Page 1 : CHAPTER, , 3, , Equilibrium, Recap Notes, Physical equilibrium : It is the process, which involves physical changes only, like, equilibrium between different states of, substances at a particular temperature., X Solid, Liquid : ice, water at 0 °C, , X, , rate of melting = rate of freezing, Liquid, Gas (vapour) :, water (liquid), water (vapours) at 100 °C, rate of evaporation = rate of condensation, Solid, Gas (vapour) :, , X, , X, , X, , CO2 (solid), X, , CO2 (vapours), , rate of sublimation = rate of condensation, Solid, Saturated solution of solid in, liquid, , rate of dissolution = rate of precipitation, Gas, Saturated solution of gas in liquid, is always exothermic and spontaneous., X Freezing point/melting point : The, temperature at which the solid-liquid, equilibrium is reached for a pure substance, under one atmospheric pressure is called, normal melting point or normal freezing, point of the substance., X Boiling point : The temperature at which, liquid-gas equilibrium is attained for a, pure substance under one atmospheric, pressure is called normal boiling point of, the substance., General characteristics of physical, equilibria :, X, , X, , X, , Equilibrium can be established only in, case of closed system, i.e., the system, should neither gain matter from the, surroundings nor lose matter to the, surroundings., , X, X, , X, , The equilibrium is always dynamic in, nature. This means that the process does, not stop but the changes take place in the, forward and backward directions with the, same speed., The measurable properties of the system, become constant at equilibrium. This, is so because the concentration of the, substances become constant., At equilibrium, there exists an expression, involving concentration of reacting, substances which becomes constant at a, given temperature. For example, for the, equilibrium,, H2O(l), H2O(g) the pressure of water, becomes constant., The magnitude of the constant value of the, concentration related expression (called, equilibrium constant) gives an indication, of the extent to which the reaction, proceeds before acquiring equilibrium., Some features of physical equilibrium, , Process, 1. H2O(l), [liquid, , H2O(g), vapour], , 2. H2O(s), [solid, , H2O(l), liquid], , 3. Sugar(s), (solution), , [Solute(s), (solution), , Conclusion, PH2O is constant, at given, temperature., Melting point, is constant at a, given pressure., , Concentration, of sugar (solute), Solute(aq)] in solution, is constant, at a given, temperature., , Sugar(aq)

Page 2 : 4. CO2(g), [gas(g), , CO2(aq), gas(aq)], , [CO2(aq) ], [CO2(g ) ], [gas(aq) ], ,  i.e.,, , [gas( g ) ] , , is constant at a, given temperature., , Chemical equilibrium : If the process, involves only chemical change, the, equilibrium is called chemical equilibrium., X Reversible reaction : A reaction in, which the reactants are formed back by, the reaction of products with each other, at the given conditions of the reaction., These reactions if carried out in a closed, vessel do not go to completion., e.g., CH3COOH + C2H5OH, , , N2(g) + 3H2(g), , CH3COOC2H5, + H2O, , 2NH3(g), , Irreversible reaction : These are the, reactions in which products do not react, back to give the reactants, i.e., reaction, cannot be retraced at any point., X State, of equilibrium : Chemical, equilibrium is that state of a reaction at, which the rate of forward reaction becomes, equal to rate of backward reaction., Dynamic nature of equilibrium : In, dynamic equilibrium, changes occur but in, opposite directions and at constant rate i.e.,, forward and backward reactions take place, even after the equilibrium is attained but, at equal speeds., X, , General characteristics of chemical, equilibrium :, X It is dynamic in nature., X The observable properties of the system, become constant at equilibrium and, remain unchanged thereafter., X The equilibrium can be approached from, either directions., X The equilibrium can be attained only if, the system is a closed one., X The free energy change at constant, pressure and temperature is zero., X Addition of a catalyst does not change, the equilibrium state, it only helps in, attaining the equilibrium faster., , Law of chemical equilibrium :, X For the reversible reaction,, aA(aq) + bB(aq), xX(aq) + yY(aq), [ X]x[Y ] y, , At equilibrium,, , = K c , Kc is called, [A]a[B]b, equilibrium constant which is constant for a, reaction at constant temperature., X Relations between equilibrium constants, for a general reaction and its multiples., , Chemical, equation, , Equilibrium, constant, , cC + dD, , aA + bB, , K, , aA + bB, , cC + dD, naA + nbB, , Kc′ = (1/Kc), K′′c = (K nc), , ncC + ndD, , Types of equilibrium :, X Homogeneous equilibrium :, Equilibrium is said to be homogeneous if, reactants and products are in same phase,, e.g.,, H2(g) + I2(g), , 2HI(g), , 2SO2(g) + O2(g), X, , 2SO3(g), , Relationship between Kp and Kc :, xX + yY, , For reaction : aA + bB, Kc =, , [ X]x[Y ] y, [A]a[B]b, , and Kp =, , PX ′ x PY ′ y, PA ′ a PB ′ b, , , (for gaseous reaction), For ideal gases PV = nRT, P, n, (concentration in mole L–1), =, RT V, n, P ′, \ Concentration of A, [A] = A = A, V, RT, nB PB ′, =, Similarly, [B] =, V RT, , or, , Substituting these values :, , Kc =, , x, , y, , a, , b, , P ′  P ′ , X, Y,  RT,   RT, , ,  , , , P ′  P ′ , A, B,  RT,   RT, , ,  , , , x, , =, , ′x, , ′y, ,  1   1 ,  RT   RT , , y, , PX .PY,  1 , = Kp, ×, a, b, a, b,  RT , ′, ′, PA .PB,  1   1 ,  RT   RT , , (x + y)−(a + b)

Page 3 : (x + y) – (a + b) = Dng (difference in number, of moles of gaseous products and reactants), ∆n,  1  g, \ Kc = K p ,  RT , or Kp = Kc (RT)Dng, If, , Dng = 0, Kp = Kc, , If, , Dng = +ve, Kp > Kc, , If, , Dng = –ve, Kp < Kc, , X, , Heterogeneous equilibrium :, , Equilibrium is said to be heterogeneous, if reactants and products are in different, phases, e.g.,, Ni(s) + 4CO(g), , Ni(CO)4(g), , CaCO3(s), X, , X, , CaO(s) + CO2(g), , Units of equilibrium constant :, Concentration of a substance is measured, in terms of moles/litre, therefore, unit of Kc, is (mol L–1)Dng. Similarly, partial pressure, is measured in terms of atmosphere,, hence, unit of Kp is (atm)Dng, – If Dn = 0, both Kc and Kp have no units, – If Dn > 0, unit of Kc = (mol L–1)Dng, unit of, Kp = (atm)Dng, – If Dn < 0, unit of Kc = (L mol–1)Dng, unit of, Kp = (atm–1)Dng, Equilibrium constants can also be, expressed as dimensionless quantities, if the standard states of reactants and, products are specified., , Applications of equilibrium constant :, X Predicting the extent of a reaction :, The magnitude of equilibrium constant K, indicates the extent to which a reaction, can go. In other words, it is a measure of, the completion of a reversible reaction., Larger the value of K, greater will be, the equilibrium concentration of the, components on the right hand side of, reaction relative to those on the left hand, side, i.e., the reaction proceeds to a greater, extent., For example, consider the reaction :, N2(g) + 2O2(g) at 298 K, , 2NO2(g), K=, , 2, , [N2 ][O2 ], 2, , = 6.7 × 10 −16, , [NO2 ], The value of K is very small which means, that the forward reaction has proceeded to, , small extent only. Thus we can say that NO2, is quite stable and decomposes only slightly., X Predicting the direction of a reaction :, The equilibrium constant helps in, predicting the direction in which a reaction, can proceed at any stage. By substituting, the concentration of substances that exist, in a reaction mixture we can calculate the, reaction quotient, Q and comparing the, value of Q with the equilibrium constant,, K, we can predict whether the reaction, will proceed towards products or towards, reactants., Case I : If Q < K, the reaction will proceed in, the forward direction., Case II : If Q > K, the reaction will proceed, in the backward direction., Case III : If Q = K, the reaction mixture is, already at equilibrium., Relation between equilibrium constant, (K) and DG° :, X Free energy change (DG) is related to DG°, as follows, DG = DG°° + 2.303 RT log Q, But at equilibrium, Q = K and DG = 0, \ DG° = –2.303 RT log K, If K > 1, then DG°= –ve, i.e., reaction is, spontaneous in forward direction., If K < 1, then DG° = +ve, i.e., reaction is, spontaneous in backward direction., X Le Chatelier’s principle : The law, states that if any kind of change in, concentration, temperature or pressure, is imposed on the system in equilibrium,, then equilibrium shifts in a direction that, tends to undo the effect of the change, imposed., X Effect of change in concentration : In, a reaction at equilibrium, if concentration, of any reactant or product is increased,, the equilibrium shifts in a direction where, it will be consumed or if concentration of, any reactant or product is decreased, the, equilibrium will shift in a direction where, it will be produced., X Effect of change in temperature : In, a reaction at equilibrium if temperature, is increased, reaction will proceed in, the direction in which some heat can, be destroyed i.e., absorbed so that

Page 4 : X, , X, , X, , temperature of the system remains, constant. It means increase in temperature, supports the endothermic reaction, where, heat is absorbed. Decrease in temperature, favours exothermic reaction in which heat, is liberated., Effect of change in pressure : In a, reaction at equilibrium if pressure is, increased, then according to Le Chatelier’s, principle, the equilibrium will shift in a, direction in which the pressure decreases., This implies that equilibrium will shift, in a direction which produces smaller, number of moles, since pressure is directly, proportional to the number of moles. On, the other hand we can say that if the, pressure on the system in equilibrium is, decreased, the equilibrium will shift in, the direction in which it is accompanied, by increase in total number of moles., Effect of catalyst : Catalyst increases, the speed of forward as well as backward, reaction to the same extent. Therefore, it, does not affect the equilibrium position., It only helps in attaining the equilibrium, state quickly without shifting it in any, direction., , –, –, –, , –, , –, , –, , –, , Effect of adding inert gas :, – If inert gas is added at constant, volume to the reaction at equilibrium,, the equilibrium will not be changed/, disturbed, because at constant volume, addition of an inert gas will not, change the molar concentration of the, reactants and products., – If inert gas is added at constant, pressure to the reaction at equilibrium,, the volume of reaction mixture, increases and equilibrium will shift in, the direction in which there is increase, in the number of moles of the gas. If, , – Effect of change in temperature :, The dissociation of HI is an endothermic, reaction hence increase of temperature, favours the formation of H2 and I2., – Effect of change in pressure : As the, number of reactant molecules is equal, to the number of product molecules,, hence pressure has no effect on the, equilibrium., – Effect of change in concentration :, Addition of HI or removal of H2 or I2, favours the forward reaction., , :, , Ionic Equilibrium :, , :, , –, , :, , number of moles on either side of, equilibrium are same, there will be, no effect of adding an inert gas on the, state of equilibrium., Application of Le Chatelier’s, principle to chemical equilibria :, For reaction, PCl5(g), PCl3(g) + Cl2(g) ;, , DH = +ve, Effect of change in temperature :, The dissociation of PCl5 is endothermic., Hence, increase, of, temperature, increases the rate of formation of PCl3, and Cl2., Effect of change in pressure :, Increase of pressure will favour the, backward direction, with decrease in, the number of gaseous moles., Effect of change in concentration, : Addition of PCl3 or Cl2 or removal, of PCl5 will shift the equilibrium in, backward direction., Effect of adding an inert gas : In, above reaction, np > nr, addition of, an inert gas has no effect at constant, volume but, at constant pressure,, the equilibrium shifts in the forward, direction., For reaction, 2HI(g), H2(g) + I2(g) ;, DH = +ve

Page 5 : X, , Conjugate acid - base pairs : These are, pairs of acids and bases which differ by a, proton, e.g.,, conjugate base, , Acid, X, , X, , X, , Relative strengths of conjugate acids, or bases depend upon their tendency to, donate a proton or to accept a proton., Stronger the acid, weaker is its conjugate, base and vice-versa, e.g.,, Weak base, +, H3O(aq), , CH3COOH(aq) + H2O(l), , +, , –, , CH3COO(aq),            , Strong base, X Relative strength of acids and bases :, This is the ratio of strengths of acids. e.g.,, for acids HA1 and HA2:, , Weak acid, , –, , 2, , H+ + A1 ; Ka1 = C1a1, +, , HA2, , H +, , A2– ;, , Ka2 =, , 2, C2a2, , Relative strength, =, , [H+] furnished by HA1, +, , [H ] furnished by HA2, , =, , C1α1, C 2α 2, ( α = K a / C ), , , =, , C1 K a /C1, 1, , C2 K a /C2, , = (K a C1) / (K a C2 ), 1, , 2, , 2, , If concentrations of acids are same, then, Relative strength =, , Ka, , Ka, , 1, 2, , Dissociation constant of weak acids, and weak bases : Let us consider the, dissociation of a weak acid HA as,, H+ + A–,   , HA, At t = 0, C     0   0, At teqm., , C(1 – a), , Ca   Ca, , Dissociation constant of acid,, Ka =, , H2PO–4, , –, , Strong acid           , , HA1, , H3PO4, , H3O+(aq) + Cl (aq), , HCl(aq) + H2O(l), , [H + ][ A − ] Cα ⋅ Cα, Cα 2, =, =, [HA], C(1 − α) (1 − α), , Similarly, for the dissociation of a weak base, +, , –, , B + OH, BOH as BOH, Dissociation constant of base,, [B + ][OH − ] Cα 2, Kb =, =, [BOH], (1 − α), , Dissociation constant for polyprotic, acids and bases : These acids involve, number of steps to ionise completely,, where number of steps is equal to the, number of replaceable hydrogen atoms., The value of ionisation constant of, each step is definite and constant e.g.,, , HPO42–, , K1, , H+ + H2PO–4, , K2, , H+ + HPO42–, , K3, , H+ + PO43–, , The overall dissociation constant (K) is given, as K = K1 × K2 × K3 where K1 > K2 > K3., pH and pH-scale : pH of a solution, is defined as the negative logarithm of, hydrogen ion concentration., pH = − log [H+] or pH = − log [H 3O + ], or, , pH = log, , 1, , [H+], , Similarly, negative logarithm of hydroxyl ion, concentration is called pOH., pOH = − log [OH − ] or pOH = log, X, , 1, , [OH − ], , pH-scale : A solution is classified as, acidic, basic or neutral based on its pH, value., – If pH = pOH = 7, then solution will be, neutral., – If pH < 7 or pOH > 7, then solution will, be acidic., – If pH > 7 or pOH < 7, then solution will, be basic., – pH scale varies from 0 to 14., , Relationship between pH and pOH, pH + pOH = 14 and pH + pOH = pKw, Relation between Ka and Kb : If Ka and, Kb are the dissociation constants of weak, acid and weak base respectively then,, pKa = – logKa and pKb = – logKb, Consider a weak acid, HA, HA, , H+ + A–, Ka =, , A– + H2O, , [H + ][A− ], [HA], , HA + OH–, Kb =, , Multiplying Ka and Kb, we get, Ka × Kb = [H+][OH –], , [OH − ][HA], [A− ]

Page 6 : But [H+][OH –] = Kw, , – Acidic buffers : These are the, solutions of a mixture of weak acid and, salt of this weak acid with strong base., – For example, CH3COOH + CH3COONa., They have pH value lesser than 7., – Basic buffers : These are the solutions, of mixture of a weak base and salt of, this weak base with strong acid. For, example, NH4OH + NH4Cl. They have, the pH value more than 7., , K a × Kb = Kw, Taking log on both sides, we get, \, , log Ka + log Kb = log Kw, or – log Ka – log Kb = – log Kw, pKa + pKb = pKw, Common ion effect in the ionisation of, acids and bases : There exists a dynamic, equilibrium between unionised molecule and, ions of a weak electrolyte : AB, A+ + B–, X To this equilibrium if a solution of a strong, electrolyte (AX or YB) is mixed, due to the, presence of a common ion, either A+ or B–,, the degree of dissociation is suppressed in, both the cases., X The decrease in dissociation is much, more in case of weak electrolytes, e.g., if a, solution of NH4OH some NH4Cl is added,, –, , NH4OH(aq), , NH+4(aq) + OH(aq), , NH4Cl(aq), , NH+4(aq), , +, , –, Cl (aq), , NH4Cl ionises almost completely leading to, +, a high concentration of NH4 (common ion)., According to Le Chatelier’s principle, the, dissociation of NH4OH is suppressed., X Thus, common ion effect can be defined as, the suppression of the degree of ionisation, of a weak electrolyte by the addition of a, strong electrolyte having an ion common, with the weak electrolyte., Buffer solutions : Buffer solution is defined, as a solution which resists the change in its, pH value when small amount of acid or base, is added to it or when the solution is diluted., Buffer solution has a definite pH value at, specific temperature and it does not change, on keeping for a long time., Buffers are classified into two categories :, X Simple buffers : These are the solutions, of salt of weak acid and weak base. For, example,, CH3COONH4, (ammonium, acetate)., X Mixed buffers : These are the mixture, of two solutions. These are further of two, types :, , Solubility product : Solubility product of, an electrolyte at a specified temperature, may be defined on the product of the molar, concentrations of its ions in a saturated, solution, each concentration raised to the, power equal to the number of ion produced, on dissociation of one molecule of electrolyte., Salt, type, AB2, , Relation between, Ksp and S, Ksp = (S)(2S)2 = 4S3, 2, , Examples, PbCl2, HgCl2, , A 2B, , Ksp = (2S) (S) = 4S, , 3, , Ag2CrO4,, Ag2C2O4,, Ag2SO4, , AB3, , Ksp = (S)(3S)3 = 27S4, , Fe(OH)3,, Al(OH)3,, Cr(OH)3, , A 3B 2, , Ca3(PO4)2,, Ksp = (3S)3(2S)2, = 108S5 Zn3(PO4)2, , AB, , Ksp = (S)(S) = S2, , AlPO4, AgCl,, AgBr, PbSO4,, BaSO4, ZnS, , Effect of common ion on solubility :, Addition of a common ion lowers the, solubility of a sparingly soluble salt. Let S, be the solubility of a salt AgCl in pure water, and suppose x g equivalent of NaCl is added, to a litre of the saturated solution. Suppose, the solubility of AgCl becomes S′., Ksp = [Ag+][Cl–] = S′(S′ + x), When the addition of a common ion or an inert, salt leads to a considerable increase in the, solubility of a salt, then it can be concluded, that a complex ion has been formed.

Page 7 : Practice Time, , 1. For the following three reactions (i), (ii) and, (iii), equilibrium constants are given, CO2(g) + H2(g); K1, (i) CO(g) + H2O(g), (ii) CH4(g) + H2O(g), CO(g) + 3H2(g); K2, (iii) CH4(g) + 2H2O(g), CO2(g) + 4H2(g); K3, Which of the following relations is correct?, (b) K1 K 2 = K 3, (a) K3 K23 = K12, (c) K2 K3 = K1, (d) K3 = K1K2, 2. If K1 and K2 are the respective equilibrium, constants for the two reactions,, XeF6(g) + H2O( g), XeOF4(g) + 2HF(g), XeO4(g)+ XeF6( g), XeOF4(g) + XeO3F2( g), The equilibrium constant for the reaction,, XeO4( g) + 2HF( g), XeO3F2( g) + H2O( g) is, (a) K1K2, (b) K1/K22, (c) K2 / K1, (d) K1 / K2, 3., , For the reversible reaction,, A(s) + B(g), , C(g) + D(g), DG° = –350 kJ,, , 7. For the reaction 2SO2(g) + O2(g), 2SO3(g), at 300 K, the value of DG° is –690.9 R. The, equilibrium constant value for the reaction at, that temperature is (R is gas constant), (b) 10 atm, (a) 10 atm–1, (c) 10, (d) 1, 8. The solubility product of aluminium, sulphate is given by the expression, (a) 4s3, (b) 6912s7, 2, (c) s, (d) 108s5, 9. In a reversible chemical reaction at, equilibrium, if the concentration of any one of, the reactants is doubled, then the equilibrium, constant will, (a) also be doubled, (b) be halved, (c) remain the same, (d) become one-fourth., 10. Identify a species which is ‘NOT’ a Bronsted, acid but a Lewis acid., (a) BF3, (b) H3O+, (c) NH3, (d) HCl, , which one of the following statements is true?, (a) The reaction is thermodynamically nonfeasible., (b) The entropy change is negative., (c) Equilibrium constant is greater than one., (d) The reaction should be instantaneous., , 11. For the system 3A + 2B, for equilibrium constant K is, , 4. In a chemical equilibrium the rate constant, of backward reaction is 3.2 × 10–2 and the, equilibrium constant is 2–5. The rate constant of, forward reaction is, (b) 2 × 10–2, (a) 1 × 10–3, –2, (c) 8 × 10, (d) 4 × 10–2, , (c), , 5. It is not possible to attain equilibrium in, (a) closed system, (b) isolated system, (c) open system, (d) none of these., , 13. The solubility of Ca3(PO4)2 in water is, y moles/litre. Its solubility product is, (a) 6y4, (b) 36y4, 5, (c) 64y, (d) 108y5, , 6. Conjugate base for Bronsted acids H2O and, HF are, (a) H3O+ and H2F+, respectively, (b) OH– and H2F+, respectively, (c) H3O+ and F–, respectively, (d) OH– and F–, respectively., , (a), , [ 3 A] × [ 2 B ], [C ], C, [ A]3 × [ B ]2, , C,the expression, , (b), , [ A]3 × [ B ], [C ], , (d), , [C ], [ 3 A] × [ 2 B ], , 12. Among the following, the one which can act, as both Bronsted acid as well as Bronsted base is, (a) H3PO4, (b) AlCl3, –, (c) CH3COO, (d) H2O, , 14. Which of the following is not a characteristic, of equilibrium?, (a) Rate is equal in both directions., (b) Measurable quantities are constant at, equilibrium.

Page 8 : (c) Equilibrium occurs in reversible condition., (d) Equilibrium occurs only in an open vessel at, constant temperature, 15. When the rate of formation of reactants is, equal to the rate of formation of products, this is, known as,, (a) chemical reaction, (b) chemical equilibrium, (c) chemical kinetics, (d) none of these., 16., for, X, (a), (c), , Which of the following options will be correct, the stage of half completion of the reaction, Y., DG° = 0, (b) DG° > 0, DG° < 0, (d) DG° = –RT ln2, , 17. In Bronsted–Lowry concept of acid-base, H+, donor is a/an, (a) acid, (b) basic, (c) exclusively amphoteric substance, (d) hydroxide acceptor., 18. In HS–, I–, R — NH2, NH3 order of proton, accepting tendency will be, (a) I – > NH3 > R — NH2 > HS–, (b) NH3 > R — NH2 > HS– > I–, (c) R — NH2 > NH3 > HS– > I–, (d) HS– > R — NH2 > NH3 > I–, 19. Equilibrium constants are given for the, following reactions. Out of the following, which, is farthest towards completion?, (a) K = 100, (b) K = 0.1, (c) K = 0.01, (d) K = 1, 20. Of the following the incorrect relation is, (a) DG = DG° + RTlnK (b) DG° = –RTlnK, (c) DG° = DG + RTlnK (d) K = e–DG°/RT, 21. Calculate Kc for the reversible process given, below if Kp = 167 and T = 800°C., CaCO3(s), (a) 1.95, (c) 1.89, , CaO(s) + CO2(g), (b) 1.85, (d) 1.60, , 22. The dissociation constants for acetic acid, and HCN at 25°C are 1.5 × 10–5 and 4.5 × 10–10, respectively. The equilibrium constant for the, equilibrium,, , CN– + CH3COOH, would be, (a) 3.0 × 10–5, (c) 3.0 × 104, , HCN + CH3COO–, (b) 3.0 × 10–4, (d) 3.0 × 105, , 23. What will be the solubility product of AX3?, (b) 4S3, (a) 27S4, 4, (c) 36S, (d) 9S3, 24. Which of the following is a characteristic of, reversible reaction?, (a) It never proceeds to completion., (b) It can be influenced by a catalyst., (c) It proceeds only in the forward direction., (d) Number of moles of reactants and products are, equal., 1, 25. For the reaction, SO2( g ) + O2( g )  SO3( g ) ,, 2, if Kp = Kc(RT)x where the symbols have usual, meaning then the value of x is (assuming, ideality), (a) 1, (b) – 1, 1, 1, (c) −, (d), 2, 2, 26. Which of these is least likely to act as a, Lewis base?, (a) BF3, (b) PF3, (c) CO, (d) F–, 27. Which of the following solutions will have pH, close to 1.0?, (a) 100 mL of M/10 HCl + 100 mL of M/10 NaOH, (b) 55 mL of M/10 HCl + 45 mL of M/10 NaOH, (c) 10 mL of M/10 HCl + 90 mL of M/10 NaOH, (d) 75 mL of M/10 HCl + 25 mL of M/10 NaOH, 28. Using the Gibbs’ energy change, DG° = +63.3 kJ,, for the following reaction,, Ag2CO3(s), 2Ag+(aq) + CO2–, 3(aq), the Ksp of Ag2CO3(s) in water at 25°C is, (R = 8.314 J K–1 mol–1), (a) 3.2 × 10–26, (c) 2.9 × 10–3, , (b) 8.0 × 10–12, (d) 7.9 × 10–2, , 29. Consider the general hypothetical reaction,, 2B(g) + 3C(g), A(s), If the concentration of C at equilibrium is doubled,, then after the equilibrium is re-established, the, concentration of B will be, (a) two times the original value, (b) one half of its original value, (c) 1 / 2 2 times the original value, (d) 2 2 times the original value.

Page 9 : 30. The pH of 10–4 M KOH solution will be, (a) 4, (b) 11, (c) 10.5, (d) 10, 31. If the pH increases from 5 to 7, then acidic, strength decreases _________ times., (a) 2, (b) 20, (c) 10, (d) 100, 32. The reaction quotient, Qc is useful in, predicting the direction of the reaction. Which of, the following is incorrect?, (a) If Qc > Kc, net reaction goes from right to left., (b) If Qc < Kc, net reaction goes from left to right., (c) If Qc = Kc, no net reaction occurs., (d) If Qc > Kc, net reaction goes from left to right., 33., , Kp, Kc, , CO(g) +, , 37. The value of DH for the reaction, X2(g) + 4Y2(g), , 2XY4(g) is less than zero., , Formation of XY4(g) will be favoured at, (a) high temperature and high pressure, (b) low pressure and low temperature, (c) high temperature and low pressure, (d) high pressure and low temperature., , for following reaction will be, , 38. In which of the following equilibrium Kc and, Kp are not equal?, , 1, O, → CO2(g), 2 2(g), , (a) 2NO(g), , (a) RT, (c), , 36. In the reaction;, Fe(OH)3(s), Fe3+(aq) + 3OH–(aq), if the, concentration of OH– ions is decreased by, 1, times, then the equilibrium concentration of, 4, Fe3+ will increase by, (a) 8 times, (b) 16 times, (c) 64 times, (d) 4 times., , 1, RT, , (b), , 1, RT, , (b) SO2(g) + NO2(g), , (d), , RT, 2, , (d) 2C(s) + O2(g), , (c) H2(g) + I2(g), , 34. Buffer solutions have constant acidity and, alkalinity because, (a) these give unionised acid or base on reaction, with added acid or alkali, (b) acids and alkalies in these solutions are, shielded from attack by other ions, (c) they have large excess of H+ or OH– ions, (d) they have fixed value of pH., 35. The equilibrium, reaction,, , constant,, , Kp, , for, , the, , PCl5, PCl3 + Cl2 is 1.6 at 200°C ., The pressure at which PCl5 will be 50%, dissociated at 200°C is, (a) 3.2 atm, (c) 2.4 atm, , N2(g) + O2(g), , (b) 4.8 atm, (d) 6.4 atm, , SO3(g) + NO(g), 2HI(g), 2CO2(g), , 39. Which will make basic buffer?, (a) 100 mL of 0.1 M HCl + 100 mL of, 0.1 M NaOH, (b) 50 mL of 0.1 M NaOH + 25 mL of, 0.1 M CH3COOH, (c) 100 mL of 0.1 M CH3COOH + 100 mL of, 0.1 M NaOH, (d) 100 mL of 0.1 M HCl + 200 mL of, 0.1 M NH4OH, 40. 4 moles of A are mixed with 4 moles of, B, when 2 moles of C and D are formed at, equilibrium according to the reaction,, C+D, A+B, the value of equilibrium constant is, (a) 4, (b) 1, (c) 1/2, (d) 1/4, , Case Based MCQs, Case I : Read the passage given below and, answer the following questions from 41 to 45., Le Chatelier’s principle is also known as the, equilibrium law, used to predict the effect of, change on a system at chemical equilibrium., , This principle states that equilibrium adjust the, forward and backward reactions in such a way, as to accept the change affecting the equilibrium, condition. When factor like concentration,, pressure, temperature, inert gas that affect

Page 10 : equilibrium are changed, the equilibrium will, shift in that direction where the effects that, caused by these changes are nullified. This, principle is also used to manipulate reversible, reaction in order to obtain suitable outcomes., 41. Which one of the following conditions will, favour maximum formation of the product in the, reaction, A2(g) + B2(g), X2(g), Dr H = –X kJ ?, (a) Low temperature and high pressure, (b) Low temperature and low pressure, (c) High temperature and high pressure, (d) High temperature and low pressure, 42. In which one of the following equilibria will, the point of equilibrium shift to left when the, pressure of the system is increased?, (a) H2(g) + I2(g), (b) 2NH3(g), , 2HI(g), N2(g) + 3H2(g), , (c) C(s) + O2(g), (d) 2H2(g) + O2(g), , CO2(g), 2H2O(g), , 43. For the reversible reaction,, N2(g) + 3H2(g), 2NH3(g) + heat, The equilibrium shifts in forward direction, (a) by increasing the concentration of NH3(g), (b) by decreasing the pressure, (c) by decreasing the concentrations of N2(g) and, H2(g), (d) by increasing pressure and decreasing, temperature., 44. Favourable conditions for manufacture of, ammonia by the reaction,, N2 + 3H2, 2NH3; DH = – 21.9 kcal are, (a), (b), (c), (d), , low temperature, low pressure and catalyst, low temperature, high pressure and catalyst, high temperature, low pressure and catalyst, high temperature, high pressure and, catalyst., , P+Q, 45. X + Y, For the above equilibrium, the reactant, concentration is doubled, what would happen, then to equilibrium constant?, (a) Remains constant, (b) Be doubled, (c) Be halved, (d) Cannot be predicted, , Case II : Read the passage given below and, answer the following questions from 46 to 50., Reactants and products coexist at equilibrium,, so that the conversion of reactant to products, is always less than 100%. Equilibrium reaction, may involve the decomposition of a covalent (nonpolar) reactant or ionization of ionic compound, into their ions in polar solvents. Ostwald dilution, law is the application of the law of mass action to, the weak electrolytes in solution., A binary electrolyte AB which dissociates into, A+ and B– ions i.e., A+ + B–, , AB, , for every weak electrolyte, Since a <<1, (1 – a) = 1, K, K = C α2 ⇒ α =, ⇒ α = KV ., C, 46. A monobasic weak acid solution has a, molarity of 0.005 M and pH of 5. What is its, percentage ionization in this solution?, (a) 2.0, (b) 0.2, (c) 0.5, (d) 0.25, 47. Calculate ionisation constant for pyridinium, hydrogen chloride. (Given that H+ ion, concentration is 3.6 × 10–4 M and its concentration, is 0.02 M.), (a) 6.48 × 10–2, (b) 6 × 10–6, (c) 1.5 × 10–9, (d) 12 × 10–8, 48. The hydrogen ion concentration of a 10–8 M, HCl aqueous solution at 298 K(Kw = 10–14) is, (a) 9.525 × 10–8 M, (c) 1.0 × 10–6 M, , (b) 1.0 × 10–8 M, (d) 1.0525 × 10–7 M, , 49. Ostwald dilution law is applicable to, (a), (b), (c), (d), , weak electrolytes, non-electrolyte, strong electrolyte, all type of electrolyte., , 50. If a is the fraction of HI dissociated at, equilibrium in the reaction :, 2HI, , H2 + I2, , then starting with 2 mol of HI, the total number, of moles of reactants and products at equilibrium, are, (a) 1, (b) 2, (c) 1 + a, (d) 2 + 2a

Page 11 : For question numbers 51-60, a statement of assertion followed by a statement of reason is given. Choose, the correct answer out of the following choices., (a) Assertion and reason both are correct statements and reason is correct explanation for assertion., (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion., (c) Assertion is correct statement but reason is wrong statement., (d) Assertion is wrong statement but reason is correct statement., 51. Assertion : At 25°C the pH of 10–7 M HCl is, 6.69., Reason : pH of acidic solution is always below 7 at, 25°C., 52. Assertion : The value of Kw at 25°C is, 1.0 × 10–14 mol2 dm–6., Reason : Kw of water changes with change in, temperature., 53. Assertion : KP can be equal to, less than or, greater than the value of Kc., Dn, , Reason : Kp = Kc(RT) where Dn is the change, in the number of moles of gaseous reactants and, products., 54. Assertion : The active mass of pure solid is, taken unity., Reason : The active mass of pure solids depends, on density and molecular mass. The density and, molecular mass of pure solids are constant., 55. Assertion : H3PO4 is a stronger acid than, H2PO–4, Reason : pKa is greater than pKa for H3PO4., 1, , 2, , 56. Assertion : For the equilibrium mixture, CO(g) + 3H2(g), CH4(g) + H2O(g) if the volume is, decreased, reaction proceeds in the forward direction., Reason : For the methanation reaction (above),, decrease in volume causes Qc > Kc., 57. Assertion : At constant temperature,, the pressure of the gas is proportional to its, concentration., Reason : KP = Kc for all reactions., 58. Assertion : For any chemical reaction at a, particular temperature, the equilibrium constant, is fixed and is a characteristic property., Reason : Equilibrium constant is independent of, temperature., 59. Assertion : On dilution, the solubility of, BaSO4 increases., Reason : On dilution, Ksp increases., 60. Assertion : The equilibrium constant does, not change on addition of catalyst., Reason : A catalyst provides an alternative path of, lower activation energy for conversion of reactants, to products., , SUBJECTIVE TYPE QUESTIONS, , Very Short Answer Type Questions (VSA), 1. A mixture of 1.57 mol of N2, 1.92 mol of H2, and 8.13 mol of NH3 is introduced into a 20 L, reaction vessel at 500 K. At this temperature,, the equilibrium constant, Kc for the reaction, N2(g) + 3H2(g), 2NH3(g) is 1.7 × 102. Is the, reaction mixture at equilibrium? If not, what is, the direction of the net reaction?, , basic strength of the following conjugate bases?, OH–, RO–, CH3COO–, Cl–, 4. Classify the following species into Lewis, acids and Lewis bases and show how these act as, Lewis acid/base : (a) OH– (b) F– (c) H+ (d) BCl3., , 2. Explain why pure liquids and solids can be, ignored while writing the equilibrium constant, expression., , 5. Calculate the pH of a solution formed by, mixing equal volumes of two solutions A and, B of a strong acid having pH = 6 and pH = 4, respectively., , 3. Conjugate acid of a weak base is always, stronger. What will be the decreasing order of, , 6. Give two important, chemical equilibrium., , characteristics, , of

Page 12 : 7. What is the equilibrium concentration of, each of the substances in the equilibrium when, the initial concentration of ICl was 0.78 M?, I2(g) + Cl2(g); Kc = 0.14, 2ICl(g), 8., , Give relation between [A] and [B] for, , the stage of half completion of the reaction, A, B., 9. The pH of a sample of vinegar is 3.76., Calculate the concentration of hydrogen ions in, it., , Short Answer Type Questions (SA-I), 10. Equilibrium constant for a reaction is 10., What will be the equilibrium constant for the, reverse reaction?, 11. Write the expression for the equilibrium, constant Kc for the following equilibrium :, 3Fe(s) + 4H2O(g), , Fe3O4(s) + 4H2(g), , 12. For the system 3A + 2B, for equilibrium constant K is?, , C,the expression, , 13. Differentiate between homogeneous and, heterogeneous equilibrium giving examples., 14. Ka , Ka and Ka are the respective ionisation, 1, , 2, , 3, , constants for the following reactions., H2S, , H+ + HS–, , HS–, , H+ + S2–, , H2S, , 2H+ + S2–, , The correct relationship between Ka , Ka and Ka, 1, 2, 3, is, 15. Write expressions for Kp and Kc for the, decomposition reaction of calcium carbonate., 16. The ionization constant of HF, HCOOH and, HCN at 298 K are 6.8 × 10–4, 1.8 × 10–4 and, 4.8 × 10–9 respectively. Calculate the ionization, constants of the corresponding conjugate base., 17. Write the relation between Qc and Kc for, reverse reaction., 18. We know that the relationship between Kc, and Kp is, Kp = Kc(RT)Dn, What would be the value of Dn for the reaction, NH3(g) + HCl(g), NH4Cl(s), 19. What will be the conjugate bases for the, Bronsted acids: HF, H2SO4 and HCO–3 ?, , Short Answer Type Questions (SA-II), 20. Calculate (a) DG° and (b) the equilibrium, constant for the formation of NO2 from NO and, O2 at 298 K., 1, NO( g ) + O2( g )  NO2( g ), 2, where Df G° (NO2) = 52.0 kJ mol–1,, Df G° (NO) = 87.0 kJ mol–1,, Df G° (O2) = 0 kJ mol–1, 21. What will be the correct order of vapour, pressure of water, acetone and ether at 30°C?, Given that among these compounds, water has, maximum boiling point and ether has minimum, boiling point?, 22. In 1 L saturated solution of AgCl, [Ksp = 1.6 × 1010], 0.1 mol of CuCl, , [Ksp = 1.0 × 10–6] is added. Find out the, resultant concentration of Ag+ in the solution., , 23. Reaction between N2 and O2 takes place as, follows :, , 2N2(g) + O2(g), , 2N2O(g), , If a mixture of 0.482 mol of N2 and 0.933 mol, of O2 is placed in a 10 L reaction vessel and, allowed to form N2O at a temperature for which, Kc = 2.0 × 10–37, determine the composition of, equilibrium mixture., 24. Equal volumes of 0.002 M solutions of sodium, iodate and copper chlorate are mixed together., Will it lead to precipitation of copper iodate? (For, copper iodate, Ksp = 7.4 × 10–8), 25. Define Le-Chatelier’s Principle. What is the, effect of :, (i) addition of H2, (ii) removal of CO, on the equilibrium :, CH3OH(g) ?, 2H2(g) + CO(g), 26. The value of DG° for the phosphorylation, of glucose in glycolysis is 13.8 kJ/mol. Find the, value of Kc at 298 K.

Page 13 : 27. For the reaction,, CO(g) + 2H2(g), , CH3OH(g); DHr° = –92 kJ/mol, , predict the direction of the reaction when, (i) pressure is doubled, (ii) temperature is doubled., 28. What is the effect of increase of temperature, on the pH of a buffer solution?, 29. Equilibrium constant, Kc for the reaction,, N2(g) + 3H2(g), , 2NH3(g) at 500 K is 0.061., , At a particular time, the analysis shows, that composition of the reaction mixture is, 3.0 mol L–1 N2, 2.0 mol L–1 H2 and 0.5 mol L–1, NH3. Is the reaction at equilibrium? If not in, which direction does the reaction tend to proceed, to reach equilibrium?, 30. (i) , Define Lewis acids and bases with, example., (ii) The value of Ksp of two sparingly soluble, salts Ni(OH)2 and AgCN are 2 × 10–15 and, 6 × 10–17. Which salt is more soluble and why?, 31. Which of the following combinations would, result in the formation of a buffer solution?, (i) NH4Cl + NH3, (ii) CH3COOH + HCl, , (iii) CH3COONa + CH3COOH, (iv) NH3 + HCl in the molar ratio of 2 : 1, (v) HCl + NaOH, 32. A sparingly soluble salt having general, formula Axp+Byq–and molar solubility S is in, equilibrium with its saturated solution. Derive a, relationship between the solubility and solubility, product for such salt., 33. On increasing the pressure, in which, direction will the gas phase reaction proceed, to re-establish equilibrium, is predicted by, applying the Le Chatelier’s principle. Consider, the reaction., N2(g) + 3H2(g), 2NH3(g), What will be the effect on K. if the total pressure, at which the equilibrium is established, is, increased without changing the temperature?, 34. The solubility product of AgCl is 1.5 × 10–10., Predict whether there will be any precipitation, by mixing 50 mL of 0.01 M NaCl and 50 mL of, 0.01 M AgNO3 solution., 35. What is the pH of 0.001 M aniline, solution? The ionisation constant of aniline is, 4.27 × 10–10. Calculate the degree of ionisation, of aniline in the solution. Also calculate the, ionisation constant of the conjugate acid of, aniline., , Long Answer Type Questions (LA), 36. (i) Point out the differences between ionic, product and solubility product., (ii) The solubility of AgCl in water at 298 K is, 1.06 × 10–5 mole per litre. Calculate its solubility, product at this temperature ., 37. (i) Derive the relationship between pKw,, pH and pOH starting from ionisation constant of, water, Kw. What is the numerical value of pKw at, 298 K?, (ii) If 0.561 g of KOH is dissolved in water to, give 200 mL of solution at 298 K. Calculate the, concentrations of K+ and OH–. What is its pH?, (Give atomic masses K = 39, O = 16, H = 1), 38. (i) What is common ion effect?, (ii) Write the Ksp expressions for Ag2CrO4 and, zirconium phosphate., (iii) Calculate the pH of 0.005 M HCl solution., , 39. At 473 K, equilibrium constant, Kc, for decomposition of PCl5 is 8.3 × 10–3. If, decomposition is depicted as :, PCl5(g), , PCl3(g) + Cl2(g); DrH° = 124.0 kJ mol–1, , (a) Write an expression for Kc for the reaction., (b) What is the value of Kc for the reverse, reaction at same temperature?, (c) What would be the effect on Kc if, (i) the pressure is increased, (ii) the temperature is increased?, 40. One mole of N2 and 3 moles of PCl5 are, placed in a 100 litre vessel heated to 227°C. The, equilibrium pressure is 2.05 atm. Assuming ideal, behaviour, calculate the degree of dissociation for, PCl5 and Kp for the reaction, PCl5(g), , PCl3(g) + Cl2(g)

Page 15 : CH3COO– + H+, , 22. (c) : Given, CH3COOH, , [CH3COO − ][H+ ], K1 =, = 1.5 × 10 −5, [CH3COOH], HCN, K2 =, , H+ + CN–, [CN− ][H+ ], = 4.5 × 10 −10, [HCN], , CN– + CH3COOH, HCN + CH3COO–, −, [HCN][CH3COO ], K=, [CN− ][CH3COOH], −5, , K, 1.5 × 10, K= 1=, ≈ 0.3 × 105 or K = 3 × 104, K 2 4.5 × 10 −10, 23. (a) : AX3, , A3+ + 3X –, S, , 3S, 3, , 10–4 M, , \, , –, , [OH ] = 10 M, pOH = –log10 [OH–] = –log10(10–4) = 4, \, pH = 14 – 4 = 10, 31. (d) : pH = 5 ⇒ [H+] = 10–5, pH = 7 ⇒ [H+] = 10–7, \, Acidic strength decreases 102 or 100 times., 32. (d), 33. (c) : Dng = np – nr = 1 −, , −1, . Hence Kp = Kc(RT)–1/2, 2, 1, 1, =, =, RT, (RT )1/2, , Kp, , 24. (a) : Reversible reaction never goes to completion, rather, attains an equilibrium point., , 34. (a), 35. (b) :, , 1, SO3(g), 25. (c) : For the reaction, SO2( g ) + O2( g ), 2, Using formula, Kp = Kc(RT)Dng, where, Dng = no. of products(g) – no. of reactants(g), , Initial moles, , 28. (b) : DG° = –2.303RT logKsp, 63.3 × 103 J = – 2.303 × 8.314 × 298 logKsp, 63.3 × 103 J = –5705.84 logKsp, 63.3 × 103, = −11.09, 5705.84, Ksp = antilog (–11.09) = 8.128 × 10–12, log K sp = −, , 29. (c) : K = [B(g)]2[C(g)]3 = x2 y3., If [C(g)] is doubled i.e. = 2y, suppose [B(g)] is z, then, 1, 1, 1, K = z 2 (2y )3 = x 2y 3 or z 2 = x 2or z =, x=, x, 8, 8, 2 2, , 3, 2, , Dng =, Kc, , 27. (d) : (a) is exact neutralisation. Hence pH = 7, (b) After neutralisation, M/10 HCl left = 10 mL, Total volume = 100 mL, Dilution = 10 times. \ [H+] = 10–2 or pH = 2, (c) After neutralisation, M/10 NaOH left = 80 mL, Total volume = 100 mL, pH > 7, (d) After neutralisation, M/10 HCl left = 50 mL, Total volume = 100 mL, Dilution = 2 times, 1, 10 −1, ∴ [H+ ] =, =, M or pH = 1.3, 2 × 10, 2, , 10–4 M, , –4, , \, , Ksp = [A3+] [X –]3 = (S)·(3S) = 27S4, , 1,  1, = 1 − 1 +  = − = x,  2, 2, 26. (a) : BF3 is Lewis acid (electron pair acceptor)., , K+ + OH–, , 30. (d) : KOH, ,   , , Moles at eqm.  , , PCl5, , 1 , 0.5 , , PCl3 + Cl2, 0 , , 0, , 0.5 , , 0.5, , Total moles at equilibrium = 1.5, 0.5, 0.5, 0.5, P, P, P, Partial pressures, 1.5, 1.5, 1.5, (where P is the total pressure.),  0.5   0.5 ,  P   P , pPCl3 × pCl2, 1.5, Kp =, or 1.6 = 1.5, pPCl,  0.5 , 5,  P , 1.5, 1, or, P = 1.6 (Given) or P = 4.8 atm, 3, 36. (c) : Kc = [x] [3x]3...(i), 1, When concentration of OH– ions is decreased by times,, 4, 3, 3x, K c = [x ′]   , ...(ii), 4, 3,  3x , 3, Equating eq. (i) and (ii), x × (3x ) = x ′   ⇒ 64x = x′,  4, 2XY4(g), 37. (d) : X2(g) + 4Y2(g), Dng = –ve and DH = –ve, The reaction is favoured in forward direction at low, temperature and high pressure, 38. (d) : Kp and Kc are related by the equation,, Kp = Kc(RT)Dng where, Dng = difference in the number of, moles of products and reactants in the gaseous state., For 2C(s) + O2(g), 2CO2(g), Dng= 2 – (1) = 1 ≠ 0, 39. (d) : Acid-base titration :, NH4Cl, HCl + NH4OH, 10 mmol  , , \, , 20 mmol, , HCl is the limiting reagent.

Page 16 : Solution contains NH4OH (weak base) and NH4Cl (salt of, strong acid and weak base). Therefore, a basic buffer will be, formed., , 50. (b) :, , 2HI(g), , Initial moles, At eqm., , 2, 2(1 – a), , 40. (b), , Total number of moles = 2 (1 – a) + a + a = 2, , 41. (a) : On increasing the pressure and decreasing the, temperature, equilibrium will shift in forward direction., , 51. (a) : Since HCl is a strong acid therefore, its pH should, be less than 7. In this case we should also consider the [H+], which comes from water i.e., [H+] = 10–7 M from H2O + [Acid], = 10–7 + 10–7 = 2 × 10–7, pH = –log [H+] = – log (2 × 10–7) = 6.69, , 42. (b) : An increase in pressure applied to a system, at equilibrium, favours the reaction in the direction that, produces smaller no. of gaseous moles. Thus, only in the, N2(g) + 3H2(g), reaction, 2NH3(g), there are smaller no. of gaseous moles on left hand side., 43. (d) : As the forward reaction is exothermic and leads, to lowering of pressure (produces lesser number of gaseous, moles) hence, according to Le Chatelier’s principle, at high, pressure and low temperature, the given reversible reaction, will shift in forward direction to form more product., 44. (b) : The formation of ammonia is favoured by high, pressure which would shift the reaction in the forward, direction. The reaction shifts in forward direction at low, temperature and more product formation occurs., Fe acts as a positive catalyst and also shifts the reaction in, forward reaction., 45. (a) : Value of equilibrium constant is independent of, initial concentration of reactants., 46. (b) : HA  H+ + A–, Ka =, , [H+ ][A − ], , pH = –log[H+], [HA ], , [H+] = 10–5 ; [H+] = [A–], 10 −5 × 10 −5, Ka, = 2 × 10 −8 ; α =, = 2 × 10 −3, 0.005, C, Percentage ionization = 0.2, Ka =, , 54. (a), 55. (c) : H3PO4  H+ + H2PO4− ;, H2PO4−, , K a2, , 56. (c) : Volume is decreased, reaction proceeds in the, forward direction because decrease in volume increases the, number of moles per unit volume. In order to undo the effect, the reaction shifts towards the direction which produces, lesser number of moles. Concentration increases, Qc must, be lesser than Kc . Suppose volume is decreased to half,, concentration becomes 2 times., [CH4 ] [H2O], , Qc =, , 49. (a), , ;, , K a1, , Hence, K a2 < K a1 and pK a2 > pK a1 ., , 1, Now, pH = − [log K w − log K a + logC ), 2, , 48. (d) : 10–8 M HCl = 10–8 M H+, Also from water, [H+] = 10–7 M, \ Total [H+] = 10–7 + 0.10 × 10–7 = 1.1 × 10–7 M, , + HPO24−, , H+ and H2PO4− are having stronger ion-pair attraction., , after decreasing Q c =, , ⇒ Ka = antilog (– 8.82) = 1.5 × 10–9, , +, , H, , pH = –log[H+] = –log(3.6 × 10–4) = 3.44, , ⇒ 6.88 = 14 + logKa + 1.70 ⇒ logKa = –8.82, , 0, a, , 53. (a) : If Dn = + ve, Kp >Kc, Dn = –ve, Kp < Kc, Dn = 0, Kp = Kc, , Kc =, , 1, 3.44 = − [–14 – logKa + log(2 × 10–2)], 2, , 0, a, , 52. (b) : The value of Kw at 25ºC is about, 1.0 × 10–14 mol 2 dm–6. Since in pure water, the, concentration of H+ and OH– ions must be equal to one, another i.e., 1.0 × 10–7 mol dm–3., , 47. (c) : Pyridinium hydrochloride is a salt of weak base and, strong acid., , ⇒, , H2(g) + I2(g), , [CO] [H2 ]3, , [2CH4 ] · [2H2O], [2CO] [2H2 ]3, , Kc, 4, , 57. (c) : Kp ≠ Kc for all reactions., Kp = Kc (RT)Dn, Dn = number of moles of gaseous products – number, of moles of gaseous reactants in the balanced chemical, equation., So, if for a reaction Dn = 0, then Kp = Kc., 58. (c) : Equilibrium constant is temperature dependent., 59. (c) : Ksp remains constant at a particular temperature.

Page 17 : 60. (b) : A catalyst increases the rate of forward and, backward reactions by same factor hence, does not change, the equilibrium constant., , SUBJECTIVE TYPE QUESTIONS, 1., , The reaction is :, N2(g) + 3H2(g), Qc =, , =, , 2NH3(g), , [NH3 ]2, , [N2 ][H2 ]3,  8.13, , mol L−1 , , , 20, , 2, , ,  1.57,   1.92, mol L−1  , mol L−1 , , ,   20, 20, , 3, , = 2.38 × 103, , As Qc ≠ Kc , the reaction mixture is not in equilibrium., As Qc > Kc , the net reaction will be in the backward direction., 2., , For the concentration of pure solid or pure liquid,, Moles of the substance, Molar conc. =, Volume of the substance, Mass/Molar mass, 1, Mass, =, =, ×, Volume, Volume Molar mass, Density, =, Molar mass, Since density of pure solid or liquid is constant at constant, temperature and molar mass is also constant therefore, their, molar concentrations are constant and are included in the, equilibrium constant., 3. Conjugate acids of given bases are H2O, ROH, CH3COOH,, HCl., Their acidic strength is in the order, HCl > CH3COOH > H2O > ROH, Hence, basic strength is in the order, RO– > OH– > CH3COO– > Cl–, –, , –, , 4. (a) : OH : OH is a Lewis base because it can donate, lone pair of electrons., (b) F– : F– is a Lewis base because it can donate lone pair of, electrons., (c) H+ : H+ is a Lewis acid because it can accept lone pair of, electrons., (d) BCl3 : BCl3 is a Lewis acid because it is electron deficient, and can accept a lone pair of electrons., 5., , pH of solution A = 6, [H+] = 10–6 mol L–1, pH of solution B = 4, [H+] = 10–4 mol L–1, On mixing one litre of each solution, Total volume = 1 L + 1 L = 2 L, , Total amount of H+ in 2 L solution formed by mixing solutions, A and B = 10–6 + 10–4 mol, 10 −4 (1 + 0.01) 1.01 × 10 −4, =, 2, 2, –5, –1, , = 5 × 10 mol L, pH = – log[H+] = – log(5 × 10–5), , = –log5 – (–5log10) = –log5 + 5, , = 5 – log5 = 5 – 0.6990 = 4.3010 = 4.3, Total [H+] =, , 6. Characteristics of chemical equilibrium are as follows :, (i) Chemical equilibrium is dynamic in nature., (ii) A catalyst does not alter the state of equilibrium., I2(g) + Cl2(g); Kc = 0.14, 7. , 2ICl(g), Initial molar conc., Eqm. molar conc., , 0.78, 0.78 – 2x, , 0, x, , 0, x, , Applying law of chemical equilibrium,, [I ][Cl ], x ⋅x, K c = 2 22 ⇒ 0.14 =, [ICl], (0.78 − 2x )2, x2 = 0.14(0.78 – 2x)2, x, = 0.14 = 0.374, or, 0.78 − 2x, or x = 0.292 – 0.748x, or 1.748x = 0.292 or x = 0.167, Hence at equilibrium, [I2] = [Cl2] = 0.167 M, [ICl] = 0.78 – 2 × 0.167 = 0.446 M, B, 8. A, At the stage of half completion of reaction [A] = [B]., 9. We know that pH = – log[H+], \ [H+] = antilog[–pH] = antilog(–3.76), , = 1.738 × 10–4 M, 10. K c′ =, , 1, 1, = = 0.1, K c 10, , [H ]4 [Fe3O4(s ) ] [H2( g ) ]4, 11. K c = 2( g ), =, [Fe(s ) ]3[H2O( g ) ]4 [H2O( g ) ]4, (, 12. K =, , concentration of solids is taken as unity), , [C ], [A ]3[B ]2, , 13. Equilibrium is said to be homogeneous if reactants and, products are in same phase, e.g.,, H2(g) + I2(g), 2HI(g), 2SO2(g) + O2(g), , 2SO3(g), , Equilibrium is said to be heterogeneous if reactants and, products are in different phases, e.g.,, CO(g) +H2(g), C(s) + H2O(g), CaCO3(s), , CaO(s) + CO2(g)

Page 18 : 14. H2S, H+ + HS– ; Ka, 1, –, HS, H+ + S2– ; Ka, 2, H2S, 2H+ + S2– ; Ka, 3, The correct relationship between Ka , Ka and Ka is, 1, 2, 3, Ka = Ka × Ka ., 3, , 1, , K sp of CuCl, , K, 16. K b = w, Ka, For F –, Kb =, , 10, , −14, , 10 −14, , For HCOO–, Kb =, For CN–, Kb =, , 1.8 × 10 −4, 10, , −14, , 4.8 × 10, , −9, , = 5.6 × 10–11, , 2.303 × 8.314 JK −1 mol−1 × 298 K, , x, , –, , y = 0.1, , y, , y, , :, , 2N2O(g), , 0.482, , 0.933, , 0, , (0.482 – x), , (0.933 – x/2), , x, , 0.482 − x, 10, , 0.933 − ( x /2), 10, , x, 10, , Kc =, , (0.1x )2, , = 2.0 × 10 −37 (given), , 2, , 0.001 M, , 0.001 M, 2+, , Cu, , 0.001 M, , + 2ClO3–, , After mixing, [IO3–] = [NaIO3] = 0.001 M, [Cu2+] = [Cu(IO3)2] = 0.001 M, Solubility equilibrium for copper iodate may be written as,, –, Cu(IO3)2(s), Cu2+, (aq) + 2IO3 (aq), Ionic product of copper iodate, = [Cu2+] [IO3–]2 = (0.001) (0.001)2 = 1 × 10–9, Since ionic product (1 × 10–9) is less than Ksp(7.4 × 10–8),, therefore, no precipitation will take place., = 6.314, , 22. Let the concentration of AgCl be x mol/litre and that of, CuCl be y mol/litre, Ag+ + Cl–, AgCl, x, , Molar conc, , O2(g), , +, , \, , 21. Higher the boiling point, lesser the vapour pressure hence, the order of V.P. is, water < acetone < ether, , Cu+ + Cl, , Moles at eqm. :, , Cu(ClO3)2, , K = 2.06 × 106, , CuCl, , Initial moles :, , 0.001 M, , ∆G °, 2.303 RT, −35 × 103 J mol−1, , 2N2(g), , (0.0482) (0.0933), On solving this we get, x ≈ 6.6 × 10–20, \ [N2O] = 0.1x = 6.6 × 10–21 mol L–1, 24. When equal volumes of sodium iodate and copper, chlorate are mixed, the molar concentrations of both the, solutes would be reduce to half i.e., 0.001 M., NaIO3, Na+ + IO3–, , = 2.08 × 10–6, , 19. Bronsted acid, Conjugate base, , HF, F–, , H2SO4, HSO4–, –, , HCO3, CO32–, 1, 20. NO( g ) + O2( g )  NO2( g ), 2, 1, ∆G ° = ∆ f G °(NO2 ) − [∆ f G °(NO) + ∆ f G °(O2 )], 2, 1, = 52.0 – 87.0 – × 0 = – 35 kJ mol–1, 2, , =−, , 23. , , \, , 17. Qc > Kc , for reverse reaction., 18. Dn = np – nr = 2 – 0 = 2, , Now, log K = −, , x, 1.6 × 10 −10 x, =, ⇒, y, 1.0 × 10 −6 0.1, , As Kc = 2.0 × 10–37 is very small, this means that the amount, of N2 and O2 reacted (x) is very small. Hence, at equilibrium,, we have, [N2] = 0.0482 mol L–1,[O2] = 0.0933 mol L–1, [N2O] = 0.1x mol L–1, , = 1.47 × 10–11, , 6.8 × 10 −4, , =, , ⇒ x = 1.6 × 10–5 mol/litre, , Taking active masses of solids to be unity,, Kc = [CO2(g)], Kp = pCO2, , or, , K sp of AgCl, , 2, , 15. CaCO3(s )  CaO(s ) + CO2( g ), [CaO(s ) ][CO2( g ) ], Kc =, [CaCO3(s ) ], , , , Ksp of AgCl = [Ag+] [Cl–] = x(x + y)...(i), Ksp of CuCl = [Cu+][Cl–] = y(x + y)...(ii), On solving (i) and (ii), we get, , 25. When an equilibrium is subjected to any kind of stress, (change in concentration, temperature or pressure) it shifts in, a direction so as to undo the effect of stress., (i) When H2 is added, the rate of forward reaction will, increase., (ii) Addition of CH3OH will lead to increase in rate of, backward reaction, 26. DG° = – RT lnKc, –13.8 × 103 = 8.314 × 298 × ln Kc, \ ln Kc = – 5.569, Kc = e–5.569 ⇒ Kc = 3.81 × 10–3

Page 19 : CH3OH(g); DHr° = – 92 kJ/mol, 27. CO(g) + 2H2(g), (i) When pressure is doubled, equilibrium will shift in the, direction where pressure decreases i.e., forward direction., (ii) As this is an exothermic reaction, so the equilibrium will, shift in backward direction when the temperature is doubled., 28. pH of a buffer changes with temperature because, concentration of H+ ions increases, thus pH decreases with, increase of temperature., 29. Q c =, , [NH3 ]2, , [N2 ][H2 ]3, , =, , (0.5)2, (3.0)(2.0)3, , = 0.0104, , Given, Kc = 0.061, As Qc ≠ Kc , reaction is not in equilibrium., As Qc < Kc , reaction will proceed in the forward direction., 30. (i) Lewis acids are those which can accept a pair of, electrons or negatively charged ions e.g., BCl3. Lewis bases, can donate a pair of electrons or negatively charged ions e.g.,, NH3., (ii) Given, Ksp of Ni(OH)2 = 2 × 10–15, Ksp of AgCN = 6 × 10–17, i.e., Ni(OH)2, Ni2+ + 2OH–, , 34. On mixing 50 mL of 0.01 M NaCl and 50 mL of 0.01 M, AgNO3 the total volume becomes 100 mL., Therefore,, 0.01 × 50, Conc. of NaCl in 100 mL =, = 0.005 M, 100, 0.01 × 50, Conc. of AgNO3 in 100 mL =, = 0.005 M, 100, Now NaCl(aq), Na+(aq) + Cl–(aq), and AgNO3(aq), , [Cl–] = [NaCl] = 0.005 M, , [Ag+] = [AgNO3] = 0.005 M, \ Ionic product of [Ag+] [Cl–] = 0.005 × 0.005, , = 2.5 × 10–5, Since, ionic product is greater than its solubility product,, precipitation will occur., C6H5NH3+ + OH –, 35. C6H5NH2 + H2O, Kb =, [OH –] =, , Ag+ + CN–, , Ksp = [Ag+] [CN–] = 6 × 10–17, , Suppose [Ag+] = [CN–] = s1 and [Ni2+] = s2, Hence, [OH–] = 2s2, , Since, s12 = 6 × 10–17 ⇒ s1 =, , 60 × 10 −18, , ⇒ s1 = 7.7 × 10–9 M, Since, s2 × (2s2)2 = 2 × 10–15 ⇒ 4s23 = 2 × 10–15, , ⇒, , \, , pH = 14 – 6.18 = 7.82, , Initial :, At. eqm. :, , \, , 3, s2 = 500 × 10 −18 = 7.9 × 10–6 M, –6, , s2 = 7.9 × 10 M, Since s2 > s1, therefore Ni(OH)2 is more soluble than AgCN., 31. (i), (iii), (iv) would result in the formation of a buffer, solution., 32. Axp +B yq −  xA p + + yB q −, S, , xS, , yS, , Ksp = [Ap+]x [Bq–]y = (xS)x (yS)y = x x · y y · S(x + y), , S(x + y) = Ksp/x x · y y, 33. N2(g) + 3H2(g), , 2NH3(g), , According to Le Chatelier’s principle, at constant temperature,, the equilibrium composition will change but K will remain, same., , K b [C6H5NH2 ], , pOH = –log(6.534 × 10–7) = 6.18, , C6H5NH2 + H2O, C, C – Ca, , C6H5NH3+ + OH–, 0, Ca, , 0, Ca, , Kb =, , C α.C α, C α2, =, = C α2 (Q 1 >>> a), C (1 − α) (1 − α), , α=, , Kb, 4.27 × 10 −10, =, = 6.53 × 10–4, C, 10 −3, , ⇒ s23 = 0.5 × 10–15 ⇒ s23 = 5 × 10–16, ⇒ s23 = 500 × 10–18, , [C6H5NH3+ ][OH− ], [OH− ]2, =, [C6H5NH2 ], [C6H5NH2 ], , = (4.27 × 10 −10 )(10 −3 ), = 6.534 × 10–7 M, , Ksp = [Ni2+] [OH–]2 = 2 × 10–15, AgCN, , Ag+(aq) + NO3–(aq), , pKb = –log(4.27 × 10–10) = 9.37, pK a + pK b = 14, (for a pair of conjugate acid and base), \ pKa = 14 – 9.37 = 4.63, i.e., –log Ka = 4.63 or, log Ka = –4.63, or, Ka = antilog (–4.63) = 2.34 × 10–5, 36. (i) The term ionic product has a broad meaning since, it is applicable to all types of solution, may be unsaturated, or saturated. On the other hand, the solubility product has, restricted meaning since it applies only to a saturated solution, in which there exists a dynamic equilibrium between the, undissolved salt and the ions present in solution. Thus, the, solubility product is, in fact the ionic product for a saturated, solution.

Page 20 : The solubility product of a salt is constant at constant, temperature whereas ionic product depends upon the, concentrations of ions in the solution., (ii) The solubility equilibrium in the saturated solution is, Ag+(aq) + Cl–(aq), AgCl(s), The solubility of AgCl is 1.06 × 10–5 mole per litre., [Ag+(aq)] = 1.06 × 10–5 mol L–1, , –1, , –5, , 38. (i) Common ion effect can be defined as the suppression, of the degree of ionisation of a weak electrolyte by the, addition of a strong electrolyte having an ion common with, the weak electrolyte., 2Ag+ + CrO42–, (ii) Ag2CrO4, s, , 2s, , Ksp = [Ag+]2 [CrO42–] = (2s)2 × s = 4 s3, , Ksp = [Zr ], , 3Zr4+ + 4PO43–, , 3s, [PO43–]4, , 4s, 3, , [PCl3 ][Cl2 ], = 8.3 × 10–3, [PCl5 ], , = (3s) × (4s)4 = 6912 s7, , 1, 8.3 × 10 −3, , = 120.48, , (i) When pressure increases, Kc remains unchanged., (ii) , Kc increases with increase in temperature because, the reaction is endothermic.,  PCl3(g) + Cl2(g), 40., PCl5(g), t=0 3, 0 0, 3(1 – x), 3x 3x, teqm, where x = degree of dissociation of PCl5(g) at 227ºC, Total no. of moles at equilibrium = [(3 – 3x) + 3x + 3x +, 1 (molar concentration of N2)] = 4 + 3x., From gas equation : PV = nRT, (4 + 3x ), Total pressure (P ) =, × 0.082 × 500 atm, 100,     = 2.05 atm (given), Thus, 4 + 3x = 5.0, or 3x = 1.0, 1 .0, Hence, x =, = 0.33, 3, (c), , 37. (i) Sorensen (1909) defined pH of a solution as negative, logarithm of the hydrogen ion concentration of the solution., 1, +, +, −pH, Thus, pH = − log[H ] = log + or [H ] = 10, [H ], Likewise, pOH of a solution, 1, pOH = − log[OH− ] = log, or [OH–] = 10–pOH, [OH− ], Kw = [H+][OH–] = Ionic product of water = 10–14 (mol/L)2, pKw = –logKw, pKw = pH + pOH = 14, 0.561 1000, ×, = 0.05 M, (ii) KOH =, 56, 200, As, KOH → K+ + OH–, [K+] = [OH–] = 0.05 M, [H+] = Kw /OH– = 10–14/0.05,   = 10–14/(5 × 10–2) = 2.0 × 10–13 m, pH = –log 2 × 10–13 = 13 – log 2 = 12.69, , 4+ 3, , Kc =, , –1, , = (1.06 × 10 mol L ) × (1.06 × 10 mol L ), = 1.12 × 10–10 mol2 L–2, , Zr3(PO4)4, , (a), , =, , Ksp = [Ag+(aq)] [Cl–(aq)], , , , PCl3(g) + Cl2(g); DrH° = 124.0 kJ mol–1, , 39. PCl5(g), , (b) Kc for the reverse reaction, 1, =, K c for the forward reaction, , [Cl–(aq)] = 1.06 × 10–5 mol L–1, –5, , (iii) pH = –log [H3O+] = –log (0.005) = 2.30, , Percentage dissociation = 33%, pPCl5 =, , 3 − 3x, (3 − 0.99) × 2.05 2.01 × 2.05, = 0.825, ×P =, =, 4 + 3x, 4 + 0.33 × 3, 4.99, , pPCl3 =, , 3x, 3 × 0.33 × 2.05 2.03, ×P =, =, = 0.4067, 4 + 3x, 4.99, 4.99, , 3x, 3 × 0.33 × 2.05 2.03, ×P =, =, = 0.4067, 4 + 3x, 4.99, 4.99, pPCl3 × pCl2 0.4067 × 0.4067, = 0.200 atm., Hence, K p =, =, pPCl5, 0.825, pCl2 =, , 