equilibrium mcq ans

7/24/2019 Equilibrium MCQ Ans

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Multiple Choice Answers

1 C Acid-alkali neutralization is irreversible. On heating strongly, NH4Cl decomposes to give NH3and

HCl. While on cooling, NH3combines with HCl to give NH4Cl again.

2 B For (1), hydrated copper(II) sulphate thermally decomposes to copper(II) sulphate and water. While

on cooling and adding water, copper(II) sulphate would be converted back to hydrated copper(II)

sulphate. For (3), ammonia is a weak alkali which only ionizes slightly in water.

3 A Rusting of iron is an irreversible reaction.

4 D

5 C

6 A

7 C (1) is a reversible reaction, while (2) and (3) are irreversible reactions.

8 D When 18O2is introduced into the system, it will react with SO 2(g) to form SO3(g). Then some SO3

formed by combining 18O2with SO2will decompose to SO2and O2. Therefore, when the reaction

mixture reaches equilibrium, all the reactants and products will contain 18O.

9 D Reversible reactions will not go to completion.

10 A Carbonic acid is a weak acid that ionizes only slightly in water. Carbonic acid molecules ionize to

form carbonate ions and hydrogen ions. At the same time, carbonates ions and hydrogen ions combine

to form carbonic acid molecules.

11 D The reaction between zinc and hydrochloric acid is not a reversible reaction.

12 C

13 D Only the reaction in D is reversible as the reaction is carried out in a closed system, while reactions

in A and C are irreversible. For the reaction in B, some materials are lost in an open system and the

system cannot achieve equilibrium.

14 B Ester is produced during the progress of the reaction, therefore a fruity smell can be detected even

though it is not at equilibrium.

15 B (1) is incorrect because the forward and backward reactions are still continuing at equilibrium. (3)

is incorrect because the concentrations of reactants and products remain unchanged but they are not

necessarily the same at equilibrium.

16 C (2) is incorrect because the reaction occurs only at about 825C.

17 C At equilibrium, the rate of forward reaction is equal to the rate of backward reaction. The

concentrations of reactants and products remain constant at equilibrium, but they are not necessarily the

same.

18 D N2O4(g) and NO2(g) are colourless and brown respectively. As time passes, N2O4(g) decomposes

to NO2(g), so the colour of the reaction mixture changes from colourless to brown. The reaction

mixture then reaches equilibrium after some time and the brown colour persists.

19 C Catalysts increase both the rates of forward and backward reactions, so the time for the reversible

reactions to reach equilibrium is shortened. Catalysts do not affect the yield of products for reversible

reactions.20 D (1) is correct because equilibrium of reversible reactions can only be established in a closed system.


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As the reaction proceeds, the rate of forward reaction decreases because the concentration of reactants

decreases while the rate of backward reaction increases because the concentration of products increases.

Therefore, (2) is incorrect.

21 D

22 D (3) is correct because the concentrations of H2S(g), H2(g) and S2(g) remain unchanged.

23 A (2) is correct because constant temperature is required to maintain the equilibrium position.

However, when N2O4decomposes, the pressure is no longer constant until an equilibrium is reached.

24 C At equilibrium, there are no observable changes of the reaction mixture because the concentrations

of reactants and products remain unchanged.

25 B (2) is incorrect because the concentrations of reactants and products remain unchanged at

equilibrium. Their concentrations are not necessarily the same at equilibrium.

26 A (2) is incorrect because the rates of forward and backward reactions are the same at equilibrium.

Therefore, the brown colour of the gaseous mixture persists. (3) is incorrect because the concentrations

of NO2(g) and N2O4(g) remain unchanged at equilibrium.

27 D (3) is correct because the rate of forward reaction is equal to the rate of the backward reaction at

equilibrium.

28 A As time passes, the concentration of N2O4decreases and the concentration of NO2increases. When

equilibrium is reached, the concentrations of N2O4and NO2remain unchanged. Thus, B and D are

incorrect. When 1 mol of N2O4decomposes, 2 mol of NO2forms. Therefore, the rate of decrease in

concentration of N2O4should be half of the rate of increase in concentration of NO 2. Thus, C is

incorrect.

29 C (1) is incorrect because equilibrium can only be established in a closed system, where no materials

can enter or leave the system. (3) is correct because both reactants and products are present and their

concentrations remain unchanged at equilibrium.

30 D At equilibrium, when 1 mol of HI(g) decomposes, 0.5 mol of I2(g) and 0.5 mol of H2(g) combine at

the same time.

31 C At chemical equilibrium, both forward and backward reactions are taking place continuously at the

same rate.

32 D Equilibrium can only be established in a closed system.

33 A (1) is correct because the rate of forward reaction is equal to the rate of backward reaction at

equilibrium. (3) is incorrect because the concentrations of reactants and products remain unchanged at

equilibrium, but not equal.

34 B Concentrated sulphuric acid increases both the rates of forward and backward reactions.

35 A B is incorrect since the equilibrium is started from the left-hand side. P is being formed. C is

incorrect since the rate of the forward reaction does not decrease linearly. The reaction above is

reversible. D is incorrect since equilibrium needs some time to achieve.

36 B (2) is incorrect. The equilibrium system above can exist in room temperature.

37 B It is a dynamic situation and both forward and backward chemical reactions take place.38 D


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39 B

40 D In dynamic equilibrium, all macroscopic properties of a reaction mixture remain constant, and this

is caused by the equal reaction rates in both forward and backward directions, which also remain

constant.

41 C Dynamic equilibrium can only be attained in a closed system.

42 A An open test tube cannot provide a closed system which is necessary for attaining equilibrium.

43 C Since the starting material is the colourless gas, N2O4, A and D must be incorrect. Also, the given

reaction is a reversible reaction, B is incorrect because it would mean that all N2O4turns into NO2.

44 B Only the change in B takes place in a closed system as pressure cooker does not allow air or liquids

to escape.

45 D For an irreversible reaction, the concentrations of reactants are decreasing while the concentrations

of products are increasing. The reaction will stop when one of the reactants is used up. There is no

backward reaction for an irreversible reaction.

46 C The concentrations of products and reactants are not necessarily the same. It depends on the

equilibrium constant.

47 D Both the rates of disappearance ofA(g) and appearance ofB(g) are decreasing as the concentration

ofA(g) is decreasing. Finally both rates remain constant.

48 C pH and opacity of the system remain unchanged at equilibrium because pH relates to the

concentration of H+(aq) and opacity relates to the concentration of chemical species that blocks the

visible light. If the system is in equilibrium, the concentrations of these species must remain unchanged.

But temperature is an external factor that can disturb a closed system.

49 D Esterification is a reversible reaction. The acid catalyst should be concentrated sulphuric acid.

50 B

51 A

52 D When calcium carbonate is heated strongly in an open system, it only undergoes an irreversible

decomposition. Therefore, no equilibrium can be established.

53 B

54 A

55 A Weak acids like ethanoic acid only slightly ionize in water. Thus, at equilibrium, the solution of

ethanoic acid contains ethanoic acid molecules, ethanoate ions and hydrogen ions.

56 C 1ststatement is incorrect. A catalyst can reduce the time for a reversible reaction to reach

equilibrium by increasing both the rates of forward and backward reactions.

57 C The stoichiometric coefficients in the balanced chemical equation are involved in the expression for

the equilibrium constant.

58

B The equilibrium constant expression for the reaction is, Kc= 2eqm2

eqm42

)]g(NO[

)]g(ON[. Therefore, the unit is

the reciprocal of mol dm3

, that is mol1

dm3

.59 C The stoichiometric coefficients of PF5, P4, and F2are 4, 1 and 10 respectively.


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60 D The equation for the equilibrium: N2(g) + 3H2(g) 2NH3(g). The above expressions except D are

possible according to the stoichiometric coefficients and the way of expression of the reversible

reaction.

61 C

62 B

The unit of Kcis 5343

6343

)dmmol()dmmol(

)dmmol()dmmol(

= mol dm3.

63 C

K =eqm2eqm2

eqmeqm2

)]g(H[)]g(CO[

)]g(CO[)]g(OH[, K1=

eqm2

eqm

)]g(CO[

)]g(CO[, K2=

eqm2

eqm2

)]g(OH[

)]g(H[

K =2

1

K

K

64 D The expression of the equilibrium constant for a reaction aA + bB cC + dD is Kc=

eqmb

eqma

eqmd

eqmc

[B][A]

[D][C].

65 C

66 D The reaction of the Haber Process is N2(g) + 3H2(g) 2NH3(g).

67

A The equilibrium constant expression for the reaction is, Kc= 4eqm2

4eqm2

)]g(H[

)]g(OH[. The concentrations of

solid reactants are constant throughout the reaction and they do not appear in the expression for Kcof

heterogeneous equilibria. The concentration units cancel out in the expression for Kc, thus Kchas no

unit.

68 B Si is a solid, so the concentration of Si in a heterogeneous equilibrium does not appear in the

expression of Kc.

69 B The concentrations of solids or liquids in heterogeneous equilibria do not appear in the expressionof Kc, so

Kc= 5eqm

5eqm2eqm2

)]g(CO[

)]g(CO[])g(I[.

70

B The unit of Kcis 23

323

)dm(mol

)dm(mol)dm(mol

= mol dm3

71 A Kc=eqm32

eqm2

eqm2

(s)]CO[Ag(g)][COO(s)][Ag is not correct because the equilibrium concentrations of solids are

so large that they are considered to be constant. The concentrations of solids in heterogeneous equilibria


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do not appear in the expression of Kc.

72

B Kc=eqm2eqm2

eqm2

(g)][I(g)][H

[HI(g)]. The reversed reaction has an equilibrium constant, Kc expression =

eqm2

eqm2eqm2

[HI(g)]

(g)][I(g)][H

which equalscK

1

.

73D Kc=

eqm3

2eqm2

eqm2

3

(g)][H(g)][N

(g)][NH; while Kc =

eqm2

3

2eqm2

1

2

eqm3

(g)][H(g)][N

(g)][NH, hence, Kc= (Kc)

2, then Kc'

= cK .

74 C Chemicals that are liquids and solids, which do not change their concentration (their densities)

significantly under constant temperature, are not counted when expressing the equilibrium constant.

75 A

76

D Since Kc=eqm2eqm2

eqm2

(g)][O(g)][N

[NO(g)], the unit of Kcis

)dm)(moldm(mol

)dm(mol3-3

23

, which has no unit.

77 A

78A Kcof the reverse process is

1.45

1 = 0.69

79 C The iron and iron(II)(III) oxide are solids, so their concentrations should be neglected in the

equilibrium constant expression.

80 B If the sum of the coefficients of product side is equal to the sum of coefficients of the reactant side,

then the powers of the concentrations in the equilibrium constant expression will be cancelled out.

81 B For the reaction: CH3COOH + CN HCN + CH3COO

Kc=eqmeqm3

eqm3eqm

][CNCOOH][CH

]COO[CH[HCN]

= K

Then for the reaction: HCN + CH3COO CH3COOH + CN

Kc =eqm3eqm

eqmeqm3

]COO[CH[HCN]

][CNCOOH][CH

=K

1

82 B Homogeneous equilibrium means all the reactants and products are in the same phase.

83 C In the equilibrium constant expression, concentrations of solids in a heterogeneous equilibrium

should be considered as constant.

84D For the reaction:A+B C, the equilibrium constant expression =

eqmeqm

eqm

][][

][

BA

C = K;then, for

the reaction: 2C 2A+ 2B, the equilibrium constant expression = 2eqm

2eqm

2

eqm

2

)1(=][

][][KC

BA = K2

85 B The species in the nominator are products in an equation, while the species in the denominator are


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reactants. The coefficients in the equation are the powers of the concentrations of the chemical species.

86 C Equilibrium constant of a chemical reaction depends on temperature only. At equilibrium, the rate

of forward reaction is equal to the rate of backward reaction no matter what the value of Kcis. If a

reaction has a large Kcvalue, it indicates that the equilibrium favours the formation of products rather

than the reactants. The Kcvalue does not give any information about the rate of the reaction.

87 B (2) is incorrect because the magnitude of Kcindicates the extent of a chemical reaction but not the

rate of the reaction.

88 B When the numerical value of Kcis much smaller than 1, the equilibrium position lies mainly to the

reactant side. Equilibrium II has the smallest Kcamong the above equilibrium systems, so it favours the

reactants most.

89 A An equilibrium constant is valid only when the reaction has attained equilibrium and takes place

under a specified temperature.

90

A Equilibrium constant =eqm2eqm

22

eqm

2

3

(g)][O(g)][SO(g)][SO =

M)10(3.70M)10(5.33M)10(5.33

323

23

= 2.70 102

M1.

For the set of concentrations in option A,

eqm2eqm2

2

eqm2

3

(g)][O(g)][SO

(g)][SO =

M)10(1.013M)10(4.72

M)10(2.47522

23

= 2.70 102M

1= Equilibrium

constant.

91 D A larger Kcmeans the equilibrium position lies mainly to the product side. i.e. there is a higher

tendency to proceed towards completion.

92

D Kc=eqm

eqmeqm+

(aq)][

(aq)][OH(aq)]H[

B

B

; for a weak base, the concentration of OH(aq) is very low, the

equilibrium position lies mainly on the left-hand side. The nominator is very small while the

denominator is very large, so Kcshould be very small.

93

B The expression for the equilibrium constant, Kc=eqm2

2eqm

(g)][CO

[CO(g)], thus Kc= 3

23

dmmol42.0

)dmmol82.0(

=

1.60 mol dm3.

94 A

Letx mol dm

3

be the change in concentration of C2H5CO2H(l).Concentration C2H5CO2H(l) + C2H5OH(l) C2H5CO2C2H5(l) + H2O(l)


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(mol dm3)

Initial 1 1 0 0

Change x x +x +x

Equilibrium 1 x 1 x 0 +x=x 0 +x=x

eqm52eqm252

eqm2eqm52252c

OH(l)]H[CH(l)]COH[CO(l)][H(l)]HCCOH[CK

2)1(4

x

x2

Solving forx,

4(1 2x+x2) =x

2

x= 0.67 orx= 2 (rejected)

Equilibrium concentration of propanoic acid = 1 0.67 mol dm3= 0.33 mol dm3

95 B

After mixing the two solutions,

initial concentration of I2(aq) = 0.2 mol dm3

500

250 = 0.1 mol dm3

initial concentration of I(aq) = 0.2 mol dm3500

250 = 0.1 mol dm3

Letx mol dm3be the change in concentration of I2(aq).

Concentration (mol dm3) I2(aq) + I(aq) I3

(aq)

Initial 0.1 0.1 0

Change x x +x

Equilibrium 0.1 x 0.1 x 0 +x=x

eqmeqm2

eqm3c

)]aq([I(aq)][I

)]aq([IK

2)1.0(75.68 x

x

Solving forx,

68.75(0.01 0.2x+x2) =x

x= 0.0684 orx= 0.146 (rejected)

96 A

The amount of N2(g) consumed = 0.60 mol dm3 0.3 = 0.18 mol dm3

Concentration

(mol dm3) N2(g) + 3H2(g) 2NH3(g)

Initial 0.6 1.8 0


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Change 0.18 3(0.18) +2(0.18)

Equilibrium 0.6 0.18 = 0.42 1.8 3(0.18) = 1.26 0 + 2(0.18) = 0.36

333

23

eqm3

2eqm2

eqm2

3c

)dmmol26.1)(dmmol42.0(

)dmmol36.0(

(g)][H(g)][N

(g)][NHK

= 0.154 mol2dm6

97 D

Kc= eqm3eqm

34

5eqm

2 )]aq(OH[)]aq(PO[)]aq(Ca[

Kc= (5 2 104mol dm3)5(3 2 104mol dm3)3(2 104mol dm3)

= 4.32 1029mol9dm27

98 D

Initial concentration of CO(g) =3dm0.3

mol4.0 = 0.133 mol dm3

Initial concentration of Cl2(g) = 3dm0.3

mol2.0 = 0.0667 mol dm3

Letx mol dm3be the change in concentration of CO(g).

Concentration (mol dm3) CO(g) + Cl2(g) COCl2(g)

Initial 0.133 0.0667 0

Change x x +x


Kc=eqm2eqm

eqm2

)]g(Cl[)]g(CO[

)]g(COCl[

0.41 =)0667.0)(133.0( xx

x

Solving forx,

0.41(0.133 x)(0.0667 x) =x

x= 3.37 103orx= 2.64 (rejected)

[COCl2(g)]eqm= 3.37 103mol dm3

99 A

Initial number of moles of SO2Cl2(g) = 1molg)25.3520.161.32(

g5.13

= 0.10 mol

Initial concentration of SO2Cl2(g) = 3dm4

mol10.0 = 0.025 mol dm3


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Equilibrium concentration of Cl2(g) = 3dm4

mol069.0 = 0.01725 mol dm3

Concentration (mol dm3) SO2Cl2(g) SO2(g) + Cl2(g)

Initial 0.025 0 0

Change 0.01725 +0.01725 +0.01725Equilibrium 0.025 0.01725

= 7.75 103

0.01725 0.01725

Kc=eqm22

eqm2eqm2

)]g(ClSO[

)]g(Cl[)]g(SO[

Kc=)dmmol10(7.75

)dmmol)(0.01725dmmol(0.0172533

33

= 0.0384 mol dm3

100 B

Initial concentration of SbCl3(g) = 3dm5.2

mol28.0 = 0.112 mol dm3

Initial concentration of Cl2(g) = 3dm5.2

mol16.0 = 0.064 mol dm3

Letx mol dm3be the change in concentration of SbCl3(g).

Concentration (mol dm3) SbCl3(g) + Cl2(g) SbCl5(g)

Initial 0.112 0.064 0Change x x +x


Kc=eqm2eqm3

eqm5

)]g(Cl[)]g(SbCl[

)]g(SbCl[

40 =)064.0)(112.0( xx

x

Solving forx,

40(0.112 x)(0.064 x) =x

x= 0.0463 orx= 0.155 (rejected)

[SbCl3(g)]eqm= 0.112 0.0463 mol dm3= 0.0657 mol dm3

101 A

Initial concentration of N2(g) = 3dm

mol

V

x

Initial concentration of H2(g) = 3dm

mol

V

y


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Equilibrium concentration of NH3(g) = 3dm

mol2

V

z

Concentration

(mol dm3)N2(g) + 3H2(g) 2NH3(g)

Initial

V

x

V

y

0

Change

V

z

V

z3 +

V

z2

Equilibrium

V

zx

V

zy 3 0 +

V

z2 =

V

z2

Kc= 3eqm2eqm2

2eqm3

)]g(H[)]g(N[

)]g(NH[ =

3

2

)3

)((

)2

(

V

zy

V

zxV

z

=

3

2

)3)((

4

zyzx

Vz2

102 D

After mixing the 2 solutions,

initial concentration of Fe3+(aq) = 1.0 mol dm3100

50 = 0.5 mol dm3

initial concentration of SCN(aq) = 1.0 mol dm3100

50 = 0.5 mol dm3

Number of moles of FeSCN2+(aq) at equilibrium =1molg0.140.121.328.55

g94.1

= 0.0170 mol

Equilibrium concentration of FeSCN2+(aq) =3dm1.0

mol0170.0 = 0.17 mol dm3

Concentration

(mol dm3)Fe3+(aq) + SCN(aq) FeSCN2+(aq)

Initial 0.5 0.5 0

Change 0.17 0.17 +0.17Equilibrium 0.5 0.17 = 0.33 0.5 0.17 = 0.33 0 + 0.17 = 0.17

Kc=eqmeqm

3

eqm2

)]aq(SCN[)]aq(Fe[

)]aq(FeSCN[

=)dmmol33.0)(dmmol33.0(

dmmol17.033

3

= 1.56 mol1dm3

103 D

Letx mol dm

3

be the change in concentration ofA(aq)

Concentration (mol dm3) A(aq) 2B(aq)


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Initial 0.97 0

Change x +2x

Equilibrium 0.97 x 0 + 2x= 2x

Kc=eqm

2eqm

)]aq([

)]aq([

A

B

180 =)97.0(

)2( 2

x

x

Solving forx,

180(0.97 x) = 4x2

x= 0.950 orx= 45.9 (rejected)

Equilibrium concentration ofB= 2 0.950 mol dm3= 1.90 mol dm3

104 A A homogeneous equilibrium means all the reactants and products are in the same phase at

equilibrium. Since Kc> 1, [C]eqm[D]eqm> [A]eqm[B]eqm. (3) is incorrect because raising the temperature

may not favour the forward reaction.

105 A The concentrations of solids and liquids in heterogeneous equilibria do not appear in the expression

of Kc.

106 B It should be known that the concentrations of solids and liquids in heterogeneous equilibria do not

appear in the expression of Kc. Thus, the answer is 1 mol dm3rather than

2

1.

107 C The equations for forward reaction and backward reaction are reversed to each other, thus the

equilibrium constants are the reciprocal of each other.

108 C For (1), the constant should be reciprocal if the equation is written in reverse. For (3), the unit of the

equilibrium constant depends on the stoichiometric coefficients and the number of species present in the

equation.

109 D

Initial concentration of SO2(g) = 0.05 mol dm3

Initial concentration of O2(g) = 0.03 mol dm3

Consider the equilibrium,

Concentration (mol dm3) 2SO2(g) + O2(g) 2SO3(g)

Initial 0.05 0.03 0

Change 0.04 0.02 +0.04

Equilibrium 0.05 0.04

= 0.01

0.03 0.02

= 0.01

0 + 0.04 = 0.04

Kc=eqm2

2eqm2

2eqm3

)]g(O[)]g(SO[

)]g(SO[=

)dmmol01.0()dmmol01.0(

)dmmol04.0(323

23

= 1600 mol1dm3


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110 A

Initial concentration of PCl5(g) = 3

3

dm25.0

mol100.1 = 4 103mol dm3

Letxmol dm3be the change in concentration of PCl5(g).

Concentration (mol dm3) PCl5(g) PCl3(g) + Cl2(g)

Initial 4 103 0 0

Change x +x +x

Equilibrium 4 103x 0 +x =x 0 +x=x

Kc=eqm5

eqm2eqm3

)]g(PCl[

)]g(Cl[)]g(PCl[

0.106 =)104( 3

2

x

x

Solving forx,

)104(106.0 32 xx

x= 3.86 103orx= 0.110 (rejected)

the equilibrium concentration of PCl5(g) = 4 1033.86 103mol dm3

= 1.4 104mol dm3

111 D

Initial concentration of HCl(g) =3dm1

mol8.0 = 0.8 mol dm3

Initial concentration of O2(g) = 3dm1

mol5.0 = 0.5 mol dm3


Concentration

(mol dm3)4HCl(g) + O2(g) 2Cl2(g) + 2H2O(g)

Initial 0.8 0.5 0 0

Change 0.4 0.1 +2(0.1) +2(0.1)Equilibrium 0.8 0.4

= 0.4

0.5 0.1

= 0.4

0 + 0.2 = 0.2 0 + 0.2 = 0.2

Kc=eqm2

4eqm

2eqm2

2eqm2

)]g(O[)]g(HCl[

)]g(OH[)]g(Cl[ =

)dmmol4.0()dmmol4.0(

)dmmol2.0()dmmol2.0(343

2323

= 0.156 mol1dm3

112 C

Initial concentration of N2(g) = 3dm5.0mol35.0 = 0.7 mol dm3


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Initial concentration of O2(g) = 3dm5.0

mol30.0 = 0.6 mol dm3

Letx mol dm3be the change in concentration of N2(g)

Concentration (mol dm3) N2(g) + O2(g) 2NO(g)

Initial 0.7 0.6 0

Change x x +2x

Equilibrium 0.7 x 0.6 x 0 + 2x= 2x

Kc=eqm2eqm2

2eqm

)]g(O[)]g(N[

)]g(NO[ =

)6.0)(7.0(

)2( 2

xx

x

Solving forx,

13 =)6.0)(7.0(

4 2

xx

x

13(0.7 x)(0.6 x) = 4x2

x= 0.415 orx= 1.46 (rejected)

equilibrium concentrations:

[N2(g)]eqm= 0.7 0.415 mol dm3= 0.285 mol dm3

[O2(g)]eqm= 0.6 0.415 mol dm3= 0.185 mol dm3

[NO(g)]eqm= 2(0.415) mol dm3= 0.83 mol dm3

113 D

After mixing the 2 solutions,

initial concentration of CH3COOH = 0.05 mol dm3

3

3

cm50

cm20 = 0.02 mol dm3

initial concentration of C2H5OH = 0.03 mol dm3

3

3

cm50

cm30 = 0.018 mol dm3


Concentration

(mol dm3)CH3COOH(l) + C2H5OH(l) CH3COOC2H5(l) + H2O(l)

Initial 0.02 0.018 0 0

Change 0.012 0.02 =

0.008

0.008 +0.008 +0.008

Equilibrium 0.012 0.018 0.008 =

0.01

0 + 0.008

= 0.008

0 + 0.008

= 0.008

114 A

Initial concentration of N2O4= 3dm2mol98.0 = 0.49 mol dm3


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Concentration (mol dm3) N2O4(g) 2NO2(g)

Initial 0.49 0

Change 0.2 +0.4

Equilibrium 0.49 0.2 = 0.29 0 + 0.4 = 0.4

Kc=eqm42

2eqm2

)]g(ON[

)]g(NO[ =

3

23

dmmol29.0

)dmmol4.0(

= 0.55 mol dm3

115 C

Concentration (mol dm3) PCl5(g) PCl3(g) + Cl2(g)

Initial 1.0 0 0

Change 0.5 0.5 +0.5

Equilibrium 0.5 0.5 0.5

Number of moles of PCl5(g) at equilibrium = 0.5 mol dm3 1 dm3= 0.5 mol

Number of moles of PCl3(g) at equilibrium = 0.5 mol dm3 1 dm3= 0.5 mol

Number of moles of Cl2(g) at equilibrium = 0.5 mol dm3 1 dm3= 0.5 mol

Total number of moles of gases at equilibrium = 1.5 mol

116 C

Concentration (mol dm3) N2O4(g) 2NO2(g)

Initial 1.0 0

Change 0.9 2 (+0.9)

Equilibrium 0.1 1.8

Kc= 3

23

dmmol0.1

)dmmol(1.8

= 32.4 mol dm3

117 D

Concentration

(mol dm3)

CH3CH2OH(l) + CH3COOH(l) CH3COOCH2CH3(l) +

2H O(l)

Initial 3.0 4.0 0 0

Change 2.2 2.2 +2.2 +2.2

Equilibrium 0.8 1.8 2.2 2.2


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Kc=)dmmol)(1.8dmmol(0.8

)dmmol)(2.2dmmol(2.233

33

= 3.36

118 D

Initial concentration of H2(g) = 3dm2

mol3 = 1.5 mol dm3

Letxmol dm3be the change in concentration of H2S(g).

Concentration (mol dm3) H2(g) + S(s) H2S(g)

Initial 1.5 / 0

Change x / +x

Equilibrium 1.5 x / x

Kc= 6.8 102=

)x5.1(

x

0.068 (1.5 x) =x

0.102 0.068x=x

x= 0.0955

Number of moles of H2S(g) at equilibrium at 90C = 0.0955 mol dm3 2 dm3= 0.191 mol

119 B

Assume the volume of the closed vessel was 1 dm3.

Concentration

(mol dm3)N2(g) + 3H2(g) 2NH3(g)

Initial 1 3 0

Change 2

90. 3

2

90

. +0.9

Equilibrium 1

( 2

90.

) 3

( 32

90

.

) 0.9

Number of moles of nitrogen at equilibrium = 1 (2

90.) = 0.55 mol

Number of moles of hydrogen at equilibrium = 3 ( 32

90

.) = 1.65 mol

120C Kc=

eqm5

eqm2eqm3

(g)][PCl

(g)][Cl(g)][PCl =

M)10(7.6

M)10(2.09M)10(2.13

22

= 0.05775 mol dm3.

121 A

Suppose

1

dm3

of

1

M

HCl(aq)

and

1

dm3

of

1

M

CH3COONa(aq)

are

mixed

together.

The

final

volumeofthenewsolutionis2dm3.Hence,theinitialconcentrationsofH

+(aq)andCH3COO

(aq)are


16/24

halved,i.e.0.5M.

Letxmoldm3

bethechangeinconcentrationofCH3COOH(aq).

Concentration

(mol dm3)CH3COOH(aq) H

+(aq) + CH3COO(aq)

Initial 0 0.5 0.5Change +x x x

Equilibrium x 0.5 x 0.5 x

The equilibrium constant for the above backward reaction iscK

1 = 31 dmmol

0.304

1 .

cK

1 =

eqm3eqm

eqm3

(aq)]COO[CH(aq)][H

COOH(aq)][CH

0.304 =x

xx )(0.5)(0.5

0.304x= 0.25 x+x2

x21.304x+ 0.25 = 0

x= 0.234 or 1.070 (rejected)

122 A Let amol dm3be the change in concentration of CO2(g).

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

Initial 1 1 0 0

Change a a +a +a

Equilibrium 1 a 1 a a a

Kc=eqm2eqm

eqm2eqm2

O(g)][H[CO(g)]

(g)][CO(g)][H

4.24 =))(1(1

aa

aa

a2= 4.24 8.48a+ 4.24a2

a= 0.673 or 1.944

However, ahas to be smaller than 1 otherwise the equilibrium concentration of CO(g) and H 2O(g) will

become negative, which is impossible to happen. Therefore 1.944 is rejected.

123B 25 =

1)(1.11)(0.1

[HI(g)] eqm2, [HI(g)]eqm= 1.75

124 B If the value of equilibrium constant for the reaction 2X(g) + Y2(g) 2Z(g) is 0.2 mol1dm3, then

the value of equilibrium constant for its reverse reaction, 2Z(g) 2X(g) + Y2(g), is 31dmmol0.2

1

.

So, the value of equilibrium constant for the reactionZ(g) X(g) +2

1Y2(g) is 3 = 2.1dmmol0.2

1

24

2

3

2

1

dmmol

.

125 A Assuming the volume of container is 1 dm3.


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Concentration

(mol dm3)W(g) + X(g) Y(g) + Z(g)

Initial 0.6 0.6 0 0

Equilibrium 0.5 0.5 0.1 0.1

Kc= 2

2

(0.5)

(0.1) = 0.04

126 B

Kc= 2

3

(2)(1)

x

6.60 =x3

x= 1.88

127A Kc=

eqmeqm2

eqm4

(aq)][(aq)][

(aq)][

BA

C. By substituting each set of concentrations into the equilibrium

constant expression, only set A gives the value of Kc= 0.1 mol dm3.

128 C

For the reaction: N2(g) + 3H2(g) 2NH3(g), Kc=eqm

3eqm

eqm2

3

(g)][H(g)][N

(g)][NH

22

= 0.105 mol2dm6;

For the reaction:2

1N2(g) +

2

3H2(g) NH3(g), Kc =

eqm2

3

2eqm2

1

2

eqm3

(g)][H(g)][N

(g)][NH, so Kc = c

=

K

62dmmol0.105 = 0.324 mol1dm3

129 C

Initial concentration of SO2(g):

=3dm30.0

mol1.00 = 0.0333 mol dm3

Initial concentration of O2(g):

=3dm30.0

mol0.80 = 0.0267 mol dm3

Equilibrium concentration of SO2(g):

=3dm30.0

mol0.40 = 0.0133 mol dm3

Concentration

(mol dm3)2SO2(g) + O2(g) 2SO3(g)

Initial 0.0333 0.0267 0

Change 0.0133 0.0333

= 0.0200 2

1 (0.0200)

+0.0200


18/24


19/24

Initial concentration of N2(g) = 3dm5.0

mol0.92 = 0.184 mol dm3

Initial concentration of O2(g) = 3dm5.0

mol0.51 = 0.102 mol dm3

Equilibrium concentration of N2(g) = 0.184 mol dm3 98.9% = 0.182 mol dm3

Concentration (mol dm3) N2(g) + O2(g) 2NO(g)

Initial 0.184 0.102 0

Change 0.182 0.184

= 0.002

0.002 2 (+0.002)

= 0.004

Equilibrium 0.182 0.102 0.002

= 0.1

0.004

Kc=eqm

2

eqm

2

eqm2

(g)][O(g)][N

[NO(g)]

=)dmmol)(0.1dmmol(0.182

)dmmol(0.00433

23

Kc= 8.79 104

134 B Fe3+(aq) cannot be oxidized by potassium dichromate. The change in the mass of the equilibrium

mixture is not significant. However, the colour intensity of the mixture varies with concentration of

FeSCN2+(aq). If we use a calibrated colorimeter, it is possible to determine the concentrations of all

chemical species in the equilibrium mixture.

135 B According to the Equilibrium Law, the equilibrium constant is a constant value for a reaction at a

given temperature, no matter what the initial concentrations of the reactants or products are. Thus, the

above two statements are facts derived from the Law, rather than having a causal relationship.

136 A

137 A

138 C The magnitude of Kcindicates the extent of a chemical reaction but not the rate of the reaction.

139 C The magnitude of equilibrium constant indicates the extent of a chemical reaction but not the rate of

the reaction.

140 B

141 C

The reaction quotient, Qc=)]g(PCl[

)]g(Cl)][g(PCl[

5

23

=3

33

dmmol45.0

)dmmol15.0)(dmmol15.0(

= 0.05 mol dm3

As Qc> Kc, more products will be consumed until Qc= Kc. Hence, the equilibrium position will shift

to the left.

142 A If QcKc, the equilibrium position shifts to the left, i.e. more reactant, X(g) is formed.

143 B

144 A (2) is incorrect. Addition of dilute NaOH(aq) removes the H+(aq) and shifts the equilibrium


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position of the reaction mixture to the right. (3) is incorrect. Addition of dilute H2SO4(aq) increases the

concentration of H+(aq) and shifts the equilibrium position of the reaction mixture to the left.

145 C Kcis always a constant at a fixed temperature.

146 B When NaOH(aq) is added to the mixture, the OH(aq) reacts with the H+(aq) in the mixture to form

H2O(l). The concentration of H+(aq) drops and therefore a shift of equilibrium position to the right

occurs.

147 A When HCl(aq) is added to the mixture, the concentration of H+(aq) increases and the equilibrium

position shifts to the left.

148 B Addition of H+(aq) increases the concentration of H+(aq) in the given reaction, so the equilibrium

position shifts to the left and the concentration of H2S(aq) increases. The addition of SO42(aq) or

Na+(aq) do not affect the position of equilibrium. Addition of OH(aq) removes the H+(aq) from the

given reaction, so the equilibrium position shifts to the right and the concentration of H 2S(aq)

decreases.

149 B (1) is incorrect. Addition of dilute hydrochloric acid increases the concentration of H+(aq) and

causes the equilibrium position of the system to shift to the right. (2) is correct. Sodium carbonate

removes the H+(aq) and causes the equilibrium position of the system to shift to the left. (3) is incorrect.

Sodium sulphite reacts with Cr2O72(aq) to give Cr3+(aq) which is green.

150 B (1) is incorrect. Sodium carbonate reduces the solubility of calcium carbonate as it shifts the

equilibrium position of the mixture to the left. (2) is correct. Since H+(aq) from nitric acid removes

CO32(aq), this shifts the equilibrium position of the system to the right. (3) is incorrect. Addition of

calcium nitrate solution increases the concentration of Ca2+(aq) and causes the equilibrium position of

the mixture to shift to the right.

151 B (1) is incorrect. Addition of dilute hydrochloric acid increases the concentration of H+(aq) and

causes the equilibrium position of the system to shift to the left. (2) is correct. Sodium hydroxide

removes the H+(aq) and causes the equilibrium position of the system to shift to the right. (3) is

incorrect. Addition of potassium bromide increases the concentration of Br(aq) and causes the

equilibrium position of the system to shift to the left.

152 C The colour of the mixture is pale yellow in equilibrium due to the presence of Br2(aq). Adding

more H+(aq) to the reaction mixture will cause the equilibrium position to shift to the left, therefore

more yellow Br2(aq) is produced. The colour will change from pale yellow to yellow.153

B The reaction quotient Qc= 222

[HI(g)]

(g)](g)][I[H =

23

53

M)10(2.50

M)10M)(4.1010(8.00

= 0.05248

As Qc(=0.05248) > Kc(=0.0202), the equilibrium position of the system will shift to the left to produce

more reactants, until the value of Q cequals Kc.

154 C When there is a change in conditions in the above reaction, no matter the concentrations of the

chemical species, temperature or pressure of the system, according to Le Chteliers Principle, the

equilibrium position will shift in a way that opposes the disturbance. Therefore, the concentrations forboth reactants and products change.

155 A The equilibrium position will shift to the left and the concentration of H+(aq) decreases, then pH


21/24

will increase.

156 C If Qcis smaller than Kc, the equilibrium will shift to the right, producing more Ag(s) and Fe3+(aq)

until Qc= Kc.

157 B The equilibrium constant is dependent on temperature only, regardless of the changes in

concentrations of reactants or products.

158 A An increase in temperature will increase both the forward and backward reaction rates but to

different extents. However, whether the Kcincreases or decreases depends on the reaction type, thus,

exothermic or endothermic.

159 C Kcdepends on the temperature only.

160 A As the forward reaction is exothermic, a decrease in temperature shifts the equilibrium to the right

giving more products. Thus, the Kcvalue increases.

161 B H3NBF3(s) is a solid. If the system is at equilibrium, adding more solid would not affect the

equilibrium position. Also, the equilibrium constant at a constant temperature does not change.

162 A (1) is correct. Since the forward reaction is an endothermic reaction, so an increase in temperature

shifts the equilibrium position of the system to the right. (2) is correct. Dilute hydrochloric acid reacts

with the OH(aq) and shifts the equilibrium position of the system to the right. (3) is wrong. Addition of

sodium hydroxide solution increases the concentration of OH(aq) and shifts the equilibrium position of

the system to the left.

163 D Upon a change in temperature, there will be a change in the value of Kc(but there could be

exceptions). Subsequently, the concentrations of reactants and products also have to change in order to

attain a new equilibrium.

164 A

165 A The concentrations of reactants and products are not necessarily the same at equilibrium but the

temperature should remain the same to avoid affecting the state of equilibrium. The amount of catalyst

does not affect the state of equilibrium but only affects the time to reach equilibrium.

166 B As the forward reaction is endothermic, an increase in temperature will shift the equilibrium

position of the system to the right to consume heat energy. As more products are formed, the value of

Kcwill increase.

167 C As the forward reaction is endothermic, an increase in temperature shifts the equilibrium position of

the system to the right. MoreA(aq) andB(aq) will be produced but less C(aq) will be remained.

168 A Either adding or removing the chemical species on reactant or product sides will cause a shift in the

equilibrium position. However, adding catalyst will not cause a shift in the equilibrium position but will

alter the rates of both forward and backward reactions.

169 D When the volume of the system is increased, the pressure of the system is decreased. Since the

number of moles of gas molecules on the product side is greater than that on the reactant side, a

decrease in pressure shifts the equilibrium position to the right.

170 B As the forward reaction is exothermic, increasing the temperature shifts the equilibrium position to

the left producing more N2(g) and H2(g), thus the value of Kcdecreases.171 D Both changes in (2) and (3) can shift the equilibrium position to the right causing an increase in the


22/24

concentration of OH(aq), thus the reaction mixture appears pink in colour.

172 B The value of equilibrium constant, Kc, depends on temperature only. Since the forward reaction is

exothermic, an increase in temperature shifts the equilibrium position to the left producing more SO 2(g)

and O2(g). Thus, the value of equilibrium constant decreases.

173 C For (1), the concentration of N2O4(g) will increase until a new equilibrium is established. (2) is

incorrect because Kcdepends on temperature only.

174 C According to Le Chteliers Principle, the equilibrium position will shift to the left, consuming

some of the added H+(aq) and producing more HCOOH(aq).

175 B Since the numbers of moles of gas molecules are equal on both sides of chemical reaction, the

change in volume or pressure does not affect the equilibrium position.

176 C (2) is incorrect because the equilibrium position should shift to the left consuming some of added

SO3(g).

177 D

178 D (3) is correct. Since the number of moles of gas molecules on the reactant side is greater than that

on the products side, an increase in pressure shifts the equilibrium position to the product side.

179 C Since both changes cause the decrease in concentration of NH3(g), the value of Qcbecomes smaller

than Kc.

180 C Since the numbers of moles of gas molecules are equal on both sides of the chemical equation, the

change in pressure does not affect the equilibrium position.

181 B Since the forward reaction is endothermic, an increase in temperature favours the endothermic

change of an equilibrium system. Thus, the equilibrium position shifts to the right and the value of Kcis

increased.

182 C (2) is incorrect. Since the backward reaction is exothermic, a decrease in temperature favours an

exothermic change. Therefore, the equilibrium position will shift to the left consuming carbon dioxide.

183 B When the volume of the container is increased, the pressure of the system is decreased. According

to Le Chteliers principle, the equilibrium system shifts to the right, raising the number of gas

molecules and bringing the pressure back up.

184 D Adding a catalyst to the equilibrium mixture speeds up both the forward and backward reactions. A

catalyst does not affect the position of equilibrium and the value of Kc.

185 D For (1), since the number of moles of gas molecules on the product side is greater than that on the

reactant side, the equilibrium position shifts to the right, raising the number of gas molecules and

bringing the pressure back up. For (2), since the forward reaction is an endothermic change, an increase

in temperature shifts the system to the right, removing a certain extent of heat added. For (3), since the

concentration of CO2(g) decreases, the equilibrium shifts to the left, forming more CO 2(g).

186A (3) is incorrect. Since Qc=

)]g(ON[

)]g(NO[

42

22 , the Qcshould be smaller than the Kc.

187 C For C, as HCl(g) reacts with NaOH(aq), the concentration of HCl(g) is decreased. Thus, the systemshifts to the right, producing more Si(s) and HCl(g).


23/24

188 A When the volume of a system is decreased, the pressure is increased. For the reaction in A, the

number of moles of gas molecules on the product side is less than that on the reactant side, so an

increase in pressure shifts the equilibrium position to the right, lowering the number of gas molecules

and bringing the pressure back down.

189 D The value of Kcdepends on the temperature only.

190 C Since the forward reaction is an exothermic change, a decrease in temperature favours the forward

reaction, thus the equilibrium position shifts to the right and the value of Kcis increased.

191 C Since the numbers of moles of gas molecules are equal on both sides of chemical equation in C, the

changes in pressure do not affect the equilibrium position and the yield of product of the reaction.

192 B Since the concentration of CoCl42(aq) decreases, the equilibrium position will shift to the right,

consuming some Co2+(aq) and Cl(aq). Thus, the concentration of Cl(aq) will decrease until a new

equilibrium is established.

193 B The forward reaction is an exothermic reaction, so decreasing the temperature shifts the equilibrium

position of the system to the right. A is incorrect. The numbers of moles of gases on both sides of the

equation are equal, changing the pressure of the system would not affect the equilibrium position. C is

incorrect. Adding catalyst to the system does not change the equilibrium position. D is incorrect.

Removing hydrogen from the system shifts the equilibrium position to the left.

194 C The forward reaction is an endothermic reaction and the numbers of moles of gases on the

right-hand side of the equation is greater. So, increasing the temperature and decreasing the pressure

would shift the equilibrium position to the right. Hence, the colour of the mixture becomes darker.

195 C A is incorrect. The numbers of moles of gases on both sides of the equation are equal, changing the

pressure of the system would not affect the equilibrium position. B is incorrect. The forward reaction is

an endothermic reaction, so a decrease in temperature of the system shifts the equilibrium position to

the left. D is incorrect. Adding catalyst to a system does not change the equilibrium position.

196 A When the forward reaction is endothermic reaction, an increase in temperature causes the

equilibrium position to shift to the right. Hence, C and D are incorrect. When the numbers of moles of

gas on the right-hand side of a chemical equation is less than that on the left-hand side, an increase in

pressure causes the equilibrium position of the system to shift to the right. Hence, B is incorrect.

197 A Since the forward reaction is an endothermic reaction, so the equilibrium position would shift to the

right when temperature increases. Adding catalyst to the reaction would not cause the shift ofequilibrium position. This reaction does not involve gases, so changing the pressure would not cause the

shift of the equilibrium position.

198 C The forward reaction is an exothermic reaction, so decreasing the temperature of the system would

shifts the equilibrium position to the right. Kc is only dependent on the temperature.

199 C When there is a change in the volume of a system, its pressure also changes. Pressure increases

when volume is reduced, and vice versa. The increasing pressure causes the system to shift the

equilibrium position to the side with fewer gas molecules. In C, the numbers of moles of gas molecules

on both sides of the equation are equal. Therefore, the equilibrium position of the system would notaffected by a volume change.

200 A Adding N2O(g) will shift the equilibrium position of the system to the right. Removing O 2(g) will


24/24

shift the equilibrium position of the system to the left. Decreasing the volume of the container will

increase the pressure of gases, therefore the system will shift the equilibrium position to the left.

201

C The reaction quotient Qc=(g)][I

[I(g)]

2

2

=

)dm2.00

mol1.00

(

)dm2.00

mol105.00(

3

2

3

3

= 1.25 105mol dm3

As Qc(=1.25 105mol dm3) < Kc(=3.76 10

5mol dm3), the equilibrium position will shift to the

right to produce more products, until the value of Qcequals Kc. However, as there is no information on

whether the reaction is exothermic or endothermic, we cannot predict whether equilibrium can be

attained by increasing temperature of the system.

202 A When the numbers of moles of gases are equal on both sides of a balanced equation, even there is a

change in pressure of the system, the equilibrium position will remain unchanged.

203 C As the reaction does not involve gases, there is no change in the equilibrium position when the

pressure changes. Dissolving sodium chloride will increase the concentration of chloride ions and will

cause a shift of the equilibrium position to the left. As the forward reaction is endothermic, the

equilibrium position will shift to the right as the temperature of the system increases.

204 B As temperature increases, the values of equilibrium constant also increases. That means more

products are formed at a higher temperature, and this is consistent with an endothermic forward

reaction.

205 A Since the forward reaction is an exothermic reaction and the numbers of moles of gases on the

right-hand side of the equation is less than that on the left-hand side, so (1) and (2) are correct. Adding

catalyst to the system does not change the equilibrium position.

206 B Qchas the same unit as Kcbecause they have the same form of expression.

207 C Kccan be calculated from concentrations at equilibrium only while Q ccan be calculated from

concentrations at any particular moment, not necessarily at equilibrium.

208 B

209 C The first statement is incorrect because the equilibrium position should shift to the reactant side.

210 C From the equation, if the reaction shifts to the right, 5 moles of gas molecules are changed to 4

moles of gas molecules. Since the pressure of a gas is proportional to the number of gas moleculespresent, fewer gas molecules result in lower pressure. Thus, when the pressure is increased, the system

shifts to the right, lowering the number of gas molecules and bringing the pressure back down. This

opposes the increase in pressure.

211 C At a given temperature, a change in concentrations of the reactants or the products can alter Qc

only.

212 C At a given temperature, a change in concentrations of the reactants or pressure cannot change the

value of Kc, but may shift the equilibrium position.

213 A

equilibrium mcq ans

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