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CHEMISTRY CHEMICAL EQUILIBRIUM

CONCEPTUAL QUESTIONS

SINGLE CHOICE QUESTIONS

1. The vapour density of a mixture containing NO2 and N2O4 is 27.6. The mole fraction of NO2

in the mixture will be(A) 0.2 (B) 0.4(C) 0.6 (D) 0.8

2. The values of Kp are 610 3atm and 4 310 atm at 298 K and 323 K respectively for thereaction

4 2 4 2CuSO .3H O(s) CuSO (s) 3H O(g)

0H for the reaction is

(A) 7.7 KJ (B) 300 KJ

(C) 147.4KJ (D) 14.7KJ

3. Which of the following will shifts the equilibrium in the forward direction for the followingequilibrium

2 2N (g) O (g) 2NO (g) 0H 181KJ

(A) Increase of pressure (B) Increase in the concentration of NO(C) Decrease in pressure (D) Increase in temperature

4. At certain temperature compound AB2(g) dissociates according to the reaction

2 22AB (g) 2AB(g) B (g)

With degree of dissociation , which is small compared with unity. The expression of Kp, interms of and initial pressure P is:

(A) 3

P2

(B) 2P

3

(C) 3

P3

(D) 2P

2

5. An aqueous solution of volume 500 ml contains the reaction

2Ag+ (aq.) + Cu(s) Cu2+ (aq.) + 2Ag(s) in equilibrium with [Cu2+(aq.)] = xM. Now500 ml of water is further added. On reset of above equilibrium [Cu2+(aq.)] will be(A) xM (B) 2xM(C) Between xM and x/2 M (D) less than x/2 M

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CHEMICAL EQUILIBRIUM CHEMISTRY

6. Kc for the synthesis of HI (g) from H2(g) and I2(g) is 49. The degree of dissociation of HI is(A) 0.10 (B) 0.14(C) 0.18 (D) 0.22

STRAIGHT OBJECTIVE TYPE QUESTIONS

LEVEL - I

Equilibrium constant Kp or Kc :

1. In the reaction, N2(g) + 3H2(g) 2NH3 (g)

the value of the equilibrium constant depends on(A) volume of the reaction vessel(B) total pressure of the system(C) the initial concentration of nitrogen and hydrogen(D) the temperature

2. In the equilibrium reaction involving the dissociation of CaCO3.

CaCO3(s) CaO (s) + CO2 (g) the equilibrium constant is given by

(A) 2

3

CaO CO

CaCO

P ×PP (B)

2

3

COCaO

CaCO

PC

C

(C) 3

CaO

CaCO

PP (D)

2COP

Le-Chatelier’s principle:

3. Consider the reaction, PCl5 (g) PCl3 (g) + Cl2(g) in a closed container at

equilibrium at a fixed temperature. What will be the effect of adding more PCl5 on theequilibrium concentration of Cl2 (g) ?(A) it decreases (B) it increases(C) it remains unaffected (D) it cannot be predicted without the value of Kp.

4. In a system A (s) 2B (g) + 3C (g). If the concentration of C at equilibrium isincreased by a factor 2, it will cause the equilibrium concentration of B to change to:(A) two times of its original value (B) one half of its original value

(C) 2 2 times of its original value (D) 221

times of its original value

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5. In which of the following does the reaction go almost to completion ?(A) K = 106 (B) K = 103

(C) K = 10–6 (D) K = 10–12.

6. 2XY dissociates as 2(g) (g) (g)XY XY Y . When the initial pressure is 600 mm of Hg, the

total pressure at equilibrium developed is 800 mm of Hg. Therefore pressure of Y at equilibriumis(A) 200 (B) 50(C) 100 (D) 150

Equilibrium Constant Kp or Kc:

7. For the equilibrium 2SO2(g) + O2(g) 2SO3(g)the concentration of each substance at

equilibrium is: [SO2] = 0.60M, [O2] = 0.82M and [SO3] = 1.90M, the Kc is:(A) 1.2229 mol L-1 (B) 12.2 mol L-1

(C) 6.1145 mol L-1 (D) 18.22 mol L-1

8. Find out the value of KC for each of the following equilibria from the value of KP :

(i) 2NOCl (g) 2NO (g) + Cl2 (g); KP= 1.8 x 10-2 at 500 K

(ii) CaCO3 (s) CaO(s) + CO2(g); Kp= 167 at 1073 K

the value of Kc are respectively(A) 4.4 × 10-4 & 1.90 (B) 8.8 ×10-4 & 3.8(C) 4.4 × 104 & 1.90 (D) 8.8 ×104 & 3.8

9. For the following equilibrium, KC = 6.3 x 1014 at 1000 K

NO(g) + O3 (g) NO2 (g) + O2(g)

Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions.What isKC, for the reverse reaction?(A) 15.9 × 10-15 (B) 1.59 × 10-15

(C) 5 × 10-15 (D) 9 × 10-15

10. If a mixture of 0.482 mol N2 and 0.933 mol of O2 is placed in a 10 L reaction vessel and allowedto formN2O at a temperature for which KC = 2.0 ×10-37. The composition of equilibrium mixture are(A) N2 = 0.482 mol/ L, O2 = 0.0933 mol /L, N2O = 6.6 × 10-21 mol/L(B) N2 = 0.482 mol/ L, O2 = 0.933 mol /L, N2O = 6.6 × 10-21 mol/L(C) N2 = 0.0482 mol/ L, O2 = 0.0933 mol /L, N2O = 6.6 × 10-21 mol/L(D) N2 = 0.0482 mol/ L, O2 = 0.0931 mol /L, N2O = 8.8 × 10-21 mol/L

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11. When 0.087 mol of NO and 0.0437 mol of Br2 are mixed in a closed container at constanttemperature, 0.0518 mol of NOBr is obtained at equilibrium. The equilibrium amount of NO andBr2 are:(A) NO = 0.352 mol, Br2 = 0.0178 mol (B) NO = 0.0352 mol, Br2 = 0.178 mol(C) NO = 0.352 mol, Br2 = 0.0178 mol (D) NO = 0.0352 mol, Br2 = 0.0178 mol

12. At 450K, KP= 2.0 ×1010/bar for the given reaction at equilibrium.

2SO2(g) + O2(g) 2SO3 (g)

What is K, at this temperature ?(A) 7.4 × 1011 L mol-1 (B) 7.4 × 10-11 L mol-1

(C) 3.7 × 10-11 L mol-1 (D) 3.7 × 1011 L mol-1

13. A sample of HI(g) is placed in flask at a pressure of 0.2 atm. At equilibrium the partial pressureof HI(g) is 0.04 atm. What is Kp for the given equilibrium ?

2HI(g) H2(g) + I2(g)

(A) 6 (B) 16(C) 4 (D) 2

Le Chatelier Principle14. A mixture of 1.57 mol of N2, 1.92 mol of H, 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?(A) not at equilibrium, forward shift (B) not at equilibrium, backward shift(C) cannot be predicted (D) in equilibrium

Equilibrium constant

15. The equilibrium constant expression for a gas reaction is,4 5

3 2C 4 6

2

[NH ] [O ]K =[NO] [H O]

The balanced chemical equation corresponding to this expression is.

(A) 4NO(g) + 6H2O (g) 4NH3(g) + 5O2(g)

(B) 4NH3(g) + 5O2(g) 4NO(g) + 6H2O (g)

(C) both of above(D) none of these

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16. One mole of H2O and one mole of CO are taken in 10 L vessel and heated to 725 K. Atequilibrium 40% of water (by mass) reacts with CO according to the equation,

H2O (g) + CO (g) H2 (g) + CO2 (g)

what is the value of equilibrium constant (Kc) for the reaction.(A) 44 (B) 4.4(C) 0.44 (D) 2.22

17. At 700 K, equilibrium constant for the reaction: H2 (g) + I2(g) 2HI (g) is 64. If 0.5

mol L-1 of HI(g) is present at equilibrium at 700 K, what are the concentration of H2(g) and I2(g)assuming that we initially started with HI(g) and allowed it to reach equilibrium at 700 K?(A) 0.0625 M (B) 0.034 M(C) 0.136 M (D) 0.68 M

18. What is the equilibrium concentration of each of the substances of RHS in the equilibrium whenthe initial concentration of ICl was 0.78 M ?

2ICl(g) I2 (g) + Cl2 (g); KC = 0.25

(A) 0.195 M (B) 0.42 M(C) 4.2 M (D) 2.1 M

19. KP = 0.04 atm at 899 K for the equilibrium shown below. What is the equilibrium concentrationof C2H6 when it is placed in a flask at 4.0 atm pressure and allowed to come to equilibrium?

C2H6 (g) C2H4 (g) + H2 (g)

(A) 7.24 atm (B) 3.62 atm(C) 1 atm (D) 1.5 atm

20. A sample of pure PCl5 was introduced into an evacuated vessel at 473 K. After equilibrium wasattained, concentration of PC15 was found to be 0.5 10-10 mol L-1. If value of KC is 8.3 10-

3, what are the concentrations of PC13 and C12 at equilibrium?

PCl5(g) PCl3(g) + Cl2(g)

(A) 0.08 M (B) 0.04M(C) 0.2M (D) 0.02 M

21. One of the reaction that takes place in producing steel from iron ore is the reduction of iron(II)

oxide by carbon monoxide to give iron metal and CO2.FeO (s) + CO (g) Fe (s) + CO2

(g); KP= 0.265 atm at 1050K. What are the equilibrium partial pressures of CO and CO2

respectively at 1050 K if the initial partial pressures are: pco= 1.4 atm and 2COp = 0.80 atm?

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(A) ]CO[P = 1.739 atm and 2COP = 0.461 atm (B) ]CO[P = 17.39 atm and

2COP = 0.461 atm

(C) ]CO[P = 1.79 atm and 2COP = 0.46 atm (D) ]CO[P = 2.739 atm and

2COP = 1.461 atm

Le-Chatelier’s principle:22. 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 and if not in which directiondoes the reaction tend to proceed to reach equilibrium?(A) at equilibrium (B) not at equilibrium, backward shift(C) not at equilibrium, forward shift (D) can not be predicted

23. Bromine monochloride, BrCl decomposes into bromine and chlorine and reaches the equilibrium:

2BrCl (g) Br2 (g) + Cl2 (g)

for which K = 32 at 500 K. If initially pure BrCl is present at a concentration of 3.3 10-3 mol L-

1, what is its molar concentration in the mixture at equilibrium?(A) 3 × 10-4 (B) 1 × 10-4

(C) 1.5 × 10-4 (D) 6 × 10-4

24. What are the values of 0G and the equilibrium constant for the formation of NO2 from NOand O2 at 298K. Given log (1.365 × 106) = 6.14

NO(g) + 12 O2 (g) NO2(g)

where f0G = 52.0 kJ/mol

f0G 87.0 / kJ mol

f0

2G ( ) 0 / O kJ mol

(A) 35 kJ mol-1 and 1.365 × 106 (B) -35 kJ mol-1 and 1.365 × 10-6

(C) -350kJ mol-1 and 1.3 × 106 (D) -35 kJ mol-1 and 1.365 × 106

25. The equilibrium constant for the following reaction is 1.6 105 at 1024K

H2(g) + Br2(g) 2HBr(g)

Find the equilibrium pressure of H2 gas if 10.0 bar of HBr is introduced into a sealed container at1024K.(A) 2.8 × 103 (B) 5.2 × 10-3

(C) 2.5 × 10-3 (D) 2 × 10-3

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LEVEL - II

Equilibrium Constant

26. X2 + X– X3– (x = iodine) This reaction is set up in aqueous medium. We start with 1 mol

of X2 and 0.5 mol of X– in 1L flask. After equilibrium is reached, excess of AgNO3 gave 0.25 molof yellow ppt. equilibrium constant is(A) 1.33 (B) 2.66(C) 2.00 (D) 3.00

Le Chatelier's Principle27. A reaction mixture containing H2, N2 and NH3 has partial pressures 2 atm, 1 atm and 3 atm

respectively at 725 K. If the value of Kp for the reaction, N2(g) + 3H2(g) 2NH3(g) is 4.28 ×

10–5 atm–2 at 725 K, in which direction the net reaction will go ?(A) Forward (B) Backward(C) No net reaction (D) Direction of reaction cannot be predicted.

Equilibrium Constant28. 2.0 mol of PCl5 were introduced in a vessel of 5.0 L capacity at a particular temperature. At

equilibrium, PCl5 was found to be 35% dissociated into PCl3 and Cl2. The value of Kc for thereaction is(A) 1.89 (B) 0.377(C) 0.75 (D) 0.075.

29. For the reaction: 2HI(g) H2(g) + I2(g)

The degree of dissociation (A) of HI(g) is related to equilibrium constant, Kp by the expression

(A) p1 2 K2

(B) p1 2K

2

(C) p

p

2K1 2K (D)

p

p

2 2K

1 2 K

30. Concentration of pure solid and liquid is not included in the expression of equilibrium constantbecause:(A) Solid and liquid concentrations are independent of their quantities(B) Solid and liquids react slowly(C) Solid and liquids at equilibrium do not interact with gaseous phase(D) The molecules of solids and liquids cannot migrate to the gaseous phase

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31. In equilibrium reaction, x moles of the reactant A decompose to give 1 mole each of C and D. Ifthe fraction of A decomposed at equilibrium is independent of initial concentration, then the valueof x would be(A) 1 (B) 2(C) 3 (D) 4

32. In the reversible reaction A + B C + D, the concentration of each C and D at equilibrium

was 0.8 mole/litre, then the equilibrium costant Kc will be (if initial concentration of A and B bewill 1 M each)(A) 6.4 (B) 0.64(C) 1.6 (D) 16.0

33. 4 moles of A are mixed with 4 moles of B. At equilibrium for the reaction

A + B C + D, 2 moles of C and D are formed. The equiibrium constant for the reaction

will be(A) 1/4 (B) 1/2(C) 1 (D) 4

34. On a given condition, the equilibrium concentration of HI, H2 are 0.80, 0.10 and 0.10 moe/litre.The

equilibrium constant for the reaction H2 + I2 2HI will be

(A) 64 (B) 12(C) 8 (D) 0.8

35. The equilibrium constant for a reaction A + B C + D is 1×10-2 at 298 K and is 2 at 273K. The chemical process resulting in the formation of C and D is:(A) exothermic (B) endothermic(C) unpredictable (D) there is no relationship between H and K

36. A reversible chemical reaction having two reactants in equilibrium. If the concentration of thereactants are doubled, then the equilibrium constant will(A) Also be doubled (B) Be halved(C) Become one-fourth (D) Remain the same

37. The equilibrium constant in reversible reaction at a given temperature(A) Depends on the initial concentration of the reactants(B) Depends on the concentration of the products at equilibrium(C) Does not depend on the initial concentrations(D) It is not characteristic of the reaction

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38. Pure ammonia is placed in a vessel at temperature where its dissociation constant (A) isappreciable. At equilibrium(A) Kp does not change with pressure(B) (A) does not change with pressure(C) Concentration of NH3 does not change with pressure(D) Concentration of H2 is less than that of N2

39. Partial pressure of A, B, C and D on the basis of gaseous system at equilibrium

A + 2B C + 3D are A = 0.20; B = 0.10; C = 0.30 and D = 0.50 atm. The numerical

value ofequilibrium constant is(A) 11.25 (B) 18.75(C) 5 (D) 3.75

40. 2 moles of PCl5 were heated in a closed vessel of 2 litre capacity. At equilibrium, 40% of PCl5is dissociated into PCl3 and Cl2. The value of equilibrium constant is(A) 0.266 (B) 0.53(C) 2.66 (D) 5.3

41. For the reaction, A(g) + 2B(g) 2C(g), the rate constants for the forward and the reverse

reactions are 1 10–4 and 2.5 10–2 respectively. The value of equilibrium constant, K for thereaction would be(A) 1 10–4 (B) 2.5 10–2 (C) 4 10–3 (D) 2.5 102.

42. At a certain temperature, the equilibrium constant K c is 16 for the reaction,

SO2(g) + NO2(g) SO3(g) + NO(g). If 1.0 mol each of all the four gases is taken in a one

litre container the concentration of NO2 at equilibrium would be(A) 1.6 mol L–1 (B) 0.8 mol L–1

(C) 0.4 mol L–1 (D) 0.6 mol L–1.

43. HI was heated in sealed tube at 400°C till the equilibrium was reached. HI was found to be 22%decomposed. The equilibrium constant for dissociationis(A) 1.99 (B) 0.0199(C) 0.0796 (D) 0.282

44. When sulphur in the form of S8 is heated at 900K, the initial pressure of one atm of S(8)falls by29% at equilibrium. This is because of conversion of some S8 to S2. Find the value of equilibriumconstant for this reaction(A) 1.16 atm3 (B) 0.71 atm3

(C) 2.55 atm3 (D) 5.1 atm3

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45. The following reactions are known to occur in the body

CO2 + H2O H2CO3 H+ + HCO3–1

If CO2 escapes from the system(A) pH will decreases (B) Hydrogen ion concentration will dimnish(C) H2CO3 concentration will remain unchanged(D) The forward reaction will be favoured

46. For which of the following reactions, the degree of dissociation cannot be calculated from thevapour density data

I. 2HI(g) H2(g) + I2(g)

II. 2NH3(g) N2(g) + 3H2(g)

III. 2NO(g) N2(g) + O2(g)

IV. PCl5(g) PCl3(g) + Cl2(g)

(A) I and III (B) II and IV(C) I and II (D) III and IV

47. For the hypothetical reactions, the equilibrium constant (K) values are given :

1 2 3A B; K 2; B C; K 4; C D; K 3

The equilibrium constant (K) for the reaction A D is :

(A) 48 (B) 6(C) 24 (D) 12

48. If, in the reaction 2 4 2N O (g) 2NO (g) , x is that part of N2O4 which dissociates, then the

number of molecules at equilibrium will be :(A) 1 (B) 3(C) (1 + x) (D) (1 + x)2

49. vant' Hoff equation shows the effect of temperature on equilibrium constants Kc and Kp. TheKp varies with temperature according to the relation :

(A) 2

1

p 1 2

p 1 2

K T THlogK 2.303R T T

(B)

2

1

p 2 1

p 1 2


(C) 2

1

p 1 2

p 1 2


(D) 2

1

p 1 2

p 2 1


50. Given :

N2(g) + 3H2(g) 2NH3(g); K1

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N2(g) + O2(g) 2NO(g), K2

H2(g) +12

O2(g) H2O(g); K3

The equilibrium constant for,

3 2 252NH (g) O (g) 2NO(g) 3H O(g)2

will be :

(A) K1K2K3 (B) 1 2

3

K KK

(C) 3

21

2

K KK (D)

32 3

1

K KK

LEVEL - IIIEquilibrium Constant

51. A(g) + B(g) C(g) + D(g)

above homogeneous reaction is carried out in a 2 litre container at a particular temperature bytaking 1 mole each of A, B, C and D respectively. If KC for the reaction is 1/4 then equilibriumconcentration of C is

(A) 13

M (B) 23

M

(C) 43

M (D) 12

M

52. If KP of an equilibrium CaCO3(s) CaO(s) + CO2(g) is 1.5 atm at 29°C then the moles of

CaO formed if excess CaCO3(s) is being placed in a 1lt of closed container containing CO2 gasat 0.5 atm pressure at same temperature is:

(A) 156 (B)

124

(C) 1

12 (D) 1

13

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53.

2 mole A2

4 mole B2

1 litre at 27°C

3 litre at 27°C

The gas A2 in the left flask is allowed to react with gas B2 presence in right flask asA2 (g) + B2 (g) 2AB (g) ; Kc = 4 at 27°C. What is the concentration of AB whenequilibrium is established ?(A) 1.33 M (B) 2.66 M(C) 0.66 M (D) 0.33 M

54. 2NOBr(g) 2NO(g) + Br2 (g). If nitrosyl bromide (NOBr) is 40% dissociated at a certaintemperature and a total pressure of 0.30 atm. Kp for the rection2NO(g) + Br2(g) 2NOBr(g) is -(A) 45 (B) 25(C) 0.022 (D) 0.25

55. A reaction system in equilibrium according to reaction 2SO2(g) + O2(g) 2SO3(g) in one litrevessel at a given temperature was found to be 0.12 mole each of SO2 and SO3 and 5 mole ofO2. In another vessel of one litre contains 32 g of SO2 at the same temperature. What massof O2 must be added to this vessel to order that at equilibrium 20% of SO2 is oxidized to SO3 ?(A) 0.4125 g (B) 11.6 g(C) 1.6 g (D) None of these

56. For the reaction A 2B 2C at equilibrium [C] = 1.4 M. [A]o = 1 M, [B]o = 2M, [C]o =

3 M. Then value of Kc is (given vol. = 1 litre, []o is initial concentration)(A) 0.084 (B) 8.4(C) 48 (D) none of these

57. Kp for a reaction at 25oC in 10 atm. Then Kc for the reaction at 40oC will be(Given: H = 8 kJ and antilog (0.067) = 0.1167)(A) 4.33 10-1 M (B) 3.33 10-2 M(C) 3.89 10-1 M (D) none of these

58. Kc for the synthesis of HI (g) from H2(g) and I2(g) is 50. The degree of dissociation of HI is(A) 0.10 (B) 0.14(C) 0.18 (D) 0.22

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59. The equilibrium constant for the reaction

H2(g) + I2(g) 2HI(g) is 50 at 500 K. In presence of suitable catalyst, equilibrium isattained four times faster. The equilibrium constant of the reaction at 500 K in the presence ofcatalyst should be(A) 200 (B) 4 104

(C) 50 (D) 83.660. Predict the nature of reaction towards forward direction

2AB(g) A2(g) + B2(g) Keq. = 2 10–2 at 35°C.Keq. = 2 10–4 at 135°C

(A) Exothermic (B) Endothermic(C) H 0 (D) G° = 0

MORE THAN ONE CHOICELEVEL - I

Le-Chatelier’s principle:1. The dissociation of ammonium carbamate may be represented by the reaction

4 2 3 2

||2g g g

O

NH C ONH NH CO

H° for the forward reaction is negative. The equilibrium will shift from right to left if there is(A) a decrease in pressure(B) an increase in temperature(C) an increase in the concentration of ammonia(D) an increase in the concentration of carbondioxide

2. The equilibrium of which of the following reactions will not be disturbed by the addition of aninert gas at constant volume

(A) 2(g) 2(g) (g)H I 2HI (B) 2 4(g) 2(g)N O 2NO

(C) (g ) 2(g) 3 (g )CO 2H CH OH (D) (s) 2 (g) (g) 2(g)C H O CO H

3. For the chemical reaction

3X (g) + Y(g) X3Y(g),

the amount of X3Y at equilibrium is affected by(A) temperature and pressure (B) temperature only(C) By adding X(g) and Y(g) (D) by adding catalyst.

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4. For which of the following statements about the reaction quotient, Q, are correct?(A) the reaction quotient, Q, and the equilibrium constant always have the same numerical value(B) Q may be >< = Keq

(C) Q (numerical value) varies as reaction proceeds(D) Q = 1 at equilibrium

5. Which of the following statements are correct ?

(A) ΔnP CK = K (RT)

(B) For the reaction N2(g) + O2(g) 2NO(g), KP = KC

(C) at equilibrium QC = KC

(D) at equilibrium QP KP

6. For the following equilibrium

4 3 2NH HS(s) NH (g) H S(g) reaction is endothermic

partial pressure of NH3 will increase:(A) if NH3 is added after equilibrium is established(B) if H2S is added after equilibrium is established(C) temperature is increased(D) volume of the flask is increased

7. Which is/are correct relation(s) for thermodynamic equilibrium constant?

(A) G° = –2.303 RT logK (B) G = G° + 2.303 RT log Q

(C) 0 0.0591E cell log Kn

(D) 0 0.0591E E log Qn

8. In presence of a catalyst, what happens to the chemical equilibrium ?(A) Energy of activation of the forward and backwards reactions are lowered by same amount(B) Equilibrium amount is not disturbed(C) Rates of forward and reverse reaction increase by the same factor(D) More product is formed.

Equilibrium Constant Kp or Kc:9. Volume of the flask in which species are transferred is double of the earlier flask. In which of the

following cases, equilibrium is affected?

(A) 2(g ) 2(g ) 3(g)N 3H 2NH (B) 2(g ) 2 (g)N O 2NO

(C) 5(g) 3(g) 2(g)PCl PCl Cl (D) 2(g) 2(g)2NO N O

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10. The following equilibrium in a closed container

N2O4 (g) 2NO2 (g)

At a fixed temperature, the volume of the reaction container is halved. For this change, which ofthe following statements holds true regarding the equilibrium constant (Kp) and degree ofdissociation ?(A) neither Kp and nor changes (B) both Kp and changes(C) changes (D) Kp does not change

Le-Chatelier’s principle:

11. N2F4 (g) 2NF2 (g); H° = 38.5 kJ. Which of the following conditions will favour the

formation of NF2?(A) Adding He to the equilibrium mixture at constant temperature and volume(B) Increasing the temperature.(C) NF2 gas is removed from the reaction mixture(D) Decreasing the pressure at constant temperature.

12. Given the equilibrium, PCI3 (g) + CI2 (g) PCI5 (g); H° = -92 kJ. Which of the following

would shift the equilibrium to the right?(A) Increasing pressure (B) Adding PCl3 to the system(C) Decreasing the temperature (D) Addition of He

13. In the equilibrium, CaCO3 (s) CaO(s) + CO2 (g), the number of moles of CO2 is increased

by(A) increasing the volume of the vessel(B) removing some CaCO3 from the mixture(C) adding a few drops of NaOH to the reaction mixture(D) adding a few drops of HCI to the reaction mixture

14. The degree of dissociation, of PCI5 at a given temperature is(A) proportional to the pressure of the gas(B) inversely proportional to the density of gas(C) inversely proportional to the square root of the pressure of the gas(D) inversely proportional to the square root of the density of the gas

15. Which of the following relationship/s describe the quantitative effect of temperature on theequilibrium constant?

(A) P2

d lnK ΔHdT RT

(B) P2

d lnK ΔEdT RT

(C) C2

d lnK ΔEdT RT

(D) C2

d lnK ΔHdT RT

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16. Which of the following statement/s is/are correct with respect to the equilibrium?

PCl5(g) PCl3(g) + Cl2(g)

(A) Decrease in pressure will increase the dissociation of PCl5.(B) The degree of dissociation of PCI5 varies inversely as the square root of the pressure of thesystem, when it is small compared to unity.(C) Decrease in volume of the system will increase the dissociation of PCI 5.(D) The value of KP = 0.041 atm at one atmospheric pressure when the degree of dissociationis 0.2.

17. Consider the following equilibria:

(1) N2 (g) + 3H2 (g) 2NH3 (g)

(2) N2(g) + O2(g) 2NO (g)

(3) PCl5(g) PCl3 (g) + CI2 (g)

Choose the correct statement/s:(A) Addition of an inert gas at constant volume has no effect on all the three equilibria.(B) Addition of an inert gas at constant pressure favours the forward reaction in (3), backwardreaction in (1) and has no effect on (2).(C) Addition of an inert gas at constant pressure has no effect on equilibrium (2), but favours theforward reaction in (1) and backward reaction in (3).(D) Addition of inert gas has no effect on all the three equilibria at constant temperatureand also at constant pressure.

Equilibrium Constant Kp or Kc:

18. If K1 and K2 are the equilibrium constants at temperature T1 and T2 and 2 1T T then

(A) when H = 0; K2 = K, (B) when H is +ve; K2 > K1

(C) when H is -ve; K2 < K1 (D) when H is +ve; K2 < K1

19. For the general reaction, A + B f

b

K

K C + D

If Kf and Kb are the equilibrium constant for the forward and backward reactions, then, which ofthe following relationships hold good?

(A) fb

1K = K (B) Kf × Kb = 1

(C) Kf - 1 = Kb (D) Kb + 1 = Kf

20. For the gas phase reaction, 2NO N2 + O2; H = - 43.5 kcal

Which of the following is/are true for N2 + O2 2NO?

(A) K is independent of T. (B) K decreases as T decreases.(C) K increases as T decreases. (D) Increase in pressure has no effect on K.

17


LEVEL - IILe Chatelier's Principle21. When NaNO3 is heated in a closed vessel, oxygen is liberated and NaNO2 is left behind. At

equilibrium,(A) addition of NaNO2 favours reverse reaction(B) addition of NaNO3 favours forward reaction(C) increasing temperature favours forward reaction(D) increasing pressure favours the forward reaction

22. For the gas phase reaction, C2H4 + H2 C2H6 (H = - 32.7 kcal) carried out in a vessel,

the equilibrium concentration of C2H4 can be increased by(A) increasing the temperature (B) decreasing the pressure(C) removing some H2 (D) adding some C2H6

23. Which of the following factors will not disturb the equilibrium state of the reaction,

N2(g) + O2 2NO (g)

(A) Change in pressure (B) Change in temperature(C) Addition of catalyst (D) Addition of N2.

24. For the reaction

2A (g) + B (g) C(g), H = +x kJ, which of the following favour the reactants ?

(A) Low pressure (B) High pressure(C) Low temperature (D) Catalyst

25. For which of the following reactions at equilibrium at constant temperature doubling the volumewill cause a shift to the right ?

(A) N2O4 (g) 2NO2 (g) (B) CaCO3 (s) CaO(s) + CO2(g)

(C) 2CO(g) + O2(g) 2CO2(g) (D) N2(g) + O2(g) 2NO (g)

26. The reaction, SO2 (g) + Cl2(g) SO2Cl2 (l) is exothermic and reversible. To the above

reaction, at equilibrium, a certain quantity of extra SO2 is introduced keeping the volume constant.Which of the following are true?(A) The pressure inside the container increases(B) The temperature will increase(C) The temperature will decreases(D) The temperature will remain same.

18


27. For the reaction PCl5 (g) PCl3(g) + Cl2(g), the forward reaction at constant temperature

is favoured by(A) introducing an inert gas at constant volume(B) introducing chlorine gas at constant volume(C) introducing an inert gas at constant pressure(D) increasing the volume of the container

28. Which of the following statement/s is/are wrong?(A) At equilibrium, concentrations of reactants and products become constant because the

reaction stops(B) Addition of catalyst speeds up the forward reaction more than the backward reaction(C) Equilibrium constant of an exothermic reaction decreases with increase of temperature(D) Kp is always greater than Kc

29. Pure ammonia is placed in avessel where its dissociation is appreciable. At equilibrium,(A) KP does not change singnificantly with pressure(B) changes with pressure(C) concentration of NH3 does not change(D) concentration of H2 is less than that of N2

30. N2O2 2NO, K1; 1/2 N2 + 1/2O2 NO, K2

2NO N2+ O2, K3; NO 1/2 N2 + 1/2 O2, K4

Correct relation between K1, K2, K3 and K4 is:

(A) K1 × K3 = 1 (B) 1K × K4 = 1

(C) 3K × K2 = 1 (D) None

LEVEL - III

31. Which is correct:(A) 2.303 log K = -DHo/RT + DSo/R (B) DGo = -2.303 RT log K(C) -2.303 log K = -DHo/RT2 + DSo/ R (D) 2.303 log K = (1/RT) (DHo + DSo)

32. For the reaction CaCO3(s) = CaO (s) + CO2(g) which is correct representation:

(A) Kp = )p(2CO (B) Kp = Kc (RT)

(C) Kc = (CO2)/1 (D) None of these

19


33. At constant temperature, the equilibrium constant (Kp) for the decompostion reactionN2O4 2NO2

Is expressed by )x1(Px4K2

2

p

Where P = pressure, x = extent of decomposition. Which one of the following statements arefalse?(A) Kp increases with increase of P(B) Kp increases with increase of x(C) Kp increases with decrease of x(D) Kp remains constant with change in P and x

34. Which one of the following oxides is most stable? The equilibrium constant are given at the sametemperature:(A) 2N2O5(g) 2N2(g) + 5O2(g); K = 1.2 ×1034

(B) 2N2O(g) 2N2(g) + O2(g); K = 3.5 ×1035

(C) 2NO(g) N2(g) + O2(g); K = 2.2 ×1030

(D) 2NO2(g) N2(g) + 2O2(g); K = 6.71 ×1016

Equilibrium Constant

35. 2 2 1 2 2 21 1N O 2NO, K ; N O NO,K ;2 2

2 2 3 2 2 41 12NO N O , K ;NO N O ,K ;2 2

Correct relation(s) between K1, K2, K3 and K4 is/are :

(A) 1 3K K 1 (B) 1 4K K 1

(C) 3 2K K 1 (D) 23

1KK

20


LINKED COMPREHENSION

Comprehension I8 moles of a mixture of SO2, O2 and SO3 in 1 : 1 : 10 molar proportions was enclosed in a

container at a total pressure of 12 atm at 25°C. If 3

0f (SO )G = 78.76 kJ mol–1,

2

0f (SO )G = 50.26 kJ

mol–1 and 2.303 298R = 5.74 kJ mol–1.1. G (kJ) of the reactions:

2 21SO O2

SO3

is equal to(A) 34.24 (B) 743.28(C) 243.26 (D) 58.70

2. The Kp of the above reaction is:(A) 1.0 10–4 (B) 2.0 10–3

(C) 4.0 10–4 (D) 1.0 10–5

3. With partial pressure of SO3 and SO2 taken to be equal to 10–5 atm and 10 atm respectivelythen what must be the partial pressure of O2 so that the reaction may remain in equilibrium?(A) 10 atm (B) 0.01 atm(C) 100 atm (D) 0.1 atm

Comprehension IILet G° be the difference in free energy of the reaction when all the reactants and products arein the standard state (1 atmospheric pressure and 298K) and KC and KP be the thermodynamicequilibrium constant of the reaction. Both are related to each other at temperature T by thefollowing relation:

G° = –2.303 RT log KC

and G° = –2.303 RT log KP (incase of ideal gas)

This equation represents one of the most important results of thermodynamics and relates to theequilibrium constant of a reaction to a thermodynamic property.It is sometimes easier to calculate the free energy in a reaction rather than to measure theequilibrium constant.Standard free energy change can be thermodynamically calculated as

G° = H° –T S°

Here H° = standard enthalpy change

S° = standard entropy change.

21


4. Which of the following statement is correct for a reversible process in state of equilibrium

(A)G = –2.303 RT log K (B) G = 2.303 RT log K

(C) G° = –2.303 RT log K (D) G° = 2.303 RT log K

5. At 490°C, the value of equilibrium constant; KP is 45.9 the reaction

H2(g) + I2(g) 2HI(g)

Calculate the value of G° for the reaction at that temperature

(A) –3.5 kcal (B) 3.5 kcal(C) 5.79 kcal (D) –5.79 kcal

6. Calculate the equilibrium concentration ratio of C to A if 2.0 mol each of A and B were allowedto come to equilibrium at 300 K

A + B C + D G° = 460 Cal

(A) 1.0 (B) 0.5(C) 0.8 (D) 0.679

Comprehension IIIIf the composition of the system does not change with time, the system is said to be in chemicalequilibrium. It is the state in which net reaction of a system is zero. in another words we can saythat in reversible reactions, a stage is reached when the rate of transformation of reactants intoproducts equals the rate of transformation of products into reactants. At this stage, the compositionof reactants and products does not change with time. This does not mean that the reaction hasceased, as both reverse and forward reactions are still taking place but with equal place. Suchequilibria are called dynamic equilirbria.Let us consider a reaction of the type

A(g) + B(g) C(g) + D(g)

BA

DCKC

where KC is equilibrium constant when the ratio of the concentrations of the product to reactants

BA

DCP PP

PPK

where KP is the equilibrium constant for the ratio of partial pressure & of products to reactants.The relation between KP and KC is as follows

KP = KC(RT) n

22


7. The mass ratio of steam and hydrogen is found to be 1 : 2 at equilibrium.

3 Fe (s) + 4 H2O(g) Fe3O4(s) + 4 H2(g)The value of equilibrium constant (KC) for the above reaction is(A) 3.05 × 103 (B) 1.05 × 105

(C) 0.75 × 102 (D) 72.42 10

8. KC for the reaction 2 2 21 1 1N (g)+ O (g)+ Br (g)2 2 2 NOBr(g) from the following

information at 298 K

( )2 gNO N2(g) + O2(g) K1 = 2.4 × 1030

(g) 2(g)1NO + Br2 NOBr(g) K2 = 1.4

(A) 3.15 × 10-9 (B) 6.35 × 5 10-18

(C) 9.03 × 10-16 (D) 17× 10-17

9. The equilibrium constant for the reaction 2 SO2 + O2 2 SO3 at 1000 K is 3.5. What wouldthe partial pressure of oxygen gas have to be to give equal moles of SO2 and SO3 ?(A) 0.29 atm (B) 3.5 atm(C) 0.53 atm (D) 1.87 atm

10. For the reaction

NH2COONH4(s) 2 NH3(g) + CO2(g)

The equilibrium constant KP = 2.9 × 10-5 atm3. The total pressure of gases at equilibrium when 1mole of reactant was heated will be(A) 0.0194 atm (B) 0.0388 atm(C) 0.0580 atm (D) 0.0667 atm

23


MATRIX MATCH TYPE

1. Match the Column I , Column II and select the answers from the given codes:Column I (Mixture) Column II (pH)(A) Keq < 1 (p) affected by temperature(B) Degree of dissociation (q) affected by pressure(C) Equilibrium constant (r) Kf < Kb

(D) Melting of ice (s) Kf > Kb

(t) Kf /Kb

2. Match the chemical reaction in equilibrium (in Column I) with the pressure dependent of degreeof dissociation of the reaction (in Column II):

Column I Column II

(A) N2(g) + 3H2 (g) 2NH3 (g) (p) x α P

(B) 2SO2(g) +O2 (g) 2SO3(g) (q) x 1P

(C) PCl5(g) PCl3 (g) + Cl2 (g) (r) x P

(D) N2 + O2 2NO (s) x P0

(t) Kp = Kc

3. Match Column I (G), Column II (direction ) :

Column I Column II(A) G > 0 (p) forward reaction

(B) G = 0 (q) backward reaction

(C) G < 0 (r)

(D) G = -ve (s) rection does not occur

(t) rate of forward reaction =rate of backward reaction

24


ASSERSION AND REASION

CODE :(A) Statement – 1 is True, Statement – 2 is True; Statement – 2 is a correct explanation for Statement – 1.(B) Statement–1 is True, Statement–2 is True; Statement–2 is NOT a correct explanation for Statement – 1.(C) Statement – 1 is True, Statement – 2 is False.

(D) Statement – 1 is False, Statement – 2 is True.

1. STATEMENT – 1: A catalyst increased the equilibrium constant at any fixed temperature.STATEMENT – 2: By the uses of catalyst the rate of the reaction increases.

2. STATEMENT – 1: In the Haber process

2 2 3N 3H 2NH ; H ve

Then concentration of NH3 increases by decreasing the temperature and increasing the pressure.STATEMENT – 2: For an exothermic reaction, equilibrium constant increase by increasing thetemperature.

3. STATEMENT–1: For the reaction 3(g) 2(g) 2(g)2NH N 3H , the unit of Kp will be atm

STATEMENT – 2: Unit of Kp is Δn(atm)

4. STATEMENT – 1: 2(g) 2(g) 3(g)1SO O SO heat2

Forward reaction is favoured at high temperature and low pressure.STATEMENT – 2: Reaction is exothermic in forward direction

5. STATEMENT-1: The graph between concentration and time is as follows

Time

Con

cent

ratio

n

graph shows a reversible reaction

STATEMENT-2: In a reversible reaction the concentration of reactants and products becomingconstant after a certain time

25


INTERGER ANSWER TYPE QUESTIONS

1. For the equilibrium N2O4 2NO2

2 4

0 1N O 298K(G ) 100kJmol and

2

0 1NO 298K(G ) 50kJmol .

When 5 mole/ litre of each in taken, calculate the value of G for the reaction at 298 K.

2. When NO and NO2 are mixed, the following equilibria are readily obtained;

2NO2 N2O4 Kp = 6.8 atm-1 and NO + NO2 N2O3

In an experiment when NO and NO2 are mixed in the ratio of 1 : 2, the find total plressure was5.05 atm and the partial pressure of N2O4 was 1.7 atm. Calculate the equilibrium partial pressureof NO in atmospheres.

3. At 250C and 1 atmospheric pressure, the partial pressures in equilibrium mixture of gaseousN2O4 and NO2 are 0.7 and 0.3 atm respectively. Calculate the partial pressrue of the N2O4 gaswhen it is in equilibrium 250C and the total pressrue of 10 atm

4. For the reaction: CO2(g) + H2(g) CO(g) + H2O(g), K is 0.63 at 700°C and 1.66 at

1000°C. What is the average H° for the temperature range considered?

5. 16 moles of H2 and 4 moles of N2 are sealed in a one litre vessel. The vessel is heated at aconstant temperature until the equilibrium is established, it is found that the pressure in the vesselhas fallen to 9/10 of its original value. Calculate newly concentration of N2 and NH3

N2(g) + 3H2(g) 2NH3(g)

6. A(g) 3B(g) 4C(g) , initial concentration of A is equal to that of B. The equilibrium

concentrations of A and C are equal. What is the value of Kc.

7. 4 2 3 2NH COONH (s) 2NH (g) CO (g) . If equilibrium pressure is 3 atm for the above reaction.

What is the value of Kp

8. A B C D . If initially the concentration of A and B are both equal but at equilibrium,concentration of D will be twice of that of A, then what will be the equilibrium constant of thereaction

9. Ammonium carbamate when heated to 200°C gives a mixture of NH3 and CO2 vapour with adensity of 13. What is the degree of dissociation of ammonium carbamate ?

10. A reaction at equilibrium involving 2 moles each of PCl5, PCl3 and Cl2 is maintained at 250°C anda total pressure of 3 atm. The value of Kp is :

26


SUBJECTIVE QUESTIONS

1. At 700 K, CO2 and H2 react to form CO and H2O. For this process K is 0.11. If a mixture of 0.45mole of CO2 and 0.45 mole of H2 is heated to 700 K, then(i) Find out the amount of each gas at equilibrium.(ii) After equilibrium is reached, another 0.34 mole of CO2 and 0.34 mole of H2 are added to thereaction mixture. Find the composition of the mixture at the new equilibrium state.

2. The equilibrium constant for the reaction H2 + I2 2HI; is found to be 64 at 450°C. If 6mole of hydrogen are mixed with 3 mole of iodine in a litre vessel at this temperature; what willbe the concentration of each of the three components, when equilibrium isattained ? If the volume of reaction vessel is reduced to half; then what will be the effect onequilibrium?

3. 5 gm of PCl5 were completely vaporized at 250°C in a vessel of 1.9 litre capacity. The mixture atequilibrium exerted a pressure of one atmosphere. Calculate the degree of dissociation KC andKp for this reaction.

4. If a mixture of N2 and H2 in the ratio 1 : 3 at 50 atmosphere and 650°C is allowed to react tillequilibrium is reached. Ammonia present at equilibrium was at 25 atm pressure. Calculate theequilibrium constant for the reaction.

N2(g) + 3H2(g) 2NH3(g)

5. A mixture of SO3, SO2 and O2 gases is maintained in a 10.0 litre flask at a temperature at whichequilibrium constant Kc for the reaction2SO2(g) + O2(g) 2SO3(g) is 100.(i) If the number of mole of SO2 and SO3 in the flask are equal how many mole of O2 arepresent?(ii) If the number of mole of SO3 in the flask is twice the number of mole of SO2,how many moleof O2 are present ?

6. The equilibrium constant Kp of the reaction 2SO2(g) + O2(g) 2SO3 (g) is 900 atm-1 at

800 K. A mixture containing SO3 and O2 having initial partial pressures of 1 atm and 2 atm,respectively, is heated at constant volume to equilibrate. Calculate the partial pressure of eachgas at 800K.

7. H2S dissociates according to the equation 2H2S (g) 2H2(g) + S2(g). If a is the degree

of dissociation of H2S, show that Kp of the reaction is given as 3

p 2

α pK =2(1-α) (1+α/2)

27


At 11250C, the value of the a is found to be 0.31 when the equilibrium pressure is 1 atm. Determinethe values of Kp and Kc.

8. COCl2 gas dissociates according to the equation COCl2 CO + Cl2. When COCl2 isheated to 724 K at 1atm, the density of gas mixture at equilibrium was found to be1.162 g L-1. Calculate (A) the degree of dissociation and (B) Kp of the reaction.

9. The density of an equilibrium mixture of N2O and NO2 at 1 atm and 348 K is 1.84 g dm-3.Calculate the equilibrium constant of the reaction

N2O4 (g) 2NO2 (g)

10. For the reaction, N2O5(g) 2NO2 (g) + 0.5O2 (g), calculate the mole fraction of N2O5(g)decomposed at constant volume and temperature, if the initial pressure is 600 mmHg and thepressure at any time is 960 mm Hg. Assuming ideal gas behaviour.

PREVIOUS YEARS IIT QUESTIONS

1. For the chemical reaction 33X(g) Y(g) X Y(g) , the amount of X3 Y at equilibrium is affected

by (1999)(A) temperature and pressure (B) temperature only(C) pressure only (D) temperature, pressure and catalyst

2. For the reversible reaction, 2 2 3N (g) 3H 2NH (g) at 500°C, the value of kp is 51.44 10when partial pressure is measured in atmosphere. The corresponding value of kc, withconcentration in mole litre-1, is (2000)

(A)

5

2

1.44 10(0.082 500) (B)

5

2

1.44 10(8.314 773)

(C)

5

2

1.44 10(0.082 773) (D)

5

2

1.44 10(0.082 773)

3. When two reactants, A & B are mixed to give products C & D, the reaction quotient Q, at theinitial stages of the reaction (2000)(A) is zero (B) decreases with time(C) is independent of time (D) increases with time

4. At constant temperature, the equilibrium constant (kp) for the decomposition reaction

2 4 2N O 2NO is expressed by kp = (4x2P)/(1-x2), where P = pressure, x = extent of

28


decomposition. Which one of the following statements is true ? (2001)(A) kp increases with increase of P (B) kp increases with increase of x(C) kp increases with decrease of x (D) kp remains constant with change in P and x

5. Consider the following equilibrium in a closed container

2 4 2N O (g) 2NO (g)

At a fixed temperature, the volume of the reaction container is halved. For this change, whichof the following statements holds true regarding the equilibrium constant (kp) and degree ofdissociation ( ) ? (2002)(A) neither kp nor changes (B) both kp and change(C) kp changes, but does not change (D) kp does not change, but changes

6. The Haber's process for the formation of NH3 at 298 K is

2 2 3N 3H 2NH ; H 46.0kJ; which of the following is the correct statement (2006)

(A) The condition for equilibrium is 2 2 3N H NHG 3G 2G where G is Gibbs free energy per mole

of the gaseous species measured at that partial pressure(B) On adding N2, the equilibrium will shift to forward direction because according to IInd law ofthermodynamics the entropy must increase in the direction of spontaneous reaction(C) The catalyst will increase the rate of forward reaction by 2 times and that of backwardreaction by 1.5 times(D) None of these

7. When 3.06 g of solid NH4HS is introduced into a two litre evacuated flask at 27°C, 30% of thesolid decomposes into gaseous ammonia and hydrogen sulphide.(i) Calculate kc and kp for the reaction at 27°C.(ii) What would happen the equilibrium when more solid NH4HS is introduced in the flask ?

(1999)

29


ANSWERS


1. (D) 2. (C) 3. (D)

4. (A) 5. (D) 6. (D)


LEVEL - I

1. (D) 2. (D) 3. (B)

4. (D) 5. (A) 6. (A)

7. (B) 8. (A) 9. (B)

10. (C) 11. (D) 12. (A)

13. (C) 14. (B) 15. (A)

16. (C) 17. (A) 18. (A)

19. (B) 20. (D) 21. (A)

22. (C) 23. (A) 24. (D)

25. (C)

LEVEL - II

26. (A) 27. (B) 28. (C)

29. (D) 30. (A) 31. (B)

32. (D) 33. (C) 34. (A)

35. (B) 36. (D) 37. (C)

38. (A) 39. (B) 40. (A)

41. (C) 42. (C) 43. (B)

44. (C) 45. (B) 46. (A)

47. (C) 48. (C) 49. (B)

50. (D)

30


LEVEL - III

51. (B) 52. (B) 53. (C)

54. (A) 55. (B) 56. (A)

57. (D) 58. (D) 59. (C)

60. (A)

MORE THAN ONE CORRECT ANSWER

LEVEL - I

1. (B, C, D) 2. (A, B, C, D)

3. (A, C) 4. (B, C) 5. (A, B, C)

6. (A, C, D) 7. (A,B,D) 8. (A, B, C)

9. (A, B, C) 10. (C, D) 11. (B, C, D)

12. (A, B, C) 13. (A, C, D) 14. (C, D)

15. (A, C) 16. (A, B, D) 17. (A, B)

18. (A, B, C) 19. (A, B) 20. (B, D)

LEVEL - II

21. (A, B, C) 22. (A, B, C, D) 23. (A, C)

24. (A, C) 25. (A, B) 26. (A, B)

27. (C, D) 28. (A,B,D) 29. (A, B)

30. (A,B,C)

LEVEL - III

31. (A, B) 32. (A, B, C) 33. (A, B, C)

34. (B) 35. (A,B,C,D)

COMPREHENSION

1. (A) 2. (D) 3. (B)

4. (C) 5. (D) 6. (D)

7. (B) 8. (C) 9. (A)

10. (C)

MATRIX MATCH TYPE

1. (A - R), (B – P, Q), (C - P, T), (D – P, Q)

31


2. (A - R), (B - P), (C - Q), (D - S, T)

3. (A –Q), (B – R, T), (C –P), (D - P)


1. (D) 2. (C) 3. (D)

4. (D) 5. (D)


1. 4 2. 2 3. 9

4. 8 5. 3 6. 8

7. 4 8. 4 9. 1 10. 1


1. [CO2 ] = [H2] = 0.594 ; [CO ] = [H2O] = 0.196

2. 5.68, 3.16

3. 0.843, 0.0565,

4. 1.677 × 10-3

5. 0.1, 0.4

6. 0.023, 2.0115, 0.977

7. 2.71 × 10-2 , 2.39 × 10-4

8. 0.231

9. 0.752, 5.206

10. 0.407


1. (A) 2. (B) 3. (D)

4. (D) 5. (D) 6. (A)

7. 5 2 2 2 28.1 10 mol ,4.90 10 atm

HINTS /SOLUTIONS


1. 2 4(g) 2(g)N O 2NO

32


Initial mole 1 0at equilibrium 1 2

D d 46 27.6(n 1)d 27.6

= 0.67

Mole fraction of 2

2NO 0.801

2.0

2 2 1

1 1 2

kp H T Tlogkp 2.303R TT

4 0

610 H 25log10 2.303 8.314 298 323

0H 2522.303 8.314 298 323

0H 147.4KJ

3. The reaction is endothermic in the forward direction. Increase in temperature shifts the equilibriumin the forward direction.

4. 2 22AB (g) 2AB(g) B (g)

P 0 0

P(1 ) P P2

2

2

P(P )2Kp ,1 1

(P(1 ))

3P

Kp2

5 2Ag+ (aq.) + Cu(s) Cu2+ (aq.) + 2Ag(s)

Concentration at equilibrium y(say) x

Kc = 2

xy

On dilution [Ag+ (aq.)] and [Cu2+ (aq.)] both will be halved

33


new concentration 2Ag+ (aq.) + Cu(s) Cu2+ (aq.) + 2Ag(s) y/2 x/2

Qc = Qc Kc and reaction will proceed backward.

6. 2 22HI(g) H (g) I (g)

2

2150 2 2

= 0.22


LEVEL - I

4. 1 2 3

2 3s g gx x x

A D C

2 34 27cK x x

5108x

2 6

2 3s g gx c x

A B C

5 2 2 3108 4 216x x c x 2108 216 4c

2 1084 216

c

; 1

2 2c

6.

2

1 1 1

600mm

Initial pressure

at equilibrium 600 -

g g gxy xy y

p p p

Total pressure at equlibrium = 600 - p1 + p1 + p1

= 600 + p1

600 + p1 = 800

34


p1 = 800 - 600 = 200 mm

7. 2SO2(g) + O2(g) 2SO3(g)

2 2-1 -13

C 2 22 2

[SO ] 1.9K = = M 12.237molL[SO ] [O ] 0.6 0.82

8. (i) KP = KC (RT) n

n = 1

24P

CK 1.8 10K = 4.4 10RT 0.0821 500

(ii) 1n

PC

K 167K = 1.90RT 0.0821 1073

9. KC(reverse) = 15

14C

1 1 1.59 10K 6.3 10

10. 2N2(g) + O2(g) 2N2O(g)

t = 0 0.482 mol 0.933 mol 0Charge -2x -x +2xt = teq (0.0482 -2x) (0.0933 - x) 2x

Kc = 2 × 10-37 = 2

2

(2 )(0.0482 2 ) (0.0933 )

xx x

Since Kc has small value, X should be low0.0482 - 2x 0.04820.0933 - x - x 0.0933

2 × 10-37 = 2

2

40.0482 0.0933

x

On solving, x = 3.3 × 10-21

[N2O] = 2x = 6.6 × 10-21

[N2] 0.0482 M ; [O2] = 0.0993 M

35


11. 2NO (g) + Br2(g) 2NOBr (g)

t=0 0.087 mol 0.043 mol 0 -2x -x 2xt = teq 0.087 - 2x 0.0437 - x 2xFrom equation,2x = 0.0518 molx = 0.0259 molNo. of mole of NO = (0.087 - 0.0518)mol = 0.0352 molNo. of mole of Br2 = (0.043 - 0.0259)mol = 0.0178 mol

12. KP = KC(RT) n

n = - 1

By putting the values and on solvingKC = 7.47 × 1011 M-1

13. 2HI(g) H2(g) + I2(g)

t = 0 0.2 atm 0 0change -2x +x +xt = teq (0.2 - 2x) x xFrom question, 0.2 - 2x = 0.04 2x = 0.2 - 0.04 = 0.16 ; x = 0.08

KP = 2

0.08 0.08 40.04

14. N2 (g) + 3H2 (g) 2NH3(g)

Q =

2

2 2

3 3

8.138.13 20201.57 1.921.57 1.92

20 20

32.38 10 CQ K , so reaction proceeds in backward direction.

15. 4NO(g) + 6H2O (g) 4NH3(g) + 5O2(g)

16. H2O (g) + CO (g) H2 (g) + CO2 (g)

t=0 1 mol 1 mol 0 0

36


-0.4 mol -0.4 mol 0.4 mol 0.4 molt = teq 0.6 0.6 0.4 0.4

KC = 0.4 0.4 4 0.440.6 0.6 9

17. H2 (g) + I2(g) 2HI (g)

t = teq x x .5M

KC = 54.8 = 2

0.5 0.5x

2 0.5 0.554.8

x

On solving, x = 0.068 M

18. 2ICl(g) I2 (g) + Cl2 (g); KC = 0.14

t = 0 0.78 M 0 0 -2x x xt=teq (0.78 - 2x) x x

0.14 = 2

2(0.78 2 )x

x

on solving, x = 0.21 M

19. C2H6 (g) C2H4 (g) + H2 (g)

t=0 4 atm 0 0 -x x xt = teq 4 - x x x

0.04 = 2

4x

x

On solving, x = 0.38 ;

2 6C HP = 3.62 atm

20. PCl5(g) PCl3(g) + Cl2(g)

t = teq 0.5 × 10-1 M x x

37


KC = 8.3 × 10-3 = 2

-1

x0.5×10

x2 = 4.15 m × 10-4

x 0.02 M

21. FeO (s) + CO (g) Fe (s) + CO2 (g)

t=0 1.4 atm 0.8 atm +x -xAt equil (1.4 +x) 0.8 - x

Q = 0.8 4

= = 0.551.4 7

Q > K, system is not at equilibrium and to attain equilibrium it will shift in backward direction

KP = 0.265 = 0.8 - x1.4 + x

On solving, x = 0.339

]CO[P = 1.739 atm and 2COP = 0.461 atm

22. N2(g) + 3H2 (g) 2NH3 (g)

t = teq 0.3M 2M 0.5M

Q = 3

0.5×0.5 1 1= =

3× 2 3×8× 4 96

Q < K, system is not at equilibrium and equilibrium will be reached by shifting in forwarddirection

23. 2BrCl (g) Br2 (g) + Cl2 (g)

t=0 3.3 ×10-3 M 0 0change -2x x xt=teq (3.3 × 10-3 -2x) x x

KC = 2

-3 2

x(3.3×10 - 2x)

on solving, x = 1.5 × 10-3

[BrCl] = 33 ×10-4 - 2x = 33 × 10-4 - 30 × 10-4

38


= 3 × 10-4

24. Δ 0G = 2

0 0NO NOΔG ΔG-

= (52 - 87) kJ mol-1 = -35 kJ mol-1

0ΔG = -2.303 RT log K

log K = -1

-1 -1

-35kJmol-8.301Jk mol ×298K

on solving K = 1.365 × 106

25. H2(g) + Br2(g) 2HBr(g)

t = 0 0 0 10 bar +x +x -2xt=teq x x 10 - 2x

KC = 16 ×10-6 = 2

2

(10 - 2x)x

4 × 10-3 = 10 - 2x

x

10x

; x = 3

104 10

= 2.5 × 10-3

LEVEL - II

26. (A)

X2 + X– 3X

1 0.5 0(1 – x) (0.5 – x) x(0.5 – x) = unreacted X–

X– Ag + 0.25 ; x = 0.25

3c

2

[X ] 0.25K 1.33[X ][X ] 0.75 0.25

27. (B) Qp =

3

2 2

22

NH 23 3

N H

P (3) 9 atm(1)(2) 8P P

= 1.125 atm–2.

Since value of Qp is larger than Kp (4.2810–5 atm–2), it indicates net reaction will proceed inbackward direction.

39


28. Moles of PCl5 dissociated = 2 35100

= 0.7

Moles of PCl5 left undissociated = 2 – 0.7 = 1.3 mol

[PCl5] = 1.35 M, [PCl3] =

0.75 M, [Cl2] =

0.75 M

3 2

5

0.7 0.7[PCl ][Cl ] 5 5K 0.075

1.3[PCl ]5

29. (D) 2HI(g) H2(g) + I2(g)

1 – 2

2

2

T

p 2 2T

P2K

(1 ) P

p2 K1 –

p

p

2 K

1 2 K

31. (B) The fraction of A decomposed at equilibrium is independent of initial concentration meansthe equilibrium constant expression is free from concentration term. The equilibrium reaction is

xA C + D

Initial conc. c 0 0

Conc. at. equil. c(1 – a) cx

cx

2

C x x 2 x

c c[C][D] (c )x xK[A] [c(1– )] x [c(1– )]

The expression of KC would be free from concentration term only when value of x is 2. Puttingx = 2, gives

40


2 2 2

C 2 2

cK4[c(1– )] 4(1– )

32. (D) Suppose 1 mole of A and B each taken then 0.8 mole/litre of C and D each formed remainingconcentration of A and B will be (1 – 0.8) = 0.2 mole/litre each.

C D 0.8 0.8 16.0A B 0.2 0.2

cK

33. (C) A + B C + D

Initial conc. 4, 4 0 0After T time conc. (4-2) (4 - 2) 2 2

quilibrium constant = C D 2 2 1A B 2 2

34. (A) H2 + I2 2HI; [HI] = 0.80, [H2] = 0.10, [I2] = 0.10

2

2 2

HI 0.80 0.80 64H I 0.10 0.10

cK

36. (D) Kc is a characteristic constant for the given reaction.

37. Equilibrium constant is independent of original concentration of reactant.

38. Kp is constant and does not change with pressure.

41. (C) Apply the formula, 4

f2

b

k 1 10Kk 2.5 10

= 4 × 10–3.

42. (C)

SO2(g) + NO2(g) SO3(g) + NO(g)

1 1 1 1 Initial conc.1 – x 1 – x 1 + x 1 + x At equilibrium

3

2 2

[SO ][NO] (1 x)(1 x)K[SO ][NO ] (1 x)(1 x)

41


2

2

(1 x)16(1 x)

(1 x) 4(1 x)

or x = 0.6

[NO2] = 1 – x = 1 – 0.6 = 0.4 mol L–1.

43. (B)2HI H2 + I2

1 0 01 – x1 – 0.22 = 0.78 0.11 0.11

x = 22% of 1 mole = 22 1 0.22

100

2 22 2

[H ][I ] 0.11 0.11K[HI] (0.78)

= 0.0199

44. (C)

8(g) 2(g)S 4S

At start 1 0At eqbm 1 – 0.29 4 0.29

= 0.71 atm = 1.16 atm

4 432

p8

[pS ] (1.16)K 2.55atm[pS ] (0.71)

45. (B) This is according to Le-Chatelier’s principle

46. (A)The degree of dissociation cannot be calculated from the vapour density data. Becaluse here thenumber of moles remains unchanged before and after reaching equilibrium

47. (C)

The reaction A D is obtained by adding the three given reactions.

1 2 3K K K K 2 4 3 24

48. (C)

2 4 2N O 2NO

42


t = 0 1 0 teq 1-x 2xTotal number of molecules at equilibrium = 1 - x + 2x = (1 + x)

49. (B)H / RT

pK Ae

o

eq.Hlog K log A

2.303RT

On applying the limit of initial and final states, we will get the relation (b).

50. (D)

2 23 2

1 2 33 1/ 22 2 2 2 2 2

[NH ] [H O][NO]K ;K ;K ;[N ][H ] [N ][O ] [H ][O ]

The equilibrium constant for

3 2 252NH (g) O (g) 2NO(g) 3H O(g)2

will be :

32 32 32

2 5/ 23 2 1

K K[NO] [H O][NH ] [O ] K

LEVEL - III

52. Use PV = nRT as extra CO2 produced to find ‘n’ which will equal to net of moles of CaO.

53. (C) A2 (g) + B2 (g) 2AB (g)

Moles at eqm 2 – x 4 – x 2x

2

c4xK

(2 x) (4 x)

x = 32 1.33 mole24

43


[AB(g)] = 2 1.33 0.66M

4

55. (B)

2SO2(g) + O2(g) 2SO3(g)

Another vessel

2SO2(g) + O2(g) 2SO3(g)

Moles at eqm 0.5 – 2x y – x 2xy = 0.3625 mole mass of O2 added = 11.6 g

57. (D)Kp = Kc(RT)n, here n = 1

c85.68 K

0.0821 313

Kc = 3.33

58. (D)

2 22HI H I

2

2150 2 2

= 0.2259. Catalyst shortens the time taken to reach equilibrium without changing equilibrium position.

MORE THAN ONE CORRECT ANSWERSLEVEL - II

23. (A, C)Pressure does not disturb equilibrium state in this reaction because np = nR or n = 0.

24. (A, C)Backward reaction is exothermic and is accompanied by increase in number of moles of gaseousspecies. Hence it is favoured by low temperature and low pressure.

25. (A, B)Doubling the volume would reduce the pressure and shift the equilibrium in the direction wherethere is more number of moles of gases. In choices (A) and (B), the equilibrium will shift to theright.

44


26. (A, B)The equilibrium shifts in the forward direction according to Le-Chatelier’s principle. The forwardreaction being exothermic, temperature increases. Pressure increases due to extra SO2.

27. (C, D, ) Apply Le-Chatelier principle.

30. (A, C) ; Pressure does not disturb equilibrium state in this reaction because np = nR or n = 0.

COMPREHENSION

1.3

2 2

SO 1/ 2p 1/ 2 1/ 2

SO O

p 10Q 10atmp p 1 1

G0 = 3 2 2

0 0 0f (SO ) f (SO ) f (O )

1G G G2

= 78.76 – 50.26 – 1 02

= 28.50 kJG = G0 + 2.303RT log Qp

= 28.50 + 5.74 log 10= 34.24 kJ

2. G0 = – 2.303RT log Kp

28.50 = – 5.7 log Kp

Kp= 10–5

5. 0 2.303 log 5.79 .PG RT K k cal

6. 0 2.303 logG RT k

log 1.665k

K = Antilog of 1.665 = 0.4625

A B C + D

Initial conc. 2 2 0 0

Eq. conc. 2 x 2 x x x

2

2 0.46252

CxK

x

45


0.682

C xA x

7. (B) Since n = 0, so Kc = Kp = Kx

Kx = 2

2

4 44 5

4 4

X (2 / 2) 18 1.05 10(1/18)

H

H OX

8. (C) 2 21 1( ) O ( ) ( )2 2

N g g NO g

Kc = 1/ 2

30

12.4 10

NO (g) + 21 ( )2

Br g NOBr (g)

K2 = 1.4

2 2 21 1 1N ( ) O ( ) Br (g) NOBr( )2 2 2

g g g

K = KC × K2 = 9.03 × 10-16

9. (A) Kp = 3

SO 22

2SO

2

PP . OP

2OP

1 1P 0.29K 3.5

atm atm

10. (C) Kp = 3 2

2 2 3. (2 ) 4pNH COP P p P

P = 1/ 3

0.06674

PK atm

MATRIX MATCH TYPE

1. (A - R), (B – P, Q), (C - P, T), (D – P, Q)2. (A - R), (B - P), (C - Q), (D - S, T)3. (A – 2-Q), (B – 3 - R, T), (C – 1 - P), (D - 1 - P)

46



1. (D)By the use of catalyst the reaction rate increase. But equilibrium constant does not change untilthe temperature is varying.

2. (C)The exothermic reaction, equilibrium constant increases by decreasing the temperature.

3. (D)

2 2

3

3 3H N 2

p 2NH

p p atm atmK (atm)p (atm)

or Dn = 4 – 2 = 2Unit of Kp = (atm)2

4. (D)reaction is exothermic, low temperature favours forward reaction. High press favours forwardreaction as it is accompanied by decrease in the number of moles.

5. (D)The graph shows irreversible reaction where as in reversible reaction concentration of reactantand product become constant at equilibrium position.


1. N2O4 2NO2 ; 2 4

0 1N OG 100kJ mol

conc. at t = 0 5 52

0 1NOG 50kJ mol

0G for reaction = 2 × NO2 2 4

0 0N OG G 2 50 100 0

0G G 2.303RTlogQ

Now, 2

3 5G 0 2.303 8.314 10 298log5

= +3.99 kJ 4kJ

2. For I equilibrium 2NO2 N2O4

47


2 4

2

'N O

P ' 2NO

PK 6.8

(P ) . . . . .(i)

2 4

'N OP = 1.7 atm By eq. (i)

2

'NOP = 0.5 atm

The equilibria are maintained using NO and NO2 in the ratio 1 : 2

For II equilibria NO + NO2 N2O3

Initial pressures P 2P 0Pressures at equi (p-x) (2P-x-3.4) x

3.4 atm of NO2 are used for I equilibrium to have 2 4

'N OP = 1.7 atm

At equilibrium (P - x) 0.5 x

( 2

'NOP is same for both the equilbria since both reactions are at equilibrium at a time)

Total pressure at equilibrium (Given 5.05 atm)

= 'NOP +

2

'NOP +

2 3

'N OP +

2 4

'N OP

= P - x + 0.5 + x + 1.75.05 = P + 2.20P = 2.85 atm2P - x - 3.4 = 0.52 × 2.85 - x - 3.4 = 0.5x = 1.80 atm 2 atm

3. N2O4 2NO2

Pressure at equilibrium 0.7 0.3

2

2 4

2NO

PN O

(P ) 0.3 0.3K 0.1286atmP 0.7

Now assume decompostion at 10atm pressure

N2O4 2NO2

Initial mole 1 0Mole at equilibrium (1-x) 2x

2

2 4

12 n 2NO

PN O

(n ) P (2x) 10Kn n (1 x) (1 x)

; x = 0.0565

2

'NOP 1.07atm 1

48


2 4

'N OP 8.93atm 9 atm

4. T1 = 700 + 273 = 973 K,T2 = 1000 + 273 = 1273 KK1 = 0.63, K2 = 1.66Using the Van’t Hoff equation

02 2 1

1 1 2

K T -TΔHlog =K 2.303R T T

97312739731273

)99.1(303.2H

63.066.1log

o

; H° = 8.0 × 103 cal = 8.0 kcal

Note: The units of R and H must be same

5. The given equilibrium is

N2(g) + 3H2(g) 2NH3(g)

At t = 0 4 16 0At equilibrium 4 – x 16 – 3x 2xTotal gaseous moles at equilibrium = 4 – x + 16 – 3x + 2x = (20 – 2x)Since, pressure has fallen to 9/10 of its original value, hence no. of mole will also fall up to thesame extent.

(20 – 2x) = 9 20 18

10

x = 1

[N2] = 4 x 4 1 3

1 1

mole/litre

[NH3] = 2x1

= 2 mole/litre

6. A(g) 3B(g) 4C(g)

t = 0 a a 0 teq. a-x a-3x 4xa - x = 4x given a = 5x

4 4

c 3 3

[C] [4x] 256K 832[A][B] [4x][2x]

49


7. 3 2

2 2p NH COK [p ] [p ] 2 1 4

8. A B C D

t = 0 a a 0 0teq. a - x a - x x x x = 2(a - x) 3x = 2a

x x x xK 4x x(a x)(a x)2 2

.

9. 2 4 3 2

O||

NH C ONH 2NH (g) CO (g)

Molecular massInitial vapour density D

2

79 392

D d 39 13 26 1

(n 1)d (3 1) 13 26

10. 5 3 2PCl PCl Cl

At equilibrium 2 2 2

5 Cl 23Cl P Cl

2 2 2P 3, p 3, p 36 6 6

3 2

5

PCl Clp

PCl

p p 1 1K 1p 1


1. i) The given equilibrium isCO2(g) + H2(g) CO(g) + H2O (g)

At t = 0 0.45 0.45 0 0At equilibrium (0.45 – x) (0.45 –x) x x

2

2

C)x45.0(

xK

50


0.11 = 2

2

)x45.0(x

x = 0.112

[CO2] = [H2] = (0.45 – x) = (0.45 – 0.112) = 0.3379 0.34 mole each[CO] = [H2O] = x = 0.112 mole each

ii) When 0.34 mole of CO2 and H2 are added in above equilibrium then following case exists.CO2(g) + H2(g) CO(g) + H2O (v)

At t = 0 (0.34 + 0.34) (0.34 + 0.34) 0.11 0.11At eqbm (0.68 – ) (0.68 – ) (0.11 +) (0.11 + )

2

2 2

[ ][ ][ ][ ]CCO H OKCO H

0.11 = 2

2

(0.11 )(0.68 )

= 0.086[CO2] = [H2] = (0.68 – 0.086) = 0.594 mole each[CO] = [H2O] = 0.11 + = 0.11 + 0.086 = 0.196 mole each

2. The given equilibrium reaction is :

H2 + I2 2HIAt t = 0 6 3 0At equilibrium (6 – x)/1 3 – x 2xconcentration at equilibrium

2

C(2x)K

(6 x)(3 x)

2

24x64

x 18 9x

On solving x = 2.84

[HI] =2x 2 2.84 5.681 1

; [H2] =

6-x 6-2.84= =3.161 1

[I2] = 6-x 6-2.84= =3.161 1

51


In the present reaction n = 0 hence, volume change will not affect the equilibrium.

3. No. of mole of PCl5 = Weight

Molecular weight = 5 0.024

208.5

The equilibrium for dissociation of PCl5 maybe represented as

PCl5 PCl3 + Cl2

At t = 0 0.024 0 0At equilibrium (0.024 - )

Total moles of gas components = 0.024 – + + = (0.024 + )

We know, PV = nRT

1 × 1.9 = (0.024 + ) 0.0821 × 523

(0.024 + ) = 1.9

0.0821 523

= 0.0202

Degree of dissociation =0.02020.024 = 0.843

[PCl5] = 0.024 0.024 0.0202 0.0038

1.9 1.9 1.9

[PCl3] = [Cl2] = 0.0202

1.9V

Kc =

2

43 2

5

0.0202[ ][ ] 4.08 101.9

0.0038[ ] 1.9 0.00381.9

PCl ClPCl

= 0.0565 mole/litre

Kp = Kc (RT) n

Where Kc = 0.0565

n = 2 – 1 = 1

4. Equilibrium of formation of NH3 may be given as,

N2 + 3H2 2NH3

At t = 0 1 3 2

52


At equilibrium (1 - x) 3(1 –x) 2x

Total moles at equilibrium = 1 – x+ 3 (1 – x) + 2x = 4 – 2x

Mole fraction of NH3 = 2x

(4 2x)

Partial pressure of NH3 = 2x p

(4 2x)

= 2x ×50

(4-2x)

25 = 2x×50(4-2x) or = 0.666

Kp = 2 2

2 416x (2-x)27p (1-x) =

2 2

2 2

16 (0.666) (2 0.666)27 (50) (0.334) = 1.677 × 10–3

5. The given equilibrium is

2SO2(g) + O2(g) 2SO3(g)

23

C 22 2

[SO ]K =

[SO ] [O ] … (i)

i) If [SO3] = [SO2]

Then C2

1K =[O ]

2C

1 1[O ]= = =0.01K 100 mole/litre

Total moles of O2 present = 0.01 × 10 = 0.1 mole

Volume of vessel is 10 litre

ii) When [SO3] = [SO2]

Then2

2C 2

2 2

[SO ]K =[SO ] [O ] from Eq. (i)

2

4100=[O ]

53


[O2] = 4/100 = 0.04 mole/litre

Total moles of O2 in vessel at equilibrium = 0.04 × 10 = 0.4 mole

6. If 2x is the partial pressure of SO3 that is decreased at equilibrium, we would have

2SO2(g) + O2(g) 2SO3(g)

t = 0 0 2 atm 1 atm

teq 2x 2 atm + x 1 atm – 2x

Hence, 3

2

2 2SO -1

p 2 2SO 2

(p ) (1atm-2x)K = = =900 atm(p ) (pO ) (2x) (2atm+x)

Assuming x << 2 atm, we get

3

2

2 2SO -1

p 2 2SO 2

(p ) (1atm-2x)K = = =900atm(p ) (pO ) (2x) (2atm+x)

or 2

2

(1atm-2x) =1800(2x) or

1atm -1=42.432x

or x = 1

2 43.43atm = 0.0115

Hence, p(SO2) = 2x = 0.023 atm; p(O2) = 2 atm + x = 2.0115 atm

and p(SO3) = 1 atm – 2x = 0.977 atm

7. We have

2H2S (g) 2H2 (g) + S2(g)

n(1 - a) na + n(a/2)

Total amount of gases = n(1 + a/2)

Hence, 2H S

1-αp = p;1+α/2 2H

αp = p;1+α/2 2S

α/2p = p1+α/2

54


Thus, 2

2

2 32 2

p 22 2H S

α α/2p p(pH ) (ps ) α p1+α/2 1+α/2K = = =

(p ) 2(1-α) (1+α/2)1-α p1+α/2

Substituting the given value of a, we get

3-2

p 2

(0.31) (1atm)K = =2.71×10 atm2(1-0.31) (1+0.31/2)

3-2

p 2

(0.31) (1atm)K = =2.71×10 atm2(1-0.31) (1+0.31/2)

For the given reaction, vg = +1

Hence, Kc = Kp(RT)-Dvg

= (2.71 × 10-2 atm) {(0.0821 L atm K-1 mol-1) (1380 K)}-1

= 2.39 × 10-4 mol L-1

8. (A) Let n be the initial amount of COCl2 and a be its degree of dissociation. We will have

COCl2 CO + Cl2

t = 0 n

teq n(1-a) na na

Total amount of gases = n(1 - a) + na + na = n(1 + a)

Now, the volume of the flask would be

n(1+α)RTV=p

The density of the mixture would be

or 2 2 2COCl COCl COClnM nM p pMρ= = =

V n(1+α)RT (1+α)RT

= 2COClpM-1

ρRTSubstituting the given values, we get

55


-1

-1 -1 -1

(1atm)(99g mol )α= 1=1.4331=0.433(1.162g L )(0.0821L atm K mol )(724K)

(B) The partial pressures of the species involved in the reactions are

2COCl1-αp = p;1+α CO

αp = p;1+α 2Cl

αp = p;1+α Hence,

2

2

2CO Cl

p 2COCl

p p αK = = pp 1-α

Which gives 2

p 2

(0.433)K = (1atm)=0.2311-(0.433)

9. Let n be the amount of N2O4 at t = 0 and let a be the degree of dissociation of N2O4 at equilibrium.Then

N2O4 (g) 2NO2 (g)

t = 0 n 0

teq n(1 - a) n(2a)

Total amount at equilibrium = n(1 + a)

Using the ideal gas equation, we get

pV = n(1 + a)RT = m (1+α)RTM (or) pM = r(1 + a)RT (or) =

pM -1ρRT

Substituting the given data, we get

= -1

-3 3 -1 -1

(1atm)(92g mol ) -1=1.752-1=0.752(1.84g dm )(0.082dm atm K mol )(348K)

Hence, 2

2 4

2 2 2 2NO

p 2 2N O

p [2αp/(1+α)] 4α 4(0.752)K = = = p= (1atm)=5.206atmp (1-α)p/(1+α) 1α 10.752

10. If p is the partial pressure of N2O5 that has decomposed,

N2O5 (g) 2NO2(g) + 212

O

600 mm Hg – p 2p p/2

Pressure at any time = (600mmHg – p) + 2p + p/2 = 600 mmHg + (3/2)p

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Equating this to 960 mmHg, we get

p = (2/3) (960 – 600) mmHg = 244 mmHg

The mole fraction of N2O5 decomposed would be 244mmHgx= =0.407600mmHg


1. (A) The given reaction will be exothermic in nature due to the formation of three X - Y bonds fromthe gaseous atmos. The reaction is also accompanied with the decrease in the gaseous species.Hence, the reaction will be affected by both temperature and pressure. The use of catalyst doesnot affect the equilibrium concentrations of the species in the chemical reaction.

2. (B)

5p

c n 2

K 1.44 10n 2 4 2 K(RT) (0.082 773) (R in L atom K-1 mole-1).

3. (D) At initial stage of reaction, concentration of each product will increase and hence Q willincrease.

4. (D) With change of pressure, x will change in such a way that Kp remains a constant.

5. (D) For the equilibria :

2 4 2N O (g) 2NO (g), Kp = Kc because here n 1

np c[k k (RT) ] Since temperature is constant so kc or kp will remain constant. Further since

volume is halved, the pressure will be doubled so will decrease so as to maintain the constancyof kc or kp.

2 4 2N O 2NO

initial 1 0

at equilibrium ( 1 ) 2

Total mole = 1 2 1

Let total pressure = P

2 2 4

1 2pNO p; pN O p1 1

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2 22

p2 4

(pNO ) 4 pkpN O (1 )(1 )

Since kp = constant, so 1p

So when volume is halved, pressure gets doubled and thus will decrease.

6. (A) In a reversible reaction, catalyst speeds up both the forward and backward reactions to thesame extent so (C) is wrong. At equilibrium

products reac tan tsG G G 0

3 2 2NH N H2G (G 3G ) 0

or 3 2 2NH N H2G G 3G

7. 4 3(g) 2 (g)NH HS(s) NH H S

Initial moles 3.0651 0 0

Moles at eq. 3.06 7051 100

3.06 3051 100

3.06 3051 100

Given V = 2 litre, T = 300 K, n 2 0 2

c 3 2

3.06 30 3.06 30k [NH ][H S]51 100 2 51 100 2

5 2 28.1 10 mol litre

Also,

n 5 2p ck k (RT) 8.1 10 (0.082 300)

2 24.90 10 atm

Addition of more NH4HS on this equilibrium will cause no effect because concentration of NH4HSis not involved in formula of kp or kc.

$RI8H7A9

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