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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
Equilibrium 0.133 x 0.0667 x 0 +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
Equilibrium 0.112 x 0.064 x 0 +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
Consider the equilibrium,
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
Consider the equilibrium,
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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Consider the equilibrium,
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
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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
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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
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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
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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).
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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
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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
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