Chemistry MCQs intermidate XI Karachi Board MCQs
Five Year
Papers
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)
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Numericals
- Chemistry XI Karachi Board Numericals
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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.
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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.
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Chemistry XI Viva Notes Volumetric Analysis
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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.
Read more: Viva Notes Volumetric Analysis - Chemistry XI http://www.friendsmania.net/forum/1st-year-chemistry-notes/20552.htm#ixzz35AOezcJ8
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.
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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
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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Chemical
Bond
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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 π bond. The sigma bond is due to sp-sp overlap whereas π
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.
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Energetics Of Chemical Reaction
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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.
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Chemical Equilibrium
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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.
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Chemical
Kinetics
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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.
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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.
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