?
(ii)
C + O2 ® CO2
DH = -94.05Kcal/mole
(iii)
H2 + ½O2 ® H2O
DH = -68.32Kcal/mole
(iv)
C2H2 + 5/2O2 ® 2CO2 + H2O
DH = -310Kcal/mole
39. Calculate the pH of a 2.356 x 10-3m HCl solution.
40. For the reaction N2 + 3H2 Û 2NH3. The equilibrium mixture contains 0.25 M nitrogen, 0.15M hydrogen gas at 25°C. Calculate the concentration of NH3 gas when Kc = 9.6. the volume of the container is 1dm3.
41. Determine the initial rate of the following reaction at 303°C in which its rate constant is 8.5 x 10-5 litre-mol-1 sec-1. Initial concentration of the reaction is 9.8 x 10-2 mole/litre. 2NO2 ® 2NO + O2.
Extra Numericals
1. 4.6gm of ethyl alcohol and 6.0gm of acetic acid kept at constant temperature until equilibrium was established. 2 gm of acid were present unused. Calculate Kc.
2. Kc for the dissociation of HI at 350°C is 0.01. If 0.2 mole of H2, 1.3 moles of I2 and 4 moles of HI are present. Predict the direction of reaction.
3. What is the solubility of PbCrO4 at 30°C when Ksp is 1.8 x 10-14.
4. 1.06m of an organic compound on combustion gave 1.49 gm of CO2 and 0.763gm H2O. It also has 23.73% N. Find its compercial formula.
5. 500 dm3 of moist O2 gas was collected over water at 27°C and 726torr pressure. Find the mass in gm. Of dry O2 gas at S.T.P. When the vapour pressure of water 27°C is 26 torr.
6. Atomic mass of phosphorus is 31. Calculate the mass of 45 atoms in a.m.u.
7. Methane burn in steam according the following reaction: CH4 + 2O2 ® CO2 + 2H2O. If 100 gm of each CH4 and O2 is taken, then what amount of CO2 liberated?
8. An organic compound containing C = 65.45%, H = 5.45% and O = 29.09%. If molecular weight of compound is 110, calculate molecular formula.
9. What mass of CO2 is produced by the complete combustion of 100g pentane. C5H12 + 3O2 ® 2CO2 + 2H2O.
10. One atom of an unknown element is found to have a mass of 67.8 x 10-23g. What is the atomic weight of the element?
11. The heat of combustion of glucose and alcohol is given below.
(i)
C6H12O6 + 6O2 ® 6CO2 + 6H2O
DH = -673Kcal/mole
(ii)
C2H5OH+ 3O2 ® 2CO2 + 3H2O
DH = -328Kcal/mole
Find DH for the fermentation given below:
C6H12O6 ® 2C2H5OH + 3CO2
12. At certain temperature, the equilibrium mixture contain 0.4 mole of H2, 0.4mole I2 and 1 mole of HI. If addition 2 mole of H2 are added. How many moles of HI will be present when the new equilibrium established. H2 + I2 ® 2HI.
13. A solution has pH of 8.4. Find concentration of H+ and OH-.
14. 180cm3 of a known gas diffuse in 15minutes, when 120 cm3 of SO2 diffuses in 20 minutes. What is the molecular mass of the unknown gas.
Chapter 1
Introduction to Fundamental Concepts
1. Calculate the moles of the following in 500gm, NH3, HCl, Na2CO3, H2SO4, MgBr2, CaCO3, Xe and C.
2. How many moles of Na are present in 5gm of Na?
3. Calculate the number of atoms in 12 gms of Mg.
4. 2gm diamond is studded in a ring. Diamond is a pure carbon. How many atoms of carbon are present in the ring?
5. Calculate the number of molecules in 9gms of H2O.
6. How many molecules are present in 25 gms of CaCO3?
7. Calculate the weight in gram of 3.01 x 1020 molecules of glucose (C6H12O6)
8. How many atoms of hydrogen are there in 2.57 x 10-6 gram of hydrogen?
9. A sample of oxygen contains 1.87 x 1027 atoms of oxygen. What would be the weight of the oxygen?
10. Find the weight of oxygen obtained from 49gm of KClO3.
2KClO3 ® 2KCl + 3O2
11. What weight of CO2 and CaO can be obtained by heating 12.5gm of Limestone (CaCO2)?
CaCO3 ® CaO + CO2
12. Calculate the weight of sodium chloride required to produce 142 gm of chlorine.
2NaCl ® 2Na + Cl2
13. Calculate the weight of carbon, required to produce 88gm of CO2.
C + O2 ® CO2
14. The action of CO on Fe2O3 can be represented by the following equation.
Fe2O3 + 3CO ® 2Fe + 3CO2
15. What weight of NH3 will be required to produce 100 gm of NO?
4NH3 + 5O2 ® 4NO + 6H2O
16. Find out the moles of CuSO4 which are obtained from 31.75 gm of Cu.
Cu + H2SO2 ® CuSO2 + H2
17. Calculate the number of N2 and H2 molecules, which are obtained from 8.5 gm of NH3.
N2 + 3H2 ® 2NH3
18. Find out the number of Cu and H2O molecules obtained from 7.95gm of CuO.
CuO + H2 ® Cu + H2O
19. 400gm of H2 was made to combine with 14200gm of Cl2. How much HCl will be produced?
20. 1kg of Limestone was heated 500gm of CaO was obtained. How much CO2 gas produced into air.
21. Find the weight of O2 obtained from 49 gm of KClO3.
2KClO3 ® 2KCl + 3O2
22. Chlorine is produced on the large scale by the electrolysis of NaCl aqueous solution. Chlorine the weight of NaCl required to produce 142 gm of Cl2.
2NaCl + 2H2O ® Cl2 + H2 + 2NaOH
23. How many grams of O2 are required to completely burn 18.0gm of C? How many grams of CO2 will be formed?
24. Calculate the weight of NH3, required to produce 100 gms of NO.
4NH3 + 5O2 ® 4NO + 6H2O
25. Find out the moles of H2 and N2 required producing 17gm of NH3.
26. Calculate the volume of H2 at S.T.P, which is obtained by the reaction of 120 gm Mg with MgSO4.
Mg + H2SO4 ® MgSO2 + H2
27. NH3 gas can be produced from ammonium chloride (NH4Cl) as follows:
CaO + 2NH4Cl ® CaCl2 + H2O + NH3
Calculate the volume of NH3 obtained at S.T.P by the reaction of 100 gm of NH4Cl.
28. 500gm of C2H4 on combustion in air gave CO2 and H2O. Calculate the volume of O2 and CO2 at S.T.P.
29. Find out the volume of O2, CO2 and SO2 gases at S.T.P react and obtained from 2 moles of CS2.
CS2 + 3O2 ® CO2 + 2SO2
30. Calculate the volume of CO2 gas at S.T.P obtained by the combustion of 20gm of CH4.
CH4 + 2O2 ® CO2 + 2H2O
31. Calculate the volume of O2 gas at S.T.P required to burn 600dm3 of H2S, also find the volume of SO2 gas produced at S.T.P.
32. Calculate the volume of O2 gas at S.T.P required to burn 50 gm of CH4.
33. What volume of H2 at S.T.P can be produced by the reaction of 6.54gm Zn with HCl?
Zn + 2HCl ® ZnCl2 + 2H2
34. Calculate the volume of O2 and H2 gases at S.T.P obtained from 9gm of H2O.
35. 0.264gm of Mg was burnt in pure O2. How much MgO will be formed?
2Mg + O2 ® 2MgO
36. How much H2 can be generated by passing 200gm of steam over hot iron.
4H2O + 3Fe ® Fe3O4 + 4H2
37. If 112dm3 of N2 react with 336 dm3 of H2, both at S.T.P. How many grams of NH3 would be obtained?
N2 + 3H2 ® 2NH3
38. An organic compound contains 12.8%C, 2.1% and 85.1% Br. If the mass of the compound is 188, find the molecular formula.
39. An organic compound contains 66.70%C, 7.41% H and 25.90% N2. The molecular mass of the compound is 108. Find out its molecular formula.
40. A compound contains 19.8%C, 2.5%H, 66.1%O and 11.6%N. Find out empirical formula of the compound.
41. 0.2475gm of a compound, containing C, H and O gave 0.4950gm CO2 and 0.2025gm H2O. If the molecular mass of the compound is 88. Find out the molecular formula.
42. An organic compound contains 32%C, 6.67%H, 18.66%N and 42.67%O. Its molecular mass is 75. Find out the molecular formula of the compound.
43. 1.367gm of a compound containing C, H and O on heating gave 3.002gm CO2 and 1.640gm H2O. Find out its molecular formula, when the molecular mass is 120.
44. A compound was found to contain 40%C and 6.7%H. Its molecular mass was 60. Find out its molecular formula.
45. An organic compound contains 75.2%C, 10.15%H and oxygen. Its molecular mass is 115. Find its molecular formula.
46. The empirical formula of a compound is CH2O. If the molecular mass 180. Find out the molecular formula.
47. An organic compound composed of C, H and O. On combustion of 0.94gm of this compound, 1.32gm CO2 and 0.568gm H2O were obtained. Its molecular mass is 180. Find its molecular formula.
48. An organic compound composed of C, H and O. 4.2gm of the compound on heating gave 6.21gm CO2 and 2.54gmH2O. Its molecular mass is 60. Find its molecular formula.
49. An organic compound contains C,H and 6.38gm of compound on combustion gave 9.06gm CO2 and 5.58gm H2O. Its molecular mass is 62. Find out its molecular formula.
50. 1gm of a hydrocarbon on combustion gave 3.03gm of CO2 and 1.55gm of H2O. If the molecular mass is 58, find its molecular formula.
51. 1.434gm of a compound on combustion gave 4.444gm CO2 and 2.0 gm H2O. Find out its empirical formula.
52. An organic compound composed of C, H and N. 0.225gm of compound on combustion gave 0.44gm CO2 and 0.315gm H2O. If the molecular mass of a compound is 90, find out its molecular formula.
53. An organic compound contains 40.68%C, 8.47%H, 23.73%N and 27.12%O. Find its empirical formula.
54. An organic compound composed of C, H and N. 0.419 gm of compound on combustion gave 0.88gm CO2 and 0.27gm H2O. Find out its empirical formula.
55. The analysis of a compound shows, C = 24.24%, H = 4.04% and Cl = 71.71%. If the molecular mass of the compound is 49.5, find its molecular formula.
56. An organic compound of molecular mass 90 has the empirical formula CH2O. What is its molecular formula?
57. The empirical formula of an organic compound is CH3NO2. If it’s molecular mass is 61. What is its molecular formula?
58. 0.638gm of an organic compound on combustion gave 0.594gm H2O and 1.452gm CO2.The compound is composed of C, H and O atoms. If the molecular mass is 116, find out its molecular formula.
59. The molecular formula of ethyl acetate is CH3COOC2H5. What is its empirical formula.
60. Find the empirical formulae of the following compounds from their percentage composition by mass:
· N = 26.17% H = 7.48% Cl = 66.35%
· Ca = 71.43% O = 28.57%
· Ag = 63.53% N = 8.23% O = 28.24%
· Na = 32.40% H = 45.07% Cl = 22.53%
61. A certain compound on analysis yielded 2.00gm C, 0.34gm H and 2.67gm O. If the relative molecular mass of the compound is 60, calculate its molecular formula.
62. What is the empirical formula of a compound, which contains 42.5% chlorine and 57.5 oxygen. If it’s formula mass is 167. What is its molecular formula?
63. What will be the weight of 5 moles of water in grams?
64. What is the mass of each of the following:
· 1.25 mole of NaCl
· 2.42 mole of NaNO3
· 1.5 mole of HCl
· 3.0 mole of NaOH
65. A piece of Aluminium metal weighs 70.0g. How many atoms are present in the piece.
66. How many atoms of carbon are present in 20-carat Diamond? (1 carat = 0.2g)
67. How many grams of oxygen have the same number of atoms as 16gm of sulphur?
68. A sample of oxygen gas at STP has a mass of 16gm. Calculate:
· The number of moles of oxygen
· The volume of the sample
· The number o molecules in the sample
69. Calculate the volume of CH4 gas at STP having a mass 32g.
70. What mass of zinc sulphate can be obtained from the reaction of 10.0gm of Zinc with an excess of dilute H2SO4?
Zn + H2SO4 ® ZnSO4 + H2*
71. Calculate what mass of sodium hydroxide you would need to neutralize a solution containing 7.3g hydrogen chloride by the reaction:
NaOH + HCl ® NaCl + H2O
72. Calculate how much sodium nitrate you need to give 126g of nitric acid by the reaction:
NaNO3 + H2SO4 ® HNO3 + NaHSO4
73. What volume of hydrogen at STP is evolved when 0.325g of zinc reacts will dilute hydrochloric acid.
Zn + 2HCl ® ZnCl2 + H2
74. What mass of oxygen is formed by the decomposition of a solution containing 120cm3 of H2O2 at STP?
2H2O2 ® 2H2O + O2
75. What is the mass of one molecule of water in grams?
76. 100cm3 of butane are burned in an excess of oxygen. Calculate:
· The volume of oxygen used
· The mass and volume of CO2 produced (assume all gases at STP)
2C4H10 + 13O2 ® 8CO2 + 10H2O
77. A cook is making a small cake. It needs 500cm3 at STP of CO2 to make the cake rise. The cook decides to add baking powder, which contains sodium bicarbonate. This generates CO2 by thermal decomposition.
2NaHCO3 ® CO2 + Na2CO3 + H2O
What mass of baking powder must the cook add to cake mixture?
78. What volume of ammonia at STP can be obtained by heating 0.25 mole of ammonium sulphate with calcium hydroxide?
(NH4)SO4 + Ca(OH)2 ® 2NH3 + CaSO4 + 2H2O
79. How many grams of SO2 are produced when 100g of H2S is reacted with 50g of oxygen.
2H2S + 3O2 ® 2H2O + 2SO2
80. How many grams of chlorobenzene will be produced when 100gm of each reactant is reacted?
C6H6 + Cl2 ® C6H5Cl + HCl
81. A car releases about 5g of NO into the air for each mile driven. How many molecules of NO are emitted per mile?
82. Simplify according to the rule of significant figures.
· 2.60 x 3.05
· 0.009 ¸ 0.3
·
·
Chapter 2
The Three States of Matter
1. 540cm3 of N2 at 400mm pressure are compressed to 300cm3 without changing the temperature. What will be the pressure of the gas?
2. A gas occupies 6dm3 at 1atm pressure keeping the temperature constant. If the pressure reduces to 600mm, what volume does the gas occupy?
3. At a certain temperature and 800mm pressure, the volume of H2 is 700cm3. If the pressure is increased to 1000mm at the same temperature, find the new volume of the gas.
4. 150ml of a gas at 27°C is heated to 77°C at constant pressure. Find the new volume of the gas.
5. 300ml of N2 are at 50° and the pressure is kept constant. If the temperature is doubled, what will be the volume of the gas?
6. A gas measures 5dm3 at 5°C under 0.5atm pressure. Calculate its volume at 25° and 5000mm pressure.
7. 2060ml of a gas is at 7°C and 860mm pressure. Find its volume at S.T,P.
8. 350ml of H2 was collected over water at 26°C. The pressure of the gas was 900mm. What volume will dry gas have at 30°C and 750mm pressure? The vapour pressure at 26°C is 25mm.
9. The volume of oxygen collected over water at 20°C and 1200mm pressure, is 200cm3. If aqueous at 20°C is 17.4mm, what will be the volume of the gas under S.T.P.
10. A 20dm3 flask contains H2 at 22°C under pressure of 1.2 atm. How many moles of H2 are present.
11. A gaseous mixture is at the pressure of 3000mm. The mixture contains 6 moles of N2, 0.5mole of CO2 and 2.5 moles of O2. Find the partial pressure of each gas.
12. A 5dm3 vessel contains 1.2 moles of H2 and 0.8 mole of N2 at 27°C. Find the total pressure of the mixture.
13. Composition of a sample of air by volume is, N2 = 76%, O2 = 20%, H2O = 2.5%, CO2 = 1.4% and He = 0.1%. If the pressure of the air is 760 mm, Calculate the partial pressure of these gases.
14. A 10dm3 container contains a mixture of He and Ne gases at 17°C. There are two moles of He gas and 3 moles of Ne gas. What is the partial pressure of the gases?
15. 10gm of H2, 96gm of O2 and 196gm of N2 are mixed together. The partial pressure of H2 is 0.6 atm. What is the partial pressure of O2 and N2?
16. A cylinder contains 1 mole of H2, 3 mole of He and 6 moles of N2. The total pressure in the cylinder is 15 atm. Calculate the partial pressure of H2, He and N2.
Chapter 5
Energetics of Chemical Reaction
1. Calculate the heat of formation of Acetic Acid from the following data:
(i)
2C + 2H2+ O2 ® CH3COOH
DHf = ?
(ii)
C + O2 ® CO2
DH = -394KJ/mole
(iii)
H2 + ½O2 ® H2O
DH = -286 KJ/mole
(iv)
CH3COOH + 2O2 ® 2CO2 + 2H2O
DH = -870KJ/mole
2. Calculate the heat of formation of Ethane from the following data:
(i)
2C + 3H2 ® C2H6
DHf = ?
(ii)
C + O2 ® CO2
DH = -394KJ/mole
(iii)
H2 + ½O2 ® H2O
DH = -286 KJ/mole
(iv)
C2H6 + 7/2O2 ® 2CO2 + 3H2O
DH = -1560KJ/mole
(v)
C2H5OH + 3O2 ® 2CO2 + 3H2O
DH = -327 KJ/mole
3. Calculate the heat of formation of Methane from the following data:
(i)
C + 2H2 ® CH4
DHf = ?
(ii)
C + O2 ® CO2
DH = -394KJ/mole
(iii)
H2 + ½O2 ® H2O
DH = -286 KJ/mole
(iv)
CH4 + 2O2 ® CO2 + 2H2O
DH = -890.3KJ/mole
4. Calculate the heat of formation of Ethyl Alcohol from the following data:
(i)
2C + 3H2 ½ O2® C2H5OH
DHf = ?
(ii)
C + O2 ® CO2
DH = -394KJ/mole
(iii)
H2 + ½O2 ® H2O
DH = -286 KJ/mole
(iv)
C2H5OH+ 3O2 ® 2CO2 + 3H2O
DH = -1369KJ/mole
5. Calculate the heat of formation of Ethane from the following data:
(i)
C2H6 + 7/2O2 ® 2CO2 + 3H2O
DHf = ?
(ii)
C + O2 ® CO2
DH = -394KJ/mole
(iii)
H2 + ½O2 ® H2O
DH = -286 KJ/mole
(iv)
C2H6 ® 2C + 3H2
DH = -84.68KJ/mole
6. Calculate the heat of formation of Methane from the following data:
(i)
C + 2H2 ® CH4
DHf = ?
(ii)
C + O2 ® CO2
DH = -94.1cal
(iii)
H2 + ½O2 ® H2O
DH = -68.3 cal
(iv)
CH4 + 2O2 ® CO2 + 2H2O
DH = -212.8 cal
7. Calculate the heat of formation of Ethene from the following data:
(i)
2C + 2H2 ® C2H4
DHf = ?
(ii)
C + O2 ® CO2
DH = -97kcal
(iii)
H2 + ½O2 ® H2O
DH = -65 kcal
(iv)
C2H4 + 3O2® 2CO2 + 2H2O
DH = 340 kcal
8. Calculate the heat of formation from the following data:
(i)
2C + 3H2 +1/2O2 ® C2H5O
DHf = ?
(ii)
C + O2 ® CO2
DH = -94.2Kcal/mole
(iii)
H2 + ½O2 ® H2O
DH = -68.5 Kcal/mole
9. Calculate the heat of formation of from the following data:
(i)
C + 2H2 + O2® CH3OH
DHf = ?
(ii)
C + O2 ® CO2
DH = -94.2Kcal/mole
(iii)
H2 + ½O2 ® H2O
DH = -68.32 Kcal/mole
(iv)
CH3OH + O2 ® CO2 + 2H2O
DH = -347.6Kcal/mole
10. Calculate the heat of formation of from the following data:
(i)
3C + 4H2 ® C3H8
DHf = ?
(ii)
C + O2 ® CO2
DH = -94.1Kcal/mole
(iii)
H2 + ½O2 ® H2O
DH = -68.3 Kcal/mole
(iv)
C3H8 + 5O2 ® 3CO2 + 4H2O
DH = -530.7Kcal/mole
11. Calculate the heat of formation of from the following data:
(i)
H2 + O2® H2O2
DHf = ?
(ii)
H2 + ½O2 ® H2O
DH = -68.32Kcal
(iii)
H2O + ½ O2 ® H2O2
DH = -23.48Kcal
12. Given:
(i)
NH3 + HCl ® NH4Cl
DH1 = 42.100Kcal
(ii)
H2O + ½ O2 ® H2O2
DH2 = 3.900cal
Find DH for the reaction,
NH3 + HCl ® NH4Cl
DHf = ?
Chapter 6
Chemical Equilibrium
1. 1.5 moles of acetic acid and 1.5 moles of ethyl alcohol were reacted at a certain temperature. At equilibrium, 1 mole of ethyl acetate was present in 1 litre of the equilibrium mixture. Calculate the equilibrium constant Kc.
CH3COOH + C2H5OH Û CH3COOC2H5 + H2O
2. 6.0 gm of hydrogen and 1016gm of iodine were heated in a sealed tube at a temperature, at which Kc is 50. The volume of the tube is 1 dm3. Calculate the concentration of HI.
H2 + I2 Û 2HI
3. At a certain temperature, an equilibrium mixture contains 0.4 mole H2, 0.4 mole I2 and 1 mole of HI. The volume of the reacting vessel is 4 dm3. Find out the equilibrium constant kc.
H2 + I2 Û 2HI
4. 3 moles of A and 2 moles of B are mixed in a 4dm3 flask, at a certain temperature. The following reaction occurs.
3A + 2B Û 4C
At equilibrium the flask contains 1 mole of B. Find the equilibrium constant kc.
5. At a certain temperature, 0.205 mole of H2 and 0.319 mole of I2 were reacted. The equilibrium mixture contains 0.314 mole of I2. Calculate the kc.
H2 + I2 Û 2HI
6. The kc for the reaction A + B Û C + D is 1/3. How many moles of A must be mixed with 3 moles of B to yield at equilibrium, 2 moles of C and D each. The volume of the vessel is 2 litre.
7. At a certain temperature the equilibrium mixture for the reaction A + B Û 2C, contains 2 moles A, 3 moles of B and 5 moles of C. Find the Kc for the reaction.
8. For the reaction 2A Û B + C, equilibrium constant kc is 1. If we start with 6 moles of A, how many moles of B will be formed.
9. 20 moles of SO2 and 10 moles of O2 are taken in a 20 litre flask. If at equilibrium 5 moles of SO3 are formed, Calculate kc.
2SO2 + O2 Û 2SO3
10. A quantity of PCl5 was heated in a 12 dm3 vessel at 250°C.
PCl5 Û PCl3 + Cl2
11. 2 moles of HI was introduced in a vessel held at constant temperature. When equilibrium was reached, it was found that 0.1 mole of I2 have been formed. Calculate the equilibrium constant.
H2 + I2 Û 2HI
12. When 1 mole of pure C2H5OH is mixed with 1 mole of CH3COOH at room temperature, the equilibrium mixture contains 2/3 moles of ester and water each.
· What will be the kc?
· How many moles of ester are formed at equilibrium when 3 moles of C2H5OH are mixed with 1 mole of CH3COOH?
CH3COOH + C2H5OH Û CH3COOC2H5 + H2O
13. PCl5 Û PCl3 + Cl2. Calculate the number of moles of Cl2 produced at equilibrium when 1 mole of PCl5 is heated at 250°C in a vessel having capacity of 10dm3. At 250°C, Kc is 0.041.
14. When 2.94 moles of iodine and 8.1 moles of Hydrogen were mixed and heated at 444°C and at constant volume, until the equilibrium was established. 5.64 moles of HI were formed. Calculate the value of kc.
H2 + I2 Û 2HI
15. What is the solubility of lead chromate in moles/dm3 at 25°C. The solubility product is 1.8 x 10-14.
PbCrO4 Û Pb++ + CrO4--
16. The solubility of Mg(OH)2 at 25°C is 0.00764 gm/dm3. What is the solubility product of Mg(OH)2?
Mg(OH)2 Û Mg++ + 2OH-
17. Find the solubility of AgCl in gm/dm3, when the solubility product is 1.25 x 10-10.
18. Calculate the solubility product of BaSO4. The solubility of the salt is 1.0 x 10-5 moles/dm3.
19. Calculate the solubility product of BaSO4 is 9.0 x 10-3 gm/dm3. Find its solubility product.
20. Predict whether there will be any precipitate formation by mixing 30cm3 of 0.01M NaCl with 60cm3 of 0.01M AgNO3 solution. Ksp of AgCl is 1.5 x 10-10.
21. A saturated solution of calcium fluoride was found to contain 0.0168 gm/dm3 of solute at 25°C. Calculate the ksp for CaF2.
22. A saturated solution of BaF2 at 25°C is 0.006M. Calculate Ksp of the salt.
Five Year Papers
1. Simplify according to the rule of significant figure .
2. The atomic mass of Zn is 65.4 a.m.u. Calculate (i) the number of moles and also the number of atoms in 10.9 gm of Zn. (ii) The mass of 1.204 x 1024 atoms of Zn in gm.
3. Adipic acid is used in the manufacture of Nylon. The acid contains 49.3%C, 6.9%H and 43.6%O by mass. The molecular mass of the acid is 146 a.m.u. Find the molecular formula of the Adipic Acid.
4. Calculate the value of R (Gas constant) with the help of Gas Equation when (i) the pressure is in atmosphere and the volume in dm3 or litre. (ii) the pressure is in Nm-2 and the volume is in cubic metre.
5. 400cm3 of helium gas effuse from a porous container in 20 seconds. How long will SO2 gas take to effuse from the same container? (Atomic Weight = S = 32, He = 4).
6. A system absorbs 200J of heat from the surroundings and does 120 J of work on the surroundings by expansions. Find the internal energy change of the system.
7. 1.2 gm of acetic acid (CH3COOH) is dissolved in water to make 200cm3 of the solution. Find the concentration of the solution in Molarity.
8. The solubility of calcium oxalate (CaC2O4) is 0.0016 g/dm3 at 25°C. Find the solubility product of calcium oxalate: CaC2O4 ® Ca2+ + C2O42-
9. Calculate H+ ion concentration of a solution whose pH = 5.6.
10. The rate constant (k) for the decomposition of nitrogen dioxide 2NO2(g) ® 2NO(g) + O2(g) is 1.8 x 103- dm3mole1-sec1-. Write down the rate expression and (i) find the initial rate when the initial concentration of NO2 is 0.75 M. (ii) Find the rate constant (k) when the initial concentration of NO2 is doubled.
11. Calculate the volume of nitrogen gas produced by heating 800 gm of ammonia at 21°C and 823 torr pressure. 2NH3 ® N2 + 3H2 (Atomic Weight = N = 14, H = 1)
12. In collection of 24 x 1025 molecules of C2H5OH. What is the number of moles. ( Atomic weight = C = 12, O = 16, H = 1)
13. Simplify using exponential notation: 43100 + 3900 + 2100.
14. A given compound contains 75. 2% carbon, 10.75% hydrogen and 14.05% oxygen. Calculate the empirical formula of the compound. (Atomic weight: C = 12, O = 16, H = 1)
15. Calculate the wave number of spectral line of hydrogen gas when an electron jumps from n = 4 to n= 2. (RH = 109678 cm-1)
16. 13.2 gm of gas occupies a volume of 0.918 dm3 at 25°C and 8 atm pressure. Calculate the molecular mass of the gas.
17. Calculate the heat of formation of benzene at 25°C when the heat of formation of CO2 and water and heat of combustion of benzene are given:
(i)
6C + 3H2 ® C6H6
DHf = ?
(ii)
C + O2 ® CO2
DH = -286KJ/mole
(iii)
H2 + ½O2 ® H2O
DH = -286KJ/mole
(iv)
C6H6 + 7.5O2 ® 6CO2 + 3H2O
DH = -3267 KJ/mole
18. The rate constant for the decomposition of nitrogen dioxide is 1.8 x 10-8 dm3 mole-1s-1. What is the initial rate when the initial concentration of NO2 is 0.50M? 2NO2 ® 2NO + O2.
19. Should AgCl precipitate from a solution prepared by mixing 400cm3 of 0.1M NaCl and 600cm3 of 0.03 M of solution of AgNO3? (Ksp for AgCl = 1.6 x 10-10 mole/dm3)
20. A sample of chlorine gas at S.T.P has a volume of 800cm3 calculate The number of moles of chlorine, the mass of the sample and the number of chlorine molecules in the sample.
21. How many atoms of carbon are present in 10 gm of coke?
22. The volume of the oxygen gas, collected over water at 24°C and 762mm pressure, is 128 ml. Calculate the mass in gm of oxygen gas obtained. The pressure of water vapour at 24°C is 22 mm.
23. Calculate the radius of orbit n = 3 for a Hydrogen atom in Armstrong unit. (h = 6.625 x 10-27 erg-sec, p = 3.14, m = 9.11 x 10-28gm, e = 4.8 x 10-10 esu)
24. For the reaction H2 + I2 ® 2HI. Kc is 49. Calculate the concentration of HI at equilibrium when initially one mole of H2 is mixed with one mole of I2 in one litre flask.
25. Determine the mass of HCl required to prepare 400 ml of 0.85M HCl solution.
26. Calculate pH value of 0.004M NaOH solution.
27. Kc for the reaction is 0.0194 and the calculated ratio of the concentration of the reactants and the product is 0.0116. Predict the direction of the reaction.
28. For the decomposition of ethyl chlorocarbonate ClCOOC2H5 ® CO2 + Cl.C2H5. Find the value of rate constant when initial concentration of Ethyl Chlorocarbonate is 0.25 M and the initial rate of the reaction is 3.25 x 10-4 mole/dm3/sec.
29. 1.0 gm of a sample of an organic substance was burnt in excess of oxygen yield 3.03 gm of CO2 and 1.55 gm of H2O. If the molecular mass of the compound is 58. Find the molecular formula.
30. Calculate the volume of the oxygen at S.T.P that may be obtained by complete decomposition of 51.3 gm of KClO3 on heating in presence of MnO2 as a catalyst. 2KClO3 ® 2KCl + 3O2. (Atomic mass of K = 39, Cl = 35.5, O = 16, Mn = 55)
31. Calculate the wave number of the Line in Lyman Series when an electron jumps from orbit 3 to orbit 1.
32. Calculate the heat of formation of ethane (C2H6) at 25°C from the following data:
(i)
2C + 3H2 ® C2H6
DHf = ?
(ii)
C + O2 ® CO2
DH = -394KJ/mole
(iii)
H2 + ½O2 ® H2O
DH = -286KJ/mole
(iv)
C2H6 + ½O2 ® 2CO2 + 3H2O
DH = -1560.632KJ/mole
33. At the equilibrium a 12 litre flask contains 0.21 mole of PCl5, 0.32 mole of Cl2 at 250°C. Find the value of Kc for the reaction. PCl5 Û PCl3 + Cl2.
34. A given compound contains C = 60%, H = 13.0% and O = 27%. Calculate its Empirical Formula.
35. How many grams of chlorine are required to prepare 7.75 dm3 of chloro benzene? The equation of the reaction is C6H6 + Cl2 ® C6H5Cl + HCl. (Atomic Number of C = 12, H = 1 and Cl = 35.5)
36. A mixture of helium and hydrogen is confined in a 12 dm3 flask at 30°C. If 0.2 mole of the helium is present, find out the partial pressure of each gas whereas the pressure of the mixture of gases is 2atm.
37. Calculate the radius by hydrogen atom by applying Bohr’s Theory. (h = 6.625 x 10-27 erg-sec, p = 3.14, m = 9.11 x 10-28gm, e = 4.8 x 10-10 esu)
38. Calculate the heat of formation of C2H2 from carbon and hydrogen from the following data:
(i)
2C + H2 ® C2H2
DHf = ?
(ii)
C + O2 ® CO2
DH = -94.05Kcal/mole
(iii)
H2 + ½O2 ® H2O
DH = -68.32Kcal/mole
(iv)
C2H2 + 5/2O2 ® 2CO2 + H2O
DH = -310Kcal/mole
39. Calculate the pH of a 2.356 x 10-3m HCl solution.
40. For the reaction N2 + 3H2 Û 2NH3. The equilibrium mixture contains 0.25 M nitrogen, 0.15M hydrogen gas at 25°C. Calculate the concentration of NH3 gas when Kc = 9.6. the volume of the container is 1dm3.
41. Determine the initial rate of the following reaction at 303°C in which its rate constant is 8.5 x 10-5 litre-mol-1 sec-1. Initial concentration of the reaction is 9.8 x 10-2 mole/litre. 2NO2 ® 2NO + O2.
Extra Numericals
1. 4.6gm of ethyl alcohol and 6.0gm of acetic acid kept at constant temperature until equilibrium was established. 2 gm of acid were present unused. Calculate Kc.
2. Kc for the dissociation of HI at 350°C is 0.01. If 0.2 mole of H2, 1.3 moles of I2 and 4 moles of HI are present. Predict the direction of reaction.
3. What is the solubility of PbCrO4 at 30°C when Ksp is 1.8 x 10-14.
4. 1.06m of an organic compound on combustion gave 1.49 gm of CO2 and 0.763gm H2O. It also has 23.73% N. Find its compercial formula.
5. 500 dm3 of moist O2 gas was collected over water at 27°C and 726torr pressure. Find the mass in gm. Of dry O2 gas at S.T.P. When the vapour pressure of water 27°C is 26 torr.
6. Atomic mass of phosphorus is 31. Calculate the mass of 45 atoms in a.m.u.
7. Methane burn in steam according the following reaction: CH4 + 2O2 ® CO2 + 2H2O. If 100 gm of each CH4 and O2 is taken, then what amount of CO2 liberated?
8. An organic compound containing C = 65.45%, H = 5.45% and O = 29.09%. If molecular weight of compound is 110, calculate molecular formula.
9. What mass of CO2 is produced by the complete combustion of 100g pentane. C5H12 + 3O2 ® 2CO2 + 2H2O.
10. One atom of an unknown element is found to have a mass of 67.8 x 10-23g. What is the atomic weight of the element?
11. The heat of combustion of glucose and alcohol is given below.
(i)
C6H12O6 + 6O2 ® 6CO2 + 6H2O
DH = -673Kcal/mole
(ii)
C2H5OH+ 3O2 ® 2CO2 + 3H2O
DH = -328Kcal/mole
Find DH for the fermentation given below:
C6H12O6 ® 2C2H5OH + 3CO2
12. At certain temperature, the equilibrium mixture contain 0.4 mole of H2, 0.4mole I2 and 1 mole of HI. If addition 2 mole of H2 are added. How many moles of HI will be present when the new equilibrium established. H2 + I2 ® 2HI.
13. A solution has pH of 8.4. Find concentration of H+ and OH-.
14. 180cm3 of a known gas diffuse in 15minutes, when 120 cm3 of SO2 diffuses in 20 minutes. What is the molecular mass of the unknown gas.
Chapter 1
Introduction to Fundamental Concepts
1. Calculate the moles of the following in 500gm, NH3, HCl, Na2CO3, H2SO4, MgBr2, CaCO3, Xe and C.
2. How many moles of Na are present in 5gm of Na?
3. Calculate the number of atoms in 12 gms of Mg.
4. 2gm diamond is studded in a ring. Diamond is a pure carbon. How many atoms of carbon are present in the ring?
5. Calculate the number of molecules in 9gms of H2O.
6. How many molecules are present in 25 gms of CaCO3?
7. Calculate the weight in gram of 3.01 x 1020 molecules of glucose (C6H12O6)
8. How many atoms of hydrogen are there in 2.57 x 10-6 gram of hydrogen?
9. A sample of oxygen contains 1.87 x 1027 atoms of oxygen. What would be the weight of the oxygen?
10. Find the weight of oxygen obtained from 49gm of KClO3.
2KClO3 ® 2KCl + 3O2
11. What weight of CO2 and CaO can be obtained by heating 12.5gm of Limestone (CaCO2)?
CaCO3 ® CaO + CO2
12. Calculate the weight of sodium chloride required to produce 142 gm of chlorine.
2NaCl ® 2Na + Cl2
13. Calculate the weight of carbon, required to produce 88gm of CO2.
C + O2 ® CO2
14. The action of CO on Fe2O3 can be represented by the following equation.
Fe2O3 + 3CO ® 2Fe + 3CO2
15. What weight of NH3 will be required to produce 100 gm of NO?
4NH3 + 5O2 ® 4NO + 6H2O
16. Find out the moles of CuSO4 which are obtained from 31.75 gm of Cu.
Cu + H2SO2 ® CuSO2 + H2
17. Calculate the number of N2 and H2 molecules, which are obtained from 8.5 gm of NH3.
N2 + 3H2 ® 2NH3
18. Find out the number of Cu and H2O molecules obtained from 7.95gm of CuO.
CuO + H2 ® Cu + H2O
19. 400gm of H2 was made to combine with 14200gm of Cl2. How much HCl will be produced?
20. 1kg of Limestone was heated 500gm of CaO was obtained. How much CO2 gas produced into air.
21. Find the weight of O2 obtained from 49 gm of KClO3.
2KClO3 ® 2KCl + 3O2
22. Chlorine is produced on the large scale by the electrolysis of NaCl aqueous solution. Chlorine the weight of NaCl required to produce 142 gm of Cl2.
2NaCl + 2H2O ® Cl2 + H2 + 2NaOH
23. How many grams of O2 are required to completely burn 18.0gm of C? How many grams of CO2 will be formed?
24. Calculate the weight of NH3, required to produce 100 gms of NO.
4NH3 + 5O2 ® 4NO + 6H2O
25. Find out the moles of H2 and N2 required producing 17gm of NH3.
26. Calculate the volume of H2 at S.T.P, which is obtained by the reaction of 120 gm Mg with MgSO4.
Mg + H2SO4 ® MgSO2 + H2
27. NH3 gas can be produced from ammonium chloride (NH4Cl) as follows:
CaO + 2NH4Cl ® CaCl2 + H2O + NH3
Calculate the volume of NH3 obtained at S.T.P by the reaction of 100 gm of NH4Cl.
28. 500gm of C2H4 on combustion in air gave CO2 and H2O. Calculate the volume of O2 and CO2 at S.T.P.
29. Find out the volume of O2, CO2 and SO2 gases at S.T.P react and obtained from 2 moles of CS2.
CS2 + 3O2 ® CO2 + 2SO2
30. Calculate the volume of CO2 gas at S.T.P obtained by the combustion of 20gm of CH4.
CH4 + 2O2 ® CO2 + 2H2O
31. Calculate the volume of O2 gas at S.T.P required to burn 600dm3 of H2S, also find the volume of SO2 gas produced at S.T.P.
32. Calculate the volume of O2 gas at S.T.P required to burn 50 gm of CH4.
33. What volume of H2 at S.T.P can be produced by the reaction of 6.54gm Zn with HCl?
Zn + 2HCl ® ZnCl2 + 2H2
34. Calculate the volume of O2 and H2 gases at S.T.P obtained from 9gm of H2O.
35. 0.264gm of Mg was burnt in pure O2. How much MgO will be formed?
2Mg + O2 ® 2MgO
36. How much H2 can be generated by passing 200gm of steam over hot iron.
4H2O + 3Fe ® Fe3O4 + 4H2
37. If 112dm3 of N2 react with 336 dm3 of H2, both at S.T.P. How many grams of NH3 would be obtained?
N2 + 3H2 ® 2NH3
38. An organic compound contains 12.8%C, 2.1% and 85.1% Br. If the mass of the compound is 188, find the molecular formula.
39. An organic compound contains 66.70%C, 7.41% H and 25.90% N2. The molecular mass of the compound is 108. Find out its molecular formula.
40. A compound contains 19.8%C, 2.5%H, 66.1%O and 11.6%N. Find out empirical formula of the compound.
41. 0.2475gm of a compound, containing C, H and O gave 0.4950gm CO2 and 0.2025gm H2O. If the molecular mass of the compound is 88. Find out the molecular formula.
42. An organic compound contains 32%C, 6.67%H, 18.66%N and 42.67%O. Its molecular mass is 75. Find out the molecular formula of the compound.
43. 1.367gm of a compound containing C, H and O on heating gave 3.002gm CO2 and 1.640gm H2O. Find out its molecular formula, when the molecular mass is 120.
44. A compound was found to contain 40%C and 6.7%H. Its molecular mass was 60. Find out its molecular formula.
45. An organic compound contains 75.2%C, 10.15%H and oxygen. Its molecular mass is 115. Find its molecular formula.
46. The empirical formula of a compound is CH2O. If the molecular mass 180. Find out the molecular formula.
47. An organic compound composed of C, H and O. On combustion of 0.94gm of this compound, 1.32gm CO2 and 0.568gm H2O were obtained. Its molecular mass is 180. Find its molecular formula.
48. An organic compound composed of C, H and O. 4.2gm of the compound on heating gave 6.21gm CO2 and 2.54gmH2O. Its molecular mass is 60. Find its molecular formula.
49. An organic compound contains C,H and 6.38gm of compound on combustion gave 9.06gm CO2 and 5.58gm H2O. Its molecular mass is 62. Find out its molecular formula.
50. 1gm of a hydrocarbon on combustion gave 3.03gm of CO2 and 1.55gm of H2O. If the molecular mass is 58, find its molecular formula.
51. 1.434gm of a compound on combustion gave 4.444gm CO2 and 2.0 gm H2O. Find out its empirical formula.
52. An organic compound composed of C, H and N. 0.225gm of compound on combustion gave 0.44gm CO2 and 0.315gm H2O. If the molecular mass of a compound is 90, find out its molecular formula.
53. An organic compound contains 40.68%C, 8.47%H, 23.73%N and 27.12%O. Find its empirical formula.
54. An organic compound composed of C, H and N. 0.419 gm of compound on combustion gave 0.88gm CO2 and 0.27gm H2O. Find out its empirical formula.
55. The analysis of a compound shows, C = 24.24%, H = 4.04% and Cl = 71.71%. If the molecular mass of the compound is 49.5, find its molecular formula.
56. An organic compound of molecular mass 90 has the empirical formula CH2O. What is its molecular formula?
57. The empirical formula of an organic compound is CH3NO2. If it’s molecular mass is 61. What is its molecular formula?
58. 0.638gm of an organic compound on combustion gave 0.594gm H2O and 1.452gm CO2.The compound is composed of C, H and O atoms. If the molecular mass is 116, find out its molecular formula.
59. The molecular formula of ethyl acetate is CH3COOC2H5. What is its empirical formula.
60. Find the empirical formulae of the following compounds from their percentage composition by mass:
· N = 26.17% H = 7.48% Cl = 66.35%
· Ca = 71.43% O = 28.57%
· Ag = 63.53% N = 8.23% O = 28.24%
· Na = 32.40% H = 45.07% Cl = 22.53%
61. A certain compound on analysis yielded 2.00gm C, 0.34gm H and 2.67gm O. If the relative molecular mass of the compound is 60, calculate its molecular formula.
62. What is the empirical formula of a compound, which contains 42.5% chlorine and 57.5 oxygen. If it’s formula mass is 167. What is its molecular formula?
63. What will be the weight of 5 moles of water in grams?
64. What is the mass of each of the following:
· 1.25 mole of NaCl
· 2.42 mole of NaNO3
· 1.5 mole of HCl
· 3.0 mole of NaOH
65. A piece of Aluminium metal weighs 70.0g. How many atoms are present in the piece.
66. How many atoms of carbon are present in 20-carat Diamond? (1 carat = 0.2g)
67. How many grams of oxygen have the same number of atoms as 16gm of sulphur?
68. A sample of oxygen gas at STP has a mass of 16gm. Calculate:
· The number of moles of oxygen
· The volume of the sample
· The number o molecules in the sample
69. Calculate the volume of CH4 gas at STP having a mass 32g.
70. What mass of zinc sulphate can be obtained from the reaction of 10.0gm of Zinc with an excess of dilute H2SO4?
Zn + H2SO4 ® ZnSO4 + H2*
71. Calculate what mass of sodium hydroxide you would need to neutralize a solution containing 7.3g hydrogen chloride by the reaction:
NaOH + HCl ® NaCl + H2O
72. Calculate how much sodium nitrate you need to give 126g of nitric acid by the reaction:
NaNO3 + H2SO4 ® HNO3 + NaHSO4
73. What volume of hydrogen at STP is evolved when 0.325g of zinc reacts will dilute hydrochloric acid.
Zn + 2HCl ® ZnCl2 + H2
74. What mass of oxygen is formed by the decomposition of a solution containing 120cm3 of H2O2 at STP?
2H2O2 ® 2H2O + O2
75. What is the mass of one molecule of water in grams?
76. 100cm3 of butane are burned in an excess of oxygen. Calculate:
· The volume of oxygen used
· The mass and volume of CO2 produced (assume all gases at STP)
2C4H10 + 13O2 ® 8CO2 + 10H2O
77. A cook is making a small cake. It needs 500cm3 at STP of CO2 to make the cake rise. The cook decides to add baking powder, which contains sodium bicarbonate. This generates CO2 by thermal decomposition.
2NaHCO3 ® CO2 + Na2CO3 + H2O
What mass of baking powder must the cook add to cake mixture?
78. What volume of ammonia at STP can be obtained by heating 0.25 mole of ammonium sulphate with calcium hydroxide?
(NH4)SO4 + Ca(OH)2 ® 2NH3 + CaSO4 + 2H2O
79. How many grams of SO2 are produced when 100g of H2S is reacted with 50g of oxygen.
2H2S + 3O2 ® 2H2O + 2SO2
80. How many grams of chlorobenzene will be produced when 100gm of each reactant is reacted?
C6H6 + Cl2 ® C6H5Cl + HCl
81. A car releases about 5g of NO into the air for each mile driven. How many molecules of NO are emitted per mile?
82. Simplify according to the rule of significant figures.
· 2.60 x 3.05
· 0.009 ¸ 0.3
·
·
Chapter 2
The Three States of Matter
1. 540cm3 of N2 at 400mm pressure are compressed to 300cm3 without changing the temperature. What will be the pressure of the gas?
2. A gas occupies 6dm3 at 1atm pressure keeping the temperature constant. If the pressure reduces to 600mm, what volume does the gas occupy?
3. At a certain temperature and 800mm pressure, the volume of H2 is 700cm3. If the pressure is increased to 1000mm at the same temperature, find the new volume of the gas.
4. 150ml of a gas at 27°C is heated to 77°C at constant pressure. Find the new volume of the gas.
5. 300ml of N2 are at 50° and the pressure is kept constant. If the temperature is doubled, what will be the volume of the gas?
6. A gas measures 5dm3 at 5°C under 0.5atm pressure. Calculate its volume at 25° and 5000mm pressure.
7. 2060ml of a gas is at 7°C and 860mm pressure. Find its volume at S.T,P.
8. 350ml of H2 was collected over water at 26°C. The pressure of the gas was 900mm. What volume will dry gas have at 30°C and 750mm pressure? The vapour pressure at 26°C is 25mm.
9. The volume of oxygen collected over water at 20°C and 1200mm pressure, is 200cm3. If aqueous at 20°C is 17.4mm, what will be the volume of the gas under S.T.P.
10. A 20dm3 flask contains H2 at 22°C under pressure of 1.2 atm. How many moles of H2 are present.
11. A gaseous mixture is at the pressure of 3000mm. The mixture contains 6 moles of N2, 0.5mole of CO2 and 2.5 moles of O2. Find the partial pressure of each gas.
12. A 5dm3 vessel contains 1.2 moles of H2 and 0.8 mole of N2 at 27°C. Find the total pressure of the mixture.
13. Composition of a sample of air by volume is, N2 = 76%, O2 = 20%, H2O = 2.5%, CO2 = 1.4% and He = 0.1%. If the pressure of the air is 760 mm, Calculate the partial pressure of these gases.
14. A 10dm3 container contains a mixture of He and Ne gases at 17°C. There are two moles of He gas and 3 moles of Ne gas. What is the partial pressure of the gases?
15. 10gm of H2, 96gm of O2 and 196gm of N2 are mixed together. The partial pressure of H2 is 0.6 atm. What is the partial pressure of O2 and N2?
16. A cylinder contains 1 mole of H2, 3 mole of He and 6 moles of N2. The total pressure in the cylinder is 15 atm. Calculate the partial pressure of H2, He and N2.
Chapter 5
Energetics of Chemical Reaction
1. Calculate the heat of formation of Acetic Acid from the following data:
(i)
2C + 2H2+ O2 ® CH3COOH
DHf = ?
(ii)
C + O2 ® CO2
DH = -394KJ/mole
(iii)
H2 + ½O2 ® H2O
DH = -286 KJ/mole
(iv)
CH3COOH + 2O2 ® 2CO2 + 2H2O
DH = -870KJ/mole
2. Calculate the heat of formation of Ethane from the following data:
(i)
2C + 3H2 ® C2H6
DHf = ?
(ii)
C + O2 ® CO2
DH = -394KJ/mole
(iii)
H2 + ½O2 ® H2O
DH = -286 KJ/mole
(iv)
C2H6 + 7/2O2 ® 2CO2 + 3H2O
DH = -1560KJ/mole
(v)
C2H5OH + 3O2 ® 2CO2 + 3H2O
DH = -327 KJ/mole
3. Calculate the heat of formation of Methane from the following data:
(i)
C + 2H2 ® CH4
DHf = ?
(ii)
C + O2 ® CO2
DH = -394KJ/mole
(iii)
H2 + ½O2 ® H2O
DH = -286 KJ/mole
(iv)
CH4 + 2O2 ® CO2 + 2H2O
DH = -890.3KJ/mole
4. Calculate the heat of formation of Ethyl Alcohol from the following data:
(i)
2C + 3H2 ½ O2® C2H5OH
DHf = ?
(ii)
C + O2 ® CO2
DH = -394KJ/mole
(iii)
H2 + ½O2 ® H2O
DH = -286 KJ/mole
(iv)
C2H5OH+ 3O2 ® 2CO2 + 3H2O
DH = -1369KJ/mole
5. Calculate the heat of formation of Ethane from the following data:
(i)
C2H6 + 7/2O2 ® 2CO2 + 3H2O
DHf = ?
(ii)
C + O2 ® CO2
DH = -394KJ/mole
(iii)
H2 + ½O2 ® H2O
DH = -286 KJ/mole
(iv)
C2H6 ® 2C + 3H2
DH = -84.68KJ/mole
6. Calculate the heat of formation of Methane from the following data:
(i)
C + 2H2 ® CH4
DHf = ?
(ii)
C + O2 ® CO2
DH = -94.1cal
(iii)
H2 + ½O2 ® H2O
DH = -68.3 cal
(iv)
CH4 + 2O2 ® CO2 + 2H2O
DH = -212.8 cal
7. Calculate the heat of formation of Ethene from the following data:
(i)
2C + 2H2 ® C2H4
DHf = ?
(ii)
C + O2 ® CO2
DH = -97kcal
(iii)
H2 + ½O2 ® H2O
DH = -65 kcal
(iv)
C2H4 + 3O2® 2CO2 + 2H2O
DH = 340 kcal
8. Calculate the heat of formation from the following data:
(i)
2C + 3H2 +1/2O2 ® C2H5O
DHf = ?
(ii)
C + O2 ® CO2
DH = -94.2Kcal/mole
(iii)
H2 + ½O2 ® H2O
DH = -68.5 Kcal/mole
9. Calculate the heat of formation of from the following data:
(i)
C + 2H2 + O2® CH3OH
DHf = ?
(ii)
C + O2 ® CO2
DH = -94.2Kcal/mole
(iii)
H2 + ½O2 ® H2O
DH = -68.32 Kcal/mole
(iv)
CH3OH + O2 ® CO2 + 2H2O
DH = -347.6Kcal/mole
10. Calculate the heat of formation of from the following data:
(i)
3C + 4H2 ® C3H8
DHf = ?
(ii)
C + O2 ® CO2
DH = -94.1Kcal/mole
(iii)
H2 + ½O2 ® H2O
DH = -68.3 Kcal/mole
(iv)
C3H8 + 5O2 ® 3CO2 + 4H2O
DH = -530.7Kcal/mole
11. Calculate the heat of formation of from the following data:
(i)
H2 + O2® H2O2
DHf = ?
(ii)
H2 + ½O2 ® H2O
DH = -68.32Kcal
(iii)
H2O + ½ O2 ® H2O2
DH = -23.48Kcal
12. Given:
(i)
NH3 + HCl ® NH4Cl
DH1 = 42.100Kcal
(ii)
H2O + ½ O2 ® H2O2
DH2 = 3.900cal
Find DH for the reaction,
NH3 + HCl ® NH4Cl
DHf = ?
Chapter 6
Chemical Equilibrium
1. 1.5 moles of acetic acid and 1.5 moles of ethyl alcohol were reacted at a certain temperature. At equilibrium, 1 mole of ethyl acetate was present in 1 litre of the equilibrium mixture. Calculate the equilibrium constant Kc.
CH3COOH + C2H5OH Û CH3COOC2H5 + H2O
2. 6.0 gm of hydrogen and 1016gm of iodine were heated in a sealed tube at a temperature, at which Kc is 50. The volume of the tube is 1 dm3. Calculate the concentration of HI.
H2 + I2 Û 2HI
3. At a certain temperature, an equilibrium mixture contains 0.4 mole H2, 0.4 mole I2 and 1 mole of HI. The volume of the reacting vessel is 4 dm3. Find out the equilibrium constant kc.
H2 + I2 Û 2HI
4. 3 moles of A and 2 moles of B are mixed in a 4dm3 flask, at a certain temperature. The following reaction occurs.
3A + 2B Û 4C
At equilibrium the flask contains 1 mole of B. Find the equilibrium constant kc.
5. At a certain temperature, 0.205 mole of H2 and 0.319 mole of I2 were reacted. The equilibrium mixture contains 0.314 mole of I2. Calculate the kc.
H2 + I2 Û 2HI
6. The kc for the reaction A + B Û C + D is 1/3. How many moles of A must be mixed with 3 moles of B to yield at equilibrium, 2 moles of C and D each. The volume of the vessel is 2 litre.
7. At a certain temperature the equilibrium mixture for the reaction A + B Û 2C, contains 2 moles A, 3 moles of B and 5 moles of C. Find the Kc for the reaction.
8. For the reaction 2A Û B + C, equilibrium constant kc is 1. If we start with 6 moles of A, how many moles of B will be formed.
9. 20 moles of SO2 and 10 moles of O2 are taken in a 20 litre flask. If at equilibrium 5 moles of SO3 are formed, Calculate kc.
2SO2 + O2 Û 2SO3
10. A quantity of PCl5 was heated in a 12 dm3 vessel at 250°C.
PCl5 Û PCl3 + Cl2
11. 2 moles of HI was introduced in a vessel held at constant temperature. When equilibrium was reached, it was found that 0.1 mole of I2 have been formed. Calculate the equilibrium constant.
H2 + I2 Û 2HI
12. When 1 mole of pure C2H5OH is mixed with 1 mole of CH3COOH at room temperature, the equilibrium mixture contains 2/3 moles of ester and water each.
· What will be the kc?
· How many moles of ester are formed at equilibrium when 3 moles of C2H5OH are mixed with 1 mole of CH3COOH?
CH3COOH + C2H5OH Û CH3COOC2H5 + H2O
13. PCl5 Û PCl3 + Cl2. Calculate the number of moles of Cl2 produced at equilibrium when 1 mole of PCl5 is heated at 250°C in a vessel having capacity of 10dm3. At 250°C, Kc is 0.041.
14. When 2.94 moles of iodine and 8.1 moles of Hydrogen were mixed and heated at 444°C and at constant volume, until the equilibrium was established. 5.64 moles of HI were formed. Calculate the value of kc.
H2 + I2 Û 2HI
15. What is the solubility of lead chromate in moles/dm3 at 25°C. The solubility product is 1.8 x 10-14.
PbCrO4 Û Pb++ + CrO4--
16. The solubility of Mg(OH)2 at 25°C is 0.00764 gm/dm3. What is the solubility product of Mg(OH)2?
Mg(OH)2 Û Mg++ + 2OH-
17. Find the solubility of AgCl in gm/dm3, when the solubility product is 1.25 x 10-10.
18. Calculate the solubility product of BaSO4. The solubility of the salt is 1.0 x 10-5 moles/dm3.
19. Calculate the solubility product of BaSO4 is 9.0 x 10-3 gm/dm3. Find its solubility product.
20. Predict whether there will be any precipitate formation by mixing 30cm3 of 0.01M NaCl with 60cm3 of 0.01M AgNO3 solution. Ksp of AgCl is 1.5 x 10-10.
21. A saturated solution of calcium fluoride was found to contain 0.0168 gm/dm3 of solute at 25°C. Calculate the ksp for CaF2.
22. A saturated solution of BaF2 at 25°C is 0.006M. Calculate Ksp of the salt.
Chemistry - Fill in the Blanks
Five Year Papers
1. The property of a crystal, which is different in different directions, is called __________.
2. 0.00051 contains __________ significant figures.
3. The oxidation number of oxygen in OF2 is __________.
4. The volume of 1 gm of hydrogen gas at S.T.P is __________.
5. The oxidation number Mn in KMnO4 is __________.
6. The product of ionic concentration in a saturated solution is called __________.
7. 16 gm of oxygen at S.T.P occupies a volume of __________ dm3.
8. The shape of the orbital for which l = 0 is __________.
9. The radius of Cl-1 is __________ than the radius of Cl0.
10. Sp2 hybridization is also known as __________.
11. The value of 1 Debye is __________.
12. The reactions catalyzed by sunlight are called __________.
13. The blue colour of CuSO4 is due to the presence of __________.
14. The force of attraction between the liquid molecules and the surface of container is called __________.
15. The heat of neutralization of a strong acid and a strong base is __________.
16. C º C triple bond is __________. C = C double bond length.
17. The ions having the same electronic configuration are called iso electronic.
18. On heating, if a solid changes directly into vapours without changing into the liquid state, the phenomenon is called __________.
19. Each orbital in an atom can be completely described by __________.
20. In a molecule of alkene, __________ restricts the rotation of the group of atoms at either end of the molecule.
21. Density, refractive index and vapour pressure are __________ properties.
22. The addition of HCl to H2 solution __________ the ionization of H2S.
23. The reaction of cation or anion (or both) with water so as to change its __________ is known as Hydrolysis.
24. A reaction with higher activation energy will start at __________ temperature.
25. 6.02 x 1023 has __________ significant figures.
26. The internal resistance in the flow of liquid is called __________.
27. A catalyst increases the velocity of a reaction but decreases the __________.
Chapter 1
Introduction to Fundamental Concepts
1. 1 mole of a gas at S.T.P occupies a volume of __________.
2. A gas occupying a volume of 22.4 dm3 at S.T.P contains __________ molecules.
3. A formula, which gives the relative number of atoms in the molecule of a compound, is called __________.
4. A formula which gives the actual number of all kinds of atoms present in the molecule of compound is termed as __________.
5. The chemical formula that not only gives the actual number of atoms but also shows the arrangement of different atoms present in the molecule is called __________.
6. Atomic weight or molecular weight expressed in grams is known as __________.
7. 2 moles of H2O contain __________ grams and __________ number of molecules.
8. Any thing that occupies space and has __________ is called matter.
9. Volume of one __________ mole of a gas at S.T.P is 22.4 cubic feet.
10. A ton mole of iron is equal to __________ tons.
11. The force with which the earth attracts a body is called the __________ of the body.
12. A pure substance contains __________ kind of molecules.
13. The smallest indivisible particle of matter is called __________.
14. The atomic number is equal to the number of __________ in nucleus.
15. The atomic mass is the total number of protons and __________ in an atom of the element.
16. The average weight of atoms of an element as compared to the weight of one atom of __________ is called the atomic mass.
17. 1.0007 contains __________ significant figures.
18. The figure 24.75 will be rounded off to __________.
19. __________ means that the readings and measurements obtained in different experiments are very close to each other.
20. __________ means that the results obtained in different experiments are very close to the accepted values.
21. The degree of a measured quantity __________ with increasing number of significant figures in it.
22. The atomic mass of sodium is __________.
23. The symbolic representation of a molecule of a compound is called __________.
24. Molecular formula of CHCl3 and its Empirical formula is __________.
25. Molecular formula of benzene is C6H6 and its empirical formula is __________.
26. 58.5 is the __________ of NaCl.
27. 4.5 gms of nitrogen will have __________ molecules.
28. 28 gms of nitrogen will have __________ molecules.
29. 2 moles of SO2 is equal to __________ gms.
30. 1000 gms of H2O is equal to __________ moles.
31. The reactions, which proceed in both directions, are called __________.
32. The reactions, which proceed in forward directions only, are called __________ reactions.
33. The __________ reactions are completed after some time.
34. 0.0006 has __________ significant figures
35. 7.40 x 108 has __________ significant figures.
36. 7 x 108 has __________ significant figures.
37. Usually Molecular formula is simple multiple of the __________.
38. 0.1 mole of H2O contains __________ molecules of H2O.
39. Mass of 3.01 x 1022 molecules of CO2 is __________.
40. __________ is the branch of science which deals with the properties, composition and structure of matter.
41. None zero digits are all __________.
42. The integer part of logarithm is called __________.
43. The decimal fraction of logarithm is called __________.
44. __________ is the amount of substance, which contains as many number of particles as there are in 12 gms of Carbon.
45. 6.02 x 1023 is called the __________.
46. The accuracy of measurement depends on the number of __________.
47. __________ is the branch of chemistry that deals with quantitative relationships among the substances undergoing chemical changes.
48. The sum of atomic weights of all the elements present in molecular formula is called the __________.
49. __________ is the sum of atomic weights of the elements represented by the Empirical formula of the compound.
50. Very small and very large quantities are expressed in terms of __________.
51. In rounding off __________ figure is dropped.
52. Mole is the quantity, which has __________ particle of the substance.
53. For three significant figures, 25.55 is rounded off to __________.
54. The S.I unit of a mass is __________.
55. Mass of 6.02 x 1023 molecules of NaCl is __________ gm.
56. 1 mole of NaOH is __________ gm of NaOH.
57. Formula weight is used for __________ substances.
58. The word S.I stands for __________.
59. 4.5 gms of water will have __________ molecules.
60. 0.0087 has __________ significant figure.
Chapter 2
The Three States of Matter
1. The intermixing of gases or liquids in a container irrespective of their densities, is called __________.
2. At constant temperature, if the pressure of a given mass of a gas is decreased, its volume will __________.
3. A volume of __________ dm3 will hold 128 gms of SO2.
4. At constant temperature of a given mass of a gas, the product of its __________ and __________ is constant.
5. The rates of diffusion of gases are __________ proportional to the square root of their densities.
6. Gases deviate from ideal behaviour more markedly at high __________.
7. Liquid diffuse __________ than gases.
8. An imaginary line passing through the centre of a crystal is called __________.
9. The temperature at which more than one crystalline forms of a substance coexist in equilibrium is called __________.
10. Two or more substances crystallizing in the same form is called __________.
11. The existence of solid substances in more than one crystalline form is known as __________.
12. Rate of diffusion of gases is __________ as compared to liquids.
13. Boiling point of a liquid __________ with the pressure.
14. Mercury in a glass tube forms __________ curvature.
15. Gases can be compressed to __________ extent.
16. Viscosity of a liquid __________ with the increase of temperature.
17. Surface tension of water __________ by adding soap solution into it.
18. The internal resistance to the flow of a liquid is called __________.
19. The rise or the fall of a liquid in a capillary tube is called __________.
20. Matter exists in __________ states.
21. The freezing point of water in Fahrenheit scale is __________.
22. Boiling point of water is __________ °K.
23. SI unit for measurement of pressure is __________.
24. The value of gas law constant R = __________ dm3 atm/°K/mole.
25. The absolute Zero is equal to __________.
26. If P is plotted against 1/V at constant temperature a __________ is obtained.
27. Gases __________ in heating.
28. The pressure of air __________ at higher altitude.
29. Standard temperature means __________.
30. Standard pressure means __________.
31. Cooling is caused by __________ of gases.
32. Rate of diffusion of O2 is __________ times more than H2.
33. H2O has __________ viscosity than CH3OH.
34. Mercury does not wet the glass surface due to its higher __________.
35. Surface tension of mercury is __________ than water.
36. Viscosity can be easily measured by an instrument called __________.
37. The pressure exerted by the vapours when these vapours are in equilibrium with the liquid is called __________.
38. Vapour pressure __________ at high temperature.
39. Boyle’s Law and Charles Law can be combined into the mathematical expression __________.
40. Equal volumes of all gases at the same temperature and pressure contain __________ number of molecules.
41. The average Kinetic energy of a gas is proportional to its __________ temperature.
42. Kinetic equation may be mathematically written as __________.
43. The temperature at which two crystalline forms of a substance can coexist in equilibrium is called __________.
44. Lighter gases diffuse __________ than heavier gases.
45. Rain drops are __________ in shape.
46. Due to surface tension, the surface area of the liquid is __________.
47. Water __________ in the capillary tube.
48. Viscosity of a solution at 10°C is __________ than at 20°C.
49. Shape of NaCl crystal is __________.
50. Gases intermix to form a mixture.
51. Pressure of a dry gas is __________ than the pressure of a moist gas.
52. 22.4 dm3 of nitrogen at S.T.P will weigh equal to __________ gm.
53. 1 mole of any gas at S.T.P is equal to __________ dm3.
54. At -273°C, volume of all gases becomes __________.
55. The gases, which strictly follows the gas Laws are called __________ gas.
56. __________ is the property that determines the direction of flow of heat.
57. __________ is defined as force per unit area.
58. __________ viscosity is defined as the viscosity of a liquid as compared to the water.
Chapter 3
Structure of Atom
1. The maximum number of electrons in 2p orbital is __________.
2. 3d orbital has __________ energy than 4s orbital.
3. __________ rays are non-material in nature.
4. Charge to mass ratio of cathode rays resembles to that of __________.
5. __________ rays are most penetrating.
6. Neutrons have mass equal to that of __________.
7. Energy is __________ when an electron jumps from higher to lower orbit.
8. Second Ionization Potential has __________ value than the First Ionization Potential.
9. Electronegativity __________ from left to right in a period of Periodic Table.
10. __________ was discovered during the course of Artificial Radioactivity.
11. The velocity of alpha rays is nearly __________ of velocity of light.
12. Natural Radioactivity is confined in __________ elements.
13. The isotopes of an element differ in their __________.
14. Two electrons with the __________ spin, can never occupy the same atomic orbital.
15. ‘Al’ has electronic configuration, 1s2, 2s2, __________.
16. In a group of Periodic Table, the ionization potential __________ from top to bottom as the size of atom increases.
17. Ionization potential values __________ from left to right in a period.
18. The energy required to remove the most loosely bond electron from an atom in gaseous state is called __________.
19. The SI unit of Ionization Potential is __________.
20. An atom of sodium possesses 11 protons and __________ neutrons.
21. The particles of Cathode rays possess __________ charge.
22. The negatively charged particles found in Cathode rays are named as __________.
23. Positive rays are emitted from __________.
24. __________ rays are also known as Canal rays.
25. __________ consists of helium ions and are doubly positively charged.
26. __________ rays consists of negatively charged particles.
27. __________ rays are light waves of very short wavelength.
28. The phenomenon in which a stable element is made radioactive by artificial disintegration is called __________.
29. The electron move around the nucleus in different circular paths called __________.
30. The maximum number of electron in a shell is determined by the formula __________.
31. A particle whose mass is equal to that of electron but carries a positive charge is called __________.
32. 2p electrons are __________ in energy that 2s electrons in the same atom.
33. Number of protons of an element also indicates its __________.
34. According to __________ Principle electrons are fed in the order of increasing orbital energy.
35. According to __________ electrons are distributed among the orbitals of a sub shell to give maximum number of unpaired electron and have same spin.
36. The specific way in which the orbitals of an atom are occupied by electrons is called __________.
37. __________ rays are stream of doubly positively charged particles.
38. Electron in the outer most shell of an atom is called __________.
39. Protons are found in the __________ of an atom and bear __________ charge.
40. The atomic number of an atom is the sum of __________ inside the nucleus.
41. __________ limits the number of electron to different shell or orbits.
42. Sir William Crookes in 1878, discovered that the cathode in high vacuum tube emit radiations what he called __________.
43. X-rays were discovered in 1895 by __________.
44. The discovery of proton was done in 1886 by __________.
45. Neutrons were discovered by __________ in 1932 by the bombardment of beryllium with alpha particles.
46. Each atom has a __________, which contains all the positive charge and practically all the mass of atom.
47. Complete the reaction: 4Be9 + 2H4 ® __________ + __________.
48. __________ have higher ionization power as compared to b-rays.
49. No dark spaces between the colours are present in __________.
50. The symbol e+ represents __________.
51. p-orbitals are __________ shaped.
52. The energy released when an electron is added to an atom in the gaseous state is called __________.
53. The power of an atom to attract a shared pair of electrons towards itself is called __________.
54. Fluorine is __________ electronegative than chlorine.
55. Lyman series of spherical lines appear in the __________ portion of spectrum.
56. The electrons with __________ spin occupy the same orbital.
57. 3d orbital has __________ energy than 4s orbital.
58. Energy and frequency are __________ proportional to each other.
59. Ionic radii of cations are __________ than the atoms from which they are formed.
60. Ionic radii of anions are __________ than the atoms from which they are formed.
Chapter 4
Chemical Bonding
1. A bond formed due to transference of electron is called __________.
2. A bond formed due to sharing of electron is called __________.
3. Sigma bond is __________ than pi bond.
4. The shape of methane molecule is __________.
5. One s and 3p orbitals overlap to produce four __________ hybrid orbitals.
6. Ethene, C2H4 is an example of __________ hybridization.
7. Water molecule has __________ structure.
8. Water molecules are inter-linked with one another due to __________.
9. Polarity of the molecule is due to the difference of __________ between the two bonded atoms.
10. A chemical bond formed between to different atoms by mutual sharing of electron is termed as __________.
11. A chemical bond formed between two similar atoms by mutual sharing of electrons is known as __________.
12. The difference between the Electronegativity values of the two atoms forming covalent bond must be __________ than 1.7.
13. When two orbitals of different atoms by hybridize with each other having their axes in the same straight lines, the bond formed is termed as __________.
14. __________ bond is formed when p-orbitals of the two atoms with their axes parallel to each other overlap with each other.
15. Melting and boiling point of ionic compounds are usually __________ than that of covalent compounds.
16. Non polar compounds are usually __________ in non polar solvent.
17. The nitrogen in NH3 is __________ hybridized.
18. A hybrid orbital is called __________ orbital.
19. Since dipole moment of CS2 is zero, it is a __________ molecule.
20. A bond formed due to the electrostatic forces of attraction between the oppositely charged ions is called __________ bond.
21. The ionic bond is formed between the atoms with low ionization potential and high __________.
22. A bond formed by the sharing of an electron pair contributed by one atom only is called a __________ bond.
23. A co-ordinate covalent bond is also known as __________ bond.
24. Polar covalent bond is __________ than a non polar covalent bond.
25. H-F bond is __________ than H-Br bond.
26. The SI unit of dipole moment is __________.
27. Commonly used unit of dipole moment is __________.
28. Dipole moment of non-polar compound is __________ D.
29. The reactions of ionic compounds are usually very __________.
30. Covalent compounds are generally __________ in nature.
31. Ionic compounds are generally __________ in nature.
32. A covalent bond is represented by a __________.
33. A co-ordinate covalent bond is represented by an __________.
34. The covalent bond between H-F is called __________ covalent bond.
35. The power of an atom to attract a shared pair of electron itself is called __________ of that atom.
36. m = d x e represents __________.
37. CO2 and SO2 molecules have __________ polar bonds.
38. NH3 molecule has __________ polar bonds.
39. A double bond has __________ bond energy than a single bond.
40. An orbital which surrounds a single nucleus is called __________ orbital.
41. An orbital which surrounds two or more atomic nuclei is called __________ orbital.
42. A molecular orbital, which is of lower energy than the atomic orbitals from which it is derived, is known as __________ orbital.
43. A molecular orbital, which has higher energy than the atomic orbitals from which it is derived, is known as __________ orbital.
44. Orbitals formed after hybridization are called __________ orbitals.
45. Bond angle in Sp3 hybridization is of __________.
46. Bond angle in Sp2 hybridization is of __________.
47. Bond angle in Sp hybridization is of __________.
48. Sp3 hybridization is also known as __________.
49. Sp2 hybridization is also known as __________.
50. Sp hybridization is also known as __________.
51. A pair of electrons residing on the central atom and which is not used in bonding is called a __________.
52. The sum of total number of electron pairs (bonding and lone pairs) is called __________ number.
53. __________ bond is usually expressed by dotted line.
54. Water molecule has dipole moment because of its __________ structure.
55. CO2 is non polar because of its __________ structure.
56. Overlapping in __________ bond is perfect.
57. Overlapping in __________ bond is not perfect.
58. H-H bond is __________ than H-Cl bond.
59. __________ hybrid orbitals are not co-planar.
60. Covalent bond is Cl2 molecule is __________.
Chapter 5
Energetics of Chemical Reaction
1. The branch of Chemistry, which deals with the heat changes that take place during chemical reaction, is called __________.
2. The branch of science which deals with energy changes accompanying physical and chemical transformation is called __________.
3. The amount of heat evolved or absorbed in a chemical reaction is called __________.
4. Such reactions in which heat is evolved are called __________ reactions.
5. Such reactions in which heat is absorbed are called __________ reactions.
6. In exothermic reactions, heat evolved is given by __________ sign of DH.
7. In endothermic reactions heat absorbed is given by __________ sign of DH.
8. The total heat change in a reaction is the same whether it takes place in one or several steps.
9. The first law of thermodynamics is also known as __________.
10. The part of universe under observation is called __________.
11. The system plus its surrounding is called __________.
12. Such properties, which give description of a system at a particular moment, is called __________.
13. The term E + PV is called __________.
14. DH represents change in __________.
15. The temperature of water is raised up when sulphuric acid is added to it. This is an __________ reaction.
16. The characteristic properties of a system which is independent of amount of material concerned is called __________ properties.
17. The characteristic properties of a system which depend on amount of substance present in it is called __________ properties.
18. Density, pressure and temperature are the examples of __________ properties.
19. Mole numbers and enthalpy are the examples of __________ properties.
20. A system, which exchange both energy and matter with its surrounding, is called __________ system.
21. A system, which only exchange energy with the surrounding but not matter is, called __________ system.
22. A system which neither exchange energy nor matter with its surrounding is called __________ system.
23. A system is __________ if it contains only one phase.
24. A system is __________ if it contains more than one phase.
25. 1 kilojoule is equal to __________ joules.
26. 1 Calorie is equal to __________ joules.
27. 1 kilo calorie is equal to __________ joules.
28. The work done (w) is mathematically denoted by __________.
29. The change in enthalpy is denoted by __________.
30. __________ law is used in calculating heat of reaction.
31. __________ is defined as the change in enthalpy when one gram mole of a compound is produced from its element.
32. Heat of formation is denoted by __________.
33. When the work is done on the system by the surrounding the sign of work done (w) is __________.
34. When the work is done by the system on surrounding the sign of work done is __________.
35. First law of Thermodynamics is mathematically represented as __________.
36. Standard enthalpies are measured at __________.
37. Hess’s Law is employed to calculate __________ of a chemical reaction.
38. Heat absorbed by the system at constant volume is completely utilize to increase the __________ of the system.
39. Heat change at constant pressure from initial to final state of the system is simply equal to the __________.
40. SI unit of measurement of heat change is __________.
Chapter 6
Chemical Equilibrium
1. The reactions, which proceed in both the directions, are called __________ reactions.
2. The reactions, which proceed to one direction only, are called __________ reactions.
3. Reversible reactions are __________ completed.
4. Irreversible reactions are __________ after some time.
5. A reversible reaction is said to be in __________ when the rate of forward reaction becomes equal to the rate of backward reaction.
6. The concentrations of reactants and products are __________ at equilibrium point.
7. The value of Kc depends upon the __________ of the reactants.
8. A increase of the value of Kc tends to move the reaction to the __________ direction.
9. A decrease of the value of Kc tends to move the reaction to the __________ direction.
10. An increase in the concentration of the reactants will move the reaction to the __________ direction.
11. A decrease in the concentration of the reactants will move the reaction to the __________ direction.
12. Equilibrium constant is denoted by __________.
13. When the equilibrium constant value is very __________, we can conclude that the forward reaction is almost completed.
14. When equilibrium constant value is very __________ we can conclude that forward reaction will occur to very little extent.
15. According to __________ principle, if system in equilibrium is subjected to a stress, the equilibrium shifts in a direction to minimize or undo the effect of the stress.
16. In exothermic reaction, the __________ of temperature favour the forward rate of reaction.
17. In endothermic reactions, the __________ of temperature favour the forward rate of reaction.
18. A __________ is a substance which effects the rate of reaction but remains unaltered at the end of the reaction.
19. A catalyst increases the velocity of the reaction by decreasing the __________.
20. The suppression of degree of ionization of a sparingly soluble weak electrolyte by the addition of a strong electrolyte containing an ion in common is called __________.
21. __________ is purified in industries by Common Ion Effect.
22. A reaction moves to the left when the concentrations of the products are __________.
23. A reaction moves to the right when the concentrations of the products are __________.
24. Increase in pressure will move the reaction in the direction of __________ volume.
25. Decrease in pressure will move the reaction in the direction of __________ volume.
26. An increase of temperature favours the formation of products in case of __________ reaction.
27. A decrease of temperature fovours the formation of products in case of __________ reaction.
28. Heating moves an endothermic reaction to the __________.
29. Cooling move an exothermic reaction to the __________.
30. The product of ionic concentration in a saturated solution is called __________ constant.
31. When HCl is added to NaCl, the concentration of __________ ion is increased.
32. Chemical reaction involving the substances in more than one phases are called __________.
33. The formation of NH3 is exothermic process hence __________ temperature will favour the formation of NH3.
34. The formation of NO from N2 and O2 is endothermic process hence __________ temperature will favour the formation of NO.
35. Chemical Equilibrium is __________ equilibrium.
36. Molar concentration is also called __________.
37. The rate at which a substance takes part in a chemical reaction depends upon its __________.
38. __________ principle is applied to all reversible reaction.
39. A common ion __________ the solubility of the salt.
40. Number of moles present per dm3 of a substance is called __________.
Chapter 7
Solutions and Electrolytes
1. A mixture of two or more substances, which are homogeneously mixed, is called a __________.
2. __________ is defined as the amount of solute dissolved in a given amount of solvent.
3. A solution is composed of two components __________ and __________.
4. A solution containing one mole of solute per dm3 of solution is called one __________ solution.
5. Molarity is denoted by __________.
6. 1M solution of NaOH contains __________ gms of it dissolved per dm3 of solution.
7. A solution containing one mole of solute dissolved by per kg of solvent is called __________ solution.
8. Molality is denoted by __________.
9. 1M solution of H2SO4 contains __________ gms of it per kg of solvent.
10. The process in which ions are surrounded by water molecules is called __________.
11. The water molecules attached with the hydrated substance are called __________.
12. Hydrated copper sulphate evolves __________ water molecules on heating.
13. The interaction between salt and water to produce acids and bases is called __________.
14. The products of ionic concentration in a saturated solution at a certain temperature are called the __________.
15. Solubility product constant expressed as __________.
16. The suppression of ionization by adding a common ion is called __________.
17. The process of dissociation of an electrolyte into ions is known as __________.
18. The chemical decomposition of a compound in a solution or in fused state brought about by a flow of electric current is known as __________.
19. Electrolysis is performed in an electrolytic cell, which is known as __________.
20. The positive electrode of a voltmeter is called __________ and negative as __________.
21. A solution, which tends to resist changes in pH is called a __________ solution.
22. A mixture of acetic acid and sodium acetate acts as a __________.
23. According to Sorenson __________ is defined as negative logarithm of the hydrogen ion concentration.
24. pH is mathematically expressed as __________.
25. The pH of a neutral solution is __________.
26. __________ substances have pH values lower than 7.
27. __________ solutions have pH values more than 7.
28. Oxidation is __________ of electron.
29. Reduction is the __________ of electron.
30. Such chemical reactions in which the oxidation number of atoms or ions is changed are called __________ reactions.
31. Oxidation number of a free element is __________.
32. Oxidation number of Oxygen in a compound is __________.
33. The sum of oxidation number of any formula of a compound is __________.
34. The oxidation number of any ion is equal to the __________ on the ion.
35. __________ is the reaction in which an acid reacts with a base to form salt and water.
36. __________ are organic compounds which change colour in accordance with the pH of the medium.
37. An indicator that changes from colourless to pink in the presence of an alkaline solution is called __________.
38. An indicator that changes from red to yellow in the presence of an alkaline solution is called __________.
39. Dissociation constant is denoted by __________.
40. According to Bronsted-Lowry Concept, __________ is the donor of proton and __________ is the acceptor of proton.
41. According to Arrhenius, acid is substance that produces __________ ions when dissolved in water.
42. According to Arrhenius, base is a substance that produces __________ ions when dissolved in water.
43. When ionic product is less than ksp, the solution will __________.
44. When ionic product is greater than ksp, the solution will __________.
45. The electrode at which oxidation takes place is called __________.
46. The electrode at which reduction takes place is called __________.
47. H3O+ ion is called __________ ion.
48. The logarithm of reciprocal of hydroxyl ion (OH)- is called __________.
49. Aqueous solution of NH4Cl is __________ while that of NaHCO3 is __________.
50. The ionic product of [H+] and [OH-] of pure water is __________.
51. An increase in the oxidation number of an element or ion during a chemical change is called __________.
52. A decrease in the oxidation number of an element or ion during a chemical change is called __________.
53. The degree of dissociation __________ with the increase in temperature.
54. The degree of dissociation __________ with the dilution of electrolytic solution.
55. A __________ consists of an electrode immersed in solution of its ion.
56. The potential difference between the electrode and the solution of its salt at equilibrium position is called __________ potential.
57. If the pH of a solution is 14, the solution is __________.
58. If the pH of a solution is 4, the solution is __________.
59. The oxidation number of Mn in KMnO4 is __________.
60. The oxidation number of Fe in FeCl3 is __________.
Chapter 8
Introduction to Chemical Kinetics
1. The branch of chemistry, which deals with the study of rates and mechanisms of chemical reactions, is known as __________.
2. Such reactions, which proceeds with very high velocities and are completed very quickly are called __________ reactions.
3. Such reactions, which take place very slowly, are called __________ reactions.
4. Reactions between silver nitrate and sodium chloride to form white precipitates of silver chloride are an example of __________ reaction.
5. Reactions of Organic compounds are slow and are called __________ reactions.
6. There are some reactions, which proceed slowly with a __________ speed.
7. The rate of __________ reaction can only be determined.
8. The amount of chemical change taking place in concentration of the per unit time is called __________ of reaction.
9. Rate of reaction is expressed in __________.
10. The rate of reaction between two specific interval of time is called __________.
11. The addition energy required to bring about a chemical reaction is called __________.
12. According to __________ theory for a chemical reaction to take place, the reacting molecules must come closed together.
13. The addition of __________ helps the reaction by lowering the energy of activation.
14. The rate of reaction __________ with the increase in concentration of the reacting molecules.
15. When the concentration of both the reacting molecules is double, the probability of collisions between them will be __________ times.
16. By __________ the surface area of the reactants, the rate of reaction is increased.
17. Rate of reaction generally __________ with the rise of temperature.
18. A __________ is a substance, which either accelerates or retards the rate of reaction without taking part in the reaction.
19. In the preparation of Oxygen from Potassium Chlorate, __________ is used as catalyst.
20. In the oxidation of SO2 to SO3 by the contact process for the manufacture of H2SO4 __________ is used as catalyst.
21. An unstable intermediate compound formed during a chemical reaction is called __________.
22. When a catalyst and the reactants are in the same phases, it is known as __________ catalyst.
23. When a catalyst and the reactants are in different phases, it is called __________.
24. When a catalyst increases the rate of reaction, it is called __________ catalyst.
25. When a catalyst retards the rate of reaction, it is called __________ catalyst.
26. A negative catalyst __________ the energy of activation, hence the rate of reaction is decreased.
27. The ratio between the rate of reaction and concentration of reactants is known as __________.
28. Velocity constant is independent of concentration but depends on __________.
29. Ionic reactions are __________ than molecular reactions.
30. The value of specific rates constant for a reaction __________ with time.
31. The sum of all exponents of concentration terms in the equation is called __________.
32. The sum of moles taking part in a chemical reaction is called __________ of the reactions.
Five Year Papers
1. The property of a crystal, which is different in different directions, is called __________.
2. 0.00051 contains __________ significant figures.
3. The oxidation number of oxygen in OF2 is __________.
4. The volume of 1 gm of hydrogen gas at S.T.P is __________.
5. The oxidation number Mn in KMnO4 is __________.
6. The product of ionic concentration in a saturated solution is called __________.
7. 16 gm of oxygen at S.T.P occupies a volume of __________ dm3.
8. The shape of the orbital for which l = 0 is __________.
9. The radius of Cl-1 is __________ than the radius of Cl0.
10. Sp2 hybridization is also known as __________.
11. The value of 1 Debye is __________.
12. The reactions catalyzed by sunlight are called __________.
13. The blue colour of CuSO4 is due to the presence of __________.
14. The force of attraction between the liquid molecules and the surface of container is called __________.
15. The heat of neutralization of a strong acid and a strong base is __________.
16. C º C triple bond is __________. C = C double bond length.
17. The ions having the same electronic configuration are called iso electronic.
18. On heating, if a solid changes directly into vapours without changing into the liquid state, the phenomenon is called __________.
19. Each orbital in an atom can be completely described by __________.
20. In a molecule of alkene, __________ restricts the rotation of the group of atoms at either end of the molecule.
21. Density, refractive index and vapour pressure are __________ properties.
22. The addition of HCl to H2 solution __________ the ionization of H2S.
23. The reaction of cation or anion (or both) with water so as to change its __________ is known as Hydrolysis.
24. A reaction with higher activation energy will start at __________ temperature.
25. 6.02 x 1023 has __________ significant figures.
26. The internal resistance in the flow of liquid is called __________.
27. A catalyst increases the velocity of a reaction but decreases the __________.
Chapter 1
Introduction to Fundamental Concepts
1. 1 mole of a gas at S.T.P occupies a volume of __________.
2. A gas occupying a volume of 22.4 dm3 at S.T.P contains __________ molecules.
3. A formula, which gives the relative number of atoms in the molecule of a compound, is called __________.
4. A formula which gives the actual number of all kinds of atoms present in the molecule of compound is termed as __________.
5. The chemical formula that not only gives the actual number of atoms but also shows the arrangement of different atoms present in the molecule is called __________.
6. Atomic weight or molecular weight expressed in grams is known as __________.
7. 2 moles of H2O contain __________ grams and __________ number of molecules.
8. Any thing that occupies space and has __________ is called matter.
9. Volume of one __________ mole of a gas at S.T.P is 22.4 cubic feet.
10. A ton mole of iron is equal to __________ tons.
11. The force with which the earth attracts a body is called the __________ of the body.
12. A pure substance contains __________ kind of molecules.
13. The smallest indivisible particle of matter is called __________.
14. The atomic number is equal to the number of __________ in nucleus.
15. The atomic mass is the total number of protons and __________ in an atom of the element.
16. The average weight of atoms of an element as compared to the weight of one atom of __________ is called the atomic mass.
17. 1.0007 contains __________ significant figures.
18. The figure 24.75 will be rounded off to __________.
19. __________ means that the readings and measurements obtained in different experiments are very close to each other.
20. __________ means that the results obtained in different experiments are very close to the accepted values.
21. The degree of a measured quantity __________ with increasing number of significant figures in it.
22. The atomic mass of sodium is __________.
23. The symbolic representation of a molecule of a compound is called __________.
24. Molecular formula of CHCl3 and its Empirical formula is __________.
25. Molecular formula of benzene is C6H6 and its empirical formula is __________.
26. 58.5 is the __________ of NaCl.
27. 4.5 gms of nitrogen will have __________ molecules.
28. 28 gms of nitrogen will have __________ molecules.
29. 2 moles of SO2 is equal to __________ gms.
30. 1000 gms of H2O is equal to __________ moles.
31. The reactions, which proceed in both directions, are called __________.
32. The reactions, which proceed in forward directions only, are called __________ reactions.
33. The __________ reactions are completed after some time.
34. 0.0006 has __________ significant figures
35. 7.40 x 108 has __________ significant figures.
36. 7 x 108 has __________ significant figures.
37. Usually Molecular formula is simple multiple of the __________.
38. 0.1 mole of H2O contains __________ molecules of H2O.
39. Mass of 3.01 x 1022 molecules of CO2 is __________.
40. __________ is the branch of science which deals with the properties, composition and structure of matter.
41. None zero digits are all __________.
42. The integer part of logarithm is called __________.
43. The decimal fraction of logarithm is called __________.
44. __________ is the amount of substance, which contains as many number of particles as there are in 12 gms of Carbon.
45. 6.02 x 1023 is called the __________.
46. The accuracy of measurement depends on the number of __________.
47. __________ is the branch of chemistry that deals with quantitative relationships among the substances undergoing chemical changes.
48. The sum of atomic weights of all the elements present in molecular formula is called the __________.
49. __________ is the sum of atomic weights of the elements represented by the Empirical formula of the compound.
50. Very small and very large quantities are expressed in terms of __________.
51. In rounding off __________ figure is dropped.
52. Mole is the quantity, which has __________ particle of the substance.
53. For three significant figures, 25.55 is rounded off to __________.
54. The S.I unit of a mass is __________.
55. Mass of 6.02 x 1023 molecules of NaCl is __________ gm.
56. 1 mole of NaOH is __________ gm of NaOH.
57. Formula weight is used for __________ substances.
58. The word S.I stands for __________.
59. 4.5 gms of water will have __________ molecules.
60. 0.0087 has __________ significant figure.
Chapter 2
The Three States of Matter
1. The intermixing of gases or liquids in a container irrespective of their densities, is called __________.
2. At constant temperature, if the pressure of a given mass of a gas is decreased, its volume will __________.
3. A volume of __________ dm3 will hold 128 gms of SO2.
4. At constant temperature of a given mass of a gas, the product of its __________ and __________ is constant.
5. The rates of diffusion of gases are __________ proportional to the square root of their densities.
6. Gases deviate from ideal behaviour more markedly at high __________.
7. Liquid diffuse __________ than gases.
8. An imaginary line passing through the centre of a crystal is called __________.
9. The temperature at which more than one crystalline forms of a substance coexist in equilibrium is called __________.
10. Two or more substances crystallizing in the same form is called __________.
11. The existence of solid substances in more than one crystalline form is known as __________.
12. Rate of diffusion of gases is __________ as compared to liquids.
13. Boiling point of a liquid __________ with the pressure.
14. Mercury in a glass tube forms __________ curvature.
15. Gases can be compressed to __________ extent.
16. Viscosity of a liquid __________ with the increase of temperature.
17. Surface tension of water __________ by adding soap solution into it.
18. The internal resistance to the flow of a liquid is called __________.
19. The rise or the fall of a liquid in a capillary tube is called __________.
20. Matter exists in __________ states.
21. The freezing point of water in Fahrenheit scale is __________.
22. Boiling point of water is __________ °K.
23. SI unit for measurement of pressure is __________.
24. The value of gas law constant R = __________ dm3 atm/°K/mole.
25. The absolute Zero is equal to __________.
26. If P is plotted against 1/V at constant temperature a __________ is obtained.
27. Gases __________ in heating.
28. The pressure of air __________ at higher altitude.
29. Standard temperature means __________.
30. Standard pressure means __________.
31. Cooling is caused by __________ of gases.
32. Rate of diffusion of O2 is __________ times more than H2.
33. H2O has __________ viscosity than CH3OH.
34. Mercury does not wet the glass surface due to its higher __________.
35. Surface tension of mercury is __________ than water.
36. Viscosity can be easily measured by an instrument called __________.
37. The pressure exerted by the vapours when these vapours are in equilibrium with the liquid is called __________.
38. Vapour pressure __________ at high temperature.
39. Boyle’s Law and Charles Law can be combined into the mathematical expression __________.
40. Equal volumes of all gases at the same temperature and pressure contain __________ number of molecules.
41. The average Kinetic energy of a gas is proportional to its __________ temperature.
42. Kinetic equation may be mathematically written as __________.
43. The temperature at which two crystalline forms of a substance can coexist in equilibrium is called __________.
44. Lighter gases diffuse __________ than heavier gases.
45. Rain drops are __________ in shape.
46. Due to surface tension, the surface area of the liquid is __________.
47. Water __________ in the capillary tube.
48. Viscosity of a solution at 10°C is __________ than at 20°C.
49. Shape of NaCl crystal is __________.
50. Gases intermix to form a mixture.
51. Pressure of a dry gas is __________ than the pressure of a moist gas.
52. 22.4 dm3 of nitrogen at S.T.P will weigh equal to __________ gm.
53. 1 mole of any gas at S.T.P is equal to __________ dm3.
54. At -273°C, volume of all gases becomes __________.
55. The gases, which strictly follows the gas Laws are called __________ gas.
56. __________ is the property that determines the direction of flow of heat.
57. __________ is defined as force per unit area.
58. __________ viscosity is defined as the viscosity of a liquid as compared to the water.
Chapter 3
Structure of Atom
1. The maximum number of electrons in 2p orbital is __________.
2. 3d orbital has __________ energy than 4s orbital.
3. __________ rays are non-material in nature.
4. Charge to mass ratio of cathode rays resembles to that of __________.
5. __________ rays are most penetrating.
6. Neutrons have mass equal to that of __________.
7. Energy is __________ when an electron jumps from higher to lower orbit.
8. Second Ionization Potential has __________ value than the First Ionization Potential.
9. Electronegativity __________ from left to right in a period of Periodic Table.
10. __________ was discovered during the course of Artificial Radioactivity.
11. The velocity of alpha rays is nearly __________ of velocity of light.
12. Natural Radioactivity is confined in __________ elements.
13. The isotopes of an element differ in their __________.
14. Two electrons with the __________ spin, can never occupy the same atomic orbital.
15. ‘Al’ has electronic configuration, 1s2, 2s2, __________.
16. In a group of Periodic Table, the ionization potential __________ from top to bottom as the size of atom increases.
17. Ionization potential values __________ from left to right in a period.
18. The energy required to remove the most loosely bond electron from an atom in gaseous state is called __________.
19. The SI unit of Ionization Potential is __________.
20. An atom of sodium possesses 11 protons and __________ neutrons.
21. The particles of Cathode rays possess __________ charge.
22. The negatively charged particles found in Cathode rays are named as __________.
23. Positive rays are emitted from __________.
24. __________ rays are also known as Canal rays.
25. __________ consists of helium ions and are doubly positively charged.
26. __________ rays consists of negatively charged particles.
27. __________ rays are light waves of very short wavelength.
28. The phenomenon in which a stable element is made radioactive by artificial disintegration is called __________.
29. The electron move around the nucleus in different circular paths called __________.
30. The maximum number of electron in a shell is determined by the formula __________.
31. A particle whose mass is equal to that of electron but carries a positive charge is called __________.
32. 2p electrons are __________ in energy that 2s electrons in the same atom.
33. Number of protons of an element also indicates its __________.
34. According to __________ Principle electrons are fed in the order of increasing orbital energy.
35. According to __________ electrons are distributed among the orbitals of a sub shell to give maximum number of unpaired electron and have same spin.
36. The specific way in which the orbitals of an atom are occupied by electrons is called __________.
37. __________ rays are stream of doubly positively charged particles.
38. Electron in the outer most shell of an atom is called __________.
39. Protons are found in the __________ of an atom and bear __________ charge.
40. The atomic number of an atom is the sum of __________ inside the nucleus.
41. __________ limits the number of electron to different shell or orbits.
42. Sir William Crookes in 1878, discovered that the cathode in high vacuum tube emit radiations what he called __________.
43. X-rays were discovered in 1895 by __________.
44. The discovery of proton was done in 1886 by __________.
45. Neutrons were discovered by __________ in 1932 by the bombardment of beryllium with alpha particles.
46. Each atom has a __________, which contains all the positive charge and practically all the mass of atom.
47. Complete the reaction: 4Be9 + 2H4 ® __________ + __________.
48. __________ have higher ionization power as compared to b-rays.
49. No dark spaces between the colours are present in __________.
50. The symbol e+ represents __________.
51. p-orbitals are __________ shaped.
52. The energy released when an electron is added to an atom in the gaseous state is called __________.
53. The power of an atom to attract a shared pair of electrons towards itself is called __________.
54. Fluorine is __________ electronegative than chlorine.
55. Lyman series of spherical lines appear in the __________ portion of spectrum.
56. The electrons with __________ spin occupy the same orbital.
57. 3d orbital has __________ energy than 4s orbital.
58. Energy and frequency are __________ proportional to each other.
59. Ionic radii of cations are __________ than the atoms from which they are formed.
60. Ionic radii of anions are __________ than the atoms from which they are formed.
Chapter 4
Chemical Bonding
1. A bond formed due to transference of electron is called __________.
2. A bond formed due to sharing of electron is called __________.
3. Sigma bond is __________ than pi bond.
4. The shape of methane molecule is __________.
5. One s and 3p orbitals overlap to produce four __________ hybrid orbitals.
6. Ethene, C2H4 is an example of __________ hybridization.
7. Water molecule has __________ structure.
8. Water molecules are inter-linked with one another due to __________.
9. Polarity of the molecule is due to the difference of __________ between the two bonded atoms.
10. A chemical bond formed between to different atoms by mutual sharing of electron is termed as __________.
11. A chemical bond formed between two similar atoms by mutual sharing of electrons is known as __________.
12. The difference between the Electronegativity values of the two atoms forming covalent bond must be __________ than 1.7.
13. When two orbitals of different atoms by hybridize with each other having their axes in the same straight lines, the bond formed is termed as __________.
14. __________ bond is formed when p-orbitals of the two atoms with their axes parallel to each other overlap with each other.
15. Melting and boiling point of ionic compounds are usually __________ than that of covalent compounds.
16. Non polar compounds are usually __________ in non polar solvent.
17. The nitrogen in NH3 is __________ hybridized.
18. A hybrid orbital is called __________ orbital.
19. Since dipole moment of CS2 is zero, it is a __________ molecule.
20. A bond formed due to the electrostatic forces of attraction between the oppositely charged ions is called __________ bond.
21. The ionic bond is formed between the atoms with low ionization potential and high __________.
22. A bond formed by the sharing of an electron pair contributed by one atom only is called a __________ bond.
23. A co-ordinate covalent bond is also known as __________ bond.
24. Polar covalent bond is __________ than a non polar covalent bond.
25. H-F bond is __________ than H-Br bond.
26. The SI unit of dipole moment is __________.
27. Commonly used unit of dipole moment is __________.
28. Dipole moment of non-polar compound is __________ D.
29. The reactions of ionic compounds are usually very __________.
30. Covalent compounds are generally __________ in nature.
31. Ionic compounds are generally __________ in nature.
32. A covalent bond is represented by a __________.
33. A co-ordinate covalent bond is represented by an __________.
34. The covalent bond between H-F is called __________ covalent bond.
35. The power of an atom to attract a shared pair of electron itself is called __________ of that atom.
36. m = d x e represents __________.
37. CO2 and SO2 molecules have __________ polar bonds.
38. NH3 molecule has __________ polar bonds.
39. A double bond has __________ bond energy than a single bond.
40. An orbital which surrounds a single nucleus is called __________ orbital.
41. An orbital which surrounds two or more atomic nuclei is called __________ orbital.
42. A molecular orbital, which is of lower energy than the atomic orbitals from which it is derived, is known as __________ orbital.
43. A molecular orbital, which has higher energy than the atomic orbitals from which it is derived, is known as __________ orbital.
44. Orbitals formed after hybridization are called __________ orbitals.
45. Bond angle in Sp3 hybridization is of __________.
46. Bond angle in Sp2 hybridization is of __________.
47. Bond angle in Sp hybridization is of __________.
48. Sp3 hybridization is also known as __________.
49. Sp2 hybridization is also known as __________.
50. Sp hybridization is also known as __________.
51. A pair of electrons residing on the central atom and which is not used in bonding is called a __________.
52. The sum of total number of electron pairs (bonding and lone pairs) is called __________ number.
53. __________ bond is usually expressed by dotted line.
54. Water molecule has dipole moment because of its __________ structure.
55. CO2 is non polar because of its __________ structure.
56. Overlapping in __________ bond is perfect.
57. Overlapping in __________ bond is not perfect.
58. H-H bond is __________ than H-Cl bond.
59. __________ hybrid orbitals are not co-planar.
60. Covalent bond is Cl2 molecule is __________.
Chapter 5
Energetics of Chemical Reaction
1. The branch of Chemistry, which deals with the heat changes that take place during chemical reaction, is called __________.
2. The branch of science which deals with energy changes accompanying physical and chemical transformation is called __________.
3. The amount of heat evolved or absorbed in a chemical reaction is called __________.
4. Such reactions in which heat is evolved are called __________ reactions.
5. Such reactions in which heat is absorbed are called __________ reactions.
6. In exothermic reactions, heat evolved is given by __________ sign of DH.
7. In endothermic reactions heat absorbed is given by __________ sign of DH.
8. The total heat change in a reaction is the same whether it takes place in one or several steps.
9. The first law of thermodynamics is also known as __________.
10. The part of universe under observation is called __________.
11. The system plus its surrounding is called __________.
12. Such properties, which give description of a system at a particular moment, is called __________.
13. The term E + PV is called __________.
14. DH represents change in __________.
15. The temperature of water is raised up when sulphuric acid is added to it. This is an __________ reaction.
16. The characteristic properties of a system which is independent of amount of material concerned is called __________ properties.
17. The characteristic properties of a system which depend on amount of substance present in it is called __________ properties.
18. Density, pressure and temperature are the examples of __________ properties.
19. Mole numbers and enthalpy are the examples of __________ properties.
20. A system, which exchange both energy and matter with its surrounding, is called __________ system.
21. A system, which only exchange energy with the surrounding but not matter is, called __________ system.
22. A system which neither exchange energy nor matter with its surrounding is called __________ system.
23. A system is __________ if it contains only one phase.
24. A system is __________ if it contains more than one phase.
25. 1 kilojoule is equal to __________ joules.
26. 1 Calorie is equal to __________ joules.
27. 1 kilo calorie is equal to __________ joules.
28. The work done (w) is mathematically denoted by __________.
29. The change in enthalpy is denoted by __________.
30. __________ law is used in calculating heat of reaction.
31. __________ is defined as the change in enthalpy when one gram mole of a compound is produced from its element.
32. Heat of formation is denoted by __________.
33. When the work is done on the system by the surrounding the sign of work done (w) is __________.
34. When the work is done by the system on surrounding the sign of work done is __________.
35. First law of Thermodynamics is mathematically represented as __________.
36. Standard enthalpies are measured at __________.
37. Hess’s Law is employed to calculate __________ of a chemical reaction.
38. Heat absorbed by the system at constant volume is completely utilize to increase the __________ of the system.
39. Heat change at constant pressure from initial to final state of the system is simply equal to the __________.
40. SI unit of measurement of heat change is __________.
Chapter 6
Chemical Equilibrium
1. The reactions, which proceed in both the directions, are called __________ reactions.
2. The reactions, which proceed to one direction only, are called __________ reactions.
3. Reversible reactions are __________ completed.
4. Irreversible reactions are __________ after some time.
5. A reversible reaction is said to be in __________ when the rate of forward reaction becomes equal to the rate of backward reaction.
6. The concentrations of reactants and products are __________ at equilibrium point.
7. The value of Kc depends upon the __________ of the reactants.
8. A increase of the value of Kc tends to move the reaction to the __________ direction.
9. A decrease of the value of Kc tends to move the reaction to the __________ direction.
10. An increase in the concentration of the reactants will move the reaction to the __________ direction.
11. A decrease in the concentration of the reactants will move the reaction to the __________ direction.
12. Equilibrium constant is denoted by __________.
13. When the equilibrium constant value is very __________, we can conclude that the forward reaction is almost completed.
14. When equilibrium constant value is very __________ we can conclude that forward reaction will occur to very little extent.
15. According to __________ principle, if system in equilibrium is subjected to a stress, the equilibrium shifts in a direction to minimize or undo the effect of the stress.
16. In exothermic reaction, the __________ of temperature favour the forward rate of reaction.
17. In endothermic reactions, the __________ of temperature favour the forward rate of reaction.
18. A __________ is a substance which effects the rate of reaction but remains unaltered at the end of the reaction.
19. A catalyst increases the velocity of the reaction by decreasing the __________.
20. The suppression of degree of ionization of a sparingly soluble weak electrolyte by the addition of a strong electrolyte containing an ion in common is called __________.
21. __________ is purified in industries by Common Ion Effect.
22. A reaction moves to the left when the concentrations of the products are __________.
23. A reaction moves to the right when the concentrations of the products are __________.
24. Increase in pressure will move the reaction in the direction of __________ volume.
25. Decrease in pressure will move the reaction in the direction of __________ volume.
26. An increase of temperature favours the formation of products in case of __________ reaction.
27. A decrease of temperature fovours the formation of products in case of __________ reaction.
28. Heating moves an endothermic reaction to the __________.
29. Cooling move an exothermic reaction to the __________.
30. The product of ionic concentration in a saturated solution is called __________ constant.
31. When HCl is added to NaCl, the concentration of __________ ion is increased.
32. Chemical reaction involving the substances in more than one phases are called __________.
33. The formation of NH3 is exothermic process hence __________ temperature will favour the formation of NH3.
34. The formation of NO from N2 and O2 is endothermic process hence __________ temperature will favour the formation of NO.
35. Chemical Equilibrium is __________ equilibrium.
36. Molar concentration is also called __________.
37. The rate at which a substance takes part in a chemical reaction depends upon its __________.
38. __________ principle is applied to all reversible reaction.
39. A common ion __________ the solubility of the salt.
40. Number of moles present per dm3 of a substance is called __________.
Chapter 7
Solutions and Electrolytes
1. A mixture of two or more substances, which are homogeneously mixed, is called a __________.
2. __________ is defined as the amount of solute dissolved in a given amount of solvent.
3. A solution is composed of two components __________ and __________.
4. A solution containing one mole of solute per dm3 of solution is called one __________ solution.
5. Molarity is denoted by __________.
6. 1M solution of NaOH contains __________ gms of it dissolved per dm3 of solution.
7. A solution containing one mole of solute dissolved by per kg of solvent is called __________ solution.
8. Molality is denoted by __________.
9. 1M solution of H2SO4 contains __________ gms of it per kg of solvent.
10. The process in which ions are surrounded by water molecules is called __________.
11. The water molecules attached with the hydrated substance are called __________.
12. Hydrated copper sulphate evolves __________ water molecules on heating.
13. The interaction between salt and water to produce acids and bases is called __________.
14. The products of ionic concentration in a saturated solution at a certain temperature are called the __________.
15. Solubility product constant expressed as __________.
16. The suppression of ionization by adding a common ion is called __________.
17. The process of dissociation of an electrolyte into ions is known as __________.
18. The chemical decomposition of a compound in a solution or in fused state brought about by a flow of electric current is known as __________.
19. Electrolysis is performed in an electrolytic cell, which is known as __________.
20. The positive electrode of a voltmeter is called __________ and negative as __________.
21. A solution, which tends to resist changes in pH is called a __________ solution.
22. A mixture of acetic acid and sodium acetate acts as a __________.
23. According to Sorenson __________ is defined as negative logarithm of the hydrogen ion concentration.
24. pH is mathematically expressed as __________.
25. The pH of a neutral solution is __________.
26. __________ substances have pH values lower than 7.
27. __________ solutions have pH values more than 7.
28. Oxidation is __________ of electron.
29. Reduction is the __________ of electron.
30. Such chemical reactions in which the oxidation number of atoms or ions is changed are called __________ reactions.
31. Oxidation number of a free element is __________.
32. Oxidation number of Oxygen in a compound is __________.
33. The sum of oxidation number of any formula of a compound is __________.
34. The oxidation number of any ion is equal to the __________ on the ion.
35. __________ is the reaction in which an acid reacts with a base to form salt and water.
36. __________ are organic compounds which change colour in accordance with the pH of the medium.
37. An indicator that changes from colourless to pink in the presence of an alkaline solution is called __________.
38. An indicator that changes from red to yellow in the presence of an alkaline solution is called __________.
39. Dissociation constant is denoted by __________.
40. According to Bronsted-Lowry Concept, __________ is the donor of proton and __________ is the acceptor of proton.
41. According to Arrhenius, acid is substance that produces __________ ions when dissolved in water.
42. According to Arrhenius, base is a substance that produces __________ ions when dissolved in water.
43. When ionic product is less than ksp, the solution will __________.
44. When ionic product is greater than ksp, the solution will __________.
45. The electrode at which oxidation takes place is called __________.
46. The electrode at which reduction takes place is called __________.
47. H3O+ ion is called __________ ion.
48. The logarithm of reciprocal of hydroxyl ion (OH)- is called __________.
49. Aqueous solution of NH4Cl is __________ while that of NaHCO3 is __________.
50. The ionic product of [H+] and [OH-] of pure water is __________.
51. An increase in the oxidation number of an element or ion during a chemical change is called __________.
52. A decrease in the oxidation number of an element or ion during a chemical change is called __________.
53. The degree of dissociation __________ with the increase in temperature.
54. The degree of dissociation __________ with the dilution of electrolytic solution.
55. A __________ consists of an electrode immersed in solution of its ion.
56. The potential difference between the electrode and the solution of its salt at equilibrium position is called __________ potential.
57. If the pH of a solution is 14, the solution is __________.
58. If the pH of a solution is 4, the solution is __________.
59. The oxidation number of Mn in KMnO4 is __________.
60. The oxidation number of Fe in FeCl3 is __________.
Chapter 8
Introduction to Chemical Kinetics
1. The branch of chemistry, which deals with the study of rates and mechanisms of chemical reactions, is known as __________.
2. Such reactions, which proceeds with very high velocities and are completed very quickly are called __________ reactions.
3. Such reactions, which take place very slowly, are called __________ reactions.
4. Reactions between silver nitrate and sodium chloride to form white precipitates of silver chloride are an example of __________ reaction.
5. Reactions of Organic compounds are slow and are called __________ reactions.
6. There are some reactions, which proceed slowly with a __________ speed.
7. The rate of __________ reaction can only be determined.
8. The amount of chemical change taking place in concentration of the per unit time is called __________ of reaction.
9. Rate of reaction is expressed in __________.
10. The rate of reaction between two specific interval of time is called __________.
11. The addition energy required to bring about a chemical reaction is called __________.
12. According to __________ theory for a chemical reaction to take place, the reacting molecules must come closed together.
13. The addition of __________ helps the reaction by lowering the energy of activation.
14. The rate of reaction __________ with the increase in concentration of the reacting molecules.
15. When the concentration of both the reacting molecules is double, the probability of collisions between them will be __________ times.
16. By __________ the surface area of the reactants, the rate of reaction is increased.
17. Rate of reaction generally __________ with the rise of temperature.
18. A __________ is a substance, which either accelerates or retards the rate of reaction without taking part in the reaction.
19. In the preparation of Oxygen from Potassium Chlorate, __________ is used as catalyst.
20. In the oxidation of SO2 to SO3 by the contact process for the manufacture of H2SO4 __________ is used as catalyst.
21. An unstable intermediate compound formed during a chemical reaction is called __________.
22. When a catalyst and the reactants are in the same phases, it is known as __________ catalyst.
23. When a catalyst and the reactants are in different phases, it is called __________.
24. When a catalyst increases the rate of reaction, it is called __________ catalyst.
25. When a catalyst retards the rate of reaction, it is called __________ catalyst.
26. A negative catalyst __________ the energy of activation, hence the rate of reaction is decreased.
27. The ratio between the rate of reaction and concentration of reactants is known as __________.
28. Velocity constant is independent of concentration but depends on __________.
29. Ionic reactions are __________ than molecular reactions.
30. The value of specific rates constant for a reaction __________ with time.
31. The sum of all exponents of concentration terms in the equation is called __________.
32. The sum of moles taking part in a chemical reaction is called __________ of the reactions.
Chemistry XI Viva Notes Volumetric Analysis
VIVA NOTES
(A) Volumetric Analysis (Titrations)
Qs.1 What is Titration?
Ans. The process of adding one solution from the burette into another in the conical flask in order to determine its volume after the completion of the chemical reaction is known as Titration.
Qs.2 What is Neutralization?
Ans. Neutralization is simply a reaction between Hydrogen ion (H+) of an acid and Hydroxide ion (OH-) of the base to form water. In this reaction another class of compound known as “Salt is also produced which remains in the solution as ions, when water is boiled off these ions re-unite to form salt.
Acid + Base → Salt + Water
Qs.3 What is an acid?
Ans. An acid is a substance which gives off proton (H+) in solution or in other words it is a donor of proton e.g. HCl, H2SO4 or HNO3. When dissolve in water ionizes to give Hydrogen ion (H+)
HCL ↔ H+ + Cl-
H2SO4 ↔ 2H+ + SO4--
HNO3 ↔ H+ + NO3-
Qs.4 What do you mean by basicity of an acid?
Ans. It is the number of ionizable Hydrogen presents in the molecule of an acid.
e.g. In the above examples basicity of HCl and HNO3 is one, while that of H2*SO4 is 2.
Qs.5 What is a base?
Ans. A base is now regarded as a molecule or an ion which furnishes OH- ion or accepts a proton given up by an acid. Thus it is proton acceptor.
e.g. NaOH, KOH, Ca (OH)2
NaOH ↔ Na+ + OH-
KOH ↔ K+ + OH-
Ca(OH)2 ↔ CA++ + 2(OH)-
Qs.6 What is meant by acidity of a base?
Ans. It is the number of hydroxyl (OH-) groups present in the molecule of a base.
e.g. In the above examples the acidity of NaOH and KOH is one while that of
Ca (OH)2 is 2.
Qs.7 Why Phenolphthalein is added into the solution of titration flask?
Ans. Phenolphthalein serves as an indicator for determining the end point. It gives pink colour in presence of an alkali and becomes colourless with slight excess of an acid.
Qs.8 While using a burette what precautions are necessary to be observed?
Ans. Burette must be washed first with ordinary water and then rinsed with the solution which is to be taken in it. It must be held vertically and air bubbles must be removed.
While taking a reading the eyes should be in level with the surface of the liquid.
Qs.9 What is a pipette?
Ans. It is an instrument which delivers a definite volume of a liquid. It consists of a glass tube with a cylindrical bulb in the middle and the lower end is drawn into a jet. There is a circular mark on the upper tube.
Qs.10 What is a standard solution?
Ans. A standard solution is a solution of known strength.
Qs.11 What do you mean by strength of a solution?
Ans. Strength of a solution is the quantity of a substance present in any known volume of the solution.
Qs.12 Define a “Normal solution”?
Ans. A standard solution which contains 1 gram equivalent of a substance per dm3 is known as a Normal solution and is denoted by 1 N.
Qs.13 What is a decinormal solution?
Ans. A decinormal solution contains 1/10 fraction of gram equivalent weight of a substance dissolved per dm3 and is denoted by 0.1 N or N/10.
Qs.14 What is the relationship between Normality and Strength per dm 3 of solution?
Ans. Normality =
Qs.15 What is “Acidimetry”?
Ans. It is an operation by which the strength of an alkali is determined by neutralizing it with an acid of known strength in presence of an indicator.
Qs.16 What is “Alkalimetry”?
Ans. It is an operation by which the strength of an alkali is determined by neutralizing it with an alkali of known strength in presence of an indicator.
Qs.17 Define “Equivalent Weight”?
Ans. It is the number of parts by weight that will combine with or displace from 1 part by weight of H2, or 8 parts by weight of O2 or 35.5 by weight of Cl2.
Qs.18 What is “Gram Equivalent Weight”?
Ans. It is the equivalent weight of a substance expressed in gram.
Qs.19 How equivalent weight of an acid is determined?
Ans. Equivalent weight of an acid =
Where basicity is the total number of replaceable Hydrogen.
Qs.20 How equivalent weight of a base is determined?
Ans. Equivalent weight of a base =
Where acidity is the total number of hydroxyl (OH-) groups.
Qs.21 What do you mean by “Standardization of a solution”?
Ans. To standardize means to determine its strength by titration against some standard solution.
Qs.22 What is the equivalent weight of NaOH?
Ans. Eq. Wt. Of a base = = = 40
Therefore, equivalent weight of NaOH is 40.
Qs.23 How a decinormal solution of NaOH is prepared?
Ans. A decinormal solution (0.1 N) of NaOH can be prepared by dissolving 1/10 fraction of its equivalent wt. i.e. 40/10 = 4 gms, in one dm3 of distilled water.
Qs.24 What do you mean by “End-point”?
Ans. It is the exact stage at which the chemical reaction of the titrating solutions is just completed.
Qs.25 Why alkali is taken in burette when phenolphthalein is used as an indicator?
Ans. The appearance of a pink colour at the end point is easily detectable. So it is a better criterion than the disappearance of colour when acid is used in the burette.
Qs.26 How Equivalent weight of Na2CO3 is calculated?
Ans. Na2CO3 is a basic salt. Its equivalent weight can be calculated from the equation of its reaction with an acid e.g. HCl.
Hence the equivalent weight of Sodium Carbonate is determined as follows:
Na2CO3 + 2HCl 2 NaCl + H2O +CO2
2x23+12 2(1+35.5)
+3x16 = 73
= 106
2 HCl = Na2CO3
1 HCl = ½ Na2CO3
Eq. Wt. of Na2CO3= = 53.
Qs.27 What is Methyl Orange?
Ans. It is Sodium salt of an azo dye. It is very good indicator for titrating strong acid against strong base or strong acid against weak base.
Qs.28 What is meant by Anhydrous salt?
Ans. A Salt without water molecule is called Anhydrous.
Qs.29 What indicator is suitable for Sodium Carbonate titration against strong acids and why?
Ans. The pH range of Methyl Orange is pH 3.0 to 4.4.
Hence, it is very suitable indicator when a weak alkali like Sodium Varbonatye is neutralized with a strong acid. In such vases the end point would be at a pH some what below 7.0.
Qs.30 Give the structural formula of Phenolphthalein?
Ans. It is as follows:
Qs.31 Which is the suitable indicator for the titration of?
(i) Weak acid against strong alkali.
(ii) Strong acid against weak alkali.
(iii) Strong acid against strong alkali.
Ans. (i) Phenolphthalein for the titration of weak acid with strong alkali.
(ii) Methyl Orange for the titration of strong acid with a weak alkali.
(iii) Phenolphthalein or Methyl orange for the titration of strong acid with strong alkali (preferably phenolphthalein)
Qs.32 Define Oxidation?
Ans. The loss of electron from an atom, ion or molecule is called Oxidation.
Fe++ ® Fe+++ + e-
Qs.33 Define Reduction?
Ans. Gain of electron or the loss of positive valence is called reduction.
Fe+++ + e- ® Fe++
Qs.34 What are “Oxidation-Reduction Titration”?
Ans. Titrations based upon the reaction between an oxidation agent and reducing agent are known as “Oxidation-Reduction Titrations”.
Qs.35 What are Oxidizing and Reducing agents?
Ans. An oxidizing agent is that substance which oxidizes the other substance e.g., KMnO4 in this titration.
A reducing agent is that which reduces the other substance e.g., Oxalic acid in this titration.
Qs.36 Why do we add equal volume of dilute Sulphuric acid in KMnO4 titration?
Ans. In presence of acid it acts as a strong oxidizing agent and liberates atomic Oxygen from KMnO4.
2 KMnO4 + 3H2SO4 ® K2SO4 + 2 MnSO4 + 3 H2O + 5 [O]
Qs.37 Why do we heat Oxalic acid solution to 60-70º C?
Ans. Oxalic acid reacts with Potassium Permanganate very slowly at room temperature. In order to facilitate the reaction, it is heated to 60º to 70ºC.
Qs.38 What indicator is used in KMnO4 titrations?
Ans. KMnO4 itself acts as an indicator. So no external indicator is required. Near end point it produces a permanent pinkish tinge.
Qs.39 Explain how the change of colour takes place near end point while titrating Oxalic acid with KMnO4 in presence of sulphuric acid?
Ans. In the presence of Sulphuric acid, Potassium permanganate reacts with reducing agent as follows.
2 KMnO4 + 3H2SO4 ® K2SO4 + 2 MnSO4 + 3 H2O + 5 [O]
As the titration proceeds, Potassium Sulphate and Manganese Sulphate are formed, both give colourless solution. As soon as KMnO4 is in excess, the solution becomes pink and so it acts as its own indicator.
Qs.40 Why upper meniscus is noted while using KMnO4 solution in the burette?
Ans. Potassium Permanganate solution is highly coloured and lower meniscus is not distinctly visible. That is why reading of upper meniscus of the liquid is noted.
Qs.41 What is the nature of FeSO4 · 7H2O in Redox titration?
Ans. Ferrous Sulphate acts as reducing agent.
Qs.42 How equivalent weight of FeSO4 · 7H2O is calculated?
Ans. According to the equation:
10 FeSO4 · 7H2O + 5H2SO4 + 5[O] → 5Fe2 (SO4)3 + 12H2O
5[O] combines with 10 FeSO4 · 7H2O
5 x 16 parts by wt. of [O] combines with 10 x 278 parts by
wt. of FeSO4 · 7H2O
8 parts by wt. of [O] combines with = 278
Hence equivalent weight of FeSO4 · 7H2O = 278
Qs.43 Why no external Indicator is required for this titration?
Ans. During titration the dark purple colour of permanganate solution disappears entirely. As soon as the reaction is completed, a single drop of permanganate produces a permanent pinkish tinge to the solution. Thus KMnO4 acts as internal indicator and no external indicator is required.
Qs.44 How equivalent weight of hydrated and Anhydrous Oxalic acid are calculated?
Ans. Mol. Wt. of hydrated Oxalic acid H2C2O4 • 2H2O
2+24+64+2x18=126
Basicity of acid =2
Eq. wt. of hydrated Oxalic acid = = = 63
Mol. wt. of Anhydrous Oxalic acid H2C2O4 = 2+24+64 = 90
Eq. wt. of Anhydrous Oxalic acid = = = 45.
Qs.45 What do you mean by Mohr’s Salt?
Ans. Ferrous Ammonium Sulphate FeSO4 (NH4)2 SO4.6H2O is commonly known as “Mohr’s salt”.
Qs.46 What is the nature of Ferrous Ammonium Sulphate in redox titration?
Ans. Ferrous Ammonium Sulphate acts as a reducing agent.
Qs.47 How can you calculate the equivalent weight of KMnO4?
Ans. KMnO4 *is an Oxidizing agent. The equivalent weight of an Oxidant is the number of parts by weight which gives 8 parts by weight of Oxygen for oxidation.
2 KMnO4 *+ 3H2SO4 → K2SO4 + 2MnSO4 + 3H2O + 5 [O]
5x16 parts by wt. of O2 comes from 2x158 parts of KMnO4
׃Ù 8 ״ ״ ״ ״ ״ ״ = 31.6
Therefore, the equivalent weight of KMnO4 is 31.6
Qs.48 Why Mohr’s salt solution is prepared in water containing dilute Sulphuric acid?
Ans. Solution of Ferrous Ammonium Sulphate (Mohr’s salt) is always prepared by dissolving it in water containing some dilute Sulphuric acid which prevents hydrolysis.
Qs.49 What happens when KMnO4 solution is added in acidified Ferrous Ammonium Sulphate solution?
Ans. When KMnO4 solution is added to Ferrous Ammonium Sulphate is presence of dulute H2SO4, the Ferrous salt is Oxidized to ferric state.
Qs.50 What is the equivalent weight of Ferrous Ammonium Sulphate ?
Ans. Equivalent weight of Ferrous Ammonium Sulphate.
FeSO4. (NH4)2SO4.6H2O = 152+36+32+64+108 = 392
(B) Melting Point & Boiling Point
Qs.1 What is meant by melting or fusion of a substance?
Ans. It is the change of a substance from solid to the liquid state.
Qs.2 Define melting point?
Ans. The temperature at which the solid substance fuses (i.e. changes into liquid) and continue to take place until the whole of the solid is converted into liquid is known as the “Melting point” of the substance.
Qs.3 Why a thin walled capillary tube and not a thick walled tube is selected to determine the Melting point?
Ans. The wall of the capillary tube should be thin otherwise the temperature of the bath may not be equal to the temperature of the substance inside the capillary tube.
Qs.4 Why it is necessary to heat the bath slowly with constant stirring?
Ans. It is necessary to heat the bath slowly with constant stirring to ensure uniformity of temperature; otherwise the rate of rise of temperature is so rapid that it will be difficult to observe the temperature at which the substance just melts.
Qs.5 Why water as a bath cannot be used for the substance having Melting point above 100˚C?
Ans. Boiling point of water is 100˚C. So it cannot be used as a bath for determination of Melting points of such substances. Instead of water, Sulphuric acid or Glycerine can be used.
Qs.6 What is “Boiling”?
Ans. It is a rapid change from the liquid to the gaseous state.
Qs.7 What is meant by the “Boiling point” of a liquid?
Ans. The Boiling point of a liquid is the temperature at which the vapour pressure of the liquid is equal to the atmospheric pressure (i.e. 760 mm.)
Qs.8 Why temperature remains constant at boiling point of a liquid?
Ans. Because the heat is utilized for converting liquid into vapours, so temperature remains constant.
Qs.9 What is the Boiling point of pure water? Is the same at all places?
Ans. The Boiling point of pure water is 100˚C at a pressure of 760 torr. The pressure of air at all places is not the same. So B.P. differs at various places.
Qs.10 What is the difference between Boiling and Evaporation?
Ans. Boiling is a rapid change and takes place through out the mass of the liquid at a definite temperature (i.e., boiling point). While Evaporation is a slow change and takes place at the surface of the liquid at all temperature.
Qs.11 What is the effect of pressure on Boiling point?
Ans. It is raised by the increase of pressure and is lowered by the decrease of pressure. (An increase or decrease of 26.7 m.m. of pressure increase or decrease the Boiling point by 1˚ C).
Qs.12 What is the effect of height on the Boiling point of a liquid?
Ans. With heights, the boiling points are reduced as the pressure decreases.
Qs.13 Why should we stir the liquid in the beaker?
Ans. We should stir it constantly because, this helps the liquid to maintain uniform temperature.
VIVA NOTES
(A) Volumetric Analysis (Titrations)
Qs.1 What is Titration?
Ans. The process of adding one solution from the burette into another in the conical flask in order to determine its volume after the completion of the chemical reaction is known as Titration.
Qs.2 What is Neutralization?
Ans. Neutralization is simply a reaction between Hydrogen ion (H+) of an acid and Hydroxide ion (OH-) of the base to form water. In this reaction another class of compound known as “Salt is also produced which remains in the solution as ions, when water is boiled off these ions re-unite to form salt.
Acid + Base → Salt + Water
Qs.3 What is an acid?
Ans. An acid is a substance which gives off proton (H+) in solution or in other words it is a donor of proton e.g. HCl, H2SO4 or HNO3. When dissolve in water ionizes to give Hydrogen ion (H+)
HCL ↔ H+ + Cl-
H2SO4 ↔ 2H+ + SO4--
HNO3 ↔ H+ + NO3-
Qs.4 What do you mean by basicity of an acid?
Ans. It is the number of ionizable Hydrogen presents in the molecule of an acid.
e.g. In the above examples basicity of HCl and HNO3 is one, while that of H2*SO4 is 2.
Qs.5 What is a base?
Ans. A base is now regarded as a molecule or an ion which furnishes OH- ion or accepts a proton given up by an acid. Thus it is proton acceptor.
e.g. NaOH, KOH, Ca (OH)2
NaOH ↔ Na+ + OH-
KOH ↔ K+ + OH-
Ca(OH)2 ↔ CA++ + 2(OH)-
Qs.6 What is meant by acidity of a base?
Ans. It is the number of hydroxyl (OH-) groups present in the molecule of a base.
e.g. In the above examples the acidity of NaOH and KOH is one while that of
Ca (OH)2 is 2.
Qs.7 Why Phenolphthalein is added into the solution of titration flask?
Ans. Phenolphthalein serves as an indicator for determining the end point. It gives pink colour in presence of an alkali and becomes colourless with slight excess of an acid.
Qs.8 While using a burette what precautions are necessary to be observed?
Ans. Burette must be washed first with ordinary water and then rinsed with the solution which is to be taken in it. It must be held vertically and air bubbles must be removed.
While taking a reading the eyes should be in level with the surface of the liquid.
Qs.9 What is a pipette?
Ans. It is an instrument which delivers a definite volume of a liquid. It consists of a glass tube with a cylindrical bulb in the middle and the lower end is drawn into a jet. There is a circular mark on the upper tube.
Qs.10 What is a standard solution?
Ans. A standard solution is a solution of known strength.
Qs.11 What do you mean by strength of a solution?
Ans. Strength of a solution is the quantity of a substance present in any known volume of the solution.
Qs.12 Define a “Normal solution”?
Ans. A standard solution which contains 1 gram equivalent of a substance per dm3 is known as a Normal solution and is denoted by 1 N.
Qs.13 What is a decinormal solution?
Ans. A decinormal solution contains 1/10 fraction of gram equivalent weight of a substance dissolved per dm3 and is denoted by 0.1 N or N/10.
Qs.14 What is the relationship between Normality and Strength per dm 3 of solution?
Ans. Normality =
Qs.15 What is “Acidimetry”?
Ans. It is an operation by which the strength of an alkali is determined by neutralizing it with an acid of known strength in presence of an indicator.
Qs.16 What is “Alkalimetry”?
Ans. It is an operation by which the strength of an alkali is determined by neutralizing it with an alkali of known strength in presence of an indicator.
Qs.17 Define “Equivalent Weight”?
Ans. It is the number of parts by weight that will combine with or displace from 1 part by weight of H2, or 8 parts by weight of O2 or 35.5 by weight of Cl2.
Qs.18 What is “Gram Equivalent Weight”?
Ans. It is the equivalent weight of a substance expressed in gram.
Qs.19 How equivalent weight of an acid is determined?
Ans. Equivalent weight of an acid =
Where basicity is the total number of replaceable Hydrogen.
Qs.20 How equivalent weight of a base is determined?
Ans. Equivalent weight of a base =
Where acidity is the total number of hydroxyl (OH-) groups.
Qs.21 What do you mean by “Standardization of a solution”?
Ans. To standardize means to determine its strength by titration against some standard solution.
Qs.22 What is the equivalent weight of NaOH?
Ans. Eq. Wt. Of a base = = = 40
Therefore, equivalent weight of NaOH is 40.
Qs.23 How a decinormal solution of NaOH is prepared?
Ans. A decinormal solution (0.1 N) of NaOH can be prepared by dissolving 1/10 fraction of its equivalent wt. i.e. 40/10 = 4 gms, in one dm3 of distilled water.
Qs.24 What do you mean by “End-point”?
Ans. It is the exact stage at which the chemical reaction of the titrating solutions is just completed.
Qs.25 Why alkali is taken in burette when phenolphthalein is used as an indicator?
Ans. The appearance of a pink colour at the end point is easily detectable. So it is a better criterion than the disappearance of colour when acid is used in the burette.
Qs.26 How Equivalent weight of Na2CO3 is calculated?
Ans. Na2CO3 is a basic salt. Its equivalent weight can be calculated from the equation of its reaction with an acid e.g. HCl.
Hence the equivalent weight of Sodium Carbonate is determined as follows:
Na2CO3 + 2HCl 2 NaCl + H2O +CO2
2x23+12 2(1+35.5)
+3x16 = 73
= 106
2 HCl = Na2CO3
1 HCl = ½ Na2CO3
Eq. Wt. of Na2CO3= = 53.
Qs.27 What is Methyl Orange?
Ans. It is Sodium salt of an azo dye. It is very good indicator for titrating strong acid against strong base or strong acid against weak base.
Qs.28 What is meant by Anhydrous salt?
Ans. A Salt without water molecule is called Anhydrous.
Qs.29 What indicator is suitable for Sodium Carbonate titration against strong acids and why?
Ans. The pH range of Methyl Orange is pH 3.0 to 4.4.
Hence, it is very suitable indicator when a weak alkali like Sodium Varbonatye is neutralized with a strong acid. In such vases the end point would be at a pH some what below 7.0.
Qs.30 Give the structural formula of Phenolphthalein?
Ans. It is as follows:
Qs.31 Which is the suitable indicator for the titration of?
(i) Weak acid against strong alkali.
(ii) Strong acid against weak alkali.
(iii) Strong acid against strong alkali.
Ans. (i) Phenolphthalein for the titration of weak acid with strong alkali.
(ii) Methyl Orange for the titration of strong acid with a weak alkali.
(iii) Phenolphthalein or Methyl orange for the titration of strong acid with strong alkali (preferably phenolphthalein)
Qs.32 Define Oxidation?
Ans. The loss of electron from an atom, ion or molecule is called Oxidation.
Fe++ ® Fe+++ + e-
Qs.33 Define Reduction?
Ans. Gain of electron or the loss of positive valence is called reduction.
Fe+++ + e- ® Fe++
Qs.34 What are “Oxidation-Reduction Titration”?
Ans. Titrations based upon the reaction between an oxidation agent and reducing agent are known as “Oxidation-Reduction Titrations”.
Qs.35 What are Oxidizing and Reducing agents?
Ans. An oxidizing agent is that substance which oxidizes the other substance e.g., KMnO4 in this titration.
A reducing agent is that which reduces the other substance e.g., Oxalic acid in this titration.
Qs.36 Why do we add equal volume of dilute Sulphuric acid in KMnO4 titration?
Ans. In presence of acid it acts as a strong oxidizing agent and liberates atomic Oxygen from KMnO4.
2 KMnO4 + 3H2SO4 ® K2SO4 + 2 MnSO4 + 3 H2O + 5 [O]
Qs.37 Why do we heat Oxalic acid solution to 60-70º C?
Ans. Oxalic acid reacts with Potassium Permanganate very slowly at room temperature. In order to facilitate the reaction, it is heated to 60º to 70ºC.
Qs.38 What indicator is used in KMnO4 titrations?
Ans. KMnO4 itself acts as an indicator. So no external indicator is required. Near end point it produces a permanent pinkish tinge.
Qs.39 Explain how the change of colour takes place near end point while titrating Oxalic acid with KMnO4 in presence of sulphuric acid?
Ans. In the presence of Sulphuric acid, Potassium permanganate reacts with reducing agent as follows.
2 KMnO4 + 3H2SO4 ® K2SO4 + 2 MnSO4 + 3 H2O + 5 [O]
As the titration proceeds, Potassium Sulphate and Manganese Sulphate are formed, both give colourless solution. As soon as KMnO4 is in excess, the solution becomes pink and so it acts as its own indicator.
Qs.40 Why upper meniscus is noted while using KMnO4 solution in the burette?
Ans. Potassium Permanganate solution is highly coloured and lower meniscus is not distinctly visible. That is why reading of upper meniscus of the liquid is noted.
Qs.41 What is the nature of FeSO4 · 7H2O in Redox titration?
Ans. Ferrous Sulphate acts as reducing agent.
Qs.42 How equivalent weight of FeSO4 · 7H2O is calculated?
Ans. According to the equation:
10 FeSO4 · 7H2O + 5H2SO4 + 5[O] → 5Fe2 (SO4)3 + 12H2O
5[O] combines with 10 FeSO4 · 7H2O
5 x 16 parts by wt. of [O] combines with 10 x 278 parts by
wt. of FeSO4 · 7H2O
8 parts by wt. of [O] combines with = 278
Hence equivalent weight of FeSO4 · 7H2O = 278
Qs.43 Why no external Indicator is required for this titration?
Ans. During titration the dark purple colour of permanganate solution disappears entirely. As soon as the reaction is completed, a single drop of permanganate produces a permanent pinkish tinge to the solution. Thus KMnO4 acts as internal indicator and no external indicator is required.
Qs.44 How equivalent weight of hydrated and Anhydrous Oxalic acid are calculated?
Ans. Mol. Wt. of hydrated Oxalic acid H2C2O4 • 2H2O
2+24+64+2x18=126
Basicity of acid =2
Eq. wt. of hydrated Oxalic acid = = = 63
Mol. wt. of Anhydrous Oxalic acid H2C2O4 = 2+24+64 = 90
Eq. wt. of Anhydrous Oxalic acid = = = 45.
Qs.45 What do you mean by Mohr’s Salt?
Ans. Ferrous Ammonium Sulphate FeSO4 (NH4)2 SO4.6H2O is commonly known as “Mohr’s salt”.
Qs.46 What is the nature of Ferrous Ammonium Sulphate in redox titration?
Ans. Ferrous Ammonium Sulphate acts as a reducing agent.
Qs.47 How can you calculate the equivalent weight of KMnO4?
Ans. KMnO4 *is an Oxidizing agent. The equivalent weight of an Oxidant is the number of parts by weight which gives 8 parts by weight of Oxygen for oxidation.
2 KMnO4 *+ 3H2SO4 → K2SO4 + 2MnSO4 + 3H2O + 5 [O]
5x16 parts by wt. of O2 comes from 2x158 parts of KMnO4
׃Ù 8 ״ ״ ״ ״ ״ ״ = 31.6
Therefore, the equivalent weight of KMnO4 is 31.6
Qs.48 Why Mohr’s salt solution is prepared in water containing dilute Sulphuric acid?
Ans. Solution of Ferrous Ammonium Sulphate (Mohr’s salt) is always prepared by dissolving it in water containing some dilute Sulphuric acid which prevents hydrolysis.
Qs.49 What happens when KMnO4 solution is added in acidified Ferrous Ammonium Sulphate solution?
Ans. When KMnO4 solution is added to Ferrous Ammonium Sulphate is presence of dulute H2SO4, the Ferrous salt is Oxidized to ferric state.
Qs.50 What is the equivalent weight of Ferrous Ammonium Sulphate ?
Ans. Equivalent weight of Ferrous Ammonium Sulphate.
FeSO4. (NH4)2SO4.6H2O = 152+36+32+64+108 = 392
(B) Melting Point & Boiling Point
Qs.1 What is meant by melting or fusion of a substance?
Ans. It is the change of a substance from solid to the liquid state.
Qs.2 Define melting point?
Ans. The temperature at which the solid substance fuses (i.e. changes into liquid) and continue to take place until the whole of the solid is converted into liquid is known as the “Melting point” of the substance.
Qs.3 Why a thin walled capillary tube and not a thick walled tube is selected to determine the Melting point?
Ans. The wall of the capillary tube should be thin otherwise the temperature of the bath may not be equal to the temperature of the substance inside the capillary tube.
Qs.4 Why it is necessary to heat the bath slowly with constant stirring?
Ans. It is necessary to heat the bath slowly with constant stirring to ensure uniformity of temperature; otherwise the rate of rise of temperature is so rapid that it will be difficult to observe the temperature at which the substance just melts.
Qs.5 Why water as a bath cannot be used for the substance having Melting point above 100˚C?
Ans. Boiling point of water is 100˚C. So it cannot be used as a bath for determination of Melting points of such substances. Instead of water, Sulphuric acid or Glycerine can be used.
Qs.6 What is “Boiling”?
Ans. It is a rapid change from the liquid to the gaseous state.
Qs.7 What is meant by the “Boiling point” of a liquid?
Ans. The Boiling point of a liquid is the temperature at which the vapour pressure of the liquid is equal to the atmospheric pressure (i.e. 760 mm.)
Qs.8 Why temperature remains constant at boiling point of a liquid?
Ans. Because the heat is utilized for converting liquid into vapours, so temperature remains constant.
Qs.9 What is the Boiling point of pure water? Is the same at all places?
Ans. The Boiling point of pure water is 100˚C at a pressure of 760 torr. The pressure of air at all places is not the same. So B.P. differs at various places.
Qs.10 What is the difference between Boiling and Evaporation?
Ans. Boiling is a rapid change and takes place through out the mass of the liquid at a definite temperature (i.e., boiling point). While Evaporation is a slow change and takes place at the surface of the liquid at all temperature.
Qs.11 What is the effect of pressure on Boiling point?
Ans. It is raised by the increase of pressure and is lowered by the decrease of pressure. (An increase or decrease of 26.7 m.m. of pressure increase or decrease the Boiling point by 1˚ C).
Qs.12 What is the effect of height on the Boiling point of a liquid?
Ans. With heights, the boiling points are reduced as the pressure decreases.
Qs.13 Why should we stir the liquid in the beaker?
Ans. We should stir it constantly because, this helps the liquid to maintain uniform temperature.
Introduction to Fundamental Concepts of Chemistry
Atom
It is the smallest particle of an element which can exist with all the properties of its own element but it cannot exist in atmosphere alone.
Molecule
When two or more than two atoms are combined with each other a molecule is formed. It can exist freely in nature.
Formula Weight
It is the sum of the weights of the atoms present in the formula of a substance.
Molecular Weight
It is the sum of the atomic masses of all the atoms present in a molecule.
Chemistry
It is a branch of science which deals with the properties, composition and the structure of matter.
Empirical Formula
Definition
It is the simplest formula of a chemical compound which represents the element present of the compound and also represent the simplest ratio between the elements of the compound.
Examples
The empirical formula of benzene is "CH". It indicates that the benzene molecule is composed of two elements carbon and hydrogen and the ratio between these two elements is 1:1.
The empirical formula of glucose is "CH2O". This formula represents that glucose molecule is composed of three elements carbon, hydrogen and oxygen. The ratio between carbon and oxygen is equal but hydrogen is double.
Determination of Empirical Formula
To determine the empirical formula of a compound following steps are required.
1. To detect the elements present in the compound.
2. To determine the masses of each element.
3. To calculate the percentage of each element.
4. Determination of mole composition of each element.
5. Determination of simplest ratio between the element of the compound.
Illustrated Example of Empirical Formula
Consider an unknown compound whose empirical formula is to be determined is given to us. Now we will use the above five steps in order to calculate the empirical formula.
Step I - Determination of the Elements
By performing test it is found that the compound contains magnesium and oxygen elements.
Step II - Determination of the Masses
Masses of the elements are experimentally determined which are given below.
Mass of Mg = 2.4 gm
Mass of Oxygen = 1.6 gm
Step III - Estimation of the Percentage
The percentage of an element may be determined by using the formula.
% of element = Mass of element / Mass of compound x 100
In the given compound two elements are present which are magnesium and oxygen, therefore mass of compound is equal to the sum of the mass of magnesium and mass of oxygen.
Mass of compound = 2.4 + 1.6 = 4.0 gm
% Mg = Mass of Mg / Mass of Compound x 100
= 2.4 / 4.0 x 100
= 60%
% O = Mass of Oxygen / Mass of Compound x 100
= 1.6 / 4.0 x 100
= 40%
Step IV - Determination of Mole Composition
Mole composition of the elements is obtained by dividing percentage of each element with its atomic mass.
Mole ratio of Mg = Percentage of Mg / Atomic Mass of Mg
= 60 / 24
= 2.5
Mole ratio of Mg = Percentage of Oxygen / Atomic Mass of Oxygen
= 40 / 16
= 2.5
Step V - Determination of Simplest Ratio
To obtain the simplest ratio of the atoms the quotients obtained in the step IV are divided by the smallest quotients.
Mg = 2.5 / 2.5 = 1
O = 2.5 / 2.5 = 1
Thus the empirical formula of the compound is MgO
Note
If the number obtained in the simplest ratio is not a whole number then multiply this number with a smallest number such that it becomes a whole number maintain their proportion.
Molecular Formula
Definition
The formula which shows the actual number of atoms of each element present in a molecule is called molecular formula.
OR
It is a formula which represents the element ratio between the elements and actual number of atoms of each type of elements present per molecule of the compound.
Examples
The molecular formula of benzene is "C6H6". It indicates that
1. Benzene molecule is composed of two elements carbon and hydrogen.
2. The ratio between carbon and hydrogen is 1:1.
3. The number of atoms present per molecule of benzene are 6 carbon and 6 hydrogen atoms.
The molecular formula of glucose is "C6H12O6". The formula represents that
1. Glucose molecule is composed of three elements carbon, hydrogen and oxygen.
2. The ratio between the atoms of carbon, hydrogen and oxygen is 1:2:1.
3. The number of atoms present per molecule of glucose are 6 carbon atoms. 12 hydrogen atoms and 6 oxygen atoms.
Determination of Molecular Formula
The molecular formula of a compound is an integral multiple of its empirical formula.
Molecular formula = (Empirical formula)n
Where n is a digit = 1, 2, 3 etc.
Hence the first step in the determination of molecular formula is to calculate its empirical formula by using the procedure as explained in empirical formula. After that the next step is to calculate the value of n
n = Molecular Mass / Empirical Formula Mass
Example
The empirical formula of a compound is CH2O and its molecular mass is 180.
To calculate the molecular formula of the compound first of all we will calculate its empirical formula mass
Empirical formula mass of CH2O = 12 + 1 x 2 + 16
= 30
n = Molecular Mass / Empirical Formula Mass
= 180 / 30
= 6
Molecular formula = (Empirical formula)n
= (CH2O)6
= C6H12O6
Molecular Mass
Definition
The sum of masses of the atoms present in a molecule is called as molecular mass.
OR
It is the comparison that how mach a molecule of a substance is heavier than 1/12th weight or mass of carbon atom.
Example
The molecular mass of CO2 may be calculated as
Molecular mass of CO2 = Mass of Carbon + 2 (Mass of Oxygen)
= 12 + 2 x 16
= 44 a.m.u
Molecular mass of H2O = (Mass of Hydrogen) x 2 + Mass of Oxygen
= 1 x 2 + 16
= 18 a.m.u
Molecular mass of HCl = Mass of Hydrogen + Mass of Chlorine
= 1 + 35.5
= 36.5 a.m.u
Gram Molecular Mass
Definition
The molecular mass of a compound expressed in gram is called gram molecular mass or mole.
Examples
1. The molecular mass of H2O is 18. If we take 18 gm H2O then it is called 1 gm molecular mass of H2O or 1 mole of water.
2. The molecular mass of HCl is 36.5. If we take 36.5 gm of HCl then it is called as 1 gm molecular mass of HCl or 1 mole of HCl.
Mole
Definition
It is defined as atomic mass of an element, molecular mass of a compound or formula mass of a substance expressed in grams is called as mole.
OR
The amount of a substance that contains as many number of particles (atoms, molecules or ions) as there are atoms contained in 12 gm of pure carbon.
Examples
1. The atomic mass of hydrogen is one. If we take 1 gm of hydrogen, it is equal to one mole of hydrogen.
2. The atomic mass of Na is 23 if we take 23 gm of Na then it is equal to one mole of Na.
3. The atomic mass of sulphur is 32. When we take 32 gm of sulphur then it is called one mole of sulphur.
From these examples we can say that atomic mass of an element expressed in grams is called mole.
Similarly molecular masses expressed in grams is also known as mole e.g.
The molecular mass of CO2 is 44. If we take 44 gm of CO2 it is called one mole of CO2 or the molecular mass of H2O is 18. If we take 18 gm of H2O it is called one mole of H2O.
When atomic mass of an element expressed in grams it is called gram atom
While
The molecular mass of a compound expressed in grams is called gram molecule.
According to the definition of mole.
One gram atom contain 6.02 x 10(23) atoms
While
One gram molecule contain 6.02 x 10(23) molecules.
Avagadro's Number
An Italian scientist, Avagadro's calculated that the number of particles (atoms, molecules) in one mole of a substance are always equal to 6.02 x 10(23). This number is known as Avogadro's number and represented as N(A).
Example
1 gm mole of Na contain 6.02 x 10(23) atoms of Na.
1 gm mole of Sulphur = 6.02 x 10(23) atoms of Sulphur.
1 gm mole of H2SO4 = 6.02 x 10(23) molecules H2SO4
1 gm mole of H2O = 6.02 x 10(23) molecules of H2O
On the basis of Avogadro's Number "mole" is also defined as
Mass of 6.02 x 10(23) molecules, atoms or ions in gram is called mole.
Determination Of The Number Of Atoms Or Molecules In The Given Mass Of A Substance
Example 1
Calculate the number of atoms in 9.2 gm of Na.
Solution
Atomic mass of Na = 23 a.m.u
If we take 23 gm of Na, it is equal to 1 mole.
23 gm of Na contain 6.02 x 10(23) atoms
1 gm of Na contain 6.02 x 10(23) / 23 atoms
9.2 gm of Na contain 9.2 x 6.02 x 10(23) /23
= 2.408 x 10(23) atoms of Na
Determination Of The Mass Of Given Number Of Atoms Or Molecules Of A Substance
Example 2
Calculate the mass in grams of 3.01 x 10(23) molecules of glucose.
Solution
Molecular mass of glucose = 180 a.m.u
So when we take 180 gm of glucose it is equal to one mole So,
6.02 x 10(23) molecules of glucose = 180 gm
1 molecule of glucose = 180 / 6.02 x 10(23) gm
3.01 x 10(23) molecules of glucose = 3.01 x 10(23) x 180 / 6.02 x 10(23)
= 90 gm
Stoichiometry
(Calculation Based On Chemical Equations)
Definition
The study of relationship between the amount of reactant and the products in chemical reactions as given by chemical equations is called stoichiometry.
In this study we always use a balanced chemical equation because a balanced chemical equation tells us the exact mass ratio of the reactants and products in the chemical reaction.
There are three relationships involved for the stoichiometric calculations from the balanced chemical equations which are
1. Mass - Mass Relationship
2. Mass - Volume Relationship
3. Volume - Volume Relationship
Mass - Mass Relationship
In this relationship we can determine the unknown mass of a reactant or product from a given mass of teh substance involved in the chemical reaction by using a balanced chemical equation.
Example
Calculate the mass of CO2 that can be obtained by heating 50 gm of limestone.
Solution
Step I - Write a Balanced Equation
CaCO3 ----> CaO + CO2
Step II - Write Down The Molecular Masses And Moles Of Reactant & Product
CaCO3 ----> CaO + CO2
Method I - MOLE METHOD
Number of moles of 50 gm of CaCO3 = 50 / 100 = 0.5 mole
According to equation
1 mole of CaCO3 gives 1 mole of CO2
0.5 mole of CaCO3 will give 0.5 mole of CO2
Mass of CO2 = Moles x Molecular Mass
= 0.5 x 44
= 22 gm
Method II - FACTOR METHOD
From equation we may write as
100 gm of CaCO3 gives 44 gm of CO2
1 gm of CaCO3 will give 44/100 gm of CO2
50 gm of CaCO3 will give 50 x 44 / 100 gm of CO2
= 22 gm of CO2
Mass - Volume Relationship
The major quantities of gases can be expressed in terms of volume as well as masses. According to Avogardro One gm mole of any gas always occupies 22.4 dm3 volume at S.T.P. So this law is applied in mass-volume relationship.
This relationship is useful in determining the unknown mass or volume of reactant or product by using a given mass or volume of some substance in a chemical reaction.
Example
Calculate the volume of CO2 gas produced at S.T.P by combustion of 20 gm of CH4.
Solution
Step I - Write a Balanced Equation
CH4 + 2 O2 ----> CO2 + 2 H2O
Step II - Write Down The Molecular Masses And Moles Of Reactant & Product
CH4 + 2 O2 ----> CO2 + 2 H2O
Method I - MOLE METHOD
Convert the given mass of CH4 in moles
Number of moles of CH4 = Given Mass of CH4 / Molar Mass of CH4
From Equation
1 mole of CH4 gives 1 moles of CO2
1.25 mole of CH4 will give 1.25 mole of CO2
No. of moles of CO2 obtained = 1.25
But 1 mole of CO2 at S.T.P occupies 22.4 dm3
1.25 mole of CO2 at S.T.P occupies 22.4 x 1.25
= 28 dm3
Method II - FACTOR METHOD
Molecular mass of CH4 = 16
Molecular mass of CO2 = 44
According to the equation
16 gm of CH4 gives 44 gm of CO2
1 gm of CH4 will give 44/16 gm of CO2
20 gm of CH4 will give 20 x 44/16 gm of CO2
= 55 gm of CO2
44 gm of CO2 at S.T.P occupy a volume 22.4 dm3
1 gm of CO2 at S.T.P occupy a volume 22.4/44 dm3
55 gm of CO2 at S.T.P occupy a volume 55 x 22.4/44
= 28 dm3
Volume - Volume Relationship
This relationship determine the unknown volumes of reactants or products from a known volume of other gas.
This relationship is based on Gay-Lussac's law of combining volume which states that gases react in the ratio of small whole number by volume under similar conditions of temperature & pressure.
Consider this equation
CH4 + 2 O2 ----> CO2 + 2 H2O
In this reaction one volume of CH4 gas reacts with two volumes of oxygen gas to give one volume of CO2 and two volumes of H2O
Examples
What volume of O2 at S.T.P is required to burn 500 litres (dm3) of C2H4 (ethylene)?
Solution
Step I - Write a Balanced Equation
C2H4 + 3 O2 ----> 2 CO2 + 2 H2O
Step II - Write Down The Moles And Volume Of Reactant & Product
C2H4 + 3 O2 ----> 2 CO2 + 2 H2O
According to Equation
1 dm3 of C2H4 requires 3 dm3 of O2
500 dm3 of C2H4 requires 3 x 500 dm3 of O2
= 1500 dm3 of O2
Limiting Reactant
In stoichiometry when more than one reactant is involved in a chemical reaction, it is not so simple to get actual result of the stoichiometric problem by making relationship between any one of the reactant and product, which are involved in the chemical reaction. As we know that when any one of the reactant is completely used or consumed the reaction is stopped no matter the other reactants are present in very large quantity. This reactant which is totally consumed during the chemical reaction due to which the reaction is stopped is called limiting reactant.
Limiting reactant help us in calculating the actual amount of product formed during the chemical reaction. To understand the concept the limiting reactant consider the following calculation.
Problem
We are provided 50 gm of H2 and 50 gm of N2. Calculate how many gm of NH3 will be formed when the reaction is irreversible.
The equation for the reaction is as follows.
N2 + 3 H2 ----> 2 NH3
Solution
In this problem moles of N2 and H2 are as follows
Moles of N2 = Mass of N2 / Mol. Mass of N2
= 50 / 28
= 1.79
Moles of H2 = Mass of H2 / Mol. Mass of H2
= 50 / 2
= 25
So, the provided moles for the reaction are
nitrogen = 1.79 moles and hydrogen = 25 moles
But in the equation of the process 1 mole of nitrogen require 3 mole of hydrogen. Therefore the provided moles of nitrogen i.e. 1.79 require 1.79 x 3 moles of hydrogen i.e. 5.37 moles although 25 moles of H2 are provided but when nitrogen is consumed the reaction will be stopped and the remaining hydrogen is useless for the reaction so in this problem N2 is a limiting reactant by which we can calculate the actual amount of product formed during the reaction.
N2 + 3 H2 ----> 2 NH3
1 mole of N2 gives 2 moles of NH3
1.79 mole of N2 gives 2 x 1.79 moles of NH3
= 3.58 moles of NH3
Mass of NH3 = Moles of NH3 x Mol. Mass
= 3.58 x 17
= 60.86 gm of NH3
Atom
It is the smallest particle of an element which can exist with all the properties of its own element but it cannot exist in atmosphere alone.
Molecule
When two or more than two atoms are combined with each other a molecule is formed. It can exist freely in nature.
Formula Weight
It is the sum of the weights of the atoms present in the formula of a substance.
Molecular Weight
It is the sum of the atomic masses of all the atoms present in a molecule.
Chemistry
It is a branch of science which deals with the properties, composition and the structure of matter.
Empirical Formula
Definition
It is the simplest formula of a chemical compound which represents the element present of the compound and also represent the simplest ratio between the elements of the compound.
Examples
The empirical formula of benzene is "CH". It indicates that the benzene molecule is composed of two elements carbon and hydrogen and the ratio between these two elements is 1:1.
The empirical formula of glucose is "CH2O". This formula represents that glucose molecule is composed of three elements carbon, hydrogen and oxygen. The ratio between carbon and oxygen is equal but hydrogen is double.
Determination of Empirical Formula
To determine the empirical formula of a compound following steps are required.
1. To detect the elements present in the compound.
2. To determine the masses of each element.
3. To calculate the percentage of each element.
4. Determination of mole composition of each element.
5. Determination of simplest ratio between the element of the compound.
Illustrated Example of Empirical Formula
Consider an unknown compound whose empirical formula is to be determined is given to us. Now we will use the above five steps in order to calculate the empirical formula.
Step I - Determination of the Elements
By performing test it is found that the compound contains magnesium and oxygen elements.
Step II - Determination of the Masses
Masses of the elements are experimentally determined which are given below.
Mass of Mg = 2.4 gm
Mass of Oxygen = 1.6 gm
Step III - Estimation of the Percentage
The percentage of an element may be determined by using the formula.
% of element = Mass of element / Mass of compound x 100
In the given compound two elements are present which are magnesium and oxygen, therefore mass of compound is equal to the sum of the mass of magnesium and mass of oxygen.
Mass of compound = 2.4 + 1.6 = 4.0 gm
% Mg = Mass of Mg / Mass of Compound x 100
= 2.4 / 4.0 x 100
= 60%
% O = Mass of Oxygen / Mass of Compound x 100
= 1.6 / 4.0 x 100
= 40%
Step IV - Determination of Mole Composition
Mole composition of the elements is obtained by dividing percentage of each element with its atomic mass.
Mole ratio of Mg = Percentage of Mg / Atomic Mass of Mg
= 60 / 24
= 2.5
Mole ratio of Mg = Percentage of Oxygen / Atomic Mass of Oxygen
= 40 / 16
= 2.5
Step V - Determination of Simplest Ratio
To obtain the simplest ratio of the atoms the quotients obtained in the step IV are divided by the smallest quotients.
Mg = 2.5 / 2.5 = 1
O = 2.5 / 2.5 = 1
Thus the empirical formula of the compound is MgO
Note
If the number obtained in the simplest ratio is not a whole number then multiply this number with a smallest number such that it becomes a whole number maintain their proportion.
Molecular Formula
Definition
The formula which shows the actual number of atoms of each element present in a molecule is called molecular formula.
OR
It is a formula which represents the element ratio between the elements and actual number of atoms of each type of elements present per molecule of the compound.
Examples
The molecular formula of benzene is "C6H6". It indicates that
1. Benzene molecule is composed of two elements carbon and hydrogen.
2. The ratio between carbon and hydrogen is 1:1.
3. The number of atoms present per molecule of benzene are 6 carbon and 6 hydrogen atoms.
The molecular formula of glucose is "C6H12O6". The formula represents that
1. Glucose molecule is composed of three elements carbon, hydrogen and oxygen.
2. The ratio between the atoms of carbon, hydrogen and oxygen is 1:2:1.
3. The number of atoms present per molecule of glucose are 6 carbon atoms. 12 hydrogen atoms and 6 oxygen atoms.
Determination of Molecular Formula
The molecular formula of a compound is an integral multiple of its empirical formula.
Molecular formula = (Empirical formula)n
Where n is a digit = 1, 2, 3 etc.
Hence the first step in the determination of molecular formula is to calculate its empirical formula by using the procedure as explained in empirical formula. After that the next step is to calculate the value of n
n = Molecular Mass / Empirical Formula Mass
Example
The empirical formula of a compound is CH2O and its molecular mass is 180.
To calculate the molecular formula of the compound first of all we will calculate its empirical formula mass
Empirical formula mass of CH2O = 12 + 1 x 2 + 16
= 30
n = Molecular Mass / Empirical Formula Mass
= 180 / 30
= 6
Molecular formula = (Empirical formula)n
= (CH2O)6
= C6H12O6
Molecular Mass
Definition
The sum of masses of the atoms present in a molecule is called as molecular mass.
OR
It is the comparison that how mach a molecule of a substance is heavier than 1/12th weight or mass of carbon atom.
Example
The molecular mass of CO2 may be calculated as
Molecular mass of CO2 = Mass of Carbon + 2 (Mass of Oxygen)
= 12 + 2 x 16
= 44 a.m.u
Molecular mass of H2O = (Mass of Hydrogen) x 2 + Mass of Oxygen
= 1 x 2 + 16
= 18 a.m.u
Molecular mass of HCl = Mass of Hydrogen + Mass of Chlorine
= 1 + 35.5
= 36.5 a.m.u
Gram Molecular Mass
Definition
The molecular mass of a compound expressed in gram is called gram molecular mass or mole.
Examples
1. The molecular mass of H2O is 18. If we take 18 gm H2O then it is called 1 gm molecular mass of H2O or 1 mole of water.
2. The molecular mass of HCl is 36.5. If we take 36.5 gm of HCl then it is called as 1 gm molecular mass of HCl or 1 mole of HCl.
Mole
Definition
It is defined as atomic mass of an element, molecular mass of a compound or formula mass of a substance expressed in grams is called as mole.
OR
The amount of a substance that contains as many number of particles (atoms, molecules or ions) as there are atoms contained in 12 gm of pure carbon.
Examples
1. The atomic mass of hydrogen is one. If we take 1 gm of hydrogen, it is equal to one mole of hydrogen.
2. The atomic mass of Na is 23 if we take 23 gm of Na then it is equal to one mole of Na.
3. The atomic mass of sulphur is 32. When we take 32 gm of sulphur then it is called one mole of sulphur.
From these examples we can say that atomic mass of an element expressed in grams is called mole.
Similarly molecular masses expressed in grams is also known as mole e.g.
The molecular mass of CO2 is 44. If we take 44 gm of CO2 it is called one mole of CO2 or the molecular mass of H2O is 18. If we take 18 gm of H2O it is called one mole of H2O.
When atomic mass of an element expressed in grams it is called gram atom
While
The molecular mass of a compound expressed in grams is called gram molecule.
According to the definition of mole.
One gram atom contain 6.02 x 10(23) atoms
While
One gram molecule contain 6.02 x 10(23) molecules.
Avagadro's Number
An Italian scientist, Avagadro's calculated that the number of particles (atoms, molecules) in one mole of a substance are always equal to 6.02 x 10(23). This number is known as Avogadro's number and represented as N(A).
Example
1 gm mole of Na contain 6.02 x 10(23) atoms of Na.
1 gm mole of Sulphur = 6.02 x 10(23) atoms of Sulphur.
1 gm mole of H2SO4 = 6.02 x 10(23) molecules H2SO4
1 gm mole of H2O = 6.02 x 10(23) molecules of H2O
On the basis of Avogadro's Number "mole" is also defined as
Mass of 6.02 x 10(23) molecules, atoms or ions in gram is called mole.
Determination Of The Number Of Atoms Or Molecules In The Given Mass Of A Substance
Example 1
Calculate the number of atoms in 9.2 gm of Na.
Solution
Atomic mass of Na = 23 a.m.u
If we take 23 gm of Na, it is equal to 1 mole.
23 gm of Na contain 6.02 x 10(23) atoms
1 gm of Na contain 6.02 x 10(23) / 23 atoms
9.2 gm of Na contain 9.2 x 6.02 x 10(23) /23
= 2.408 x 10(23) atoms of Na
Determination Of The Mass Of Given Number Of Atoms Or Molecules Of A Substance
Example 2
Calculate the mass in grams of 3.01 x 10(23) molecules of glucose.
Solution
Molecular mass of glucose = 180 a.m.u
So when we take 180 gm of glucose it is equal to one mole So,
6.02 x 10(23) molecules of glucose = 180 gm
1 molecule of glucose = 180 / 6.02 x 10(23) gm
3.01 x 10(23) molecules of glucose = 3.01 x 10(23) x 180 / 6.02 x 10(23)
= 90 gm
Stoichiometry
(Calculation Based On Chemical Equations)
Definition
The study of relationship between the amount of reactant and the products in chemical reactions as given by chemical equations is called stoichiometry.
In this study we always use a balanced chemical equation because a balanced chemical equation tells us the exact mass ratio of the reactants and products in the chemical reaction.
There are three relationships involved for the stoichiometric calculations from the balanced chemical equations which are
1. Mass - Mass Relationship
2. Mass - Volume Relationship
3. Volume - Volume Relationship
Mass - Mass Relationship
In this relationship we can determine the unknown mass of a reactant or product from a given mass of teh substance involved in the chemical reaction by using a balanced chemical equation.
Example
Calculate the mass of CO2 that can be obtained by heating 50 gm of limestone.
Solution
Step I - Write a Balanced Equation
CaCO3 ----> CaO + CO2
Step II - Write Down The Molecular Masses And Moles Of Reactant & Product
CaCO3 ----> CaO + CO2
Method I - MOLE METHOD
Number of moles of 50 gm of CaCO3 = 50 / 100 = 0.5 mole
According to equation
1 mole of CaCO3 gives 1 mole of CO2
0.5 mole of CaCO3 will give 0.5 mole of CO2
Mass of CO2 = Moles x Molecular Mass
= 0.5 x 44
= 22 gm
Method II - FACTOR METHOD
From equation we may write as
100 gm of CaCO3 gives 44 gm of CO2
1 gm of CaCO3 will give 44/100 gm of CO2
50 gm of CaCO3 will give 50 x 44 / 100 gm of CO2
= 22 gm of CO2
Mass - Volume Relationship
The major quantities of gases can be expressed in terms of volume as well as masses. According to Avogardro One gm mole of any gas always occupies 22.4 dm3 volume at S.T.P. So this law is applied in mass-volume relationship.
This relationship is useful in determining the unknown mass or volume of reactant or product by using a given mass or volume of some substance in a chemical reaction.
Example
Calculate the volume of CO2 gas produced at S.T.P by combustion of 20 gm of CH4.
Solution
Step I - Write a Balanced Equation
CH4 + 2 O2 ----> CO2 + 2 H2O
Step II - Write Down The Molecular Masses And Moles Of Reactant & Product
CH4 + 2 O2 ----> CO2 + 2 H2O
Method I - MOLE METHOD
Convert the given mass of CH4 in moles
Number of moles of CH4 = Given Mass of CH4 / Molar Mass of CH4
From Equation
1 mole of CH4 gives 1 moles of CO2
1.25 mole of CH4 will give 1.25 mole of CO2
No. of moles of CO2 obtained = 1.25
But 1 mole of CO2 at S.T.P occupies 22.4 dm3
1.25 mole of CO2 at S.T.P occupies 22.4 x 1.25
= 28 dm3
Method II - FACTOR METHOD
Molecular mass of CH4 = 16
Molecular mass of CO2 = 44
According to the equation
16 gm of CH4 gives 44 gm of CO2
1 gm of CH4 will give 44/16 gm of CO2
20 gm of CH4 will give 20 x 44/16 gm of CO2
= 55 gm of CO2
44 gm of CO2 at S.T.P occupy a volume 22.4 dm3
1 gm of CO2 at S.T.P occupy a volume 22.4/44 dm3
55 gm of CO2 at S.T.P occupy a volume 55 x 22.4/44
= 28 dm3
Volume - Volume Relationship
This relationship determine the unknown volumes of reactants or products from a known volume of other gas.
This relationship is based on Gay-Lussac's law of combining volume which states that gases react in the ratio of small whole number by volume under similar conditions of temperature & pressure.
Consider this equation
CH4 + 2 O2 ----> CO2 + 2 H2O
In this reaction one volume of CH4 gas reacts with two volumes of oxygen gas to give one volume of CO2 and two volumes of H2O
Examples
What volume of O2 at S.T.P is required to burn 500 litres (dm3) of C2H4 (ethylene)?
Solution
Step I - Write a Balanced Equation
C2H4 + 3 O2 ----> 2 CO2 + 2 H2O
Step II - Write Down The Moles And Volume Of Reactant & Product
C2H4 + 3 O2 ----> 2 CO2 + 2 H2O
According to Equation
1 dm3 of C2H4 requires 3 dm3 of O2
500 dm3 of C2H4 requires 3 x 500 dm3 of O2
= 1500 dm3 of O2
Limiting Reactant
In stoichiometry when more than one reactant is involved in a chemical reaction, it is not so simple to get actual result of the stoichiometric problem by making relationship between any one of the reactant and product, which are involved in the chemical reaction. As we know that when any one of the reactant is completely used or consumed the reaction is stopped no matter the other reactants are present in very large quantity. This reactant which is totally consumed during the chemical reaction due to which the reaction is stopped is called limiting reactant.
Limiting reactant help us in calculating the actual amount of product formed during the chemical reaction. To understand the concept the limiting reactant consider the following calculation.
Problem
We are provided 50 gm of H2 and 50 gm of N2. Calculate how many gm of NH3 will be formed when the reaction is irreversible.
The equation for the reaction is as follows.
N2 + 3 H2 ----> 2 NH3
Solution
In this problem moles of N2 and H2 are as follows
Moles of N2 = Mass of N2 / Mol. Mass of N2
= 50 / 28
= 1.79
Moles of H2 = Mass of H2 / Mol. Mass of H2
= 50 / 2
= 25
So, the provided moles for the reaction are
nitrogen = 1.79 moles and hydrogen = 25 moles
But in the equation of the process 1 mole of nitrogen require 3 mole of hydrogen. Therefore the provided moles of nitrogen i.e. 1.79 require 1.79 x 3 moles of hydrogen i.e. 5.37 moles although 25 moles of H2 are provided but when nitrogen is consumed the reaction will be stopped and the remaining hydrogen is useless for the reaction so in this problem N2 is a limiting reactant by which we can calculate the actual amount of product formed during the reaction.
N2 + 3 H2 ----> 2 NH3
1 mole of N2 gives 2 moles of NH3
1.79 mole of N2 gives 2 x 1.79 moles of NH3
= 3.58 moles of NH3
Mass of NH3 = Moles of NH3 x Mol. Mass
= 3.58 x 17
= 60.86 gm of NH3
Three States Of Matter
Matter
It is defined as any thing which has mass and occupies space is called matter.
Matter is composed of small and tiny particles called Atoms or molecules. It exist in three different states which are gaseous, liquid & solid.
Properties of Gas
1. It has no definite shape.
2. It has no definite volume, so it can be compressed or expanded.
3. A gas may diffuse with the other gas.
4. The molecules of a gas are in continuous motion.
Properties of Liquids
1. A liquid has no definite shape.
2. It has a fixed volume.
3. The diffusion of a liquid into the other liquid is possible if both of the liquids are polar or non-polar.
4. It can be compressed to a negligible.
Properties of Solids
1. A solid has a definite shape.
2. It has a fixed volume.
3. The rate of diffusion of solid with each other is very slow.
4. It cannot be compressed easily.
Kinetic Theory of Gases
It was an idea of some scientist like Maxwell & Bolzmann that the properties of gases are due to their molecular motion. This motion of the molecules is related with the kinetic energy, so the postulates give by the scientist about the behaviour of gases are collectively known as kinetic molecular theory of gases.
The postulates of kinetic molecular theory are as follows.
1. All gases consists of very large number of tiny particles called molecules.
2. These molecules are widely separated from each other and are so small that they are invisible.
3. The size of the molecules is very small as compared to the distance between them.
4. There is no attractive or repulsive force between molecules so they can move freely.
5. The molecules are very hard and perfectly elastic so when they collide no loss of energy takes place.
6. The gas molecules are in continuous motion they move in a straight path until they collide. The distance between two continuous collision is called Mean Free Path.
7. During their motion these molecules are collided with one another and with the walls of the container.
8. The collision of the molecules are perfectly elastic. When molecules collide they rebound with perfect elasticity and without loss or gain of energy.
9. The pressure of the gas is the result of collision of molecules on the walls of the container.
10. The average kinetic energy of gas molecules depends upon the absolute temperature. At any given temperature the molecules of all gases have the same average kinetic energy (1/2 mv2).
Kinetic Theory of Liquids
This theory is bases on the following assumptions.
1. The particles of a liquid are very close to each other due to which a liquid has fixed volume.
2. The particles in a liquid are free to move so they have no definite shape.
3. During the motion these molecules collides with each other and with the walls of the container.
4. These molecules possess kinetic which is directly proportional to its absolute temperature.
Kinetic Theory of Solids
The assumptions of kinetic theory for solids are as follows.
1. The particles in a solid are very closely packed due to strong attractive forces between the molecules.
2. These molecules are present at a fixed position and are unable to move.
3. They have definite shape because the particles are arranged in a fixed pattern.
4. They possess only vibrational energy.
Mean Free Path
The distance which a molecule of a gas travels before its collision with the other molecule is called free path. This distance between the collision of the molecules changes constantly so the average distance which a molecule travels before its collision is called mean free path.
Boyle's Law
A relationship of volume with external pressure was given by Boyle's in the form of law. This law is known as Boyle's Law which states,
For a given mass of a gas the volume of the gas is inversely proportional to its pressure provided the temperature is kept constant.
Mathematically it may be written as
V ∞ 1 / P
Or V = K / P
Or PV = K
On the bases of the relation, Boyle's law can also be stated as
The product of the pressure and volume of a given mass of a gas is always constant at constant temperature.
Explanation
Consider for a given mass a gas having volume V1 at pressure P1, so according to Boyle's Law we may write as
P1V1 = K1 (constant)
If the pressure of the above system is changed from P1 to P2 then the volume of the gas will also change from V1 to V2. For this new condition of the gas we can write as,
P2V2 = K2 (constant)
But for the same mass of the gas.
K1 = K2
P1V1 = P2V2
This equation is known as Boyle's Equation.
Charle's Law
We know that everything expand on heating and contract cooling. This change in volume is small in liquids and solids but gases exhibit enormous changes due to the presence of large intermolecular spaces.
Change of volume of a gas with the change of temperature at constant pressure was studied by Charles and was given in the form of a law. which states,
Statement
For a given mass of a gas the volume of the gas is directly proportional to its absolute temperature provided the pressure is kept constant.
Mathematically this law may be written as
V ∞ T
V = K T
OR
V / T = K
This relation shows that the ratio of volume of a given mass of a gas to its absolute temperature is always constant provided the pressure is kept constant. On this bases Charles Law may also be defined as,
If the pressure remains constant for each 1ºC change of temperature the volume of the gas changes to 1/273 of its original volume.
On the bases of this statement
V1 / T = K & V2 / T2 = K
V1 / T1 = V2 / T2
This equation is known as Charle's equation.
The volume temperature relationship can be represented graphically. When volume of a given mass of gas is plotted against temperature, a straight line is obtained.
Graph Coming Soon
Absolute Scale Of Temperature
There are different scales for the measurement of temperature such as Celsius ºC and Fahrenheit ºC. Similarly another scale known as absolute scale or Kelvin scale is determined on the basis of Charle's law.
On the basis of Charle's law we known that the volume of the gas changes to 1/273 times of its original volume for each 1 ºC change of temperature. It suggests that the volume of a gas would theoretically be zero at -273ºC. But this temperature has never been achieved for any gas because all the gases condense to liquid at a temperature above this point. So the minimum possible temperature for a gaseous system is to be -273ºC. This temperature is referred as absolute zero or zero degree of the absolute scale or Kelvin scale.
To form an absolute scale thermometer if the equally spaced divisions of centigrade thermometer are extended below zero and when the point -273ºC is maked then this point is called as absolute zero and the scale is called as absolute scale. It shows that for the conversion of centigrade scale into Kelvin scale 273 is added to the degrees on the centigrade scale.
K = 273 + ºC
Avogadro's Law
In 1811, a scientist Avogadro's established a relationship between the volume and number of molecules of the gas, which is known as Avogadro's law.
Statement
Equal volume of all gases contains equal number of molecules under the same condition of temperature & pressure.
Mathematically it may be represented as
V ∞ n
OR
V = K n
On the basis of the above statement we can say that
1 dm3 of O2 gas will contain the same number of molecules as 1 dm3 of H2 or N2 or any other gas at same temperature and pressure.
It was also observed that 22.4 dm3 of any gas at S.T.P contain 1 mole of that gas, so 22.4 dm3 volume at S.T.P is called as molar volume or the volume of 1 mole of the gas and the mass present in 22.4 dm3 of any gas will be equal to its molar mass or molecular mass. It can also be explained on the basis of following figures.
Determination of Unknown Molecular Mass of a Gas With the Help of Avogadro's Law
Suppose we have two gases (i) Oxygen (ii) CO
The volume of these two gases are equal which are 1 dm3.
The mass of 1 dm3 of oxygen is 1.43 gm
The mass of 1 dm3 of Co is 1.25 gm
According to Avogadro's law we know that 1 dm3 of CO at S.T.P contain the same number of molecules as 1 dm3 of O2 under similar condition. Hence a molecule of CO has 1.25 / 1.43 times as much as a molecule of O2 and we know that the molecular mass of oxygen is 32 so the molecular mass of CO would be
1.25 / 1.43 x 32 = 28 g / mole
General Gas Equation (Ideal Gas Equation)
To give a relation between the volume, pressure and number of moles of n gas, Boyle's law, Charle's law and Avogadro's law are used.
According to Boyle's law | V ∞ 1 / P
According to Charle's law | V ∞ T
According to Avogadro's law | V ∞ n
By combining these laws we get
V ∞ 1 / P x T x n
OR
V = R x 1 / P x T x P
OR
P V = n R T
This equation is known as general gas equation n is also known as equation of state because when we specify the four variables = pressure, temperature, volume and number of moles we define the state for a gas.
In this equation "R" is a constant known as gas constant.
Value of R
1. When Pressure is Expressed in Atmosphere and Volume in Litres or dm3
According to general gas equation
P V = n R T
OR
R = PV / nT
For 1 mole of a gas at S.T.P we know that
V = 22.4 dm3 or litres
T = 273 K (standard temperature)
P = 1 atm (standard pressure)
So,
R = PV / nT
= 1 atm x 22.4 dm3 / 1 mole x 273 K
= 0.0821 dm3 K-1 mol0-1
2. When Pressure is Expressed in Newtons Per Square Metre and Volume in Cubic Metres
For 1 mole of a gas at S.T.P
V = 0.0224 m3 .......... ( 1 dm3 = 10-3 m3)
n = 1 mole
T = 273 K
P = 101200 Nm-2
So,
R = PV / nT
= 101300 Nm-2 x 0.0224 m3 / 1 mole x 273 K
= 8.3143 Nm K-1 mole-1
= 8.3143 J K-1 mol-1
Derivation of Gas Equation
According to general gas equation
P V = n R T
For 1 mole of a gas n = 1
P V = R T
OR
P V / T = R
Consider for a known mass of a gas the volume of the gas is V1 at a temperature T1 and pressure P1. Therefore for this gas we can write as
P1 V1 / T1 = R
If this gas is heated to a temperature T2 due to which the pressure is changed to P2 and volume is changed to V2. For this condition we may write as
P2 V2 / T2 = R
P1 V1 / T1 = P2 V2 / T2 = R
P1 V1 / T1 = P2 V2 / T2
This equation is known as gas equation.
Graham's Law of Diffusion
We know that gas molecules are constantly moving in haphazard direction, therefore when two gases are placed separated by a porous membrane, they diffuse through the membrane and intermix with each other. The phenomenon of mixing of molecules of different gases is called diffusion.
In 1881, Graham established a relationship between the rates of diffusion of gases and their densities which is known as Graham's law of diffusion.
Statement
The rate of diffusion of any gas is inversely proportional to the square root of its density.
Mathematically it can be represented as
r ∞ 1 / √d
r = K / √d
Graham also studied the comparative rates of diffusion of two gases. On this basis the law os defined as
The comparative rates of diffusion of two gases under same condition of temperature and pressure are inversely proportional to the square root of their densities.
If the rate of diffusion of gas A is r1 and its density is d1 then according to Graham's law
r1 ∞ 1 / √d1
OR
r1 = K / √d1
Similarly the rate of diffusion of gas B is r2 and its density is d2 then
r2 ∞ 1 / √d2
OR
r2 = K / √d2
Comparing the two rates
r1 / r2 = (K / √d1) / (K / √d2)
r1 / r2 = √d2 / d1 ................... (A)
But density d = mass / volume
Therefore,
For d1 we may write as
d1 = m1 / v1
And for d2
d2 = m2 / v2
Substituting these values of d1 & d2 in equation (A)
r1 / r2 = √(m2 / v2) / (m1 / v1)
But v1 = v2 because both gases are diffusing in the same volume.
Therefore,
r1 / r2 = √m2 / m1
Hence Graham's law can also be stated as,
The comparative rates of diffusion of two gases are inversely proportional to the square root of their masses under the same condition of temperature and pressure.
It means that a lighter gas will diffuse faster than the heavier gas. For example compare the rate of diffusion of hydrogen and oxygen.
Rate of diffusion of H2 / Rate of diffusion of O2 = √Mass of O2 / Mass of H2 = √32/ 2 = √16 = 4
It shows that H2 gas which is lighter gas than O2 will diffuse four times faster than O2.
Dalton's Law of Partial Pressures
Partial Pressure
In a gaseous mixture the individual pressure oxerted by a gas is known as partial pressure.
When two or more gases which do not react chemically are mixed in the same container each gas will exert the same pressure as it would exert if it alone occupy the same volume.
John Dalton in 1801 formulated a law which is known as Dalton's Law of partial pressure and stated as.
Statement
The total pressure of a gaseous system is equal to the sum of the partial pressures of all the gases present in the system.
Suppose in a system three gases A, B & C are present. The partial pressure of these gases are
PA = Partial pressure of gas A
PB = Partial pressure of gas B
PC = Partial pressure of gas C
Then Dalton's law may be mathematically written as
PT = PA + PB + PC
Where PT is the total pressure of the system.
To calculate the individual pressures of gases in the above example suppose the number of moles of A, B & C in the container are nA, nB and nC. So the total number of moles in the container will be
n = nA + nB + nC
Apply the general gas equation
P V = n R T
PT = n R T / V
Since R, T and V are same for gases A, B and C, therefore the partial pressure of these gases are as follows.
Partial pressure of gas A | PA = n(A)RT / V ......... (2)
Partial pressure of gas B | PB = n(B)RT / V ......... (3)
Partial pressure of gas C | PC = n(C)RT / V ......... (4)
Now divide equation (2) by (1)
PA / PT = (nA RT/V) / (nRT/V)
OR
PA / PT = nA / nT
Therefore,
P(gas) = P1 x n(gas) / n(total)
Application of Dalton's Law
In an inert mixture of gases the individual gas exerts its own pressure due to collision of its molecules with the walls of the container but the total pressure produced on the container wall will be the sum of pressure of all the individual gases of the mixture.
On this basis the number of moles formed during a chemical reaction can be measured. For this purpose a gas produced in a chemical reaction is collected over water. The gas also contains some of water vapours. So the pressure exerted by the gas would be the pressure of pure gas and the pressure of water vapours.
Therefore the pressure of the system may be represented as
P(moist) = P(dry) + P(water vapour)
So,
P(dry) = P(moist) - P(water vapour)
In this way we can obtain the pressure of the gas and by using general gas equation we can calculate the number of moles of the prepared gas.
Ideal Gas
A gas which obeys all the gas laws at all temperatures and pressures is known as ideal gas.
It means that the product of pressure and volume must be constant at all pressures.
Similarly the rate of V/T will remain constant for an ideal gas.
But there is no gas which is perfectly ideal because of the presence of the force of attraction or repulsion between the molecules.
Gas Laws on the Basis of Kinetic Theory
Boyle's Law
According to Boyle's law the volume of a given mass of a gas is inversely proportional to its pressure at constant temperature.
It means that when the volume of the gas is decreased the pressure of the gas will increase.
According to kinetic molecular theory of gases the pressure exerted by a gas is due to the collisions of the molecules with the walls of the container. If the volume of a gas is reduced at constant temperature, the average velocity of the gas molecules remains constant so they collide more frequently wit the walls which causes higher pressure.
Charle's Law
According to Charles law the volume of a given mass of a gas is directly proportional to its absolute temperature at constant pressure.
According to kinetic molecular theory the average kinetic energy of gas molecules is directly proportional to its absolute temperature so if the temperature of the gas is increased the average kinetic energy of the gas molecules is also increased due to which the sample of the gas expanded to keep the pressure constant. It is accordance with the law.
Graham's Law
According to Graham's Law
r1 / r2 = √m2 / m1
The rate of diffusion of a gas is directly proportional to the velocity of the molecules so,
v1 / v2 = √m2 / m1
Liquefaction
According to kinetic theory, the kinetic energy of the molecules is low for lower temperature. These slower moving molecules become subject to inter molecular attraction. At a sufficiently low temperature these attractive forces are capable of holding the molecules with one another so the gas is changed into liquid and the process is called liquefaction.
Liquid State
It is one of the state of matter. In this state, the kinetic energy of the molecule is very high due to which the molecules of the liquid are able to move but due to compact nature liquids are not compressible. On this basis we can say that the volume of a liquid is always constant but its shape can be changed.
Behaviour of Liquids
The main properties of liquids are as follows.
Diffusibility
The diffusion of one liquid into another liquid is possible but its rate is slow as compared with the rate of diffusion of gases. Example of diffusion of liquids is mixing of alcohol in water.
Explanation of Diffusion in Terms of Kinetic Energy
As the molecular of a liquid are in cluster form they are very close to each other but these molecules are movable so they can mix with the other molecules. Since the intermolecular distance are smaller due to which the rate of diffusion of liquids is slow.
Compressibility
The space between liquid molecules are very small due to strong Van der Waals forces. When the pressure is applied, they can be compressed but to a very little extent.
Expansion
When a liquid is heated, the kinetic energy of its molecules also increases so the attraction between the molecules becomes weaker due to which they go further apart and hence the liquid expands.
Contraction
When a liquid is cooled its kinetic energy is lowered and the attraction among the molecules becomes stronger so they comes close to each other and hence the liquid contract.
Viscosity
Definition
The internal resistance in the flow of a liquid is called viscosity.
Liquids have the ability to flow, but different liquids have different rates of flow. Some liquids like honey mobil oil etc. flow slowly and are called viscous liquids while ether, gasoline etc. which flow quickly are called less viscous.
Explanation
The viscosity of liquid can be understood by considering a liquid in a tube, a liquid in a tube is considered as made up of a series of molecular layer. The layer of the liquid in contact with the walls of the tube remains stationary and the layer in the center of the tube has highest velocity as shown.
Each layer exerts a drag on the next layer and causes resistance to flow.
Factors on Which Viscosity Depends
1. Size of Molecules
The viscosity of a liquid depends upon the size of its molecules. If the size of the molecules is bigger the viscosity of the liquid is high.
2. Shape of Molecules
Shape of the molecules affects the viscosity. If the shapes of the molecules are spherical they can move easily but if the shapes of the molecules are irregular such as linear or trigonal then the molecules will move slowly and its viscosity will be high.
3. Intermolecular Attraction
If the force of attraction between the molecules of a liquid is greater the viscosity of the liquid is also greater.
4. Temperature
Viscosity of a liquid decreases with the increase of temperature.
Units of Viscosity
Viscosity of a liquid is measured in poise, centipoise or millipoise & S.I unit.
1 poise = 1 N.s.m(-2)
1 centipoise = 10(-2) N.s.m(-2)
Surface Tension
Definition
The force acting per unit length on the surface of a liquid at right angle direction is called surface tension.
Explanation
Consider a liquid is present in a beaker. The molecules inside the liquid are surrounded by the other molecules of the liquid. So the force of attraction on a molecule is balanced from all direction. But the force of attraction acting on the molecules of the surface from the lower layer molecules is not balanced.
The molecules lying on the surface are attracted by the molecules present below the surface Due to this downward pull the surface of the liquid behave as a membrane which tends to contract to a smaller area and causes a tension on the surface of the liquid known as surface tension.
Diagram Coming Soon
Factors on Which Surface Tension Depends
1. Molecular Structure of the Liquid
If the force of attraction between the molecules is greater, the surface tension of the liquid is also greater. Those liquids in which hydrogen bond formation take place will have more surface tension.
2. Temperature
Surface tension of a liquid is inversely proportional to the temperature.
Units
1. Dynes / cm
2. Ergs / cm2
Capillary Action
The fall or rise of a liquid in a capillary tube is called capillary action.
When a capillary tube is dipped in a liquid which wets the wall of the tube, the liquid will rise in the capillary tube, to decrease the surface area due to surface tension. The liquid will rise in the capillary tube until the upward force due to surface tension is just balanced by the downward gravitational pull. This is called capillary action.
Vapour Pressre
Definition
The pressure exerted by the vapours of a liquid in its equilibrium state with the pure liquid at a given temperature is called vapour pressure.
Explanation
Consider a liquid is present in a bottle as shown.
Diagram Coming Soon
In the beginning the atmosphere above the surface of liquid is unsaturated but due to continuous evaporation the molecule of the liquid are trapped in the bottle and the air present above the surface of the liquid is becomes saturated and after it the molecules present in the vapour state may hit the liquid again and rejoin it by condensing into liquid. Thus in this closed vessel two process are going on simultaneously which are evaporation and condensation of vapours. When the rates of these two processes becomes equal at this point the pressure exerted by vapours is called vapour pressure.
Units of Vapour Pressure
The units for vapour pressure are
1. Millimeter of Hg
2. Atmosphere
3. Torr
4. Newton / m(2)
Factors for Vapour Pressure
1. Nature of Liquid
Vapour pressure of a liquid depends upon the nature of the liquid. Low boiling liquid exert more vapour pressure at a given temperature.
2. Temperature
Vapour pressure of a liquid also depends upon temperature. The vapour pressure of the liquid increases with the increase of temperature due to the increase of average of kinetic energy.
3. Intermolecular Forces
Those liquids in which the intermolecular forces are weak shows high vapour pressure.
Explanation of Evaporation on the Basis of Kinetic Theory
According to this theory the molecules of a liquid collide with each other during their motion. Due to these collisions some of the molecules acquire greater energy than Van der Walls forces which binds the molecules of the liquid together so these molecules of higher energy escapes from the surface into the air in the form of vapours.
Evaporation is a Cooling Process
In liquids, due to collision between molecules some molecules acquire higher energy and escapes from the surface of the liquid in the form of vapours. The kinetic energy of the remaining molecules decreases due to which the temperature of the liquid also decreases and hence we can say that evaporation is a cooling process.
Boiling Point
Definition
The temperature at which the vapour pressure of a liquid becomes equal to the atmospheric pressure is called boiling point.
When a liquid is heated the rate of evaporation of the molecules also increases with the increase in temperature. When the pressure of the vapours becomes equal to the atmospheric pressure the liquid starts boiling and this temperature is known as boiling point.
If the external pressure on a liquid is changed the boiling point of the liquid also change. The increase in external pressure on a liquid increases the boiling point while the decreases of external pressure decrease the boiling point.
Solid State
It is a state of matter which posses both definite shape and definite volume. In solids the particles are very close to each and tightly packed with a greater force of attraction.
Properties of Solids
1. Diffusibility
Diffusion also occurs in solids but its rate is very slow. If a polished piece of zinc is clamped with a piece of copper for a long time. After few years we will see that some particles of zinc are penetrated into copper and some particles of copper are penetrated into zinc. It shows that the diffusion in solids is possible but it occurs with a slow rate.
2. Compressibility
In solids the molecules are close to each other so it is not easy to compress a solid. In other words we can say that the effect of pressure on solids is negligible.
3. Sublimation
It is a property of some solids that on heating these solids are directly converted into vapours without liquification. This property of solids is known as sublimation.
4. Melting
When solids are heated, they are changed into liquids and the property is called melting of the solids.
5. Deformity
Solids may be deformed by high pressure. When a high pressure is applied on solids due to which some particles are dislocated the force of attraction is so strong that the rearranged atoms are held equally well with their new neighbours and hence the solid is deformed.
Classification of Solids
Solids are classified into two main classes.
1. Crystalline
2. Amorphous
1. Crystalline Solids
In a solid if the atoms are attached with each other with a definite arrangement and it also possesses a definite geometrical shape. This type of solid is called crystalline solid.
e.g. NaCl, NiSO4 are crystalline solids.
2. Amorphous Solids
In these solids there is no definite arrangement of the particles so they do not have a definite shape. The particles of such solids have a random three dimensional arrangement. Examples of amorphous solids are glass, rubber, plastic etc.
The properties of crystalline and amorphous solids are quite different from each other. These differences in properties are given below.
Difference of Geometry
1. Crystalline Solids
In crystalline solids particles are arranged in a definite order due to which it possesses a definite structure.
2. Amorphous Solids
In amorphous solids particles are present without any definite arrangement so they do not have definite shape.
Difference of Melting Point
1. Crystalline Solids
Crystalline solids have sharp melting point due to uniform arrangement.
2. Amorphous Solids
Amorphous solids melts over a wide range of temperature.
Cleavage and Cleavage Plane
1. Crystalline Solids
When a big crystal is broken down into smaller pieces the shape of the smaller crystals is identical with the bigger crystal. This property of crystalline solids is called cleavage and the plane from where a big crystal is broken is called cleavage plane.
2. Amorphous Solids
Amorphous solids do not break up into smaller pieces with an identical shape.
Anisotropy & Isotropy
1. Crystalline Solids
It is a property of crystalline solid that they show different physical properties in different direction. For example graphite can conduct electric current only through the plane which is parallel to its layers. This property is called anisotropy.
2. In amorphous solids the physical properties are same in all directions. This property of solids is called isotropy.
Symmetry in Structure
1. Crystalline solids are symmetric in their structure when they are rotated about an axis, their appearance remains same so they are symmetric in structure.
2. Amorphous Solids
Amorphous solids are not symmetric.
Types of Crystals
There are four types of crystals.
1. Atomic crystals
2. Ionic crystals
3. Covalent crystals
4. Molecular crystal
1. Atomic Crystals
Metals are composed of atoms. These atoms are combined with each other by metallic bond and the valency electrons in metals can move freely throughout the crystal lattice. This type of solid is called atomic crystal.
The properties of atomic crystals are
1. High melting point.
2. Electrical and thermal conductivity.
3. These are converted into sheets so these are malleable.
4. These are used as wire so these are ductile.
2. Ionic Crystals
Those solids which consists of negativity and positively charged ions held together by strong electrostatic force of attraction are called ionic crystals. Ionic crystalline solids possesses the following properties.
1. The melting and boiling point of ionic crystals is high.
2. They conduct electricity in molten state.
3. Ionic crystals are very hard.
4. Indefinite growth of crystals is also a property of ionic crystals.
3. Covalent Crystals
In covalent solids, the atoms or molecules are attached with each other by sharing of electrons. Such type of solids are called covalent solids e.g. diamond is a covalent solid in which carbon atoms are attached with each other by covalent bond. The other examples of covalent crystals are sulphur, graphite etc.
Covalent crystals possesses the following properties.
1. High melting point.
2. High refractive index.
3. Low density.
4. Molecular Crystals
Those solid in which molecules are held together due to intermolecular forces to form a crystal lattice are called molecular crystals e.g. iodine and solid CO2 are molecular crystals. The general properties of molecular crystals are as follows.
1. Low melting and boiling point.
2. Non - conductor of heat and electricity.
Isomorphism
When two different substance have same crystalline structure, they are said to be isomorphous and the phenomenon is called isomorphism.
e.g. ZnSO4 and NiSO4 are two different substances but both are orthorhombic similarly the structure of CaCO3 and NaNO3 is frigonal.
Polymorphism
If a substance exist in more than one crystalline form it is called polymorphous and the phenomenon is known as polymorphism. E.g. sulphur exist in rhombic and monoclinic form similarly CaCO3 exist in trigonal and orthorhombic form.
Unit Cell
The basic structural unit of a crystalline solid which when repeated in three dimensions generates the crystal structure is called a unit cell.
A unit cell of any crystalline solid has a definite geometric shape and distinguish from other crystals on the basis of length of the edges and angle between the edges.
Crystal Lattice
In crystalline solids atoms, ions or molecules are arranged in a definite order and form a three dimensional array of particles which is known as crystal lattice
Matter
It is defined as any thing which has mass and occupies space is called matter.
Matter is composed of small and tiny particles called Atoms or molecules. It exist in three different states which are gaseous, liquid & solid.
Properties of Gas
1. It has no definite shape.
2. It has no definite volume, so it can be compressed or expanded.
3. A gas may diffuse with the other gas.
4. The molecules of a gas are in continuous motion.
Properties of Liquids
1. A liquid has no definite shape.
2. It has a fixed volume.
3. The diffusion of a liquid into the other liquid is possible if both of the liquids are polar or non-polar.
4. It can be compressed to a negligible.
Properties of Solids
1. A solid has a definite shape.
2. It has a fixed volume.
3. The rate of diffusion of solid with each other is very slow.
4. It cannot be compressed easily.
Kinetic Theory of Gases
It was an idea of some scientist like Maxwell & Bolzmann that the properties of gases are due to their molecular motion. This motion of the molecules is related with the kinetic energy, so the postulates give by the scientist about the behaviour of gases are collectively known as kinetic molecular theory of gases.
The postulates of kinetic molecular theory are as follows.
1. All gases consists of very large number of tiny particles called molecules.
2. These molecules are widely separated from each other and are so small that they are invisible.
3. The size of the molecules is very small as compared to the distance between them.
4. There is no attractive or repulsive force between molecules so they can move freely.
5. The molecules are very hard and perfectly elastic so when they collide no loss of energy takes place.
6. The gas molecules are in continuous motion they move in a straight path until they collide. The distance between two continuous collision is called Mean Free Path.
7. During their motion these molecules are collided with one another and with the walls of the container.
8. The collision of the molecules are perfectly elastic. When molecules collide they rebound with perfect elasticity and without loss or gain of energy.
9. The pressure of the gas is the result of collision of molecules on the walls of the container.
10. The average kinetic energy of gas molecules depends upon the absolute temperature. At any given temperature the molecules of all gases have the same average kinetic energy (1/2 mv2).
Kinetic Theory of Liquids
This theory is bases on the following assumptions.
1. The particles of a liquid are very close to each other due to which a liquid has fixed volume.
2. The particles in a liquid are free to move so they have no definite shape.
3. During the motion these molecules collides with each other and with the walls of the container.
4. These molecules possess kinetic which is directly proportional to its absolute temperature.
Kinetic Theory of Solids
The assumptions of kinetic theory for solids are as follows.
1. The particles in a solid are very closely packed due to strong attractive forces between the molecules.
2. These molecules are present at a fixed position and are unable to move.
3. They have definite shape because the particles are arranged in a fixed pattern.
4. They possess only vibrational energy.
Mean Free Path
The distance which a molecule of a gas travels before its collision with the other molecule is called free path. This distance between the collision of the molecules changes constantly so the average distance which a molecule travels before its collision is called mean free path.
Boyle's Law
A relationship of volume with external pressure was given by Boyle's in the form of law. This law is known as Boyle's Law which states,
For a given mass of a gas the volume of the gas is inversely proportional to its pressure provided the temperature is kept constant.
Mathematically it may be written as
V ∞ 1 / P
Or V = K / P
Or PV = K
On the bases of the relation, Boyle's law can also be stated as
The product of the pressure and volume of a given mass of a gas is always constant at constant temperature.
Explanation
Consider for a given mass a gas having volume V1 at pressure P1, so according to Boyle's Law we may write as
P1V1 = K1 (constant)
If the pressure of the above system is changed from P1 to P2 then the volume of the gas will also change from V1 to V2. For this new condition of the gas we can write as,
P2V2 = K2 (constant)
But for the same mass of the gas.
K1 = K2
P1V1 = P2V2
This equation is known as Boyle's Equation.
Charle's Law
We know that everything expand on heating and contract cooling. This change in volume is small in liquids and solids but gases exhibit enormous changes due to the presence of large intermolecular spaces.
Change of volume of a gas with the change of temperature at constant pressure was studied by Charles and was given in the form of a law. which states,
Statement
For a given mass of a gas the volume of the gas is directly proportional to its absolute temperature provided the pressure is kept constant.
Mathematically this law may be written as
V ∞ T
V = K T
OR
V / T = K
This relation shows that the ratio of volume of a given mass of a gas to its absolute temperature is always constant provided the pressure is kept constant. On this bases Charles Law may also be defined as,
If the pressure remains constant for each 1ºC change of temperature the volume of the gas changes to 1/273 of its original volume.
On the bases of this statement
V1 / T = K & V2 / T2 = K
V1 / T1 = V2 / T2
This equation is known as Charle's equation.
The volume temperature relationship can be represented graphically. When volume of a given mass of gas is plotted against temperature, a straight line is obtained.
Graph Coming Soon
Absolute Scale Of Temperature
There are different scales for the measurement of temperature such as Celsius ºC and Fahrenheit ºC. Similarly another scale known as absolute scale or Kelvin scale is determined on the basis of Charle's law.
On the basis of Charle's law we known that the volume of the gas changes to 1/273 times of its original volume for each 1 ºC change of temperature. It suggests that the volume of a gas would theoretically be zero at -273ºC. But this temperature has never been achieved for any gas because all the gases condense to liquid at a temperature above this point. So the minimum possible temperature for a gaseous system is to be -273ºC. This temperature is referred as absolute zero or zero degree of the absolute scale or Kelvin scale.
To form an absolute scale thermometer if the equally spaced divisions of centigrade thermometer are extended below zero and when the point -273ºC is maked then this point is called as absolute zero and the scale is called as absolute scale. It shows that for the conversion of centigrade scale into Kelvin scale 273 is added to the degrees on the centigrade scale.
K = 273 + ºC
Avogadro's Law
In 1811, a scientist Avogadro's established a relationship between the volume and number of molecules of the gas, which is known as Avogadro's law.
Statement
Equal volume of all gases contains equal number of molecules under the same condition of temperature & pressure.
Mathematically it may be represented as
V ∞ n
OR
V = K n
On the basis of the above statement we can say that
1 dm3 of O2 gas will contain the same number of molecules as 1 dm3 of H2 or N2 or any other gas at same temperature and pressure.
It was also observed that 22.4 dm3 of any gas at S.T.P contain 1 mole of that gas, so 22.4 dm3 volume at S.T.P is called as molar volume or the volume of 1 mole of the gas and the mass present in 22.4 dm3 of any gas will be equal to its molar mass or molecular mass. It can also be explained on the basis of following figures.
Determination of Unknown Molecular Mass of a Gas With the Help of Avogadro's Law
Suppose we have two gases (i) Oxygen (ii) CO
The volume of these two gases are equal which are 1 dm3.
The mass of 1 dm3 of oxygen is 1.43 gm
The mass of 1 dm3 of Co is 1.25 gm
According to Avogadro's law we know that 1 dm3 of CO at S.T.P contain the same number of molecules as 1 dm3 of O2 under similar condition. Hence a molecule of CO has 1.25 / 1.43 times as much as a molecule of O2 and we know that the molecular mass of oxygen is 32 so the molecular mass of CO would be
1.25 / 1.43 x 32 = 28 g / mole
General Gas Equation (Ideal Gas Equation)
To give a relation between the volume, pressure and number of moles of n gas, Boyle's law, Charle's law and Avogadro's law are used.
According to Boyle's law | V ∞ 1 / P
According to Charle's law | V ∞ T
According to Avogadro's law | V ∞ n
By combining these laws we get
V ∞ 1 / P x T x n
OR
V = R x 1 / P x T x P
OR
P V = n R T
This equation is known as general gas equation n is also known as equation of state because when we specify the four variables = pressure, temperature, volume and number of moles we define the state for a gas.
In this equation "R" is a constant known as gas constant.
Value of R
1. When Pressure is Expressed in Atmosphere and Volume in Litres or dm3
According to general gas equation
P V = n R T
OR
R = PV / nT
For 1 mole of a gas at S.T.P we know that
V = 22.4 dm3 or litres
T = 273 K (standard temperature)
P = 1 atm (standard pressure)
So,
R = PV / nT
= 1 atm x 22.4 dm3 / 1 mole x 273 K
= 0.0821 dm3 K-1 mol0-1
2. When Pressure is Expressed in Newtons Per Square Metre and Volume in Cubic Metres
For 1 mole of a gas at S.T.P
V = 0.0224 m3 .......... ( 1 dm3 = 10-3 m3)
n = 1 mole
T = 273 K
P = 101200 Nm-2
So,
R = PV / nT
= 101300 Nm-2 x 0.0224 m3 / 1 mole x 273 K
= 8.3143 Nm K-1 mole-1
= 8.3143 J K-1 mol-1
Derivation of Gas Equation
According to general gas equation
P V = n R T
For 1 mole of a gas n = 1
P V = R T
OR
P V / T = R
Consider for a known mass of a gas the volume of the gas is V1 at a temperature T1 and pressure P1. Therefore for this gas we can write as
P1 V1 / T1 = R
If this gas is heated to a temperature T2 due to which the pressure is changed to P2 and volume is changed to V2. For this condition we may write as
P2 V2 / T2 = R
P1 V1 / T1 = P2 V2 / T2 = R
P1 V1 / T1 = P2 V2 / T2
This equation is known as gas equation.
Graham's Law of Diffusion
We know that gas molecules are constantly moving in haphazard direction, therefore when two gases are placed separated by a porous membrane, they diffuse through the membrane and intermix with each other. The phenomenon of mixing of molecules of different gases is called diffusion.
In 1881, Graham established a relationship between the rates of diffusion of gases and their densities which is known as Graham's law of diffusion.
Statement
The rate of diffusion of any gas is inversely proportional to the square root of its density.
Mathematically it can be represented as
r ∞ 1 / √d
r = K / √d
Graham also studied the comparative rates of diffusion of two gases. On this basis the law os defined as
The comparative rates of diffusion of two gases under same condition of temperature and pressure are inversely proportional to the square root of their densities.
If the rate of diffusion of gas A is r1 and its density is d1 then according to Graham's law
r1 ∞ 1 / √d1
OR
r1 = K / √d1
Similarly the rate of diffusion of gas B is r2 and its density is d2 then
r2 ∞ 1 / √d2
OR
r2 = K / √d2
Comparing the two rates
r1 / r2 = (K / √d1) / (K / √d2)
r1 / r2 = √d2 / d1 ................... (A)
But density d = mass / volume
Therefore,
For d1 we may write as
d1 = m1 / v1
And for d2
d2 = m2 / v2
Substituting these values of d1 & d2 in equation (A)
r1 / r2 = √(m2 / v2) / (m1 / v1)
But v1 = v2 because both gases are diffusing in the same volume.
Therefore,
r1 / r2 = √m2 / m1
Hence Graham's law can also be stated as,
The comparative rates of diffusion of two gases are inversely proportional to the square root of their masses under the same condition of temperature and pressure.
It means that a lighter gas will diffuse faster than the heavier gas. For example compare the rate of diffusion of hydrogen and oxygen.
Rate of diffusion of H2 / Rate of diffusion of O2 = √Mass of O2 / Mass of H2 = √32/ 2 = √16 = 4
It shows that H2 gas which is lighter gas than O2 will diffuse four times faster than O2.
Dalton's Law of Partial Pressures
Partial Pressure
In a gaseous mixture the individual pressure oxerted by a gas is known as partial pressure.
When two or more gases which do not react chemically are mixed in the same container each gas will exert the same pressure as it would exert if it alone occupy the same volume.
John Dalton in 1801 formulated a law which is known as Dalton's Law of partial pressure and stated as.
Statement
The total pressure of a gaseous system is equal to the sum of the partial pressures of all the gases present in the system.
Suppose in a system three gases A, B & C are present. The partial pressure of these gases are
PA = Partial pressure of gas A
PB = Partial pressure of gas B
PC = Partial pressure of gas C
Then Dalton's law may be mathematically written as
PT = PA + PB + PC
Where PT is the total pressure of the system.
To calculate the individual pressures of gases in the above example suppose the number of moles of A, B & C in the container are nA, nB and nC. So the total number of moles in the container will be
n = nA + nB + nC
Apply the general gas equation
P V = n R T
PT = n R T / V
Since R, T and V are same for gases A, B and C, therefore the partial pressure of these gases are as follows.
Partial pressure of gas A | PA = n(A)RT / V ......... (2)
Partial pressure of gas B | PB = n(B)RT / V ......... (3)
Partial pressure of gas C | PC = n(C)RT / V ......... (4)
Now divide equation (2) by (1)
PA / PT = (nA RT/V) / (nRT/V)
OR
PA / PT = nA / nT
Therefore,
P(gas) = P1 x n(gas) / n(total)
Application of Dalton's Law
In an inert mixture of gases the individual gas exerts its own pressure due to collision of its molecules with the walls of the container but the total pressure produced on the container wall will be the sum of pressure of all the individual gases of the mixture.
On this basis the number of moles formed during a chemical reaction can be measured. For this purpose a gas produced in a chemical reaction is collected over water. The gas also contains some of water vapours. So the pressure exerted by the gas would be the pressure of pure gas and the pressure of water vapours.
Therefore the pressure of the system may be represented as
P(moist) = P(dry) + P(water vapour)
So,
P(dry) = P(moist) - P(water vapour)
In this way we can obtain the pressure of the gas and by using general gas equation we can calculate the number of moles of the prepared gas.
Ideal Gas
A gas which obeys all the gas laws at all temperatures and pressures is known as ideal gas.
It means that the product of pressure and volume must be constant at all pressures.
Similarly the rate of V/T will remain constant for an ideal gas.
But there is no gas which is perfectly ideal because of the presence of the force of attraction or repulsion between the molecules.
Gas Laws on the Basis of Kinetic Theory
Boyle's Law
According to Boyle's law the volume of a given mass of a gas is inversely proportional to its pressure at constant temperature.
It means that when the volume of the gas is decreased the pressure of the gas will increase.
According to kinetic molecular theory of gases the pressure exerted by a gas is due to the collisions of the molecules with the walls of the container. If the volume of a gas is reduced at constant temperature, the average velocity of the gas molecules remains constant so they collide more frequently wit the walls which causes higher pressure.
Charle's Law
According to Charles law the volume of a given mass of a gas is directly proportional to its absolute temperature at constant pressure.
According to kinetic molecular theory the average kinetic energy of gas molecules is directly proportional to its absolute temperature so if the temperature of the gas is increased the average kinetic energy of the gas molecules is also increased due to which the sample of the gas expanded to keep the pressure constant. It is accordance with the law.
Graham's Law
According to Graham's Law
r1 / r2 = √m2 / m1
The rate of diffusion of a gas is directly proportional to the velocity of the molecules so,
v1 / v2 = √m2 / m1
Liquefaction
According to kinetic theory, the kinetic energy of the molecules is low for lower temperature. These slower moving molecules become subject to inter molecular attraction. At a sufficiently low temperature these attractive forces are capable of holding the molecules with one another so the gas is changed into liquid and the process is called liquefaction.
Liquid State
It is one of the state of matter. In this state, the kinetic energy of the molecule is very high due to which the molecules of the liquid are able to move but due to compact nature liquids are not compressible. On this basis we can say that the volume of a liquid is always constant but its shape can be changed.
Behaviour of Liquids
The main properties of liquids are as follows.
Diffusibility
The diffusion of one liquid into another liquid is possible but its rate is slow as compared with the rate of diffusion of gases. Example of diffusion of liquids is mixing of alcohol in water.
Explanation of Diffusion in Terms of Kinetic Energy
As the molecular of a liquid are in cluster form they are very close to each other but these molecules are movable so they can mix with the other molecules. Since the intermolecular distance are smaller due to which the rate of diffusion of liquids is slow.
Compressibility
The space between liquid molecules are very small due to strong Van der Waals forces. When the pressure is applied, they can be compressed but to a very little extent.
Expansion
When a liquid is heated, the kinetic energy of its molecules also increases so the attraction between the molecules becomes weaker due to which they go further apart and hence the liquid expands.
Contraction
When a liquid is cooled its kinetic energy is lowered and the attraction among the molecules becomes stronger so they comes close to each other and hence the liquid contract.
Viscosity
Definition
The internal resistance in the flow of a liquid is called viscosity.
Liquids have the ability to flow, but different liquids have different rates of flow. Some liquids like honey mobil oil etc. flow slowly and are called viscous liquids while ether, gasoline etc. which flow quickly are called less viscous.
Explanation
The viscosity of liquid can be understood by considering a liquid in a tube, a liquid in a tube is considered as made up of a series of molecular layer. The layer of the liquid in contact with the walls of the tube remains stationary and the layer in the center of the tube has highest velocity as shown.
Each layer exerts a drag on the next layer and causes resistance to flow.
Factors on Which Viscosity Depends
1. Size of Molecules
The viscosity of a liquid depends upon the size of its molecules. If the size of the molecules is bigger the viscosity of the liquid is high.
2. Shape of Molecules
Shape of the molecules affects the viscosity. If the shapes of the molecules are spherical they can move easily but if the shapes of the molecules are irregular such as linear or trigonal then the molecules will move slowly and its viscosity will be high.
3. Intermolecular Attraction
If the force of attraction between the molecules of a liquid is greater the viscosity of the liquid is also greater.
4. Temperature
Viscosity of a liquid decreases with the increase of temperature.
Units of Viscosity
Viscosity of a liquid is measured in poise, centipoise or millipoise & S.I unit.
1 poise = 1 N.s.m(-2)
1 centipoise = 10(-2) N.s.m(-2)
Surface Tension
Definition
The force acting per unit length on the surface of a liquid at right angle direction is called surface tension.
Explanation
Consider a liquid is present in a beaker. The molecules inside the liquid are surrounded by the other molecules of the liquid. So the force of attraction on a molecule is balanced from all direction. But the force of attraction acting on the molecules of the surface from the lower layer molecules is not balanced.
The molecules lying on the surface are attracted by the molecules present below the surface Due to this downward pull the surface of the liquid behave as a membrane which tends to contract to a smaller area and causes a tension on the surface of the liquid known as surface tension.
Diagram Coming Soon
Factors on Which Surface Tension Depends
1. Molecular Structure of the Liquid
If the force of attraction between the molecules is greater, the surface tension of the liquid is also greater. Those liquids in which hydrogen bond formation take place will have more surface tension.
2. Temperature
Surface tension of a liquid is inversely proportional to the temperature.
Units
1. Dynes / cm
2. Ergs / cm2
Capillary Action
The fall or rise of a liquid in a capillary tube is called capillary action.
When a capillary tube is dipped in a liquid which wets the wall of the tube, the liquid will rise in the capillary tube, to decrease the surface area due to surface tension. The liquid will rise in the capillary tube until the upward force due to surface tension is just balanced by the downward gravitational pull. This is called capillary action.
Vapour Pressre
Definition
The pressure exerted by the vapours of a liquid in its equilibrium state with the pure liquid at a given temperature is called vapour pressure.
Explanation
Consider a liquid is present in a bottle as shown.
Diagram Coming Soon
In the beginning the atmosphere above the surface of liquid is unsaturated but due to continuous evaporation the molecule of the liquid are trapped in the bottle and the air present above the surface of the liquid is becomes saturated and after it the molecules present in the vapour state may hit the liquid again and rejoin it by condensing into liquid. Thus in this closed vessel two process are going on simultaneously which are evaporation and condensation of vapours. When the rates of these two processes becomes equal at this point the pressure exerted by vapours is called vapour pressure.
Units of Vapour Pressure
The units for vapour pressure are
1. Millimeter of Hg
2. Atmosphere
3. Torr
4. Newton / m(2)
Factors for Vapour Pressure
1. Nature of Liquid
Vapour pressure of a liquid depends upon the nature of the liquid. Low boiling liquid exert more vapour pressure at a given temperature.
2. Temperature
Vapour pressure of a liquid also depends upon temperature. The vapour pressure of the liquid increases with the increase of temperature due to the increase of average of kinetic energy.
3. Intermolecular Forces
Those liquids in which the intermolecular forces are weak shows high vapour pressure.
Explanation of Evaporation on the Basis of Kinetic Theory
According to this theory the molecules of a liquid collide with each other during their motion. Due to these collisions some of the molecules acquire greater energy than Van der Walls forces which binds the molecules of the liquid together so these molecules of higher energy escapes from the surface into the air in the form of vapours.
Evaporation is a Cooling Process
In liquids, due to collision between molecules some molecules acquire higher energy and escapes from the surface of the liquid in the form of vapours. The kinetic energy of the remaining molecules decreases due to which the temperature of the liquid also decreases and hence we can say that evaporation is a cooling process.
Boiling Point
Definition
The temperature at which the vapour pressure of a liquid becomes equal to the atmospheric pressure is called boiling point.
When a liquid is heated the rate of evaporation of the molecules also increases with the increase in temperature. When the pressure of the vapours becomes equal to the atmospheric pressure the liquid starts boiling and this temperature is known as boiling point.
If the external pressure on a liquid is changed the boiling point of the liquid also change. The increase in external pressure on a liquid increases the boiling point while the decreases of external pressure decrease the boiling point.
Solid State
It is a state of matter which posses both definite shape and definite volume. In solids the particles are very close to each and tightly packed with a greater force of attraction.
Properties of Solids
1. Diffusibility
Diffusion also occurs in solids but its rate is very slow. If a polished piece of zinc is clamped with a piece of copper for a long time. After few years we will see that some particles of zinc are penetrated into copper and some particles of copper are penetrated into zinc. It shows that the diffusion in solids is possible but it occurs with a slow rate.
2. Compressibility
In solids the molecules are close to each other so it is not easy to compress a solid. In other words we can say that the effect of pressure on solids is negligible.
3. Sublimation
It is a property of some solids that on heating these solids are directly converted into vapours without liquification. This property of solids is known as sublimation.
4. Melting
When solids are heated, they are changed into liquids and the property is called melting of the solids.
5. Deformity
Solids may be deformed by high pressure. When a high pressure is applied on solids due to which some particles are dislocated the force of attraction is so strong that the rearranged atoms are held equally well with their new neighbours and hence the solid is deformed.
Classification of Solids
Solids are classified into two main classes.
1. Crystalline
2. Amorphous
1. Crystalline Solids
In a solid if the atoms are attached with each other with a definite arrangement and it also possesses a definite geometrical shape. This type of solid is called crystalline solid.
e.g. NaCl, NiSO4 are crystalline solids.
2. Amorphous Solids
In these solids there is no definite arrangement of the particles so they do not have a definite shape. The particles of such solids have a random three dimensional arrangement. Examples of amorphous solids are glass, rubber, plastic etc.
The properties of crystalline and amorphous solids are quite different from each other. These differences in properties are given below.
Difference of Geometry
1. Crystalline Solids
In crystalline solids particles are arranged in a definite order due to which it possesses a definite structure.
2. Amorphous Solids
In amorphous solids particles are present without any definite arrangement so they do not have definite shape.
Difference of Melting Point
1. Crystalline Solids
Crystalline solids have sharp melting point due to uniform arrangement.
2. Amorphous Solids
Amorphous solids melts over a wide range of temperature.
Cleavage and Cleavage Plane
1. Crystalline Solids
When a big crystal is broken down into smaller pieces the shape of the smaller crystals is identical with the bigger crystal. This property of crystalline solids is called cleavage and the plane from where a big crystal is broken is called cleavage plane.
2. Amorphous Solids
Amorphous solids do not break up into smaller pieces with an identical shape.
Anisotropy & Isotropy
1. Crystalline Solids
It is a property of crystalline solid that they show different physical properties in different direction. For example graphite can conduct electric current only through the plane which is parallel to its layers. This property is called anisotropy.
2. In amorphous solids the physical properties are same in all directions. This property of solids is called isotropy.
Symmetry in Structure
1. Crystalline solids are symmetric in their structure when they are rotated about an axis, their appearance remains same so they are symmetric in structure.
2. Amorphous Solids
Amorphous solids are not symmetric.
Types of Crystals
There are four types of crystals.
1. Atomic crystals
2. Ionic crystals
3. Covalent crystals
4. Molecular crystal
1. Atomic Crystals
Metals are composed of atoms. These atoms are combined with each other by metallic bond and the valency electrons in metals can move freely throughout the crystal lattice. This type of solid is called atomic crystal.
The properties of atomic crystals are
1. High melting point.
2. Electrical and thermal conductivity.
3. These are converted into sheets so these are malleable.
4. These are used as wire so these are ductile.
2. Ionic Crystals
Those solids which consists of negativity and positively charged ions held together by strong electrostatic force of attraction are called ionic crystals. Ionic crystalline solids possesses the following properties.
1. The melting and boiling point of ionic crystals is high.
2. They conduct electricity in molten state.
3. Ionic crystals are very hard.
4. Indefinite growth of crystals is also a property of ionic crystals.
3. Covalent Crystals
In covalent solids, the atoms or molecules are attached with each other by sharing of electrons. Such type of solids are called covalent solids e.g. diamond is a covalent solid in which carbon atoms are attached with each other by covalent bond. The other examples of covalent crystals are sulphur, graphite etc.
Covalent crystals possesses the following properties.
1. High melting point.
2. High refractive index.
3. Low density.
4. Molecular Crystals
Those solid in which molecules are held together due to intermolecular forces to form a crystal lattice are called molecular crystals e.g. iodine and solid CO2 are molecular crystals. The general properties of molecular crystals are as follows.
1. Low melting and boiling point.
2. Non - conductor of heat and electricity.
Isomorphism
When two different substance have same crystalline structure, they are said to be isomorphous and the phenomenon is called isomorphism.
e.g. ZnSO4 and NiSO4 are two different substances but both are orthorhombic similarly the structure of CaCO3 and NaNO3 is frigonal.
Polymorphism
If a substance exist in more than one crystalline form it is called polymorphous and the phenomenon is known as polymorphism. E.g. sulphur exist in rhombic and monoclinic form similarly CaCO3 exist in trigonal and orthorhombic form.
Unit Cell
The basic structural unit of a crystalline solid which when repeated in three dimensions generates the crystal structure is called a unit cell.
A unit cell of any crystalline solid has a definite geometric shape and distinguish from other crystals on the basis of length of the edges and angle between the edges.
Crystal Lattice
In crystalline solids atoms, ions or molecules are arranged in a definite order and form a three dimensional array of particles which is known as crystal lattice
Atomic Structure
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Introduction
About the structure of atom a theory was put on by John Dalton in 1808. According to this theory matter was made from small indivisible particles called atoms.
But after several experiments many particles have been discovered with in the atom which are electrons, protons, neutrons, positrons etc. For the discovery of these fundamental particles the experiments are as follows.
1. Faraday's experiment indicates the existence of electron.
2. Crook's tube experiment explains the discovery of electron and proton.
3. Radioactivity also confirms the presence of electrons and protons.
4. Chadwick's experiment shows the presence of neutrons.
The details of these experiments are given below.
Faraday's Experiment
Passage of Electricity Through Solution
In this experiment Faraday passed the electricity through an electrolytic solution. He observed that when two metal plates called electrodes are placed in an electrolytic solution and electricity is passed through his solution the ions present in the solution are moves towards their respective electrodes. In other words these ions are moves towards the oppositely charge electrodes to give up their charge and liberated as a neutral particles.
Faraday also determined the charges of different ions and the amount of elements liberated from the electrolytic solution. Due to this experiment presence of charge particles in the structure of atoms is discovered. The basic unit of electric charge was later named as electron by Stoney in 1891.
Diagram Coming Soon
Crook's Tube Or Discharge Tube Experiment
Passage of Electricity Through Gases Under Low Pressure
Introduction
The first of the subatomic particles to be discovered was electron. The knowledge about the electron was derived as a result of the study of the electric discharge in the discharge tube by J.J. Thomson in 1896. This work was later extended by W. Crooke
Working of Discharge Tube
When a very high voltage about 10,000 volts is applied between the two electrodes, no electric discharge occurs until the part of the air has been pumped out of the tube. When the pressure of the gas inside the tube is less than 1 mm, a dark space appears near the cathode and thread like lines are observed in the rest of 0.01 mm Hg it fills the whole tube. The electric discharge passes between the electrodes and the residual gas in the tube begins to glow. These rays which proceed from the cathode and move away from it at right angle in straight lines are called cathode rays.
Properties of Cathode Rays
1. They travel in straight lines away from the cathode and produce shadow of the object placed in their path.
2. The rays carry a negative charge.
3. These rays can also be easily deflected by an electrostatic field.
4. The rays can exert mechanical pressure showing that these consist of material particle which are moving with kinetic energy.
5. The produce fluorescence when they strike the glass wall of the discharge tube.
6. Cathode rays produce x-rays when they strike a metallic plate.
7. These rays consists of material particle whose e/m resembles with electron.
8. These rays emerge normally from the cathode and can be focused by using a concave cathode.
Positive Rays
In 1890 Goldstein used a discharge tube with a hole in the cathode. He observed that while cathode rays were emitting away from the cathode, there were coloured rays produced simultaneously which passed through the perforated cathode and caused a glow on the wall opposite to the anode. Thomson studied these rays and showed that they consisted of particles carrying a positive charge. He called them positive rays.
Properties of Positive Rays
1. These rays travel in a straight line in a direction opposite to the cathode.
2. These are deflected by electric as well as magnetic field in the way indicating that they are positively charged.
3. The charge to mass ratio (e/m) of positive particles varies with the nature of the gas placed in the discharge tube.
4. Positive rays are produced from the ionization of gas and not from anode electrode.
5. Positive rays are deflected in electric field. This deflection shows that these are positively charged so these are named as protons.
The Information Obtained From Discharge Tube Experiment
The negatively charge particles electrons and the positively charge particles protons are the fundamental particle of every atom.
Radioactivity
In 1895, Henry Becqueral observed that uranium and its compounds spontaneously emitted certain type of radiation which affected a photographic plate in the dark and were able to penetrate solid matter. He called these rays as radioactivity rays and a substance which possessed the property of emitting these radioactivity rays was said to be radioactivity element and the phenomenon was called radioactivity.
On further investigation by Maric Curic, it was found that the radiation emitted from the element uranium as well as its salts is independent of temperature and the source of the mineral but depend upon the mineral but depend upon the quantity of uranium present e.g. Pitchblende U3O8 was found to be about four times more radioactive than uranium.
Radioactive Rays
Soon after the discovery of radium it was suspected that the rays given out by radium and other radioactive substance were not of one kind. Rutherford in 1902 devised an ingenious method for separating these rays from each other by passing them between two oppositely charged plate. It was observed that the radioactive rays were of three kinds, the one bending towards the negative plate obviously carrying positive charge were called α-rays and those deflected to the positive plate and carrying -ve charge were named as β-rays. The third type gamma rays, pass unaffected and carry no charge.
Properties of α - RAYS
1. These rays consists of positively charged particles.
2. These particles are fast moving helium nuclei.
3. The velocity of α-particles is approximately equal to 1/10th of the velocity of light.
4. Being relatively large in size, the penetrating power of α-rays is very low.
5. They ionize air and their ionization power is high.
Properties of β - RAYS
1. These rays consists of negatively charged particles.
2. These particles are fast moving electron.
3. The velocity of β-particles is approximately equal to the velocity of light.
4. The penetrating power of β-rays is much greater than α-rays.
5. These rays ionizes gases to lesser extent.
Properties of γ - RAYS
1. Gamma rays do not consist of particles. These are electromagnetic radiations.
2. They carry no charge so they are not deflected by electric or magnetic field.
3. Their speed is equal to that of light.
4. These are weak ionizer of gases.
5. Due to high speed and non-material nature they have great power of penetration.
Chadwick Experiment (Discovery of Neutron)
When a light element is bombarded by α-particles, these α-particles leaves the nucleus in an unstable disturbed state which on settling down to stable condition sends out radioactivity rays. The phenomenon is known as "Artificial Radioactivity".
In 1933, Chadwick identified a new particle obtained from the bombardment of beryllium by α-particles. It had a unit mass and carried no charge. It was named "Neutron".
Spectroscopic Experiment
After the discovery of fundamental particles which are electrons, protons & neutron, the next question concerned with electronic structure of atom.
The electronic structure of the atom was explained by the spectroscopic studies. In this connection Plank's Quantum theory has great impact on the development of the theory of structure of atom.
Planck's Quantum Theory
In 1900, Max Planck studied the spectral lines obtained from hot body radiations at different temperatures. According to him,
When atoms or molecules absorb or emit radiant energy, they do so in separate units of waves called Quanta or Photons.
Thus light radiations obtained from excited atoms consists of a stream of photons and not continuous waves.
The energy E of a quantum or photon is given by the relation
E = h v
Where v is the frequency of the emitted radiation and h the Planck's constant. The value of h = 6.62 x 10(-27) erg. sec.
The main point of this theory is that the amount of energy gained or lost is quantized which means that energy change occurs in small packets or multiple of those packets, hv, 2 hv, 3 hv and so on.
Spectra
A spectrum is an energy of waves or particles spread out according to the increasing or decreasing of some property. E.g. when a beam of light is allowed to pass through a prism it splits into seven colours. This phenomenon is called dispersion and the band of colours is called spectrum. This spectrum is also known as emission spectrum. Emission spectra are of two types.
1. Continuous Spectrum
2. Line Spectrum
1. Continuous Spectrum
When a beam of white light is passed through a prism, different wave lengths are refracted through different angles. When received on a screen these form a continuous series of colours bands: violet, indigo, blue, green, yellow and red (VIBGYOR). The colours of this spectrum are so mixed up that there is no line of demarcation between different colours. This series of bands that form a continuous rainbow of colours is called continuous spectrum.
Diagram Coming Soon
2. Line Spectrum
When light emitted from a gas source passes through a prism a different kind of spectrum may be obtained.
If the emitted from the discharge tube is allowed to pass through a prism some discrete sharp lines on a completely dark back ground are obtained. Such spectrum is known as line spectrum. In this spectrum each line corresponds to a definite wave length.
Diagram Coming Soon
Identification of Element By Spectrum
Each element produces a characteristics set of lines, so line spectra came to serve as "finger prints" for the identification of element. It is possible because same element always emit the same wave length of radiation. Under normal condition only certain wave lengths are emitted by an element.
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Rutherford's Atomic Model
Evidence for Nucleus and Arrangement of Particles
Having known that atom contain electrons and a positive ion, Rutherford and Marsden performed their historic "Alpha particle scattering experiment" in 1909 to know how and where these fundamental particles were located in the structure of atom.
Rutherford took a thin of gold with thickness 0.0004 cm and bombarded in with α-particles. He observed that most of the α-particles passed straight through the gold foil and thus produced a flash on the screen behind it. This indicated that old atoms had a structure with plenty of empty space but some flashes were also seen on portion of the screen. This showed that gold atoms deflected or scattered α-particles through large angles so much so that some of these bounced back to the source.
Based on these observations Rutherford proposed a model of the atom which is known as Rutherford's atomic model.
Diagram Coming Soon
Assumption Drawn From the Model
1. Atom has a tiny dense central core or the nucleus which contains practically the entire mass of the atom leaving the rest of the atom almost empty.
2. The entire positive charge of the atom is located on the nucleus. While electrons were distributed in vacant space around it.
3. The electrons were moving in orbits or closed circular paths around the nucleus like planets around the sun.
4. The greater part of the atomic volume comprises of empty space in which electrons revolve and spin.
Weakness of Rutherford Atomic Model
According to the classical electromagnetic theory if a charged particle accelerate around an oppositely charge particle it will radiate energy. If an electron radiates energy, its speed will decrease and it will go into spiral motion finally falling into the nucleus. Similarly if an electron moving through orbitals of ever decreasing radii would give rise to radiations of all possible frequencies. In other words it would given rise to a continuous spectrum. In actual practise, atom gives discontinuous spectrum.
X-Rays and Atomic Number
In 1895, W.Roentgen discovered that when high energy electrons from cathode collide with the anode in the Crook's tube, very penetrating rays are produced. These rays were named as X-rays.
Explanation
When an electron coming from the cathode strike with the anode in the crook's tube, it can remove an electron from the inner shell of the atom. Due to removal of t his electron the electronic configuration of this ion is unstable and an electron from an orbital of higher energy drops into the inner orbital by emitting energy in form of a photon. This photon corresponds to electromagnetic radiations in the x-rays region.
Relationship Between Wave Length and Nuclear Charge
In 1911, Mosley stablished a relationship between the wave length and nuclear charge. He found that when cathode rays struck elements used as anode targets in the discharge tube, characteristic x-rays were emitted. The wave length of the x-rays emitted decreases regularly with the increase of atomic mass. On careful examination of his data Mosely found that the number of positive charges on the nucleus increases from atom to atom by single electronic unit. He called the number of positive charges as the atomic number.
Diagram Coming Soon
Bohr's Theory
Rutherford's model of atom fails to explain the stability of atom and appearance of the line spectra. Bohr in 1913 was the first to present a simple model of the atom which explained the appearance of line spectra.
Some of the postulates of Bohr's theory are given below.
1. An atom has a number of stable orbits or stationary states in which an electron can reside without emission or absorption of energy.
2. An electron may pass from one of these non-radiating states to another of lower energy with the emission of radiations whose energy equals the energy difference between the initial and final states.
3. In any of these states the electrons move in a circular path about the nucleus.
4. The motion of the electron in these states is governed by the ordinary laws of mechanics and electrostatic provided its angular momentum is an integral multiple of h/2Ï€
It can be written as
mvr = nh / 2Ï€
Here mvr becomes the angular momentum of the electron. Thus Bohr's first condition defining the stationary states could be stated as
"Only those orbits were possible in which the angular momentum of the electrons would be an integral multiple of h/2Ï€". These stationary states correspond to energy levels in the atom.
Calculation of Radius of Orbits
Consider an electrons of charge e revolving.
Atomic number and e the charge on a proton.
Let m be the mass of the electro, r the radius of the orbit and v the tangential velocity of the revolving electron.
The electrostatic force of attraction between the nucleus and the electron according to Coulomb's law
= Z e x e / r2
Diagram Coming Soon The centrifugal force acting on the electron.
= mv2 / r
Bohr assumed that these two opposing forces must be balanced each other exactly to keep the electron in an orbit.
Therefore
Ze2 / r2 = m v2 / r
Multiply both sides by r
r x Ze2 / r2 = r x m v2 / r
Ze2 / r = m v2
OR
r = Ze2 / m v2 .................. (1)
The Bohr's postulate states that only those orbits are possible in which
mvr = nh / 2Ï€
Therefore,
V = nh / 2Ï€mr
Substituting the value of V in eq (1)
r = Ze2 / m(nh/2Ï€mr)2
or
r = Ze2 x 4Ï€2 mr2/n2h2
or
1/r = 4Ï€2mZe2/n2h2
cr
r = n2h2 / 4Ï€2mZe2 ............... (2)
This equation gives the radii of all the possible stationary states. The values of constants present in this equation are as follows.
H = 6.625 x 10(-27) ergs sec OR 6.625 x 10(-37) J.s
Me = 9.11 x 10(-28) gm OR 9.11 x 10(-31) kg
E = 4.802 x 10(-10) e.s.u OR 1.601 x 10(-19) C
By substituting these values we get for first shell of H atom
r = 0.529 x 10(-8) m OR 0.529
The above equation may also be written as
r = n2 (h2 / 4Ï€2mZe2) x n2 a0 .................... (3)
For the first orbit n = 1 and r = 0.529. This is the value of the terms in the brackets sometimes written as a0 called Bohr's Radius. For the second shell n = 2 and for 3rd orbit n = 3 and so on.
Hydrogen Atom Spectrum
Balmer Series
The simplest element is hydrogen which contain only one electron in its valence shell.
Balmer in 1885 studied the spectrum of hydrogen. For this purpose he used hydrogen gas in the discharge tube. Balmer observed that hydrogen atom spectrum consisted of a series of lines called Balmer Series. Balmer determined the wave number of each of the lines in the series and found that the series could be derived by a simple formula.
Lyman Series
Lyman series is obtained when the electron returns to the ground state i.e. n = 1 from higher energy level n(2) = 2, 3, 4, 5, etc. This series of lines belongs to the ultraviolet region of spectrum.
Paschen Series
Paschen series is obtained when the electron returns to the 3rd shell i.e. n = 3 from the higher energy levels n2 = 4, 5, 6 etc. This series belongs to infrared region.
Bracket Series
This series is obtained when an electron jumps from higher energy levels to 4th energy level.
Heisenberg Uncertainty Principle
According to Bohr's theory an electron was considered to be a particle but electron also behaves as a wave according to be Broglie.
Due to this dual nature of electron in 1925 Heisenberg gave a principle known as Heisenberg Uncertainty Principle which is stated as,
It is impossible to calculate the position and momentum of a moving electron simultaneously.
It means that if one was known exactly it would be impossible to known the other exactly. Therefore if the uncertainty in the determination of momentum is Δpx and the uncertainty in position is Δx then according to this principle the product of these two uncertainties may written as
Δpx . Δx ≈ h
So if one of these uncertainties is known exactly then the uncertainty in its determination is zero and the other uncertainty will become infinite which is according to the principle.
Energy Levels and Sub-Levels
According to Bohr's atomic theory, electrons are revolving around the nucleus in circular orbits which are present at definite distance from the nucleus. These orbits are associated with definite energy of the electron increasing outwards from the nucleus, so these orbits are referred as Energy Levels or Shells.
These shells or energy levels are designated as 1, 2, 3, 4 etc K, L, M, N etc.
The spectral lines which correspond to the transition of an electron from one energy level to another consists of several separate close lying lines as doublets, triplets and so on. It indicates that some of the electrons of the given energy level have different energies or the electrons belonging to same energy level may differ in their energy. So the energy levels are accordingly divided into sub energy levels which are denoted by letters s, p, f (sharp, principle, diffuse & fundamental).
The number of sub levels in a given energy level or shell is equal to its value of n.
e.g. in third shell where n = 3 three sub levels s, p, d are possible.
Quantum Numbers
There are four quantum numbers which describe the electron in an atom.
1. Principle Quantum Number
It is represented by "n" which describe the size of orbital or energy level.
The energy level K, L, M, N, O etc correspond to n = 1, 2, 3, 4, 5 etc.
If
n = 1 the electron is in K shell
n = 2 the electron is in L shell
n = 3 the electron is in M shell
2. Azimuthal Quantum Number
This quantum number is represented by "l" which describes the shape of the orbit. The value of Azimuthal Quantum number may be calculated by a relation.
l = 0 ----> n - 1
So for different shell the value of l are as
n = 1 K Shell l = 0
n = 2 L Shell l = 0, 1
n = 3 M Shell l = 0, 1, 2
n = 4 N Shell l = 0, 1, 2, 3
when l = 0 the orbit is s
when l = 1 the orbit is p
when l = 2 the orbit is d
when l = 3 the orbit is f
3. Magnetic Quantum Number
It is represented by "m" and explains the magnetic properties of an electron. The value of m depends upon the value of l. It is given by
m = + l ----> 0 ----> l
when l = 1, m has three values (+1, 0, -1) which corresponds to p orbital. Similarly when l = 2, m has five values which corresponds to d orbital.
4. Spin Quantum Number
It is represented by "s" which represents spin of a moving electron. This spin may be either clockwise or anticlockwise so the values for s may be +1/2 or -1/2.
Pauli's Exclusion Principle
According to this principle
No two electrons in the same atom can have the same four quantum number.
Consider an electron is present in 1s orbital. For this electron n = 1, l = 0, m = 0. Suppose the spin of this electron is s = +1/2 which will be indicated by an upward arrow ↑. Now if another electron is put in the same orbital (1s) for that electron n = 1, l = 0, m = 0. It can occupy this orbital only if the direction of its spin is opposite to that of the first electron so s = -1/2 which is symbolized by downward arrow ↓. From this example, we can observe the application of Pauli's exclusion principle on the electronic structure of atom.
Electronic Configuration
The distribution of electrons in the available orbitals is proceeded according to these rules.
1. Pauli Exclusion Principle
2. Aufbau Principle
3. (n + l) Rule
4. Hund's Rule
The detail of these rules and principles is given below.
1. Aufbau Principle
It is states as
The orbitals are filled up with electrons in the increasing order of their energy.
It means that the orbitals are fulled with the electrons according to their energy level. The orbitals of minimum energy are filled up first and after it the orbitals of higher energy are filled.
2. Hund's Rule
If orbitals of equal energy are provided to electron then electron will go to different orbitals and having their parallel spin.
In other words we can say that electrons are distributed among the orbitals of a sub shell in such a way as to give the maximum number of unpaired electrons and have the same direction of spin.
3. (n + l) Rule
According to this rule
The orbital with the lowest value of (n + l) fills first but when the two orbitals have the same value of (n + l) the orbital with the lower value of n fills first.
For the electronic configuration the order of the orbital is as follows.
1s, 2s, 2p, 3s, 4s, 3d, 4p, 5s, 4d, 5p, 6s etc.
Atomic Radius
For homonuclear diatomic molecules the atomic radius may be defined as
The half of the distance between the two nuclei present in a homonuclear diatomic molecules is called atomic radius.
It may be shown as
In case of hetronuclear molecular like AB, the bond length is calculated which is (rA + rB) and if radii of any one is known the other can be calculated.
For the elements present in periodic table the atomic radius decreases from left to right due to the more attraction on the valence shell but it increases down the group with the increase of number of shells.
Ionic Radius
Ionic radius is defined as
The distance between nucleus of an ion and the point up to which nucleus has influence of its electron cloud.
When an electron is removed from a neutral atom the atom is left with an excess of positive charge called a cation e.g
Na ----> Na+ + c-
But when an electron is added in a neutral atom a negative ion or anion is formed.
Cl + e- ----> Cl-
As the atomic radius, the ionic radii are known from x-ray analysis. The value of ionic radius depends upon the ions that surround it.
Ionic radii of cations have smaller radii than the neutral atom because when an electron is removed. The effective charge on the nucleus increases and pulls the remaining electrons with a greater force.
Ionic radii of anions have a large radii than the neutral atom because an excess of negative charge results in greater electron repulsion.
Radius of Na atom = 1.57
Radius of Na+ atom = 0.95 (smaller than neutral atom)
Radius of Cl atom = 0.99
Radius of Cl- atom = 1.81 (larger than neutral atom)
Ionization Potential
Definition
The amount of energy required to remove most loosely bounded electron from the outermost shell of an atom in its gaseous state is called is called ionization potential energy.
It is represented as
M(gas) ----> M+(gas) + e- ................... ΔE = I.P
The energy required to remove first electron is called first I.P. The energy required to remove 2nd or 3rd electron is called 2nd I.P or 3rd I.P
M(gas) ----> M+(gas) + e- ................... ΔE = 1st I.P
M+(gas) ----> M++(gas) + e- ................ΔE = 2nd I.P
M++(gas) ----> M+++(gas) + e- ............ ΔE = 3rd I.P
The units of I.P is kilo-Joule per mole.
Factors on which I.P Depends
1. Size of the Atom
If the size of an atom is bigger the I.P of the atom is low, but if the size of the atom is small then the I.P will be high, due to fact if we move down the group in the periodic table. The I.P value decreases down the group.
2. Magnitude of Nuclear Charge
If the nuclear charge of atom is greater than the force of attraction on the valence electron is also greater so the I.P value for the atom is high therefore as we move from left to right in the periodic table the I.P is increased.
3. Screening Effect
The shell present between the nucleus and valence electrons also decreases the force of attraction due to which I.P will be low for such elements.
Electron Affinity
Definition
The amount of energy liberated by an atom when an electron is added in it is called electron affinity.
It shows that this process is an exothermic change which is represented as
Cl + e- ----> Cl- ............ ΔH = -348 kJ / mole
Factors on which Electron Affinity Depends
1. Size of the Atom
If the size of atom is small, the force of attraction from the nucleus on the valence electron will be high and hence the E.A for the element will also be high but if the size of the atoms is larger the E.A for these atoms will be low.
2. Magnitude of the Nuclear Charge
Due to greater nuclear charge the force of attraction on the added electron is greater so the E.A of the atom is also high.
3. Electronic Configuration
The atoms with the stable configuration has no tendency to gain an electron so the E.A of such elements is zero. The stable configuration may exist in the following cases.
1. Inert gas configuration
2. Fully filled orbital
3. Half filled orbital
Electronegativity
Definition
The force of attraction by which an atom attract a shared pair of electrons is called electronegativity.
Application of Electronegativity
1. Nature of Chemical Bond
If the difference of electronegativity between the two combining atoms is more than 1.7 eV, the nature of the bond between these atoms is ionic but if the difference of electronegativity is less than 1.7 eV then the bond will be covalent.
2. Metallic Character
If an element possesses high electronegativity value then this element is a non-metal but if an element exist with less electronegativity, it will be a metal.
Factors for Electronegativity
1. Size of the Atom
If the size of the atom is greater the electronegativity of the atom is low due to the large distance between the nucleus and valence electron.
2. Number of Valence Electrons
If the electrons present in the valence shell are greater in number, the electronegativity of the element is high
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Introduction
About the structure of atom a theory was put on by John Dalton in 1808. According to this theory matter was made from small indivisible particles called atoms.
But after several experiments many particles have been discovered with in the atom which are electrons, protons, neutrons, positrons etc. For the discovery of these fundamental particles the experiments are as follows.
1. Faraday's experiment indicates the existence of electron.
2. Crook's tube experiment explains the discovery of electron and proton.
3. Radioactivity also confirms the presence of electrons and protons.
4. Chadwick's experiment shows the presence of neutrons.
The details of these experiments are given below.
Faraday's Experiment
Passage of Electricity Through Solution
In this experiment Faraday passed the electricity through an electrolytic solution. He observed that when two metal plates called electrodes are placed in an electrolytic solution and electricity is passed through his solution the ions present in the solution are moves towards their respective electrodes. In other words these ions are moves towards the oppositely charge electrodes to give up their charge and liberated as a neutral particles.
Faraday also determined the charges of different ions and the amount of elements liberated from the electrolytic solution. Due to this experiment presence of charge particles in the structure of atoms is discovered. The basic unit of electric charge was later named as electron by Stoney in 1891.
Diagram Coming Soon
Crook's Tube Or Discharge Tube Experiment
Passage of Electricity Through Gases Under Low Pressure
Introduction
The first of the subatomic particles to be discovered was electron. The knowledge about the electron was derived as a result of the study of the electric discharge in the discharge tube by J.J. Thomson in 1896. This work was later extended by W. Crooke
Working of Discharge Tube
When a very high voltage about 10,000 volts is applied between the two electrodes, no electric discharge occurs until the part of the air has been pumped out of the tube. When the pressure of the gas inside the tube is less than 1 mm, a dark space appears near the cathode and thread like lines are observed in the rest of 0.01 mm Hg it fills the whole tube. The electric discharge passes between the electrodes and the residual gas in the tube begins to glow. These rays which proceed from the cathode and move away from it at right angle in straight lines are called cathode rays.
Properties of Cathode Rays
1. They travel in straight lines away from the cathode and produce shadow of the object placed in their path.
2. The rays carry a negative charge.
3. These rays can also be easily deflected by an electrostatic field.
4. The rays can exert mechanical pressure showing that these consist of material particle which are moving with kinetic energy.
5. The produce fluorescence when they strike the glass wall of the discharge tube.
6. Cathode rays produce x-rays when they strike a metallic plate.
7. These rays consists of material particle whose e/m resembles with electron.
8. These rays emerge normally from the cathode and can be focused by using a concave cathode.
Positive Rays
In 1890 Goldstein used a discharge tube with a hole in the cathode. He observed that while cathode rays were emitting away from the cathode, there were coloured rays produced simultaneously which passed through the perforated cathode and caused a glow on the wall opposite to the anode. Thomson studied these rays and showed that they consisted of particles carrying a positive charge. He called them positive rays.
Properties of Positive Rays
1. These rays travel in a straight line in a direction opposite to the cathode.
2. These are deflected by electric as well as magnetic field in the way indicating that they are positively charged.
3. The charge to mass ratio (e/m) of positive particles varies with the nature of the gas placed in the discharge tube.
4. Positive rays are produced from the ionization of gas and not from anode electrode.
5. Positive rays are deflected in electric field. This deflection shows that these are positively charged so these are named as protons.
The Information Obtained From Discharge Tube Experiment
The negatively charge particles electrons and the positively charge particles protons are the fundamental particle of every atom.
Radioactivity
In 1895, Henry Becqueral observed that uranium and its compounds spontaneously emitted certain type of radiation which affected a photographic plate in the dark and were able to penetrate solid matter. He called these rays as radioactivity rays and a substance which possessed the property of emitting these radioactivity rays was said to be radioactivity element and the phenomenon was called radioactivity.
On further investigation by Maric Curic, it was found that the radiation emitted from the element uranium as well as its salts is independent of temperature and the source of the mineral but depend upon the mineral but depend upon the quantity of uranium present e.g. Pitchblende U3O8 was found to be about four times more radioactive than uranium.
Radioactive Rays
Soon after the discovery of radium it was suspected that the rays given out by radium and other radioactive substance were not of one kind. Rutherford in 1902 devised an ingenious method for separating these rays from each other by passing them between two oppositely charged plate. It was observed that the radioactive rays were of three kinds, the one bending towards the negative plate obviously carrying positive charge were called α-rays and those deflected to the positive plate and carrying -ve charge were named as β-rays. The third type gamma rays, pass unaffected and carry no charge.
Properties of α - RAYS
1. These rays consists of positively charged particles.
2. These particles are fast moving helium nuclei.
3. The velocity of α-particles is approximately equal to 1/10th of the velocity of light.
4. Being relatively large in size, the penetrating power of α-rays is very low.
5. They ionize air and their ionization power is high.
Properties of β - RAYS
1. These rays consists of negatively charged particles.
2. These particles are fast moving electron.
3. The velocity of β-particles is approximately equal to the velocity of light.
4. The penetrating power of β-rays is much greater than α-rays.
5. These rays ionizes gases to lesser extent.
Properties of γ - RAYS
1. Gamma rays do not consist of particles. These are electromagnetic radiations.
2. They carry no charge so they are not deflected by electric or magnetic field.
3. Their speed is equal to that of light.
4. These are weak ionizer of gases.
5. Due to high speed and non-material nature they have great power of penetration.
Chadwick Experiment (Discovery of Neutron)
When a light element is bombarded by α-particles, these α-particles leaves the nucleus in an unstable disturbed state which on settling down to stable condition sends out radioactivity rays. The phenomenon is known as "Artificial Radioactivity".
In 1933, Chadwick identified a new particle obtained from the bombardment of beryllium by α-particles. It had a unit mass and carried no charge. It was named "Neutron".
Spectroscopic Experiment
After the discovery of fundamental particles which are electrons, protons & neutron, the next question concerned with electronic structure of atom.
The electronic structure of the atom was explained by the spectroscopic studies. In this connection Plank's Quantum theory has great impact on the development of the theory of structure of atom.
Planck's Quantum Theory
In 1900, Max Planck studied the spectral lines obtained from hot body radiations at different temperatures. According to him,
When atoms or molecules absorb or emit radiant energy, they do so in separate units of waves called Quanta or Photons.
Thus light radiations obtained from excited atoms consists of a stream of photons and not continuous waves.
The energy E of a quantum or photon is given by the relation
E = h v
Where v is the frequency of the emitted radiation and h the Planck's constant. The value of h = 6.62 x 10(-27) erg. sec.
The main point of this theory is that the amount of energy gained or lost is quantized which means that energy change occurs in small packets or multiple of those packets, hv, 2 hv, 3 hv and so on.
Spectra
A spectrum is an energy of waves or particles spread out according to the increasing or decreasing of some property. E.g. when a beam of light is allowed to pass through a prism it splits into seven colours. This phenomenon is called dispersion and the band of colours is called spectrum. This spectrum is also known as emission spectrum. Emission spectra are of two types.
1. Continuous Spectrum
2. Line Spectrum
1. Continuous Spectrum
When a beam of white light is passed through a prism, different wave lengths are refracted through different angles. When received on a screen these form a continuous series of colours bands: violet, indigo, blue, green, yellow and red (VIBGYOR). The colours of this spectrum are so mixed up that there is no line of demarcation between different colours. This series of bands that form a continuous rainbow of colours is called continuous spectrum.
Diagram Coming Soon
2. Line Spectrum
When light emitted from a gas source passes through a prism a different kind of spectrum may be obtained.
If the emitted from the discharge tube is allowed to pass through a prism some discrete sharp lines on a completely dark back ground are obtained. Such spectrum is known as line spectrum. In this spectrum each line corresponds to a definite wave length.
Diagram Coming Soon
Identification of Element By Spectrum
Each element produces a characteristics set of lines, so line spectra came to serve as "finger prints" for the identification of element. It is possible because same element always emit the same wave length of radiation. Under normal condition only certain wave lengths are emitted by an element.
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Rutherford's Atomic Model
Evidence for Nucleus and Arrangement of Particles
Having known that atom contain electrons and a positive ion, Rutherford and Marsden performed their historic "Alpha particle scattering experiment" in 1909 to know how and where these fundamental particles were located in the structure of atom.
Rutherford took a thin of gold with thickness 0.0004 cm and bombarded in with α-particles. He observed that most of the α-particles passed straight through the gold foil and thus produced a flash on the screen behind it. This indicated that old atoms had a structure with plenty of empty space but some flashes were also seen on portion of the screen. This showed that gold atoms deflected or scattered α-particles through large angles so much so that some of these bounced back to the source.
Based on these observations Rutherford proposed a model of the atom which is known as Rutherford's atomic model.
Diagram Coming Soon
Assumption Drawn From the Model
1. Atom has a tiny dense central core or the nucleus which contains practically the entire mass of the atom leaving the rest of the atom almost empty.
2. The entire positive charge of the atom is located on the nucleus. While electrons were distributed in vacant space around it.
3. The electrons were moving in orbits or closed circular paths around the nucleus like planets around the sun.
4. The greater part of the atomic volume comprises of empty space in which electrons revolve and spin.
Weakness of Rutherford Atomic Model
According to the classical electromagnetic theory if a charged particle accelerate around an oppositely charge particle it will radiate energy. If an electron radiates energy, its speed will decrease and it will go into spiral motion finally falling into the nucleus. Similarly if an electron moving through orbitals of ever decreasing radii would give rise to radiations of all possible frequencies. In other words it would given rise to a continuous spectrum. In actual practise, atom gives discontinuous spectrum.
X-Rays and Atomic Number
In 1895, W.Roentgen discovered that when high energy electrons from cathode collide with the anode in the Crook's tube, very penetrating rays are produced. These rays were named as X-rays.
Explanation
When an electron coming from the cathode strike with the anode in the crook's tube, it can remove an electron from the inner shell of the atom. Due to removal of t his electron the electronic configuration of this ion is unstable and an electron from an orbital of higher energy drops into the inner orbital by emitting energy in form of a photon. This photon corresponds to electromagnetic radiations in the x-rays region.
Relationship Between Wave Length and Nuclear Charge
In 1911, Mosley stablished a relationship between the wave length and nuclear charge. He found that when cathode rays struck elements used as anode targets in the discharge tube, characteristic x-rays were emitted. The wave length of the x-rays emitted decreases regularly with the increase of atomic mass. On careful examination of his data Mosely found that the number of positive charges on the nucleus increases from atom to atom by single electronic unit. He called the number of positive charges as the atomic number.
Diagram Coming Soon
Bohr's Theory
Rutherford's model of atom fails to explain the stability of atom and appearance of the line spectra. Bohr in 1913 was the first to present a simple model of the atom which explained the appearance of line spectra.
Some of the postulates of Bohr's theory are given below.
1. An atom has a number of stable orbits or stationary states in which an electron can reside without emission or absorption of energy.
2. An electron may pass from one of these non-radiating states to another of lower energy with the emission of radiations whose energy equals the energy difference between the initial and final states.
3. In any of these states the electrons move in a circular path about the nucleus.
4. The motion of the electron in these states is governed by the ordinary laws of mechanics and electrostatic provided its angular momentum is an integral multiple of h/2Ï€
It can be written as
mvr = nh / 2Ï€
Here mvr becomes the angular momentum of the electron. Thus Bohr's first condition defining the stationary states could be stated as
"Only those orbits were possible in which the angular momentum of the electrons would be an integral multiple of h/2Ï€". These stationary states correspond to energy levels in the atom.
Calculation of Radius of Orbits
Consider an electrons of charge e revolving.
Atomic number and e the charge on a proton.
Let m be the mass of the electro, r the radius of the orbit and v the tangential velocity of the revolving electron.
The electrostatic force of attraction between the nucleus and the electron according to Coulomb's law
= Z e x e / r2
Diagram Coming Soon The centrifugal force acting on the electron.
= mv2 / r
Bohr assumed that these two opposing forces must be balanced each other exactly to keep the electron in an orbit.
Therefore
Ze2 / r2 = m v2 / r
Multiply both sides by r
r x Ze2 / r2 = r x m v2 / r
Ze2 / r = m v2
OR
r = Ze2 / m v2 .................. (1)
The Bohr's postulate states that only those orbits are possible in which
mvr = nh / 2Ï€
Therefore,
V = nh / 2Ï€mr
Substituting the value of V in eq (1)
r = Ze2 / m(nh/2Ï€mr)2
or
r = Ze2 x 4Ï€2 mr2/n2h2
or
1/r = 4Ï€2mZe2/n2h2
cr
r = n2h2 / 4Ï€2mZe2 ............... (2)
This equation gives the radii of all the possible stationary states. The values of constants present in this equation are as follows.
H = 6.625 x 10(-27) ergs sec OR 6.625 x 10(-37) J.s
Me = 9.11 x 10(-28) gm OR 9.11 x 10(-31) kg
E = 4.802 x 10(-10) e.s.u OR 1.601 x 10(-19) C
By substituting these values we get for first shell of H atom
r = 0.529 x 10(-8) m OR 0.529
The above equation may also be written as
r = n2 (h2 / 4Ï€2mZe2) x n2 a0 .................... (3)
For the first orbit n = 1 and r = 0.529. This is the value of the terms in the brackets sometimes written as a0 called Bohr's Radius. For the second shell n = 2 and for 3rd orbit n = 3 and so on.
Hydrogen Atom Spectrum
Balmer Series
The simplest element is hydrogen which contain only one electron in its valence shell.
Balmer in 1885 studied the spectrum of hydrogen. For this purpose he used hydrogen gas in the discharge tube. Balmer observed that hydrogen atom spectrum consisted of a series of lines called Balmer Series. Balmer determined the wave number of each of the lines in the series and found that the series could be derived by a simple formula.
Lyman Series
Lyman series is obtained when the electron returns to the ground state i.e. n = 1 from higher energy level n(2) = 2, 3, 4, 5, etc. This series of lines belongs to the ultraviolet region of spectrum.
Paschen Series
Paschen series is obtained when the electron returns to the 3rd shell i.e. n = 3 from the higher energy levels n2 = 4, 5, 6 etc. This series belongs to infrared region.
Bracket Series
This series is obtained when an electron jumps from higher energy levels to 4th energy level.
Heisenberg Uncertainty Principle
According to Bohr's theory an electron was considered to be a particle but electron also behaves as a wave according to be Broglie.
Due to this dual nature of electron in 1925 Heisenberg gave a principle known as Heisenberg Uncertainty Principle which is stated as,
It is impossible to calculate the position and momentum of a moving electron simultaneously.
It means that if one was known exactly it would be impossible to known the other exactly. Therefore if the uncertainty in the determination of momentum is Δpx and the uncertainty in position is Δx then according to this principle the product of these two uncertainties may written as
Δpx . Δx ≈ h
So if one of these uncertainties is known exactly then the uncertainty in its determination is zero and the other uncertainty will become infinite which is according to the principle.
Energy Levels and Sub-Levels
According to Bohr's atomic theory, electrons are revolving around the nucleus in circular orbits which are present at definite distance from the nucleus. These orbits are associated with definite energy of the electron increasing outwards from the nucleus, so these orbits are referred as Energy Levels or Shells.
These shells or energy levels are designated as 1, 2, 3, 4 etc K, L, M, N etc.
The spectral lines which correspond to the transition of an electron from one energy level to another consists of several separate close lying lines as doublets, triplets and so on. It indicates that some of the electrons of the given energy level have different energies or the electrons belonging to same energy level may differ in their energy. So the energy levels are accordingly divided into sub energy levels which are denoted by letters s, p, f (sharp, principle, diffuse & fundamental).
The number of sub levels in a given energy level or shell is equal to its value of n.
e.g. in third shell where n = 3 three sub levels s, p, d are possible.
Quantum Numbers
There are four quantum numbers which describe the electron in an atom.
1. Principle Quantum Number
It is represented by "n" which describe the size of orbital or energy level.
The energy level K, L, M, N, O etc correspond to n = 1, 2, 3, 4, 5 etc.
If
n = 1 the electron is in K shell
n = 2 the electron is in L shell
n = 3 the electron is in M shell
2. Azimuthal Quantum Number
This quantum number is represented by "l" which describes the shape of the orbit. The value of Azimuthal Quantum number may be calculated by a relation.
l = 0 ----> n - 1
So for different shell the value of l are as
n = 1 K Shell l = 0
n = 2 L Shell l = 0, 1
n = 3 M Shell l = 0, 1, 2
n = 4 N Shell l = 0, 1, 2, 3
when l = 0 the orbit is s
when l = 1 the orbit is p
when l = 2 the orbit is d
when l = 3 the orbit is f
3. Magnetic Quantum Number
It is represented by "m" and explains the magnetic properties of an electron. The value of m depends upon the value of l. It is given by
m = + l ----> 0 ----> l
when l = 1, m has three values (+1, 0, -1) which corresponds to p orbital. Similarly when l = 2, m has five values which corresponds to d orbital.
4. Spin Quantum Number
It is represented by "s" which represents spin of a moving electron. This spin may be either clockwise or anticlockwise so the values for s may be +1/2 or -1/2.
Pauli's Exclusion Principle
According to this principle
No two electrons in the same atom can have the same four quantum number.
Consider an electron is present in 1s orbital. For this electron n = 1, l = 0, m = 0. Suppose the spin of this electron is s = +1/2 which will be indicated by an upward arrow ↑. Now if another electron is put in the same orbital (1s) for that electron n = 1, l = 0, m = 0. It can occupy this orbital only if the direction of its spin is opposite to that of the first electron so s = -1/2 which is symbolized by downward arrow ↓. From this example, we can observe the application of Pauli's exclusion principle on the electronic structure of atom.
Electronic Configuration
The distribution of electrons in the available orbitals is proceeded according to these rules.
1. Pauli Exclusion Principle
2. Aufbau Principle
3. (n + l) Rule
4. Hund's Rule
The detail of these rules and principles is given below.
1. Aufbau Principle
It is states as
The orbitals are filled up with electrons in the increasing order of their energy.
It means that the orbitals are fulled with the electrons according to their energy level. The orbitals of minimum energy are filled up first and after it the orbitals of higher energy are filled.
2. Hund's Rule
If orbitals of equal energy are provided to electron then electron will go to different orbitals and having their parallel spin.
In other words we can say that electrons are distributed among the orbitals of a sub shell in such a way as to give the maximum number of unpaired electrons and have the same direction of spin.
3. (n + l) Rule
According to this rule
The orbital with the lowest value of (n + l) fills first but when the two orbitals have the same value of (n + l) the orbital with the lower value of n fills first.
For the electronic configuration the order of the orbital is as follows.
1s, 2s, 2p, 3s, 4s, 3d, 4p, 5s, 4d, 5p, 6s etc.
Atomic Radius
For homonuclear diatomic molecules the atomic radius may be defined as
The half of the distance between the two nuclei present in a homonuclear diatomic molecules is called atomic radius.
It may be shown as
In case of hetronuclear molecular like AB, the bond length is calculated which is (rA + rB) and if radii of any one is known the other can be calculated.
For the elements present in periodic table the atomic radius decreases from left to right due to the more attraction on the valence shell but it increases down the group with the increase of number of shells.
Ionic Radius
Ionic radius is defined as
The distance between nucleus of an ion and the point up to which nucleus has influence of its electron cloud.
When an electron is removed from a neutral atom the atom is left with an excess of positive charge called a cation e.g
Na ----> Na+ + c-
But when an electron is added in a neutral atom a negative ion or anion is formed.
Cl + e- ----> Cl-
As the atomic radius, the ionic radii are known from x-ray analysis. The value of ionic radius depends upon the ions that surround it.
Ionic radii of cations have smaller radii than the neutral atom because when an electron is removed. The effective charge on the nucleus increases and pulls the remaining electrons with a greater force.
Ionic radii of anions have a large radii than the neutral atom because an excess of negative charge results in greater electron repulsion.
Radius of Na atom = 1.57
Radius of Na+ atom = 0.95 (smaller than neutral atom)
Radius of Cl atom = 0.99
Radius of Cl- atom = 1.81 (larger than neutral atom)
Ionization Potential
Definition
The amount of energy required to remove most loosely bounded electron from the outermost shell of an atom in its gaseous state is called is called ionization potential energy.
It is represented as
M(gas) ----> M+(gas) + e- ................... ΔE = I.P
The energy required to remove first electron is called first I.P. The energy required to remove 2nd or 3rd electron is called 2nd I.P or 3rd I.P
M(gas) ----> M+(gas) + e- ................... ΔE = 1st I.P
M+(gas) ----> M++(gas) + e- ................ΔE = 2nd I.P
M++(gas) ----> M+++(gas) + e- ............ ΔE = 3rd I.P
The units of I.P is kilo-Joule per mole.
Factors on which I.P Depends
1. Size of the Atom
If the size of an atom is bigger the I.P of the atom is low, but if the size of the atom is small then the I.P will be high, due to fact if we move down the group in the periodic table. The I.P value decreases down the group.
2. Magnitude of Nuclear Charge
If the nuclear charge of atom is greater than the force of attraction on the valence electron is also greater so the I.P value for the atom is high therefore as we move from left to right in the periodic table the I.P is increased.
3. Screening Effect
The shell present between the nucleus and valence electrons also decreases the force of attraction due to which I.P will be low for such elements.
Electron Affinity
Definition
The amount of energy liberated by an atom when an electron is added in it is called electron affinity.
It shows that this process is an exothermic change which is represented as
Cl + e- ----> Cl- ............ ΔH = -348 kJ / mole
Factors on which Electron Affinity Depends
1. Size of the Atom
If the size of atom is small, the force of attraction from the nucleus on the valence electron will be high and hence the E.A for the element will also be high but if the size of the atoms is larger the E.A for these atoms will be low.
2. Magnitude of the Nuclear Charge
Due to greater nuclear charge the force of attraction on the added electron is greater so the E.A of the atom is also high.
3. Electronic Configuration
The atoms with the stable configuration has no tendency to gain an electron so the E.A of such elements is zero. The stable configuration may exist in the following cases.
1. Inert gas configuration
2. Fully filled orbital
3. Half filled orbital
Electronegativity
Definition
The force of attraction by which an atom attract a shared pair of electrons is called electronegativity.
Application of Electronegativity
1. Nature of Chemical Bond
If the difference of electronegativity between the two combining atoms is more than 1.7 eV, the nature of the bond between these atoms is ionic but if the difference of electronegativity is less than 1.7 eV then the bond will be covalent.
2. Metallic Character
If an element possesses high electronegativity value then this element is a non-metal but if an element exist with less electronegativity, it will be a metal.
Factors for Electronegativity
1. Size of the Atom
If the size of the atom is greater the electronegativity of the atom is low due to the large distance between the nucleus and valence electron.
2. Number of Valence Electrons
If the electrons present in the valence shell are greater in number, the electronegativity of the element is high
Chemical Bond
Introduction
Atoms of all the elements except noble gases have incomplete outermost orbits and tends to complete them by chemical combination with the other atoms.
In 1916, W Kossel described the ionic bond which is formed by the transfer of electron from one atom to another and also in 1916 G.N Lewis described about the formation of covalent bond which is formed by the mutual sharing of electrons between two atoms.
Both these scientists based their ideas on the fact that atoms greatest stability when they acquire an inert gas electronic configuration.
Definition
When two or more than two atoms are combined with each other in order to complete their octet a link between them is produced which is known as chemical bond.
OR
The force of attraction which holds atoms together in the molecule of a compound is called chemical bond.
Types of Chemical Bond
There are three main types of chemical bond.
1. Ionic bond or electrovalent bond
2. Covalent bond
3. Co-ordinate covalent bond or Dative covalent bond
Ionic Bond OR Electrovalent Bond
Definition
A chemical bond which is formed by the complete shifting of electron between two atoms is called ionic bond or electrovalent bond.
OR
The electrostatic attraction between positive and negative ions is called ionic bond.
Conditions for the Ionic Bond Formation
1. Electronegativity
Ionic bond is formed between the element having a difference of electronegativity more than 1.7 or equal to 1.7 eV.
Therefore ionic bond is generally formed between metals (low electronegative) and non-metal (high electronegative) elements.
2. Ionization Potential
We know that ionic bond is formed by the transference of electron from one atom to another, so in the formation of ionic bond an element is required which can lose its electrons from the outer most shell. It is possible to remove electron from the outermost shell of metals because of their low ionization potential values.
3. Electron Affinity
In the formation of ionic bond an element is also required which can gain an element is also required which can gain electron, since non-metals can attract electrons with a greater force due to high electronegativity. So a non-metal is also involved in the formation of ionic bond due to high electron affinity.
Example of Ionic Bond
In order to understand ionic bond consider the example of NaCl. During the formation of Ionic bond between Na and Cl2, Sodium loses one electron to form Na+ ion while chlorine atom gains this electron to form Cl- ion. When Na+ ion and Cl- ion attract to each other NaCl is formed. The stability of NaCl is due to the decrease in the energy. These energy change which are involved in the formation of ionic bond between Na and Cl are as follows.
i. Sodium has one valence electron. In order to complete its octet Na loses its valence electron. The loss of the valence electron required 495 kJ/mole.
Na ----> Na+ + e- ....................... ΔH = 495 kJ/mole
ii. Chlorine atom has seven electrons in its valence shell. It require only one electron to complete its octet, so chlorine gains this electron of sodium and release 348 kJ/mole energy.
Cl + e- ----> Cl- ...................... ΔH = -348 kJ/mole
Here the energy difference is 147 kJ/mole (495 - 348 = 147). This loss of energy is balanced when oppositely charged ions are associated to form a crystal lattice.
iii. In third step, positively charged Na+ ion and negatively charged Cl- ion attract to each other and a crystal lattice is formed with a definite pattern.
Na+(g) + Cl-(g) ----> Na+Cl- ........... ΔH = - 788 kJ/mole
This energy which is released when one mole of gaseous ions arrange themselves in definite pattern to form lattice is called lattice energy.
From this example, we can conclude that it is essential for the formation of ionic bond that the sum of energies released in the second and third steps must be greater than the energy required for the first step.
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Characteristics of Ionic Compounds
1. An ionic compounds, the oppositely charged ions are tightly packed with each other, so these compounds exist in solid state.
2. Due to strong attractive forces between ions a larger amount of energy is required to melt or to boil the compound and hence the melting and boiling point of the ionic compound are generally high.
3. Ionic compounds are soluble in water but insoluble in organic solvents like benzene, CCl4. etc.
4. In the aqueous solution, the ionic compounds are good electrolytes, because in water the interionic forces are so weakened that the ions are separated and free to move under the influence of electric current. Due to this free movement of ions, the ionic compounds conduct electricity in their solutions.
Covalent Bond
Definition
A link which is formed by the mutual sharing of electrons between two atoms is called covalent bond.
Explanation
In the formation of covalent bond, mutual sharing of electron takes place. This mutual sharing is possible in non-metals, therefore covalent bond is generally formed between the atoms of non-metals. For example
In Cl2 molecule, two atoms of chlorine are combined with each other to form Cl2 molecule. Each atom of chlorine having seven electrons in its valencies shell. These atoms are united with each other by sharing one of its valence electron as shown.
Cl Cl: ----> :Cl :Cl OR Cl - Cl In this molecule, one shared pair of electrons forms a single covalent bond between two chlorine the atoms. With the formation of a covalent bond the energy of the system is also decreased.
Cl + Cl ----> Cl - Cl .............. ΔH = - 242 kJ / mole
This released energy lowered the energy of the molecule and the stability of the compound is also increased.
Types of Covalent Bond
There are three main types of covalent bond.
1. Single Covalent Bond
When a covalent bond is formed by sharing of one electron from each atom, that it is called single covalent bond and denoted by (-) single line between the two bonded atoms e.g.
Cl - Cl, H - H, H - Br etc.
2. Double Covalent Bond
In a covalent bond, if two electrons are shared from each of the bonded atom then this covalent bond is called double covalent bond and denoted by (=) two lines e.g.
O = O, O : : O
3. Triple Covalent Bond
When a covalent bond is formed by sharing of three electrons from each atom then this type of covalent bond is called triple covalent bond, and denoted by (≡) three lines between the two bonded atoms e.g.
N : : N :, N ≡ N The bond distance of multiple bonds are shorter and the bond energies are higher.
Characteristics of Covalent Compounds
The main characteristics properties of covalent compounds are as follows
1. The covalent compounds exist as separate covalent molecules, because the particles are electrically neutral so they passes solid, liquid or gaseous state. This intermolecular force of attraction among the molecules.
2. Since the covalent compound exist in all the three states of matter so their melting points and boiling point may be high or low.
3. Covalent compounds are non-electrolytes so they do not conduct electricity from their aqueous solution.
4. Covalent compounds are generally insoluble in water and similar polar solvent but soluble in the organic solvents.
Co-Ordinate OR Dative Covalent Bond
Definition
It is a type of covalent bond in which both the shared electrons are donated only be one atom, this type is called co-ordinate covalent bond.
The ∞ ordinate covalent bond between two atoms is denoted by an arrow (→). The atom which donates an electron pair is called as a donor of electron and the other atom involved in this bond is called acceptor. E.g.
A + B ----> A : B OR A → B
Dipole Moment
Definition
The product of the charge and the distance present in a polar molecules is called dipole moment and represented by μ.
OR
The extent of tendency of a molecule to be oriented under the influence of an electric field is called dipole moment.
Mathematical Representation of Dipole Moment
Suppose the charge present on a polar molecule is denoted by e and the separation between the two oppositely charged poles of the molecules is d, then the product of these two may be written as
e x d = μ
Where μ is dipole moment.
Dipole Moment in Diatomic Molecules
The diatomic molecules which are made up of similar atoms will be non-polar and their dipole moment is zero but the diatomic molecules made up of two different atoms e.g. HCl or Hl are polar and have some dipole moment. The value of the dipole moment depends upon the difference of electronegativities of the two bonded atom. If the difference of electronegativity between the atoms is greater, the polarity and also the dipole moment of the molecule is greater e.g.
The dipole moment of HCl = 1.03 debye
Whereas dipole moment of HF = 1.90 debye
Dipole Moment of Poly Atomic Molecules
In poly atomic molecules, the dipole moment of molecules depends upon the polarity of the bond as well as the geometry of the molecule.
Ionic Character of Covalent Bond
In homonuclear diatomic molecules like Cl2, O2, l2, H2 both the atoms are identical so the shared electrons are equally attracted due to identical electronegativities and hence the molecules are non-polar.
When two dissimilar atoms are linked by a covalent bond the shared electrons are not attracted equally by the two bonded atoms. Due to unsymmetrical distribution of electrons one end of the molecules acquire partial positive charge and the other end acquire a partial negative charge. This character of a covalent bond is called Ionic character of a covalent bond.
The ionic character of a covalent bond depends upon the difference of electronegativity of the two dissimilar atoms joined with each other in a covalent bond. E.g., the H-F bond is 43% ionic whereas the H-Cl bond is 17% ionic. The ionic character greatly affects the properties of a molecules e.g., melting point, boiling point of polar molecules are high and they are soluble in polar solvent like H2O. Similarly the presence of partial polar character shortens the covalent bond and increases the bond energies.
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Bond Energy
Definition
The amount of energy required to break a bond between two atoms in a diatomic molecule is known as Bond Energy.
OR
The energy released in forming a bond from the free atoms is also known as Bond Energy.
It is expressed in kilo Joules per mole or kCal/mole.
Examples
i. The bond energy for hydrogen molecule is
H - H(g) ----> 2 H(g) .......................... ΔH = 435 kJ/mole
OR
H(g) + H(g) ----> H - H ....................... ΔH = 435 kJ/mole
It can be observed from this example that the breaking of bond is endothermic whereas the formation of the bond is exothermic.
ii. The bond energy for oxygen molecule is
O = O(g) ----> 2 O(g) ........................ ΔH = 498 kJ/mole
OR
O(g) + O(g) ----> O = O .................... ΔH = -498 kJ/mole
Bond energy of a molecule also measure the strength of the bond. Generally bond energies of polar bond are greater than pure covalent bond.
E.g.
Cl - Cl ----> 2 Cl ........................ ΔH = 244 kJ/mole
H - Cl ----> H+ + Cl- ................... ΔH = 431 kJ/mole
The value of bond energy e.g., triple bonds are usually shorter than the double bond therefore the bond energy for triple bond is greater than double bond.
Sigma & PI Bond
Sigma Bond Definition
When the two orbitals which are involved in a covalent bond are symmetric about an axis, then the bond formed between these orbitals is called Sigma Bond.
OR
A bond which is formed by head to head overlap of atomic orbitals is called Sigma Bond.
Explanation
In the formation of a sigma bond the atomic orbital lies on the same axis and the overlapping of these orbital is maximum therefore, all such bonds, in which regions of highest density around the bond axis are termed as sigma bond.
Types of Overlapping in Sigma Bond
There are three types of overlapping in the formation of sigma bond.
1. s-s orbitals overlapping
2. s-p orbitals overlapping
3. p-p orbitals overlapping
In all the three types, when the two atomic orbitals are overlapped with each other two molecular orbitals are formed. In these two molecular orbitals the energy of one orbital is greater than the the atomic orbitals which is known as sigma antibonding orbital while the energy of the other orbital is less than the atomic orbital this orbital of lower energy is called sigma bonding orbital and the shared electron are always present in the sigma bonding orbitals.
1. s-s Orbitals Overlapping
In order to explain s-s overlapping consider the example of H2 molecule. In this molecule is orbital of one hydrogen overlaps with is orbital of other hydrogen to form sigma bonding orbitals. Due to this bonding a single covalent bond is formed between the two hydrogen atoms.
Diagram Coming Soon
2. s-p Orbitals Overlapping
This type of overlapping takes place in H-Cl molecule. 1s orbital of hydrogen overlaps with 1p orbital of chlorine to form a single covalent bond. In this overlapping two molecular orbitals are formed, one of the lower energy while the other orbital is of higher energy. The shapes of these orbitals are as follows.
Diagram Coming Soon
3. p-p Orbitals Overlapping
This type of overlapping takes place in fluorine molecule. In this mole 1p orbital of a fluorine atom is overlapped with 1p orbital of the other fluorine atom. The molecular orbitals formed in this overlapping are given in figure
Diagram Coming Soon
PI Bond
When the two atomic orbital involved in a covalent bond are parallel to each other then the bond formed between them is called pi bond.
In this overlapping, two molecular orbitals are also formed. The lower energy molecular orbitals is called π bonding orbital while the higher energy molecular orbital is called π antibonding orbital. The shape of these molecular orbitals are as follows.
Diagram Coming Soon
Hybridization
Definition
The process in which atomic orbitals of different energy and shape are mixed together to form new set of equivalent orbitals of the same energy and same shape.
There are many different types of orbital hybridization but we will discuss here only three main types.
1. sp3 Hybridization
The mixing of one s and three p orbitals to form four equivalent sp3 hybrid orbitals is called sp3 hybridization. These sp3 orbitals are directed from the center of a regular tetrahedron to its four corners. The angles between tetrahedrally arranged orbitals are 109.5º.
It has two partially filled 2p orbitals which indicate that it is divalent, but carbon behaves as tetravalent in most of its compounds. It is only possible if one electron from 2s orbital is promoted to an empty 2pz orbital to get four equivalent sp3 hybridized orbitals.
Diagram Coming Soon The four sp3 hybrid orbitals of the carbon atom overlap with 1s orbitals of four hydrogen atoms to form a methane CH4 molecule.
The methane molecule contains four sigma bonds and each H-C-H bond angle is 109.5º.
2. sp2 Hybridization
The mixing of one s and two p orbitals to form three orbitals of equal energy is called sp2 or 3sp2 hybridization. Each sp2 orbital consists of s and p in the ratio of 1:2. These three orbitals are co-planar and at 120º angle as shown
Diagram Coming Soon A typical example of this type of hybridization is of ethane molecule. In ethylene, two sp2 hybrid orbitals of each carbon atom share and overlap with 1s orbitals of two hydrogen atoms to form two σ bonds. While the remaining sp2 orbital on each carbon atom overlaps to form a σ bond. The remaining two unhybridized p orbitals (one of each) are parallel and perpendicular to the axis joining the two carbon nuclei. These generates a parallel overlap and results in the formation of 2 π orbitals. Thus a molecule of ethylene contain five σ bonds and one π bond.
Diagram Coming Soon
3. sp Hybridization
When one s and one p orbitals combine to give two hybrid orbitals the process is called sp hybridization. The sp hybrid orbitals has two lobes, one with greater extension in shape than the other and the lobes are at an angle of 180º from each other. It means that the axis of the two orbitals form a single straight line as shown.
Now consider the formation of acetylene molecule HC ≡ CH. The two C-H σ bonds are formed due to sp-s overlap and a triple bond between two carbon atoms consist of a σ bond and two Ï€ bond. The sigma bond is due to sp-sp overlap whereas Ï€ bonds are formed as a result of parallel overlap between the unhybridized four 2p orbitals of the two carbon.
Diagram Coming Soon
Valence Shell Electron Pair Repulsion Theory
The covalent bonds are directed in space to give definite shapes to the molecules. The electrons pairs forming the bonds are distributed in space around the central atom along definite directions. The shared electron pairs as well as the lone pair of electrons are responsible for the shape of molecules.
Sidwick and Powell in 1940 pointed out that the shapes of the molecules could be explained on the basis of electron pairs present in the outermost shell of the central atom. Pairs of electrons around the central atom are arranged in space in such a way so that the distances between them are maximum and coulombie repulsion of electronic cloud are minimized.
The known geometries of many molecules based upon measurement of bond angles shows that lone pairs of electrons occupy more space than bonding pairs. The repulsion between electronic pairs in valence shell, decreases in the following order.
Lone Pair - Lone Pair > Lone Pair - Bond Pair > Bond Pair - Bond Pair
When we apply this theory we can see the variation of angle in the molecular structures.
Consider the molecular structures of NH3, OH & H2O.
Diagram Coming Soon Variation from ideal bond angles are caused by multiple covalent bonds and lone electron pairs both of which require more space than single covalent bonds and therefore cause compression of surrounding bond angles.
Thus the number of pairs of electrons in the valency shell determine the overall molecular shape.
Structure of BeCl2
The two bond pairs of electrons in BeCl2 arrange themselves as far apart as possible in order to minimize the repulsion between them.
Structure of BF3 OR BCl3
In this molecule three bond pair are present around boron to arrange themselves as far apart as possible a trigonal structure is formed.
Hydrogen Bond
When hydrogen is bonded with a highly electronegative element such as nitrogen oxygen, fluorine, the molecule will be polarized and a dipole is produced. The slightly positive hydrogen atom is attracted by the slightly negatively charged electronegative atom. An electrostatic attraction between the neighbouring molecules is set up when the positive pole of one molecule attracts the negative pole of the neighbouring molecule. This type of attractive force which involves hydrogen is known as hydrogen bonding.
Energetics Of Chemical Reaction
Introduction
Atoms of all the elements except noble gases have incomplete outermost orbits and tends to complete them by chemical combination with the other atoms.
In 1916, W Kossel described the ionic bond which is formed by the transfer of electron from one atom to another and also in 1916 G.N Lewis described about the formation of covalent bond which is formed by the mutual sharing of electrons between two atoms.
Both these scientists based their ideas on the fact that atoms greatest stability when they acquire an inert gas electronic configuration.
Definition
When two or more than two atoms are combined with each other in order to complete their octet a link between them is produced which is known as chemical bond.
OR
The force of attraction which holds atoms together in the molecule of a compound is called chemical bond.
Types of Chemical Bond
There are three main types of chemical bond.
1. Ionic bond or electrovalent bond
2. Covalent bond
3. Co-ordinate covalent bond or Dative covalent bond
Ionic Bond OR Electrovalent Bond
Definition
A chemical bond which is formed by the complete shifting of electron between two atoms is called ionic bond or electrovalent bond.
OR
The electrostatic attraction between positive and negative ions is called ionic bond.
Conditions for the Ionic Bond Formation
1. Electronegativity
Ionic bond is formed between the element having a difference of electronegativity more than 1.7 or equal to 1.7 eV.
Therefore ionic bond is generally formed between metals (low electronegative) and non-metal (high electronegative) elements.
2. Ionization Potential
We know that ionic bond is formed by the transference of electron from one atom to another, so in the formation of ionic bond an element is required which can lose its electrons from the outer most shell. It is possible to remove electron from the outermost shell of metals because of their low ionization potential values.
3. Electron Affinity
In the formation of ionic bond an element is also required which can gain an element is also required which can gain electron, since non-metals can attract electrons with a greater force due to high electronegativity. So a non-metal is also involved in the formation of ionic bond due to high electron affinity.
Example of Ionic Bond
In order to understand ionic bond consider the example of NaCl. During the formation of Ionic bond between Na and Cl2, Sodium loses one electron to form Na+ ion while chlorine atom gains this electron to form Cl- ion. When Na+ ion and Cl- ion attract to each other NaCl is formed. The stability of NaCl is due to the decrease in the energy. These energy change which are involved in the formation of ionic bond between Na and Cl are as follows.
i. Sodium has one valence electron. In order to complete its octet Na loses its valence electron. The loss of the valence electron required 495 kJ/mole.
Na ----> Na+ + e- ....................... ΔH = 495 kJ/mole
ii. Chlorine atom has seven electrons in its valence shell. It require only one electron to complete its octet, so chlorine gains this electron of sodium and release 348 kJ/mole energy.
Cl + e- ----> Cl- ...................... ΔH = -348 kJ/mole
Here the energy difference is 147 kJ/mole (495 - 348 = 147). This loss of energy is balanced when oppositely charged ions are associated to form a crystal lattice.
iii. In third step, positively charged Na+ ion and negatively charged Cl- ion attract to each other and a crystal lattice is formed with a definite pattern.
Na+(g) + Cl-(g) ----> Na+Cl- ........... ΔH = - 788 kJ/mole
This energy which is released when one mole of gaseous ions arrange themselves in definite pattern to form lattice is called lattice energy.
From this example, we can conclude that it is essential for the formation of ionic bond that the sum of energies released in the second and third steps must be greater than the energy required for the first step.
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Characteristics of Ionic Compounds
1. An ionic compounds, the oppositely charged ions are tightly packed with each other, so these compounds exist in solid state.
2. Due to strong attractive forces between ions a larger amount of energy is required to melt or to boil the compound and hence the melting and boiling point of the ionic compound are generally high.
3. Ionic compounds are soluble in water but insoluble in organic solvents like benzene, CCl4. etc.
4. In the aqueous solution, the ionic compounds are good electrolytes, because in water the interionic forces are so weakened that the ions are separated and free to move under the influence of electric current. Due to this free movement of ions, the ionic compounds conduct electricity in their solutions.
Covalent Bond
Definition
A link which is formed by the mutual sharing of electrons between two atoms is called covalent bond.
Explanation
In the formation of covalent bond, mutual sharing of electron takes place. This mutual sharing is possible in non-metals, therefore covalent bond is generally formed between the atoms of non-metals. For example
In Cl2 molecule, two atoms of chlorine are combined with each other to form Cl2 molecule. Each atom of chlorine having seven electrons in its valencies shell. These atoms are united with each other by sharing one of its valence electron as shown.
Cl Cl: ----> :Cl :Cl OR Cl - Cl In this molecule, one shared pair of electrons forms a single covalent bond between two chlorine the atoms. With the formation of a covalent bond the energy of the system is also decreased.
Cl + Cl ----> Cl - Cl .............. ΔH = - 242 kJ / mole
This released energy lowered the energy of the molecule and the stability of the compound is also increased.
Types of Covalent Bond
There are three main types of covalent bond.
1. Single Covalent Bond
When a covalent bond is formed by sharing of one electron from each atom, that it is called single covalent bond and denoted by (-) single line between the two bonded atoms e.g.
Cl - Cl, H - H, H - Br etc.
2. Double Covalent Bond
In a covalent bond, if two electrons are shared from each of the bonded atom then this covalent bond is called double covalent bond and denoted by (=) two lines e.g.
O = O, O : : O
3. Triple Covalent Bond
When a covalent bond is formed by sharing of three electrons from each atom then this type of covalent bond is called triple covalent bond, and denoted by (≡) three lines between the two bonded atoms e.g.
N : : N :, N ≡ N The bond distance of multiple bonds are shorter and the bond energies are higher.
Characteristics of Covalent Compounds
The main characteristics properties of covalent compounds are as follows
1. The covalent compounds exist as separate covalent molecules, because the particles are electrically neutral so they passes solid, liquid or gaseous state. This intermolecular force of attraction among the molecules.
2. Since the covalent compound exist in all the three states of matter so their melting points and boiling point may be high or low.
3. Covalent compounds are non-electrolytes so they do not conduct electricity from their aqueous solution.
4. Covalent compounds are generally insoluble in water and similar polar solvent but soluble in the organic solvents.
Co-Ordinate OR Dative Covalent Bond
Definition
It is a type of covalent bond in which both the shared electrons are donated only be one atom, this type is called co-ordinate covalent bond.
The ∞ ordinate covalent bond between two atoms is denoted by an arrow (→). The atom which donates an electron pair is called as a donor of electron and the other atom involved in this bond is called acceptor. E.g.
A + B ----> A : B OR A → B
Dipole Moment
Definition
The product of the charge and the distance present in a polar molecules is called dipole moment and represented by μ.
OR
The extent of tendency of a molecule to be oriented under the influence of an electric field is called dipole moment.
Mathematical Representation of Dipole Moment
Suppose the charge present on a polar molecule is denoted by e and the separation between the two oppositely charged poles of the molecules is d, then the product of these two may be written as
e x d = μ
Where μ is dipole moment.
Dipole Moment in Diatomic Molecules
The diatomic molecules which are made up of similar atoms will be non-polar and their dipole moment is zero but the diatomic molecules made up of two different atoms e.g. HCl or Hl are polar and have some dipole moment. The value of the dipole moment depends upon the difference of electronegativities of the two bonded atom. If the difference of electronegativity between the atoms is greater, the polarity and also the dipole moment of the molecule is greater e.g.
The dipole moment of HCl = 1.03 debye
Whereas dipole moment of HF = 1.90 debye
Dipole Moment of Poly Atomic Molecules
In poly atomic molecules, the dipole moment of molecules depends upon the polarity of the bond as well as the geometry of the molecule.
Ionic Character of Covalent Bond
In homonuclear diatomic molecules like Cl2, O2, l2, H2 both the atoms are identical so the shared electrons are equally attracted due to identical electronegativities and hence the molecules are non-polar.
When two dissimilar atoms are linked by a covalent bond the shared electrons are not attracted equally by the two bonded atoms. Due to unsymmetrical distribution of electrons one end of the molecules acquire partial positive charge and the other end acquire a partial negative charge. This character of a covalent bond is called Ionic character of a covalent bond.
The ionic character of a covalent bond depends upon the difference of electronegativity of the two dissimilar atoms joined with each other in a covalent bond. E.g., the H-F bond is 43% ionic whereas the H-Cl bond is 17% ionic. The ionic character greatly affects the properties of a molecules e.g., melting point, boiling point of polar molecules are high and they are soluble in polar solvent like H2O. Similarly the presence of partial polar character shortens the covalent bond and increases the bond energies.
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Bond Energy
Definition
The amount of energy required to break a bond between two atoms in a diatomic molecule is known as Bond Energy.
OR
The energy released in forming a bond from the free atoms is also known as Bond Energy.
It is expressed in kilo Joules per mole or kCal/mole.
Examples
i. The bond energy for hydrogen molecule is
H - H(g) ----> 2 H(g) .......................... ΔH = 435 kJ/mole
OR
H(g) + H(g) ----> H - H ....................... ΔH = 435 kJ/mole
It can be observed from this example that the breaking of bond is endothermic whereas the formation of the bond is exothermic.
ii. The bond energy for oxygen molecule is
O = O(g) ----> 2 O(g) ........................ ΔH = 498 kJ/mole
OR
O(g) + O(g) ----> O = O .................... ΔH = -498 kJ/mole
Bond energy of a molecule also measure the strength of the bond. Generally bond energies of polar bond are greater than pure covalent bond.
E.g.
Cl - Cl ----> 2 Cl ........................ ΔH = 244 kJ/mole
H - Cl ----> H+ + Cl- ................... ΔH = 431 kJ/mole
The value of bond energy e.g., triple bonds are usually shorter than the double bond therefore the bond energy for triple bond is greater than double bond.
Sigma & PI Bond
Sigma Bond Definition
When the two orbitals which are involved in a covalent bond are symmetric about an axis, then the bond formed between these orbitals is called Sigma Bond.
OR
A bond which is formed by head to head overlap of atomic orbitals is called Sigma Bond.
Explanation
In the formation of a sigma bond the atomic orbital lies on the same axis and the overlapping of these orbital is maximum therefore, all such bonds, in which regions of highest density around the bond axis are termed as sigma bond.
Types of Overlapping in Sigma Bond
There are three types of overlapping in the formation of sigma bond.
1. s-s orbitals overlapping
2. s-p orbitals overlapping
3. p-p orbitals overlapping
In all the three types, when the two atomic orbitals are overlapped with each other two molecular orbitals are formed. In these two molecular orbitals the energy of one orbital is greater than the the atomic orbitals which is known as sigma antibonding orbital while the energy of the other orbital is less than the atomic orbital this orbital of lower energy is called sigma bonding orbital and the shared electron are always present in the sigma bonding orbitals.
1. s-s Orbitals Overlapping
In order to explain s-s overlapping consider the example of H2 molecule. In this molecule is orbital of one hydrogen overlaps with is orbital of other hydrogen to form sigma bonding orbitals. Due to this bonding a single covalent bond is formed between the two hydrogen atoms.
Diagram Coming Soon
2. s-p Orbitals Overlapping
This type of overlapping takes place in H-Cl molecule. 1s orbital of hydrogen overlaps with 1p orbital of chlorine to form a single covalent bond. In this overlapping two molecular orbitals are formed, one of the lower energy while the other orbital is of higher energy. The shapes of these orbitals are as follows.
Diagram Coming Soon
3. p-p Orbitals Overlapping
This type of overlapping takes place in fluorine molecule. In this mole 1p orbital of a fluorine atom is overlapped with 1p orbital of the other fluorine atom. The molecular orbitals formed in this overlapping are given in figure
Diagram Coming Soon
PI Bond
When the two atomic orbital involved in a covalent bond are parallel to each other then the bond formed between them is called pi bond.
In this overlapping, two molecular orbitals are also formed. The lower energy molecular orbitals is called π bonding orbital while the higher energy molecular orbital is called π antibonding orbital. The shape of these molecular orbitals are as follows.
Diagram Coming Soon
Hybridization
Definition
The process in which atomic orbitals of different energy and shape are mixed together to form new set of equivalent orbitals of the same energy and same shape.
There are many different types of orbital hybridization but we will discuss here only three main types.
1. sp3 Hybridization
The mixing of one s and three p orbitals to form four equivalent sp3 hybrid orbitals is called sp3 hybridization. These sp3 orbitals are directed from the center of a regular tetrahedron to its four corners. The angles between tetrahedrally arranged orbitals are 109.5º.
It has two partially filled 2p orbitals which indicate that it is divalent, but carbon behaves as tetravalent in most of its compounds. It is only possible if one electron from 2s orbital is promoted to an empty 2pz orbital to get four equivalent sp3 hybridized orbitals.
Diagram Coming Soon The four sp3 hybrid orbitals of the carbon atom overlap with 1s orbitals of four hydrogen atoms to form a methane CH4 molecule.
The methane molecule contains four sigma bonds and each H-C-H bond angle is 109.5º.
2. sp2 Hybridization
The mixing of one s and two p orbitals to form three orbitals of equal energy is called sp2 or 3sp2 hybridization. Each sp2 orbital consists of s and p in the ratio of 1:2. These three orbitals are co-planar and at 120º angle as shown
Diagram Coming Soon A typical example of this type of hybridization is of ethane molecule. In ethylene, two sp2 hybrid orbitals of each carbon atom share and overlap with 1s orbitals of two hydrogen atoms to form two σ bonds. While the remaining sp2 orbital on each carbon atom overlaps to form a σ bond. The remaining two unhybridized p orbitals (one of each) are parallel and perpendicular to the axis joining the two carbon nuclei. These generates a parallel overlap and results in the formation of 2 π orbitals. Thus a molecule of ethylene contain five σ bonds and one π bond.
Diagram Coming Soon
3. sp Hybridization
When one s and one p orbitals combine to give two hybrid orbitals the process is called sp hybridization. The sp hybrid orbitals has two lobes, one with greater extension in shape than the other and the lobes are at an angle of 180º from each other. It means that the axis of the two orbitals form a single straight line as shown.
Now consider the formation of acetylene molecule HC ≡ CH. The two C-H σ bonds are formed due to sp-s overlap and a triple bond between two carbon atoms consist of a σ bond and two Ï€ bond. The sigma bond is due to sp-sp overlap whereas Ï€ bonds are formed as a result of parallel overlap between the unhybridized four 2p orbitals of the two carbon.
Diagram Coming Soon
Valence Shell Electron Pair Repulsion Theory
The covalent bonds are directed in space to give definite shapes to the molecules. The electrons pairs forming the bonds are distributed in space around the central atom along definite directions. The shared electron pairs as well as the lone pair of electrons are responsible for the shape of molecules.
Sidwick and Powell in 1940 pointed out that the shapes of the molecules could be explained on the basis of electron pairs present in the outermost shell of the central atom. Pairs of electrons around the central atom are arranged in space in such a way so that the distances between them are maximum and coulombie repulsion of electronic cloud are minimized.
The known geometries of many molecules based upon measurement of bond angles shows that lone pairs of electrons occupy more space than bonding pairs. The repulsion between electronic pairs in valence shell, decreases in the following order.
Lone Pair - Lone Pair > Lone Pair - Bond Pair > Bond Pair - Bond Pair
When we apply this theory we can see the variation of angle in the molecular structures.
Consider the molecular structures of NH3, OH & H2O.
Diagram Coming Soon Variation from ideal bond angles are caused by multiple covalent bonds and lone electron pairs both of which require more space than single covalent bonds and therefore cause compression of surrounding bond angles.
Thus the number of pairs of electrons in the valency shell determine the overall molecular shape.
Structure of BeCl2
The two bond pairs of electrons in BeCl2 arrange themselves as far apart as possible in order to minimize the repulsion between them.
Structure of BF3 OR BCl3
In this molecule three bond pair are present around boron to arrange themselves as far apart as possible a trigonal structure is formed.
Hydrogen Bond
When hydrogen is bonded with a highly electronegative element such as nitrogen oxygen, fluorine, the molecule will be polarized and a dipole is produced. The slightly positive hydrogen atom is attracted by the slightly negatively charged electronegative atom. An electrostatic attraction between the neighbouring molecules is set up when the positive pole of one molecule attracts the negative pole of the neighbouring molecule. This type of attractive force which involves hydrogen is known as hydrogen bonding.
Energetics Of Chemical Reaction
Thermodynamics
Definition
It is branch of chemistry which deals with the heat energy change during a chemical reaction.
Types of Thermochemical Reactions
Thermo-chemical reactions are of two types.
1. Exothermic Reactions
2. Endothermic Reactions
1. Exothermic Reaction
A chemical reaction in which heat energy is evolved with the formation of product is known as Exothermic Reaction.
An exothermic process is generally represented as
Reactants ----> Products + Heat
2. Endothermic Reaction
A chemical reaction in which heat energy is absorbed during the formation of product is known as endothermic reaction.
Endothermic reaction is generally represented as
Reactants + Heat ----> Products
Thermodynamic Terms
1. System
Any real or imaginary portion of the universe which is under consideration is called system.
2. Surroundings
All the remaining portion of the universe which is present around a system is called surroundings.
3. State
The state of a system is described by the properties such as temperature, pressure and volume when a system undergoes a change of state, it means that the final description of the system is different from the initial description of temperature, pressure or volume.
Properties of System
The properties of a system may be divided into two main types.
1. Intensive Properties
Those properties which are independent of the quantity of matter are called intensive properties.
e.g. melting point, boiling point, density, viscosity, surface, tension, refractive index etc.
2. Extensive Properties
Those properties which depends upon the quantity of matter are called extensive properties.
e.g. mass, volume, enthalpy, entropy etc.
First Law of Thermodynamics
This law was given by Helmheltz in 1847. According to this law
Energy can neither be created nor destroyed but it can be changed from one form to another.
In other words the total energy of a system and surroundings must remain constant.
Mathematical Derivation of First Law of Thermodynamics
Consider a gas is present in a cylinder which contain a frictionless piston as shown.
Diagram Coming Soon
Let a quantity of heat q is provided to the system from the surrounding. Suppose the internal energy of the system is E1 and after absorption of q amount of heat it changes to E2. Due to the increase of this internal energy the collisions offered by the molecules also increases or in other words the internal pressure of the system is increased after the addition of q amount of heat. With the increase of internal pressure the piston of the cylinder moves in the upward direction to maintain the pressure constant so a work is also done by the system.
Therefore if we apply first law of thermodynamics on this system we can write
q = E2 - E1 + W
OR
q = ΔE + W
OR
ΔE = q - W
This is the mathematical representation of first law of thermodynamics.
Pressure - Volume Work
Consider a cylinder of a gas which contain a frictionless and weightless piston, as shown above. Let the area of cross-section of the piston = a
Pressure on the piston = P
The initial volume of the gases = V1
And the final volume of the gases = V2
The distance through which piston moves = 1
So the change in volume = ΔV = V2 - V1
OR ΔV = a x 1
The word done by the system W = force x distance
W = Pressure x area x distance
W = P x a x 1
W = P Δ V
By substituting the value of work the first law of thermodynamics may be written as
q = ΔE + P ΔV
The absorption or evolution of heat during chemical reaction may take place in two ways.
1. Process at Constant Volume
Let qv be the amount of heat absorbed at constant volume.
According to first law qv = ΔE + P ΔV
But for constant volume ΔV = O
Therefore,
P ΔV = P x O = O
So,
qv = ΔE + 0
Or
qv = ΔE
Thus in the process carried at constant volume the heat absorbed or evolved is equal to the energy ΔE.
2. Process at Constant Pressure
Let qp is the amount of heat energy provided to a system at constant pressure. Due to this addition of heat the internal energy of the gas is increased from E1 to E2 and volume is changed from V1 to V2, so according to first law.
qp = E2 - E1 + P(V2 - V1)
Or
qp = E2 - E1 + PV2 - PV1
Or
qp = E2 + PV2- E1 - PV1
Or
qp = (E2 + PV2) - (E1 - PV1)
But we known that
H = E + PV
So
E1 + PV1 = H1
And
E2 + PV2 = H2
Therefore the above equation may be written as
qp = H2 - H1
Or
qp = Δ H
This relation indicates that the amount of heat absorbed at constant pressure is used in the enthalpy change.
Sign of ΔH
ΔH represent the change of enthalpy. It is a characteristic property of a system which depends upon the initial and final state of the system.
For all exothermic processes ΔH is negative and for all endothermic reactions ΔH is positive.
Thermochemistry
It is a branch of chemistry which deals with the measurement of heat evolved or absorbed during a chemical reaction.
The unit of heat energy which are generally used are Calorie and kilo Calorie or Joules and kilo Joules.
1 Cal = 4.184 J
OR
1 Joule = 0.239 Cal
Hess's Law of Constant Heat Summation
Statement
If a chemical reaction is completed in a single step or in several steps the total enthalpy change for the reaction is always constant.
OR
The amount of heat absorbed or evolved during a chemical reaction must be independent of the particular manner in which the reaction takes place.
Explanation
Suppose in a chemical reactant A changes to the product D in a single step with the enthalpy change ΔH
Diagram Coming Soon
This reaction may proceed through different intermediate stages i.e., A first changes to B with enthalpy change ΔH1 then B changes to C with enthalpy change ΔH2 and finally C changes to D with enthalpy ΔH3.
According to Hess's law
ΔH = ΔH1 + ΔH2 + ΔH3
Verification of Hess's Law
When CO2 reacts with excess of NaOH sodium carbonate is formed with the enthalpy change of 90 kJ/mole. This reaction may take place in two steps via sodium bicarbonate.
In the first step for the formation of NaHCO3 the enthalpy change is -49 kJ/mole and in the second step the enthalpy change is -41 kJ/mole.
According to Hess's Law
ΔH = ΔH1 + ΔH2
ΔH = -41 -49 = -90 kJ/mole
The total enthalpy change when the reaction is completed in a single step is -90 kJ/mole which is equal to the enthalpy change when the reaction is completed into two steps. Thus the Hess's law is verified from this example.
Chemical Equilibrium
INDEX
* 1 Chemical Equilibrium
o 1.1 Reversible Reactions
o 1.2 Irreversible Reactions
o 1.3 Equilibrium State
o 1.4 Law of Mass Action
o 1.5 Le Chatelier's Principle
o 1.6 Contact Process
o 1.7 Common Ion Effect
o 1.8 Solubility Product
o 1.9 Hydration
o 1.10 Hydrolysis
o 1.11 Theory of Ionization
Chemical Equilibrium
Reversible Reactions
Those chemical reactions which take place in both the directions and never proceed to completion are called Reversible reaction.
For these type of reaction both the forward and reverse reaction occur at the same time so these reaction are generally represented as
Reactant □ Product
The double arrow □ indicates that the reaction is reversible and that both the forward and reverse reaction can occur simultaneously.
Some examples of reversible reactions are given below
1. 2Hl □ H2 + l2
2. N2 + 2 H2 □ 2 NH3
Irreversible Reactions
Those reactions in which reactants are completely converted into product are called Irreversible reaction.
These reaction proceed only in one direction. Examples of such type of reaction are given below
1. NaCl + AgNO3 ----> AgCl + NaNO3
2. Cu + H2SO4 ----> CuSO4 + H2
Equilibrium State
The state at which the rate of forward reaction becomes equal to the rate of reverse reaction is called Equilibrium state.
Explanation
Consider the following reaction
A + B □ C + D
It is a reversible reaction. In this reaction both the changes (i.e. forward & backward) occur simultaneously. At initial stage reactant A & B are separated from each other therefore the concentration of C and D is zero.
When the reaction is started and the molecules of A and B react with each other the concentration of reactant is decreased while the concentration of product is increased. With the formation of product, the rate of forward reaction decreased with time but the rate of reverse reaction is increased with the formation of product C & D.
Ultimately a stage reaches when the number of reacting molecules in the forward reaction equalizes the number of reacting molecules in the reverse direction, so this state at which the rate of forward reaction becomes equal to the rate of reverse reaction is called equilibrium state.
Law of Mass Action
Statement
The rate at which a substance reacts is proportional to its active mass and the rate of a chemical reaction is proportional to the product of the active masses of the reactant.
The term "active mass" means the concentration in terms of moles/dm3.
Derivation of Equilibrium Constant Expression
Consider in a reversible reaction "m" mole of A and "n" moles of B reacts to give "x" moles of C and "y" moles of D as shown in equation.
mA + nB □ xC + yD
In this process
The rate of forward reaction ∞ [A]m [B]n
Or
The rate of forward reactin = Kf [A]m [B]n
&
The rate of reverse reaction ∞ [C]x [D]y
Or
The rate of reverse reaction = Kf [C]x [D]y
But at equilibrium state
Rate of forward reaction = Rate of reverse reaction
Therefore,
Kf [A]m [B]n = Kf [C]x [D]y
Or
Kf / Kr = [C]x [D]y / [A]m [B]n
Or
Ke = [C]x [D]y / [A]m [B]n
This is the expression for equilibrium constant which is denoted by Ke and defined as
The ratio of multiplication of active masses of the products to the product of active masses of reactant is called equilibrium constant.
Equilibrium Constant for a Gaseous System
Consider in a reversible process, the reactants and product are gases as shown
A(g) + B(g) □ C(g) + D(g)
When the reactants and products are in gaseous state, their partial pressures are used instead of their concentration, so according to law of mass action.
Determination of Equilibrium Constant
The value of equilibrium constant K(C) does not depend upon the initial concentration of reactants. In order to find out the value of K(C) we have to find out the equilibrium concentration of reactant and product.
1. Ethyl Acetate Equilibrium
Acetic acid reacts with ethyl alcohol to form ethyl acetate and water as shown
CH3COOH + C2H5OH □ CH3COOC2H5 + H2O
Suppose 'a' moles of acetic acid and 'b' moles of alcohol are mixed in this reaction. After some time when the state of equilibrium is established suppose 'x' moles of H2O and 'x' moles of ethyl acetate are formed while the number of moles of acetic acid and alcohol are a-x and b-x respectively at equilibrium.
According to law of mass action
K(C) = [CH3COOC2H5] [H2O] / [CH3COOH] [C2H5OH]
K(C) = [x/V] [x/V] / [a-x/V] [b-x/V]
K(C) = (x) (x) / (a-x) (b-x)
K(C) = x2 / (a-x) (b-x)
2. Hydrogen Iodide Equilibrium
For the reaction between hydrogen and iodine suppose a mole of hydrogen and 'b' moles of iodine are mixed in a scaled bulb at 444ºC in the boiling sulphur for some time. The equilibrium mixture is then cooled and the bulbs are opened in the solution of NaOH. Let the amount of hydrogen consumed at equilibrium be 'x' moles which means that the amount of hydrogen left at equilibrium is a-x moles. Since 1 mole of hydrogen reacts with 1 mole of iodine 'o' form two moles of hydrogen iodide hence the amount of iodine used is also x moles so its moles at equilibrium are b-x and the moles of hydrogen iodide at equilibrium are 2x.
According to law of mass action
K(C) = [Hl]2 / [H2] [l2]
K(C) = [2x/V]2 / [a-x/V] [b-x/V]
K(C) = 4x2 / (a-x) (b-x)
Applications of Law of Mass Action
There are two important applications of equilibrium constant.
1. It is used to predict the direction of reaction.
2. K(C) is also used to predict the extent of reaction.
To Predict the Direction of Reaction
The value of equilibrium constant K(C) is used to predict the direction of reaction. For a reversible process.
Reactant □ Product
With respect to the ratio of initial concentration of the reagent.
There are three possibilities for the value of K
1. It is greater than K(C)
2. It is less than K(C)
3. It is equal to K(C)
Case I
If [Reactant]initial / [Product]initial > K(C) the reaction will shift towards the reverse direction.
Case II
If [Reactant]initial / [Product]initial > K(C) the reaction will shift towards the forward direction.
Case III
If [Reactant]initial / [Product]initial > K(C) this is equilibrium state for the reaction.
To Predict the Extent of Reaction
From the value of K(C) we can predict the extent of the reaction.
If the value of K(C) is very large e.g.
For 2 O3 □ 3 O2 ........... K(C) = 10(55)
From this large value of K(C) it is predicted that the forward reaction is almost complete.
When the value of K(C) is very low e.g.,
2 HF □ H2 + F2 ........... K(C) = 10(-13)
From this value it is predicted that the forward reaction proceeds with negligible speed.
But if the value of K(C) is moderate, the reaction occurs in both the direction and equilibrium will be attained after certain period of time e.g., K(C) for
N2 + 3 H2 □ 2 NH3 ............. is 10
So the reaction occurs in both the direction.
Le Chatelier's Principle
Statement
When a stress is applied to a system at equilibrium the equilibrium position changes so as to minimize the effect of applied stress.
The equilibrium state of a chemical reaction is altered by changing concentration pressure or temperature. The effect of these changes is explained by Le Chatelier.
Effect of Concentration
By changing the concentration of any substance present in the equilibrium mixture, the balance of chemical equilibrium is disturbed. For the reaction,
A + B □ C + D
K(C) = [C][D] / [A][B]
If the concentration of a reactant A or B is increased the equilibrium state shifts tc right and yield of products increases.
But if the concentration of C or D is increased then the reaction proceed in the backward direction with a greater rate and more A & B are formed.
Effect of Temperature
The effect of temperature is different for different type of reaction.
For an exothermic reaction the value of K(C) decreased with the increase of temperature so the concentration of products decreases.
For a endothermic reaction heat is absorbed for the conversion of reactant into product so if temperature during the reaction is increased then the reaction will proceed with a greater rate in forward direction.
ENDOTHERMIC REACTION
Temperature increase ----> More products are formed
Temperature decrease ----> More reactants are formed
EXOTHERMIC REACTION
Temperature increase ----> More reactants are formed
Temperature decrease ----> More products are formed
Effect of Pressure
The state of equilibrium of gaseous reaction is distributed by the change of pressure. There are three types of reactions which show the effect of pressure change.
1. When the Number of Moles of Product are Greater
In a reaction such as
PCl5 <----> PCl3 + Cl2
The increase of pressure shifts the equilibrium towards reactant side.
2. When the Number of Moles of Reactant are Greater
In a reaction such as
N2 + 3H2 <----> 2NH3
The increase of pressure shifts the equilibrium towards product side because the no. of moles of product are less than the no. of moles of reactant.
3. When Number of Moles of Reactants and Products are Equal
In these reactions where the number of moles of reactant are equal to the number of moles of product the change of pressure does not change the equilibrium state e.g.,
H2 + l2 □ 2 Hl
Since the number of moles of reactants and products are equal in this reaction so the increase of pressure does not affect the yield of Hl.
Important Industrial Application of Le Chatelier's Principle
Haber's Process
This process is used for the production of NH3 by the reaction of nitrogen and hydrogen. In this process 1 volume of nitrogen is mixed with three volumes of hydrogen at 500ºC and 200 to 1000 atm pressure in presence of a catalyst
N2 + 3 H2 □ 2 NH3 ............... ΔH = -46.2 kJ/mole
1. Effect of Concentration
The value of K(C) for this reaction is
K(C) = [NH3]2 / [N2] [H2]3
Increase in concentration of reactants which are nitrogen and hydrogen the equilibrium of the process shifts towards the right so as to keep the value of K(C) constant. Hence the formation of NH3 increases with the increase of the concentration of N2 or hydrogen.
2. Effect of Temperature
It is an exothermic process, so heat is liberated with the formation of product. Therefore, according to Le Chatelier's principle at low temperature the equilibrium shifts towards right to balance the equilibrium state so low temperature favours the formation of NH3
3. Effect of Pressure
The formation of NH3 proceeds with the decrease in volume, therefore, the reaction is carried out under high pressure or in other words high pressure is favourable for the production of NH3.
Contact Process
The process is used to manufacture H2SO4 on large scale. In this process the most important step is the oxidation of SO2 to SO3 in presence of a catalyst vanadium pentoxide.
2 SO2 + O2 □ 2 SO3 ................... ΔH = - 395 kJ/mole
1. Effect of Concentration
The value of K(C) for this reaction is
K(C) = [SO3]2 / [SO2]2 [O2]
Increase in concentration of SO2 or O2 shifts the equilibrium towards the right and more SO3 is formed.
2. Effect of Temperature
Since the process is exothermic, so low temperature will favour the formation of SO3. The optimum temperature for this reaction is 400 to 450ºC.
3. Effect of Pressure
In this reaction decrease in volume takes place so high pressure is favourable for the formation of SO3.
Common Ion Effect
Statement
The process in which precipitation of an electrolyte is caused by lowering the degree of ionization of a weak electrolyte when a common ion is added is known as common ion effect.
Explanation
In the solution of an electrolyte in water, there exist an equilibrium between the ions and the undissociated molecules to which the law of mass action can be applied.
Considering the dissociation of an electrolyte AB we have
AB □ A+ + B-
And
[A+][B-] / [AB] = K (dissociation constant)
If now another electrolyte yielding A+ or B- ions be added to the above solution, it will result in the increase of concentration of the ions A+ or B- and in order that K may remain the same, the concentration AB must evidently increase. In other words the degree of dissociation of an electrolyte is suppressed by the addition of another electrolyte containing a common ion. This phenomenon is known as common ion effect.
Application of Common Ion Effect in Salt Analysis
An electrolyte is precipitated from its solution only when the concentration of its ions exceed from the solubility product. The precipitates are obtained when the concentration of any one ion is increased. Thus by adding the common ion, the solubility product can be exceeded.
In this solution Ou(OH)2 is a weak base while H2SO3 is a strong acid so the pH of the solution is changed towards acidic medium.
When Na2CO3 is dissolved in water, it reacts with water such as
Na2CO3 + 2 H2O □ 2 NaOH + H2CO3
In this solution H2CO3 which is weak acid an NaOH which is a strong base are formed. Due to presence of strong base the medium is changed towards basic nature.
Solubility Product
When a slightly soluble ionic solid such as silver chloride is dissolved in water, it decompose into its ions
AgCl □ Ag+ + Cl-
These Ag+ and Cl- ions from solid phase pass into solution till the solution becomes saturated. Now there exists an equilibrium between the ions present in the saturated solution and the ions present in the solid phase, thus
AgCl □ Ag+ + Cl-
Applying the law of mass action
K(C) = [Ag+][Cl-] / [AgCl]
Since the concentration of solid AgCl in the solid phase is fixed, no matter how much solid is present in contact with solution, so we can write.
K(C) = [Ag+][Cl-] / K
Or
K(C) x K = [Ag+][Cl-]
Or
K(S.P) = [Ag+][Cl-]
Where K(S.P) is known as solubility product and defined as
The product of the concentration of ions in the saturated solution of a sparingly soluble salt is called solubility product.
the value of solubility product is constant for a given temperature.
Calculation of Solubility Product From Solubility
The mass of a solute present in a saturated solution with a fixed volume of solvent is called solubility, which is generally represented in the unit of gm/dm3. With the help of solubility we can calculate the solubility product of a substance e.g., the solubility of Mg(OH)2 at 25ºC is 0.00764 gm/dm3. To calculate the K(S.P) of Mg(OH)2, first of all we will calculate the concentration of Mg(OH)2 present in the solution.
Mass of Mg(OH)2 = 0.00764 gm/dm3
Moles of Mg(OH)2 = 0.00764 / 58 moles / dm3
= 1.31 x 10(-4) moles/dm3
The ionization of Mg(OH)2 in the solution is as follows.
Mg(OH)2 □ Mg(+2) + 2 OH-
And the solubility product for Mg(OH)2 may be written as,
K(S.P) = [Mg(+2)] [OH-]2
Since in one mole of Mg(OH2) solution one mole of Mg++ ions are present while two moles of OH- ions are present, therefore in 1.31 x 10(-4) mole/dm3 solution of Mg(OH)2, the concentration of Mg(+2) is 1.31 x 10(-4) moles/dm3 while the concentration of OH- is 2. 62 x 10(-8) moles/dm3. By substituting these values
K(S.P) = [Mg(+2)][OH-]2
= [1.31 x 10(-4)] [2.62 x 10(-4)]2
= 9.0 x 10(-12) mole3 / dm9
So in this way the solubility product of a substance may be calculated with the help of solubility.
Calculation of Solubility from Solubility Product
If we know the value of solubility product, we can calculate the solubility of the salt.
For example, the solubility of PbCrO4at 25ºC is 2.8 x 10(-13) moles/dm3.
m = n2 / w1 in kg
m = (w2 / m2) / (w1 / 1000)
m = w2 / m2) x (1000 / w1)
Hydration
Addition of water or association of water molecules with a substance without dissociation is called Hydration.
Water is a good solvent and its polar nature plays very important part in dissolving substances. It dissolves ionic compounds readily.
When an ionic compound is dissolved in water, the partial negatively charged oxygen of water molecule is attracted towards the cation ion similarly the partial positively charged hydrogen of water molecule is attracted towards the anions so hydrated ions are formed.
Diagram Coming Soon
In solution, the number of water molecules which surround the ions is indefinite, but when an aqueous solution of a salt is evaporated the salt crystallizes with a definite number of water molecules which is called as water of crystallization E.g., when CuSO4 recrystallized from its solution the crystallized salt has the composition CuSO4. 5H2O. Similarly when magnesium chloride is recrystallized from the solution, it has the composition MgCl2.6H2O. This composition indicates that each magnesium ion in the crystal is surrounded by six molecules. This type of salts is called hydrated salts.
It is observed experimentally that the oxygen atom of water molecule is attached with the cation of salt through co-ordinate covalent bond so it is more better to write the molecular formulas of the hydrated salts as given below.
[Cu(H2O)5]SO4 ................. [Mg(H2O)6]Cl2
It is also observed that these compound exist with a definite geometrical structure e.g., the structure of [Mg(H2O)6]Cl2 is octahedral and [Cu(H2O)4]+2 is a square planar.
Diagram Coming Soon
Factors for Hydration
The ability of hydration of an ion depend upon its charge density.
For example the charge density of Na+ is greater than K+ because of its smaller size, so the ability of hydration for Na+ is greater than K+ ion. Similarly small positive ions with multiple charges such as Cu(+2), Al(+3), Cr(+3) posses great attraction for water molecules.
Hydrolysis
Addition of water with a substance with dissociation into ions is called Hydrolysis.
OR
The reaction of cation or anion with water so as to change its pH is known as Hydrolysis.
Theoritically it is expected that the solution of salts like CuSO4 or Na2CO3 are neutral because these solutions contain neither H+ ion nor OH-, but it is experimentally observed that the solution of CuSO4 is acidic while the solution of Na2CO3 is basic. This acidic or basic nature of solution indicate but H+ ions or OH- ions are present in their solutions which can be produced only by the dissociation of water molecules.
Theory of Ionization
1n 1880, a Swedish chemist Svante August Arrhenius put forward a theory known as theory of ionization, in order to account for the conductivity of electrolytes, electrolysis and certain properties of electrolytic solutions. According to this theory.
1. Acids, Bases and Salts when dissolved in water yield two kinds of ions, one carry positive charge and the other carry negative charge. The positively charged ions are called cations which are derived from metals or it may be H+ ion but the negatively charged ions which are known as anions are derived from non-metals
NaCl ----> Na+ + Cl-
H2SO4 ----> 2 H+ + SO4(-2)
KOH ----> K+ + OH-
2. Ions in the solution also recombine with each other to form neutral molecules and this process continues till an equilibrium state between an ionized and unionized solid is attained.
INDEX
* 1 Chemical Equilibrium
o 1.1 Reversible Reactions
o 1.2 Irreversible Reactions
o 1.3 Equilibrium State
o 1.4 Law of Mass Action
o 1.5 Le Chatelier's Principle
o 1.6 Contact Process
o 1.7 Common Ion Effect
o 1.8 Solubility Product
o 1.9 Hydration
o 1.10 Hydrolysis
o 1.11 Theory of Ionization
Chemical Equilibrium
Reversible Reactions
Those chemical reactions which take place in both the directions and never proceed to completion are called Reversible reaction.
For these type of reaction both the forward and reverse reaction occur at the same time so these reaction are generally represented as
Reactant □ Product
The double arrow □ indicates that the reaction is reversible and that both the forward and reverse reaction can occur simultaneously.
Some examples of reversible reactions are given below
1. 2Hl □ H2 + l2
2. N2 + 2 H2 □ 2 NH3
Irreversible Reactions
Those reactions in which reactants are completely converted into product are called Irreversible reaction.
These reaction proceed only in one direction. Examples of such type of reaction are given below
1. NaCl + AgNO3 ----> AgCl + NaNO3
2. Cu + H2SO4 ----> CuSO4 + H2
Equilibrium State
The state at which the rate of forward reaction becomes equal to the rate of reverse reaction is called Equilibrium state.
Explanation
Consider the following reaction
A + B □ C + D
It is a reversible reaction. In this reaction both the changes (i.e. forward & backward) occur simultaneously. At initial stage reactant A & B are separated from each other therefore the concentration of C and D is zero.
When the reaction is started and the molecules of A and B react with each other the concentration of reactant is decreased while the concentration of product is increased. With the formation of product, the rate of forward reaction decreased with time but the rate of reverse reaction is increased with the formation of product C & D.
Ultimately a stage reaches when the number of reacting molecules in the forward reaction equalizes the number of reacting molecules in the reverse direction, so this state at which the rate of forward reaction becomes equal to the rate of reverse reaction is called equilibrium state.
Law of Mass Action
Statement
The rate at which a substance reacts is proportional to its active mass and the rate of a chemical reaction is proportional to the product of the active masses of the reactant.
The term "active mass" means the concentration in terms of moles/dm3.
Derivation of Equilibrium Constant Expression
Consider in a reversible reaction "m" mole of A and "n" moles of B reacts to give "x" moles of C and "y" moles of D as shown in equation.
mA + nB □ xC + yD
In this process
The rate of forward reaction ∞ [A]m [B]n
Or
The rate of forward reactin = Kf [A]m [B]n
&
The rate of reverse reaction ∞ [C]x [D]y
Or
The rate of reverse reaction = Kf [C]x [D]y
But at equilibrium state
Rate of forward reaction = Rate of reverse reaction
Therefore,
Kf [A]m [B]n = Kf [C]x [D]y
Or
Kf / Kr = [C]x [D]y / [A]m [B]n
Or
Ke = [C]x [D]y / [A]m [B]n
This is the expression for equilibrium constant which is denoted by Ke and defined as
The ratio of multiplication of active masses of the products to the product of active masses of reactant is called equilibrium constant.
Equilibrium Constant for a Gaseous System
Consider in a reversible process, the reactants and product are gases as shown
A(g) + B(g) □ C(g) + D(g)
When the reactants and products are in gaseous state, their partial pressures are used instead of their concentration, so according to law of mass action.
Determination of Equilibrium Constant
The value of equilibrium constant K(C) does not depend upon the initial concentration of reactants. In order to find out the value of K(C) we have to find out the equilibrium concentration of reactant and product.
1. Ethyl Acetate Equilibrium
Acetic acid reacts with ethyl alcohol to form ethyl acetate and water as shown
CH3COOH + C2H5OH □ CH3COOC2H5 + H2O
Suppose 'a' moles of acetic acid and 'b' moles of alcohol are mixed in this reaction. After some time when the state of equilibrium is established suppose 'x' moles of H2O and 'x' moles of ethyl acetate are formed while the number of moles of acetic acid and alcohol are a-x and b-x respectively at equilibrium.
According to law of mass action
K(C) = [CH3COOC2H5] [H2O] / [CH3COOH] [C2H5OH]
K(C) = [x/V] [x/V] / [a-x/V] [b-x/V]
K(C) = (x) (x) / (a-x) (b-x)
K(C) = x2 / (a-x) (b-x)
2. Hydrogen Iodide Equilibrium
For the reaction between hydrogen and iodine suppose a mole of hydrogen and 'b' moles of iodine are mixed in a scaled bulb at 444ºC in the boiling sulphur for some time. The equilibrium mixture is then cooled and the bulbs are opened in the solution of NaOH. Let the amount of hydrogen consumed at equilibrium be 'x' moles which means that the amount of hydrogen left at equilibrium is a-x moles. Since 1 mole of hydrogen reacts with 1 mole of iodine 'o' form two moles of hydrogen iodide hence the amount of iodine used is also x moles so its moles at equilibrium are b-x and the moles of hydrogen iodide at equilibrium are 2x.
According to law of mass action
K(C) = [Hl]2 / [H2] [l2]
K(C) = [2x/V]2 / [a-x/V] [b-x/V]
K(C) = 4x2 / (a-x) (b-x)
Applications of Law of Mass Action
There are two important applications of equilibrium constant.
1. It is used to predict the direction of reaction.
2. K(C) is also used to predict the extent of reaction.
To Predict the Direction of Reaction
The value of equilibrium constant K(C) is used to predict the direction of reaction. For a reversible process.
Reactant □ Product
With respect to the ratio of initial concentration of the reagent.
There are three possibilities for the value of K
1. It is greater than K(C)
2. It is less than K(C)
3. It is equal to K(C)
Case I
If [Reactant]initial / [Product]initial > K(C) the reaction will shift towards the reverse direction.
Case II
If [Reactant]initial / [Product]initial > K(C) the reaction will shift towards the forward direction.
Case III
If [Reactant]initial / [Product]initial > K(C) this is equilibrium state for the reaction.
To Predict the Extent of Reaction
From the value of K(C) we can predict the extent of the reaction.
If the value of K(C) is very large e.g.
For 2 O3 □ 3 O2 ........... K(C) = 10(55)
From this large value of K(C) it is predicted that the forward reaction is almost complete.
When the value of K(C) is very low e.g.,
2 HF □ H2 + F2 ........... K(C) = 10(-13)
From this value it is predicted that the forward reaction proceeds with negligible speed.
But if the value of K(C) is moderate, the reaction occurs in both the direction and equilibrium will be attained after certain period of time e.g., K(C) for
N2 + 3 H2 □ 2 NH3 ............. is 10
So the reaction occurs in both the direction.
Le Chatelier's Principle
Statement
When a stress is applied to a system at equilibrium the equilibrium position changes so as to minimize the effect of applied stress.
The equilibrium state of a chemical reaction is altered by changing concentration pressure or temperature. The effect of these changes is explained by Le Chatelier.
Effect of Concentration
By changing the concentration of any substance present in the equilibrium mixture, the balance of chemical equilibrium is disturbed. For the reaction,
A + B □ C + D
K(C) = [C][D] / [A][B]
If the concentration of a reactant A or B is increased the equilibrium state shifts tc right and yield of products increases.
But if the concentration of C or D is increased then the reaction proceed in the backward direction with a greater rate and more A & B are formed.
Effect of Temperature
The effect of temperature is different for different type of reaction.
For an exothermic reaction the value of K(C) decreased with the increase of temperature so the concentration of products decreases.
For a endothermic reaction heat is absorbed for the conversion of reactant into product so if temperature during the reaction is increased then the reaction will proceed with a greater rate in forward direction.
ENDOTHERMIC REACTION
Temperature increase ----> More products are formed
Temperature decrease ----> More reactants are formed
EXOTHERMIC REACTION
Temperature increase ----> More reactants are formed
Temperature decrease ----> More products are formed
Effect of Pressure
The state of equilibrium of gaseous reaction is distributed by the change of pressure. There are three types of reactions which show the effect of pressure change.
1. When the Number of Moles of Product are Greater
In a reaction such as
PCl5 <----> PCl3 + Cl2
The increase of pressure shifts the equilibrium towards reactant side.
2. When the Number of Moles of Reactant are Greater
In a reaction such as
N2 + 3H2 <----> 2NH3
The increase of pressure shifts the equilibrium towards product side because the no. of moles of product are less than the no. of moles of reactant.
3. When Number of Moles of Reactants and Products are Equal
In these reactions where the number of moles of reactant are equal to the number of moles of product the change of pressure does not change the equilibrium state e.g.,
H2 + l2 □ 2 Hl
Since the number of moles of reactants and products are equal in this reaction so the increase of pressure does not affect the yield of Hl.
Important Industrial Application of Le Chatelier's Principle
Haber's Process
This process is used for the production of NH3 by the reaction of nitrogen and hydrogen. In this process 1 volume of nitrogen is mixed with three volumes of hydrogen at 500ºC and 200 to 1000 atm pressure in presence of a catalyst
N2 + 3 H2 □ 2 NH3 ............... ΔH = -46.2 kJ/mole
1. Effect of Concentration
The value of K(C) for this reaction is
K(C) = [NH3]2 / [N2] [H2]3
Increase in concentration of reactants which are nitrogen and hydrogen the equilibrium of the process shifts towards the right so as to keep the value of K(C) constant. Hence the formation of NH3 increases with the increase of the concentration of N2 or hydrogen.
2. Effect of Temperature
It is an exothermic process, so heat is liberated with the formation of product. Therefore, according to Le Chatelier's principle at low temperature the equilibrium shifts towards right to balance the equilibrium state so low temperature favours the formation of NH3
3. Effect of Pressure
The formation of NH3 proceeds with the decrease in volume, therefore, the reaction is carried out under high pressure or in other words high pressure is favourable for the production of NH3.
Contact Process
The process is used to manufacture H2SO4 on large scale. In this process the most important step is the oxidation of SO2 to SO3 in presence of a catalyst vanadium pentoxide.
2 SO2 + O2 □ 2 SO3 ................... ΔH = - 395 kJ/mole
1. Effect of Concentration
The value of K(C) for this reaction is
K(C) = [SO3]2 / [SO2]2 [O2]
Increase in concentration of SO2 or O2 shifts the equilibrium towards the right and more SO3 is formed.
2. Effect of Temperature
Since the process is exothermic, so low temperature will favour the formation of SO3. The optimum temperature for this reaction is 400 to 450ºC.
3. Effect of Pressure
In this reaction decrease in volume takes place so high pressure is favourable for the formation of SO3.
Common Ion Effect
Statement
The process in which precipitation of an electrolyte is caused by lowering the degree of ionization of a weak electrolyte when a common ion is added is known as common ion effect.
Explanation
In the solution of an electrolyte in water, there exist an equilibrium between the ions and the undissociated molecules to which the law of mass action can be applied.
Considering the dissociation of an electrolyte AB we have
AB □ A+ + B-
And
[A+][B-] / [AB] = K (dissociation constant)
If now another electrolyte yielding A+ or B- ions be added to the above solution, it will result in the increase of concentration of the ions A+ or B- and in order that K may remain the same, the concentration AB must evidently increase. In other words the degree of dissociation of an electrolyte is suppressed by the addition of another electrolyte containing a common ion. This phenomenon is known as common ion effect.
Application of Common Ion Effect in Salt Analysis
An electrolyte is precipitated from its solution only when the concentration of its ions exceed from the solubility product. The precipitates are obtained when the concentration of any one ion is increased. Thus by adding the common ion, the solubility product can be exceeded.
In this solution Ou(OH)2 is a weak base while H2SO3 is a strong acid so the pH of the solution is changed towards acidic medium.
When Na2CO3 is dissolved in water, it reacts with water such as
Na2CO3 + 2 H2O □ 2 NaOH + H2CO3
In this solution H2CO3 which is weak acid an NaOH which is a strong base are formed. Due to presence of strong base the medium is changed towards basic nature.
Solubility Product
When a slightly soluble ionic solid such as silver chloride is dissolved in water, it decompose into its ions
AgCl □ Ag+ + Cl-
These Ag+ and Cl- ions from solid phase pass into solution till the solution becomes saturated. Now there exists an equilibrium between the ions present in the saturated solution and the ions present in the solid phase, thus
AgCl □ Ag+ + Cl-
Applying the law of mass action
K(C) = [Ag+][Cl-] / [AgCl]
Since the concentration of solid AgCl in the solid phase is fixed, no matter how much solid is present in contact with solution, so we can write.
K(C) = [Ag+][Cl-] / K
Or
K(C) x K = [Ag+][Cl-]
Or
K(S.P) = [Ag+][Cl-]
Where K(S.P) is known as solubility product and defined as
The product of the concentration of ions in the saturated solution of a sparingly soluble salt is called solubility product.
the value of solubility product is constant for a given temperature.
Calculation of Solubility Product From Solubility
The mass of a solute present in a saturated solution with a fixed volume of solvent is called solubility, which is generally represented in the unit of gm/dm3. With the help of solubility we can calculate the solubility product of a substance e.g., the solubility of Mg(OH)2 at 25ºC is 0.00764 gm/dm3. To calculate the K(S.P) of Mg(OH)2, first of all we will calculate the concentration of Mg(OH)2 present in the solution.
Mass of Mg(OH)2 = 0.00764 gm/dm3
Moles of Mg(OH)2 = 0.00764 / 58 moles / dm3
= 1.31 x 10(-4) moles/dm3
The ionization of Mg(OH)2 in the solution is as follows.
Mg(OH)2 □ Mg(+2) + 2 OH-
And the solubility product for Mg(OH)2 may be written as,
K(S.P) = [Mg(+2)] [OH-]2
Since in one mole of Mg(OH2) solution one mole of Mg++ ions are present while two moles of OH- ions are present, therefore in 1.31 x 10(-4) mole/dm3 solution of Mg(OH)2, the concentration of Mg(+2) is 1.31 x 10(-4) moles/dm3 while the concentration of OH- is 2. 62 x 10(-8) moles/dm3. By substituting these values
K(S.P) = [Mg(+2)][OH-]2
= [1.31 x 10(-4)] [2.62 x 10(-4)]2
= 9.0 x 10(-12) mole3 / dm9
So in this way the solubility product of a substance may be calculated with the help of solubility.
Calculation of Solubility from Solubility Product
If we know the value of solubility product, we can calculate the solubility of the salt.
For example, the solubility of PbCrO4at 25ºC is 2.8 x 10(-13) moles/dm3.
m = n2 / w1 in kg
m = (w2 / m2) / (w1 / 1000)
m = w2 / m2) x (1000 / w1)
Hydration
Addition of water or association of water molecules with a substance without dissociation is called Hydration.
Water is a good solvent and its polar nature plays very important part in dissolving substances. It dissolves ionic compounds readily.
When an ionic compound is dissolved in water, the partial negatively charged oxygen of water molecule is attracted towards the cation ion similarly the partial positively charged hydrogen of water molecule is attracted towards the anions so hydrated ions are formed.
Diagram Coming Soon
In solution, the number of water molecules which surround the ions is indefinite, but when an aqueous solution of a salt is evaporated the salt crystallizes with a definite number of water molecules which is called as water of crystallization E.g., when CuSO4 recrystallized from its solution the crystallized salt has the composition CuSO4. 5H2O. Similarly when magnesium chloride is recrystallized from the solution, it has the composition MgCl2.6H2O. This composition indicates that each magnesium ion in the crystal is surrounded by six molecules. This type of salts is called hydrated salts.
It is observed experimentally that the oxygen atom of water molecule is attached with the cation of salt through co-ordinate covalent bond so it is more better to write the molecular formulas of the hydrated salts as given below.
[Cu(H2O)5]SO4 ................. [Mg(H2O)6]Cl2
It is also observed that these compound exist with a definite geometrical structure e.g., the structure of [Mg(H2O)6]Cl2 is octahedral and [Cu(H2O)4]+2 is a square planar.
Diagram Coming Soon
Factors for Hydration
The ability of hydration of an ion depend upon its charge density.
For example the charge density of Na+ is greater than K+ because of its smaller size, so the ability of hydration for Na+ is greater than K+ ion. Similarly small positive ions with multiple charges such as Cu(+2), Al(+3), Cr(+3) posses great attraction for water molecules.
Hydrolysis
Addition of water with a substance with dissociation into ions is called Hydrolysis.
OR
The reaction of cation or anion with water so as to change its pH is known as Hydrolysis.
Theoritically it is expected that the solution of salts like CuSO4 or Na2CO3 are neutral because these solutions contain neither H+ ion nor OH-, but it is experimentally observed that the solution of CuSO4 is acidic while the solution of Na2CO3 is basic. This acidic or basic nature of solution indicate but H+ ions or OH- ions are present in their solutions which can be produced only by the dissociation of water molecules.
Theory of Ionization
1n 1880, a Swedish chemist Svante August Arrhenius put forward a theory known as theory of ionization, in order to account for the conductivity of electrolytes, electrolysis and certain properties of electrolytic solutions. According to this theory.
1. Acids, Bases and Salts when dissolved in water yield two kinds of ions, one carry positive charge and the other carry negative charge. The positively charged ions are called cations which are derived from metals or it may be H+ ion but the negatively charged ions which are known as anions are derived from non-metals
NaCl ----> Na+ + Cl-
H2SO4 ----> 2 H+ + SO4(-2)
KOH ----> K+ + OH-
2. Ions in the solution also recombine with each other to form neutral molecules and this process continues till an equilibrium state between an ionized and unionized solid is attained.
Chemical Kinetics
Introduction
The branch of physical chemistry which deals with the speed or rate at which a reaction occurs is called chemical kinetics.
The study of chemical kinetics, therefore includes the rate of a chemical reaction and also the rate of chemical reaction and also the factors which influence its rate.
Slow and Fast Reaction
Those reactions for which short time is required to convert a reactant into product are called fast reaction but if more time is required for the formation of a product then the reactions are called slow reactions.
Usually ionic reactions which involve oppositely charged ions in aqueous medium are very fast. For example, reaction between aqueous solution of NaCl and AgNO3 gives white precipitates of AgCl instantaneously.
AgNO3 + NaCl ----> AgCl + NaNO3
Such reactions are very fast and these are completed in fractions of seconds.
But those reactions which involve covalent molecules take place very slowly. For example, conversion of SO2 into SO3
2 SO2 + O2 ----> 2 SO3
It is a slow reaction and required more time for the formation of a product.
Rate Or Velocity of a Reaction
Definition
It is the change in concentration of a reactant or product per unit time.
Mathematically it is represented as
Rate of reaction = Change in concentration of reactant or product / Time taken for the change
The determination of the rate of a reaction is not so simple because the rate of a given reaction is never uniform. It falls off gradually with time as the reactants are used up. Hence we can not get the velocity or rate of reaction simply by dividing the amount of substance transformed by the time taken for such transformation. For this reason we take a very small interval of time "dt" during which it is assumed that velocity of reaction remains constant. If "dx" is the amount of substance transformed during that small interval of time "dt" then the velocity of reaction is expressed as
Velocity of a reaction = dx / dt
Thus with the velocity of a chemical reaction we mean the velocity at the given moment or given instant.
The Rate Constant
Definition
The proportionality constant present in the rate equation is called rate constant.
According to law of mass action we know that the rate of chemical reaction is directly proportional to the molar concentration of the reactants. For example
R ----> P
The rate of reaction ∞ [R]
Or
dx / dt = K [R]
Where K is known as rate constant.
Specific Rate Constant
When the concentration and temperature both are specified, the rate constant is known as specific rate constant.
When the concentration of each reactant is 1 mole per dm3 at given temperature, the specific rate constant numerically equals to the velocity of the reaction.
dx / dt = V = K [R]
Or
K = V / [R]
When R = 1 mole/dm3
K = V
But when different reactant are reacting with different number of moles then the value of K may be calculated as
2 SO2 + O2 ----> 2 SO3
= dx / dt = K [SO2]2 [O2]
Or
K = V / [SO2]2 [O2]
Determination of Rate of Reaction
There are two method for the determination of rate of a chemical reaction.
1. Physical Method
When the rate of a chemical reaction is determined by using physical properties such as colour change, volume change, state change the method known as physical method.
2. Chemical Method
In the method the change in concentration of reactant or product is noted and with the help of this change rate of reaction is determined e.g.,
For the reaction R ----> P
Velocity of reaction = - d[R] / dt = + d[P] / dt
The negative sign indicates a decrease in concentration of the reactant while positive sign indicates an increase in the concentration of product.
Ionization is thus a reversible process. To this process, the law of mass action can be applied as
K(C) = [Na+] [Cl-] / [NaCl]
3. The number of positive and negative charges on the ions must be equal so that the solution as a whole remains neutral.
4. The degree of ionization of an electrolyte depends upon (a) the nature of electrolyte, (b) dilution of the solution (c) the temperature
5. When an electric current passes through the solution of an electrolyte the positive ions i.e., the cations move towards the cathode and the anions move towards the anode. This movement of ions is responsible for the conductance of electric current through the solution.
6. The electrical conductivity of the solution of an electrolyte depends upon the number of ions present in the solution. On reaching the electrodes, the ions lose their charge and change into neutral atoms or molecules by the gain or loss of electrons.
Applications of Arrhenius Theory
This theory explain many peculiarities in the behaviour of electrolytic solutions.
For example, the elevation in boiling point of 1 molal solution of glucose is 0.52ºC while this elevation in 1 molal solution of NaCl is 1.04ºC. This difference in elevation of boiling point can be explained on the basis of Arrhenius theory.
In one molal solution of glucose the number of (molecules) particles are 6.02 x 10(23) per dm3 of solution while in 1 molal solution of NaCl 6.02 x 10(23) ions of Na+ and 6.02 x 10(23) ions of Cl- are present because NaCl is an ionic compound. Since the number of particle are double in NaCl solution, therefore the elevation in boiling point is also double than the solution of glucose.
Similarly the other collegative properties such as lowering in vapour pressure, depression in freezing point and osmosis are explained on the basis of this theory.
Note
Collegative properties are those properties which depends upon the number of particles.
Conductance of Electric Current Through Solutions
The ability of a solution to conduct electric current depends upon the ions present in the solution. The conductance of a solution is increased when
1. The solution is diluted
2. The degree of dissociation of the electrolyte is high
3. The temperature of the solution is high
4. The velocity of the ions is high
But in a concentrated solution, the number of ions per unit volume of solution increases and the distance between ions decreases causing strong interionic attraction. As a result, migration of ions becomes more difficult and the conductance decreases with increase in concentration. As the conductance is related with the movement of ions, so conductance increase with the increase of absolute velocity of ions in the solution.
The conductance of an electrolyte also depends upon the degree of ionization. The degree of ionization is denoted by α and calculated as
α = No. of dissociated molecules / Total molecules dissovled
Electrolysis
Electrolyte
A chemical substance which can conduct electric current in molten form or in its aqueous solution with a chemical change is called electrolyte.
Electrolysis
The movement of anions and cations towards their respective electrodes with all accompanying chemical changes in an electrolytic solution under the influence of electric current is known as electrolysis.
Explanation
To explain the phenomenon of electrolysis consider the example of CuCl2 solution. the ionization of CuCl2 in the solution may be represented as
CuCl2 <----> Cu+2 + 2 Cl-
When electric current is passed through this solution, the movement of these ions begins to take place Cu+2 ions migrate towards cathode and Cl- ions towards anode. At cathode Cu+2 ions are discharged as copper atoms by the gain of electrons (reduction)
Cu+2 + 2 e- ----> Cu(M) ........ Reduction at Cathode
At anode Cl- ions are discharged as Cl2 by the loss of electrons (oxidation)
2 Cl- - 2 e- ----> Cl2(g0 ...... Oxidation at Anode
The overall reaction of the electrolysis may be written as
Cu+2 + 2 e- ----> Cu(M)
2 Cl- - 2 e- ----> Cl2(g)
Cu+2 + 2 Cl- ----> Cu(M) + Cl2(g)
OR
CuCl2 ----> Cu(M) + Cl2(g)
When all the ions present in the solution have been changed to neutral particles, the flow of current is stopped.
Introduction
The branch of physical chemistry which deals with the speed or rate at which a reaction occurs is called chemical kinetics.
The study of chemical kinetics, therefore includes the rate of a chemical reaction and also the rate of chemical reaction and also the factors which influence its rate.
Slow and Fast Reaction
Those reactions for which short time is required to convert a reactant into product are called fast reaction but if more time is required for the formation of a product then the reactions are called slow reactions.
Usually ionic reactions which involve oppositely charged ions in aqueous medium are very fast. For example, reaction between aqueous solution of NaCl and AgNO3 gives white precipitates of AgCl instantaneously.
AgNO3 + NaCl ----> AgCl + NaNO3
Such reactions are very fast and these are completed in fractions of seconds.
But those reactions which involve covalent molecules take place very slowly. For example, conversion of SO2 into SO3
2 SO2 + O2 ----> 2 SO3
It is a slow reaction and required more time for the formation of a product.
Rate Or Velocity of a Reaction
Definition
It is the change in concentration of a reactant or product per unit time.
Mathematically it is represented as
Rate of reaction = Change in concentration of reactant or product / Time taken for the change
The determination of the rate of a reaction is not so simple because the rate of a given reaction is never uniform. It falls off gradually with time as the reactants are used up. Hence we can not get the velocity or rate of reaction simply by dividing the amount of substance transformed by the time taken for such transformation. For this reason we take a very small interval of time "dt" during which it is assumed that velocity of reaction remains constant. If "dx" is the amount of substance transformed during that small interval of time "dt" then the velocity of reaction is expressed as
Velocity of a reaction = dx / dt
Thus with the velocity of a chemical reaction we mean the velocity at the given moment or given instant.
The Rate Constant
Definition
The proportionality constant present in the rate equation is called rate constant.
According to law of mass action we know that the rate of chemical reaction is directly proportional to the molar concentration of the reactants. For example
R ----> P
The rate of reaction ∞ [R]
Or
dx / dt = K [R]
Where K is known as rate constant.
Specific Rate Constant
When the concentration and temperature both are specified, the rate constant is known as specific rate constant.
When the concentration of each reactant is 1 mole per dm3 at given temperature, the specific rate constant numerically equals to the velocity of the reaction.
dx / dt = V = K [R]
Or
K = V / [R]
When R = 1 mole/dm3
K = V
But when different reactant are reacting with different number of moles then the value of K may be calculated as
2 SO2 + O2 ----> 2 SO3
= dx / dt = K [SO2]2 [O2]
Or
K = V / [SO2]2 [O2]
Determination of Rate of Reaction
There are two method for the determination of rate of a chemical reaction.
1. Physical Method
When the rate of a chemical reaction is determined by using physical properties such as colour change, volume change, state change the method known as physical method.
2. Chemical Method
In the method the change in concentration of reactant or product is noted and with the help of this change rate of reaction is determined e.g.,
For the reaction R ----> P
Velocity of reaction = - d[R] / dt = + d[P] / dt
The negative sign indicates a decrease in concentration of the reactant while positive sign indicates an increase in the concentration of product.
Ionization is thus a reversible process. To this process, the law of mass action can be applied as
K(C) = [Na+] [Cl-] / [NaCl]
3. The number of positive and negative charges on the ions must be equal so that the solution as a whole remains neutral.
4. The degree of ionization of an electrolyte depends upon (a) the nature of electrolyte, (b) dilution of the solution (c) the temperature
5. When an electric current passes through the solution of an electrolyte the positive ions i.e., the cations move towards the cathode and the anions move towards the anode. This movement of ions is responsible for the conductance of electric current through the solution.
6. The electrical conductivity of the solution of an electrolyte depends upon the number of ions present in the solution. On reaching the electrodes, the ions lose their charge and change into neutral atoms or molecules by the gain or loss of electrons.
Applications of Arrhenius Theory
This theory explain many peculiarities in the behaviour of electrolytic solutions.
For example, the elevation in boiling point of 1 molal solution of glucose is 0.52ºC while this elevation in 1 molal solution of NaCl is 1.04ºC. This difference in elevation of boiling point can be explained on the basis of Arrhenius theory.
In one molal solution of glucose the number of (molecules) particles are 6.02 x 10(23) per dm3 of solution while in 1 molal solution of NaCl 6.02 x 10(23) ions of Na+ and 6.02 x 10(23) ions of Cl- are present because NaCl is an ionic compound. Since the number of particle are double in NaCl solution, therefore the elevation in boiling point is also double than the solution of glucose.
Similarly the other collegative properties such as lowering in vapour pressure, depression in freezing point and osmosis are explained on the basis of this theory.
Note
Collegative properties are those properties which depends upon the number of particles.
Conductance of Electric Current Through Solutions
The ability of a solution to conduct electric current depends upon the ions present in the solution. The conductance of a solution is increased when
1. The solution is diluted
2. The degree of dissociation of the electrolyte is high
3. The temperature of the solution is high
4. The velocity of the ions is high
But in a concentrated solution, the number of ions per unit volume of solution increases and the distance between ions decreases causing strong interionic attraction. As a result, migration of ions becomes more difficult and the conductance decreases with increase in concentration. As the conductance is related with the movement of ions, so conductance increase with the increase of absolute velocity of ions in the solution.
The conductance of an electrolyte also depends upon the degree of ionization. The degree of ionization is denoted by α and calculated as
α = No. of dissociated molecules / Total molecules dissovled
Electrolysis
Electrolyte
A chemical substance which can conduct electric current in molten form or in its aqueous solution with a chemical change is called electrolyte.
Electrolysis
The movement of anions and cations towards their respective electrodes with all accompanying chemical changes in an electrolytic solution under the influence of electric current is known as electrolysis.
Explanation
To explain the phenomenon of electrolysis consider the example of CuCl2 solution. the ionization of CuCl2 in the solution may be represented as
CuCl2 <----> Cu+2 + 2 Cl-
When electric current is passed through this solution, the movement of these ions begins to take place Cu+2 ions migrate towards cathode and Cl- ions towards anode. At cathode Cu+2 ions are discharged as copper atoms by the gain of electrons (reduction)
Cu+2 + 2 e- ----> Cu(M) ........ Reduction at Cathode
At anode Cl- ions are discharged as Cl2 by the loss of electrons (oxidation)
2 Cl- - 2 e- ----> Cl2(g0 ...... Oxidation at Anode
The overall reaction of the electrolysis may be written as
Cu+2 + 2 e- ----> Cu(M)
2 Cl- - 2 e- ----> Cl2(g)
Cu+2 + 2 Cl- ----> Cu(M) + Cl2(g)
OR
CuCl2 ----> Cu(M) + Cl2(g)
When all the ions present in the solution have been changed to neutral particles, the flow of current is stopped.