equlibrium sq ans

Structured Question Answers

1 (a) NH4Cl(s) NH3(g) + HCl(g) [1]

(b) NH3(g) can turn a moist red litmus paper blue but HCl(g) cannot. [1]/

HCl(g) can turn a moist blue litmus paper red but NH3(g) cannot. [1]

(c) When a substance undergoes sublimation, it changes directly from solid to gaseous state on

heating. [1] When the vapour is cooled, it changes back to solid directly. [1]

2 (a) Propyl ethanoate [1]

(b) Concentrated sulphuric acid [1]

(c) (i) These two reactions are:

CH3CH2CH2OH(l) + CH3COOH(l) CH3COOCH2CH2CH3(l) + H2O(l) [1]

] CH3COOCH2CH2CH3(l) + H2O(l) CH3CH2CH2OH(l) + CH3COOH(l) [1

(ii) The rates of these two reactions are the same. [1]

3 (a) (i) The reaction is reversible. [1]

(ii) NH3(aq) + H2O(l) NH4+(aq) + OH(aq) [1]

(b) (i) The reaction is irreversible. [1]

(ii) FeSO4(aq) + 2KOH(aq) Fe(OH)2(s) + K2SO4(aq) [1]

(c) (i) The reaction is irreversible. [1]

(ii) 2AgNO3(aq) + MgCl2(aq) 2AgCl(s) + Mg(NO3)2(aq) [1]

4 (a) It is a reversible reaction. [1]

The reaction does not go to completion [1] because the ester formed reacts with water to

regenerate ethanoic acid and ethanol at the same time. [1]

(b) CH3COOH(l) + CH3CH2OH(l) CH3COOCH2CH3(l) + H2O(l) [1]

(c) Ethyl ethanoate [1]

(d) Concentrated sulphuric acid [1]

(e) Any TWO of the following:

It can only be established in a closed system. [1]/

Equilibrium can be reached from either forward or backward direction of the reversible reaction.

[1]/

At equilibrium, the rate of forward reaction is equal to the rate of backward reaction. [1]/

At equilibrium, the concentrations of products and reactants remain unchanged. [1]

5 (a) CH3COOH(aq) ions, CH3COO(aq) ions and H+(aq) ions [1]

(b) (i) The two reactions are:

CH3COOH(aq) CH3COO(aq) + H+(aq) [1]

CH3COO(aq) + H+(aq) CH3COOH(aq) [1]

(ii) The rates of these two reactions are the same at equilibrium. [1]

(c) Ethanoic acid is a weak acid that ionizes slightly in water while hydrochloric acid is a strong acid

that ionizes completely in water. [1] Therefore, the concentration of H+(aq) in hydrochloric acid is

higher and the pH value is lower. [1]

(d) The newly added small amount of H+ reacted with CH3COO(aq) in the reaction mixture, forming

CH3COOH(aq). [1] Therefore, the H+ concentration which determines the pH value did not change

much. [1]

6 (a) At the beginning, the concentration of N2O4(g) is the highest, so the rate of forward reaction is the

highest. [1] On the other hand, the concentration of NO2(g) is zero, so the rate of backward

reaction is zero. [1]

(b) At equilibrium, the rates of forward and backward reactions are the same. [2]

(c) If a hole is found in the vessel, both the reactant and product will escape. [1] As a result,

equilibrium cannot be established. [1]

7 (a) Forward reaction: 2NO(g) + O2(g) 2NO2(g) [1]

Backward reaction: 2NO2(g) 2NO(g) + O2(g) [1]

(b)

Correct drawing [2]

Correct labelling [2]

(c) His statement is incorrect. [1]

At equilibrium, the concentrations of NO(g), O2(g) and NO2(g) remain unchanged, so the pressure

should be constant. [1]

8 (a) (i) The rates of forward and backward reactions are the same at equilibrium. [1]

(ii) The rate of backward reaction at equilibrium can be represented by decomposing 2 mol of

NO(g) per second. [1]

(b)

Rea

ctio

n ra

te

Forward reaction

Backward reaction

Time

NO(g)

Con

cent

rati

on

N (g)/O (g)2 2

Time

Correct drawing [2]


9 (a) CaCO3(s) CaO(s) + CO2(g) [1]

(b) No, it would not. [1] Because the reaction system is not a sealed/closed one. [1]

(c) His statement is correct. [1]

CO2 gas will be produced when calcium carbonate reacts with acid in the stomach, [1] which

would cause discomfort to the patients. [1]

10 (a) (i) 2NH4I(s) 2NH3(g) + H2(g) + I2(g) [1]

2NH3(g) + H2(g) + I2(g) 2NH4I(s) [1]

(ii) The rates of two reactions are the same. [1]

(b) 2NH4I(s) 2NH3(g) + H2(g) + I2(g) [1]

(c)

Correct drawing [2]


11 (a) H2(g) + I2(g) 2HI(g) [1]

(b) (i) The initial concentration of H2(g) = L1

mol4.0 = 0.4 mol L1

The initial concentration of I2(g) = L1

mol3.0 = 0.3 mol L1

The concentration of HI(g) at equilibirum = L1

mol5.0 = 0.5 mol L1 [1]

According to the equation, the mole ratio of H2 : I2 : HI = 1 : 1 : 2

No. of moles of H2(g) and I2(g) reacted = 2

mol5.0 = 0.25 mol [1]

∴ the concentration of H2(g) at equilibrium = 0.4 mol L1 0.25 mol L1

= 0.15 mol L1 [1]

the concentration of I2(g) at equilibrium = 0.3 mol L1 0.25 mol L1

= 0.05 mol L1 [1]

(ii)

H2/I2 NH3

NH4I

Time

Con

cent

rati

on

Correct drawing [3]


12 (a) PCl5(g) PCl3(g) + Cl2(g) [1]

(b) The forward reaction is PCl5(g) PCl3(g) + Cl2(g) [1]

The backward reaction is PCl3(g) + Cl2(g) PCl5(g) [1]

(c)

Correct drawing [2]


13 (a) (i) The two reactions are:

2SO3(g) 2SO2(g) + O2(g) [1]

2SO2(g) + O2(g) 2SO3(g) [1]

(ii) The rates of these two reactions are the same. [1]

(b) The concentrations of SO3(g), SO2(g) and O2(g) remain unchanged at equilibrium. [1]

(c) Any TWO of the following:

It can only be established in a closed system. [1]/

Equilibrium can be reached from either forward or backward direction of the reversible reaction.

[1]/

At equilibrium, the rate of forward reaction is equal to the rate of backward reaction. [1]/

At equilibrium, the concentrations of products and reactants remain unchanged. [1]

14 (a) Any THREE:

Equilibrium can only be established in a closed system, where no materials can enter or leave the

system. [1]

Equilibrium can be reached from either the forward or backward direction of a reversible reaction.

H2 I2

HI

Con

cent

rati

on

Time

Backward reaction

Forward reaction

Time

Rea

ctio

n ra

te

[1]

At equilibrium, the rate of forward reaction is equal to the rate of backward reaction. [1]

At equilibrium, both reactants and products are present and their concentrations remain unchanged.

Hence, there are no observable changes of the reaction mixture. [1]

(b) (i) (3 0.5 × 3) mol = 1.5 mol [1]

(ii) 0.5 × 2 mol = 1.0 mol [1]

(c)

For correct labelling of axes [1]

For correct shape of diagram [1]

15 (a) Dynamic equilibrium refers to a reversible reaction [1] in which the rate of forward reaction is

equal to the rate of backward reaction. [1]

(b) (i)

[1]

It is because the two curves become horizontal and the concentrations of reactants and

product remain unchanged at time t. [1]

(ii)

NH3(g)

H2(g)

N2(g)

Time

Am

ount

(m

ol)

t Time

[A] or [B]

[C]

Con

cent

rati

on



16 (a) No. [1] As the reaction proceeds to a certain point, equilibrium is established. [1] At the end there

is still some CaCO3 solid left, no matter how long it is heated.

(b) When there is a tiny hole on the surface, after heating the CaCO3 solid for some time, all of it

decomposes into CaO and CO2. No CaCO3 solid will be left behind at the end. [1] This is because

the system is now an open system [1] and fails to attain equilibrium.

17 (a) The brown colour of the gas gradually fades and it finally becomes pale brown. [1]

(b) The concentration of NO2 drops throughout the period of time. Initially the drop is very rapid, and

then it gradually slows down. [1] Finally the concentration of NO2 stops dropping and the state of

equilibrium is reached. [1]

(c) At first, the number of NO2 molecules is large. Therefore the forward reaction occurs quickly, and

N2O4 molecules are rapidly produced. [1] As time goes by, the number of NO2 molecules

decreases and the number of N2O4 molecules increases. The forward reaction slows down and the

backward reaction speeds up. [1] Finally when the rate of formation of N2O4 from NO2 equals the

rate of formation of NO2 from N2O4, an equilibrium state is established. [1]

18 (a) Ethanoic acid and ethanol. [1]

(b) Ethyl ethanoate and water. [1]

(c) A diagram showing the changes in rates of forward and backward reactions.

backward reaction

forward reaction R

ate

Time



19 (a) (1) Equilibrium can only be established in a closed system, where no materials can enter or leave

the system. [1]

(2) Equilibrium can be reached from either the forward or the backward direction of the

reversible reaction. [1]

(3) At equilibrium, the rate of forward reaction is equal to the rate of backward reaction. [1]

(4) At equilibrium, both reactants and products are present and their concentrations remain

unchanged. [1]

(b) At equilibrium, the forward reaction CH3COOH(aq) CH3COO(aq) + H+(aq) [1] and the

backward reaction CH3COO(aq) + H+(aq) CH3COOH(aq) are still continuing, [1] even though

there is no observable change in the system [1] when it is in equilibrium.

20 (a) SbCl5(g), SbCl3(g) and Cl2(g) [2]

(b) Forward reaction: SbCl5(g) SbCl3(g) + Cl2(g) [1]

Backward reaction: SbCl3(g) + Cl2(g) SbCl5(g) [1]

(c) The rate of forward reaction decreases as SbCl5(g) is decomposed, [1] until the rate of forward

reaction becomes constant at equilibrium. [1] The rate of backward reaction increases from zero,

[1] until the rate of backward reaction becomes constant [1] and is essentially equal to the rate of

forward reaction at equilibrium. [1]

21 (a) PCl5(g) PCl3(g) + Cl2(g) [1]

(b) Kc = eqm5

eqm2eqm3

)]g(PCl[

)]g(Cl[)]g(PCl[ [1]

(c) Initial concentration of PCl5(g)

= 3

3

dm25.0

mol1000.1 = 4 × 103 mol dm3 [1]

Time

Rat

e

backward reaction

forward reaction

Equilibrium concentration of Cl2(g)

= 3

4

dm25.0

mol1065.9 = 3.86 × 103 mol dm3 [1]

Consider the equilibrium,

Concentration

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

Initial 4 × 103 0 0

Change 3.86 × 103 +3.86 × 103 +3.86 × 103

Equilibrium 4 × 103

3.86 × 103

= 1.4 × 104 [1]

0 + 3.86 × 103

= 3.86 × 103 [1]

0 + 3.86 × 103

= 3.86 × 103 [1]

Kc = ]dmmol104.1[

]dmmol1086.3][dmmol1086.3[34

3333

= 0.106 mol dm3 [1]

22 (a) Number of moles of H2(g) = 1molg20.1

g0.2

= 1 mol [1]

Number of moles of I2(g) = 1molg2127

g508

= 2 mol [1]

(b) Initial concentration of H2(g) = 3dm5

mol1 = 0.2 mol dm3 [1]

Initial concentration of I2(g) = 3dm5

mol2 = 0.4 mol dm3 [1]

Let x mol dm3 be the change in concentration of H2(g).

Concentration (mol dm3) H2(g) + I2(g) 2HI(g)

Initial 0.2 0.4 0

Change x x +2x

Equilibrium 0.2 x 0.4 x 0 + 2x = 2x

eqm2eqm2

2eqm

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

[HI(g)]K [1]

)4.0)(2.0(

)2(48

2

xx

x

[1]

Solving for x,

48(0.08 0.6x + x2) = 4x2

x = 0.186 or x = 0.468 (rejected)

∴ equilibrium concentrations:

[H2(g)]eqm = 0.2 0.186 mol dm3 = 0.014 mol dm3 [1]

[I2(g)]eqm = 0.4 0.186 mol dm3 = 0.214 mol dm3 [1]

[HI(g)]eqm = 2(0.186) mol dm3 = 0.372 mol dm3 [1]

23

(a) Kc = 2eqm2

eqm22eqm2

)]g(SH[

)]g(S[)]g(H[ [1]

(b) 2.25 × 104 = 23

32eqm2

)1042.2(

)10165.1()]g(H[

[1]

[H2(g)]eqm = 1.06 × 103 mol dm3

Therefore, the equilibrium concentration of H2(g) is 1.06 × 103 mol dm3. [1]

24 (a) Initial concentration of PCl3(g) = 3dm5.2

mol55.0 = 0.22 mol dm3 [1]

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

mol55.0 = 0.22 mol dm3 [1]

(b) Let x mol dm3 be the change in concentration of PCl3(g).

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

Initial 0.22 0.22 0

Change x x +x

Equilibrium 0.22 x 0.22 x 0 + x = x

The equilibrium constant for PCl3(g) + Cl2(g) PCl5(g)

= cK

1 = Kc = 26.3 mol1 dm3 [1]

Kc = eqm2eqm3

eqm5

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

)]g(PCl[ [1]

26.3 = )22.0)(22.0( xx

x

Solving for x,

26.3(0.22 x )(0.22 x ) = x

x = 0.146 or x = 0.332 (rejected)

∴ the equilibrium concentration of PCl3(g) = 0.22 0.146 mol dm3

= 0.074 mol dm3 [1]

the equilibrium concentration of Cl2(g) = 0.22 0.146 mol dm3

= 0.074 mol dm3 [1]

the equilibrium concentration of PCl5(g) = 0.146 mol dm3 [1]

25 (a) Initial concentration of SO2(g) =

V

x2 mol dm3 [1]

Initial concentration of O2(g) = V

y mol dm3 [1]

Equilibrium concentration of SO3(g) = V

z2 mol dm3 [1]

Concentration

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

Initial V

x2

V

y 0

Change V

z2

V

z +

V

z2

Equilibrium V

zx 22

V

zy 0 +

V

z2 =

V

z2

Kc = eqm2

2eqm2

2eqm3

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

)]g(SO[ [1]

= )()

22(

)2

(

2

2

V

zy

V

zxV

z

=

))(484(

422

2

zyzxzx

Vz

[1]

(b) Kc = eqm2

2eqm2

2eqm3

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

)]g(SO[

= )dmmol5.0()dmmol71.0(

)dmmol00.1(323

23

= 3.97 mol1 dm3 [1]

26 (a) It is a homogeneous equilibrium [1] because all the reactants and products are in the same phase.

[1]

(b) Initial concentration of CO2(g) = 3dm1

mol10.0 = 0.10 mol dm3 [1]

Initial concentration of H2(g) = 3dm1

mol12.0 = 0.12 mol dm3 [1]

Let x mol dm3 be the change in concentration of CO2(g).

Concentration

(mol dm3) CO2(g) + H2(g) CO(g) + H2O(g)

Initial 0.10 0.12 0 0

Change x x +x +x

Equilibrium 0.10 x 0.12 x 0 + x = x 0 + x = x

Kc = eqm2eqm2

eqm2eqm

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

)]g(OH[)]g(CO[ [1]

0.63 = )12.0)(10.0(

2

xx

x

[1]

Solving for x,

0.63(0.012 0.22x + x2) = x2

x = 0.0483 or x = 0.423 (rejected)


[CO2(g)]eqm = 0.10 0.0483 mol dm3 = 0.0517 mol dm3 [1]

[H2(g)]eqm = 0.12 0.0483 mol dm3 = 0.0717 mol dm3 [1]

[CO(g)]eqm = 0.0483 mol dm3 [1]

[H2O(g)]eqm = 0.0483 mol dm3 [1]

27 (a) (i) Kc1 = 2

eqm

eqm2eqm2

)]g(NO[

)]g(O[)]g(N[ [1]

(ii) Kc2 = eqm2

2eqm

2eqm

)]g(Br[)]g(NO[

)]g(NOBr[ [1]

(iii) Kc = eqm2eqm2eqm2

2eqm

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

)]g(NOBr[ [1]

= 1c

2c

K

K [1]

(b) Kc = 18

31

104.2

dmmol4.1

= 5.83 × 1017 mol1 dm3 [1]

28 (a) It is incorrect [1] because the equilibrium concentration of each species is dependent on the

equilibrium constant of the reaction. [1]

(b) It is correct. [1]

(c) It is incorrect [1] because the rates of forward and backward reactions are the same at equilibrium.

[1]

29 (a) (i) Forward reaction: 2X(g) + Y(g) 2Z(g) [1]

Backward reaction: 2Z(g) 2X(g) + Y(g) [1]

(ii) The unit of Kc for forward reaction = )dmmol()dmmol(

)dmmol(323

23

= mol1 dm3 [1]

The unit of Kc for the backward reaction = the unit for cK

1 mol dm3 [1]

(b) (i) The initial concentration of NOCl = 3dm1

mol2 = 2 mol dm3 [1]

(ii) Consider the equilibrium,

Concentration

(mol dm3) 2NOCl(g) 2NO(g) + Cl2(g)

Initial 2 0 0

Change 0.66 +0.66 +0.33

Equilibrium 2 0.66 = 1.34 0 + 0.66 = 0.66 0 + 0.33 = 0.33

The equilibrium concentration of NOCl = 1.34 mol dm3 [1]

The equilibrium concentration of Cl2 = 0.33 mol dm3 [1]

(iii) Kc = 2eqm

eqm22eqm

)]g(NOCl[

)]g(Cl[)]g(NO[ [1]

= 23

323

)dmmol34.1(

)dmmol33.0()dmmol66.0(

= 0.0801 mol dm3 [1]

30

(a) K1 = 3eqm2

2eqm

2eqm3

)]g(Cl[)]g(P[

)]g(PCl[ [1]

K2 = eqm2eqm3

eqm5

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

)]g(PCl[ [1]

(b) K = 5eqm2

2eqm

2eqm5

)]g(Cl[)]g(P[

)]g(PCl[ [1]

= K1 × K22 [1]

(c) Equilibrium concentration of PCl3(g) = 3

3

dm5.2

mol106.1 = 6.4 × 104 mol dm3 [1]

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

mol03.0 = 0.012 mol dm3 [1]

Equilibrium concentration of PCl5(g) = 3

4

dm5.2

mol1004.5 = 2.016 × 104 mol dm3 [1]

Kc = eqm2eqm3

eqm5

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

)]g(PCl[ =

)dmmol012.0)(dmmol104.6(

)dmmol10016.2(334

34

[1]

= 26.25 mol1 dm3 [1]

31 (a) Kc indicates the extent of a chemical reaction but not the rate of the reaction. [1]

(b) Kc = 2eqm2eqm4

4eqm2eqm2

)]g(SH[)]g(CH[

)]g(H[)]g(CS[ [1]

Kc = 233

433



= 5.86 mol2 dm6 [1]

Since the calculated value of Kc is greater than the given value, the reaction mixture is not at

equilibrium. [1]

(c) (i) Number of moles of ClF3(g) = 1molg30.195.35

g25.9

= 0.1 mol

The initial concentration of ClF3(g) = 3dm2

mol1.0 = 0.05 mol dm3 [1]


Concentration

(mol dm3) ClF3(g) ClF(g) + F2(g)

Initial 0.05 0 0

Change 0.05 × 0.215 =

0.01075 +0.01075 +0.01075

Equilibrium 0.05 0.01075 =

0.03925

0 + 0.01075 =

0.01075

0 + 0.01075 =

0.01075

Kc = eqm3

eqm2eqm

)]g(ClF[

)]g(F[)]g(ClF[ [1]

= )dmmol03925.0(

)dmmol01075.0)(dmmol01075.0(3

33

= 2.94 × 103 mol dm3 [1]

(ii) Number of moles of ClF3(g) = 1molg30.195.35

g4.39

= 0.426 mol

The initial concentration of ClF3(g) = 3dm2

mol426.0 = 0.213 mol dm3 [1]

Let x mol dm3 be the change in concentration of ClF3(g).

Concentration

(mol dm3) ClF3(g) ClF(g) + F2(g)

Initial 0.213 0 0

Change x +x +x

Equilibrium 0.213 x 0 + x = x 0 + x = x

Kc = x

x

213.0

2

= 2.94 × 103

Solving for x,

x2 = 2.94 × 103 (0.213 x)

x = 0.0236 or x = 0.0265 (rejected)


[ClF3(g)]eqm = 0.213 0.0236 mol dm3 = 0.189 mol dm3 [1]

[ClF(g)]eqm = 0.0236 mol dm3 [1]

[F2(g)]eqm = 0.0236 mol dm3 [1]

32 (a) N2(g) + 3H2(g) 2NH3(g) [1]

(b) Initial concentration of N2(g) = 3dm100

mol10 0.1 mol dm3 [1]

Initial concentration of H2(g) = 3dm100

mol28 = 0.28 mol dm3 [1]

Equilibrium concentration of NH3(g) = 3dm100

mol6 0.06 mol dm3 [1]


Concentration

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

Initial 0.1 0.28 0

Change 0.03 3(0.03) +0.06


0.07

0.28 0.09 =

0.19

0 + 0.06 =

0.06

Kc = 3eqm2eqm2

2eqm3

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

)]g(NH[ [1]

= 333

23


)dmmol06.0(

= 7.50 mol2 dm6 [1]

33 (a) H2(g) + I2(g) 2HI(g) [1]

(b) Initial concentration of H2(g) = 3dm3

mol15.0 = 0.05 mol dm3 [1]

Initial concentration of I2(g) = 3dm3

mol15.0 = 0.05 mol dm3 [1]

Equilibrium concentration of HI(g) = 3dm3

mol24.0 = 0.08 mol dm3 [1]


Concentration

(mol dm3) H2(g) + I2(g) 2HI(g)

Initial 0.05 0.05 0

Change 0.04 0.04 +0.08


0.01

0.05 0.04 =

0.01

0 + 0.08 =

0.08

Kc = eqm2eqm2

2eqm

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

)]g(HI[ [1]

= )dmmol01.0)(dmmol01.0(

)dmmol08.0(33

23

= 64

∴ the equilibrium constant for the reaction is 64. [1]

(c) The equilibrium position of the reaction lies mainly to the product side [1] because the equilibrium

constant is much larger than 1. [1]

34 (a) Initial number of moles of PCl5(g) =

1mol g 35.5)5(31.0

g 2.5

= 0.012 mol [1]

Initial concentration of PCl5(g) = 3dm2

mol 0.012 = 0.006 mol dm3

Let x mol dm3 be the change in concentration of PCl3(g).

Concentration


Initial 0.006 0 0

Equilibrium 0.006 x x x

Kc = )0060(

2

x.

x

[1]

0.19 × (0.006 x) = x2

0 = x2 + 0.19x 0.00114

x = 0.00582 or 0.1958 (rejected) [1]

[PCl5(g)]eqm = 0.00018 mol dm3 [1]

[PCl3(g)]eqm = 0.00582 mol dm3 [1]

[Cl2(g)]eqm = 0.00582 mol dm3 [1]

(b) Initial concentration of PCl5(g) = 3dm 5

mol 0.012 = 0.0024 mol dm3

Let y mol dm3 be the change in concentration of PCl3(g).

Concentration


Initial 0.0024 0 0

Equilibrium 0.0024 y y y

Kc = y

y

0.0024

2

[1]

0.19 × (0.0024 y) = y2

0 = y2 + 0.19y 0.000456

y = 0.002370 or 0.1924 (rejected) [1]

[PCl5(g)]eqm = 3 × 105 mol dm3 [1]

[PCl3(g)]eqm = 0.002370 mol dm3 [1]

[Cl2(g)]eqm = 0.002370 mol dm3 [1]

35 (a) Kc =

eqm2

2

eqm2eqm2

2

S(g)][H

(g)][S(g)][H [1]

(b) Initial concentration of H2S(g) = 13

3

mol dm 24.0

dm 6.0 ×

3dm 10

1 = 0.025 mol dm3 [1]

Equilibrium concentration of H2(g) = 13

3

mol dm 24.0

dm 4.2 ×

3dm 10

1 = 0.0175 mol dm3

Concentration

(mol dm3) 2H2S(g) 2H2(g) + S2(g)

Initial 0.025 0 0

Change 0.0175 +0.0175 +2

01750.

Equilibrium 0.0075 0.0175 0.00875

[2]

Kc = 23

323

)dm mol (0.0075

)dm mol (0.00875)dm mol (0.0175

[1]

= 0.04764 mol dm3 [1]

36 (a) (i) The gas gradually changed from colourless to violet. [1]

(ii) The reaction mixture was put into potassium iodide solution in order to dissolve iodine in an

aqueous solution. [1]

(b) (i) I2(aq) + 2S2O32(aq) 2I(aq) + S4O6

2(aq) [1]

(ii) Number of moles of sodium thiosulphate used = 0.1 mol dm3 × 3dm1000

22.2 = 0.00222 mol

[1]

From the equation, the mole ratio of I2 : S2O32 = 1 : 2

Number of moles of iodine at equilibrium = 0.00111 mol [1]

(c) (i) Kc = eqm

2

eqm2eqm2

[HI(g)]

(aq)][I(g)][H [1]

(ii) Assume the volume of the sealed vessel is 1 dm3.

Initial number of moles of HI(aq) = 1mol g 127.0)(1.0

g 2.56

= 0.02 mol

Initial concentration of HI(aq) = 3dm 1

mol 0.02 = 0.02 mol dm3

Concentration

(mol dm3) 2HI(g) H2(g) + I2(g)

Initial 0.02 0 0

Change 0.00222 +0.00111 +0.00111

Equilibrium 0.01778 0.00111 0.00111

[2]

Kc = 2

2

(0.01778)

(0.00111) [1]

= 3.897 103 [1]

37 (a) Since the concentration of a solid or a liquid, i.e. density, is so large that it is considered to be

constant at a fixed temperature, [1] it is not included in the equilibrium constant expression.

(b) Kc = [CO2(g)]eqm. [1] The unit of the equilibrium constant expression is mol dm3. [1]

(c) From (b), equilibrium constant = 2.80 mol dm3 = [CO2(g)]eqm

Hence, the equilibrium concentration of CO2(g) is 2.8 mol dm3. [1]

38 (a) Kc =

eqm2eqm2

eqm2

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

[HI(g)] [1]

(b) Trial 1: eqm2eqm2

eqm2

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

[HI(g)] =

M) 10M)(0.27 10(0.27

M) 10(1.4722

22

= 29.6

Trial 2: eqm2eqm2

eqm2

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

[HI(g)] =

M) 10M)(0.10 10(1.10

M) 10(1.8022

22

= 29.5

Trial 3: eqm2eqm2

eqm2

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

[HI(g)] =

M) 10M)(0.04 10(3.04

M) 10(1.9222

22

= 30.3

[1]

Average value of equilibrium constants at this temperature = 3

30.329.529.6 = 29.8 [1]

(c) Let a × 102 M be the change in concentration of H2(g).

Concentration

( × 102 M) H2(g) + I2(g) 2HI(g)

Initial 3 1 0

Change a a +2a

Equilibrium 3 a 1 a 2a

[2]

29.8 = ))(1(3

)(2 2

aa

a

[1]

29.8 × (3 4a + a2) = 4a2

25.8a2 119.2a + 89.4 = 0

a = 0.942 or 3.678 (rejected) [1]

Hence, [H2(g)]eqm = 2.058 × 102 M

[I2(g)]eqm = 0.058 × 102 M

[HI(g)]eqm = 1.884 × 102 M [3]

39 (a) Initial mass of CH3COOH present = 1.049 g cm3 × 25 cm3 = 26.225 g [1]

Initial mass of C2H5OH present = 0.789 g cm3 × 25 cm3 = 19.725 g [1]

Molar mass of CH3COOH = (12 × 2 + 16 × 2 + 1 × 4) g mol1 = 60 g mol1

Molar mass of C2H5OH = (12 × 2 + 16 × 1 + 1 × 6) g mol1 = 46 g mol1

Therefore,

initial number of moles of CH3COOH present = 1mol g 60

g 26.225 = 0.437 mol

initial number of moles of C2H5OH present = 1mol g 46

g 19.725 = 0.429 mol

initial concentration of CH3COOH present =

33

3

dm cm 1000

cm 50

mol 0.437

= 8.74 M [1]

initial concentration of C2H5OH present =

33

3

dm cm 1000

cm 50

mol 0.429

= 8.58 M [1]

(b) Let a M be the change in concentration of CH3COOC2H5.

Concentration

(M)

CH3COOH + C2H5OH CH3COOC2H5 +

H2O

Initial 8.74 8.58 0 0

Change a a +a +a

Equilibrium 8.74 a 8.58 a a a

[1]

CH3COOH + NaOH CH3COONa + H2O

Number of moles of NaOH in 3 M NaOH(aq)

= 3 mol dm3 × 33

3

dm cm 1000

cm 10.4 = 0.0312 mol [1]

From the equation, mole ratio of CH3COOH : NaOH = 1 : 1

Hence, number of moles of CH3COOH remaining in 10 cm3 sample = 0.0312 mol [1]

Thus,

[CH3COOH]eqm =

33

3

dm cm 1000

cm 10

mol 0.0312

= 3.12 M [1]

8.74 a = 3.12

a = 5.62

[C2H5OH]eqm = (8.58 a) = (8.58 5.62) = 2.96 M [1]

[CH3COOC2H5]eqm = 5.62 M [1]

[H2O]eqm = 5.62 M [1]

(c) Kc = eqm52eqm3

eqm2eqm523

OH(l)]H[CCOOH(l)][CH

O(l)][H(l)]HCOOC[CH [1]

(d) Kc = M 2.96 M 3.12

M 5.62 M 5.62

= 3.42 [1]

40 Initial concentration of SO3(g):

= 3dm 10

mol 1 = 0.1 mol dm3

Initial concentration of NO(g):

= 3dm 10

mol 1 = 0.1 mol dm3

Let x mol dm3 be the change in concentration of SO2(g).

Concentration

(mol dm3) SO3(g) + NO(g) SO2(g) + NO2(g)

Initial 0.1 0.1 0 0

Change x x +x +x

Equilibrium 0.1 x 0.1 x x x

[2]

Kc = eqmeqm3

eqm2eqm2

][NO(g)](g)[SO

(g)][NO(g)][SO

Kc = 2

2

)(0.1

)(

x

x

[1]

0.5 = x.

x

10

0.7071(0.1 x) = x

x = 0.0414 mol dm3 [1]

∴ The concentration of SO2(g) at equilibrium is 0.0414 mol dm3.

41 (a) Kc =

eqm23eqm

eqm2eqm32

OH(l)]CH[CH[HCOOH(l)]

O(l)][H(l)]CH[HCOOCH [1]

(b) Mass of HCOOH = 1.22 g cm3 × 11.31 cm3 = 13.798 g

Mass of CH3CH2OH = 0.789 g cm3 × 36.44 cm3 = 28.751 g

Number of moles of HCOOH before reaction = 1mol g 46

g 13.798 = 0.30 mol [1]

Number of moles of CH3CH2OH before reaction = 1mol g 46

g 28.751 = 0.625 mol [1]

(c) HCOOH + NaOH HCOONa + H2O

Number of moles of NaOH in 25 cm3 of 0.25 M NaOH(aq)

= 0.25 mol dm3 × 1000

25 dm3 = 0.00625 mol [1]

From the equation, mole ratio of HCOOH : NaOH = 1 : 1

Hence, number of moles of HCOOH in 5.0 cm3 sample = 0.00625 mol [1]

(d) Let x mol be the number of moles of HCOOH in the equilibrium mixture.

0.00625

x =

5

36.44)(11.31

x = 0.0597 [1]

Let the volume of container used be 1 dm3.

Concentration

(mol dm3) HCOOH + CH3CH2OH HCOOCH2CH3 + H2O

Initial 0.30 0.625 0 0

Change

0.0597

0.30

= 0.2403

0.2403 +0.2403 +0.2403

Equilibrium 0.0597 0.3847 0.2403 0.2403

[2]

Kc = .3847)(0.0597)(0

.2403)(0.2403)(0 [1]

= 2.51 [1]

42 (a) Point M represents the rate of disappearance of B(g) at time t1. [1]

Point N represents the rate of formation of A2B(g) at time t1. [1]

(b) At time t2. [1] The concentrations of all chemical species do not change after time t2. [1]

(c) [A(g)]eqm = 3dm 0.5

mol 0.15 = 0.3 mol dm3 [1]

[B(g)]eqm = 3dm 0.5

mol 0.575 = 1.15 mol dm3 [1]

[A2B(g)]eqm = 3dm 0.5

mol 0.425 = 0.85 mol dm3 [1]

(d) Kc = eqmeqm

2

eqm2

(g)][(g)][

(g)][

BA

BA [1]

= )dm mol (1.15)dm mol (0.3

)dm mol (0.85323

3

= 8.21 mol2 dm6 [1]

The unit of Kc is mol2 dm6 [1]

43 Let x mol dm3 be the change in the concentration of PCl3(g).

Concentration


Initial ? 0 0

Change x +x +x

Equilibrium 5.00 × 104 x x

[2]

Kc = eqm5

eqm2eqm3

(aq)][PCl

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

8.00 × 103 = 4

2

10 5.00 x

[1]

x = 2.00 × 103 [1]

∴ Initial concentration of PCl5(g) = 5.00 × 104 + 2.00 × 103

= 2.5 × 103 mol dm3 [1]

44 Initial concentration of XO(g):

= 3dm 2.00

mol 1.00 = 0.50 mol dm3

Initial concentration of O2(g):

= 3dm 2.00

mol 2.00 = 1.00 mol dm3

Let x mol dm3 be the change in concentration of O2(g).

Concentration

(mol dm3) 2XO(g) + O2(g) 2XO2(g)

Initial 0.50 1.00 0

Change 2x x +2x

Equilibrium 0.50 2x 1.00 x 2x

[2]

Kc = eqm2eqm

2

eqm2

2

(g)][OO(g)][

(g)]O[

X

X

1.00 × 104 = )(1.00)2 (0.50

)(22

2

xx

x

[1]

Assume x << 1.00,

1.00 × 104 = 2

2

4 2 0.25

4

xx

x

[1]

2.5 × 105 2.00 × 104x + 4 × 104x2 = 4x2

2.5 × 105 2.00 × 104x 3.9996x2 = 0

x = 2.48 × 103 or x = 2.53 × 103 (rejected) [1]

∴ [XO(g)]eqm = 0.50 2 × 2.48 × 103 = 0.495 mol dm3 [1]

[O2(g)]eqm = 1.00 2.48 × 103 = 0.998 mol dm3 [1]

[XO2(g)]eqm = 2 × 2.48 × 103 = 4.96 × 103 mol dm–3 [1]

45 Initial concentration of SO2Cl2(g) =

3dm 100.0

mol 0.030 = 0.00030 mol dm3

Initial concentration of SO2(g) = 3dm 100.0

mol 2.00 = 0.020 mol dm3

Initial concentration of Cl2(g) = 3dm 100.0

mol 1.00 = 0.010 mol dm3

Let x mol dm3 be the change in the concentration of SO2(g).

Concentration

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

Initial 0.00030 0.020 0.010

Change x +x +x

Equilibrium 0.00030 x 0.020 + x 0.010 + x

[2]

Kc = eqm22

eqm2eqm2

(g)]Cl[SO

(g)][Cl(g)][SO [1]

0.0810 = ) 000300(

)0100)(0200(

x.

x.x.

2.43 × 105 0.081x = 2 × 104 + 0.03x + x2

x2 + 0.111x + 1.757 × 104 = 0

x = 0.001606 or 0.1094 (rejected) [1]

∴ [SO2Cl2(g)]eqm = 0.00030 (0.001606) = 1.91 × 103 mol dm3 [1]

[SO2(g)]eqm = 0.020 + (0.001606) = 1.84 × 102 mol dm3 [1]

[Cl2(g)]eqm = 0.010 + (0.001606) = 8.39 × 103 mol dm3 [1]

46 (a) CH3CH2OH(l) + CH3CH2COOH(l) CH3CH2COOCH2CH3(l) + H2O(l) [1]

(b) Concentrated sulphuric acid. [1]

(c) 1. Allow the mixture to reach an equilibrium and withdraw a known amount of equilibrium

mixture by a pipette. [1]

2. Quench the reaction in the small portion of equilibrium mixture by cooling it in an ice bath.

[1]

3. Determine the equilibrium concentration of propanoic acid by titrating the equilibrium

mixture against standard sodium hydroxide solution. [1]

4. Calculate the equilibrium concentrations of ethanol, ethyl propanoate and water based on the

balanced equation. [1]

5. Substitute the equilibrium concentrations in the expression for Kc.

Kc = eqm23eqm23

eqm2eqm3223

])l(COOHCH[CH])l(OHCH[CH

])l(O[H])l(CHCOOCHCH[CH [1]

6. Calculate the average Kc by repeating steps 15 two more times. [1]

47 (a) The mixture is allowed to stand for two hours in order to ensure the equilibrium is reached. [1]

(b) When the end point is reached, a blood red colour would be observed. [1] Excess SCN(aq) reacts

with Fe3+(aq) to give the blood red colour. [1]

(c) (i) Kc = eqmeqm

2

eqm3

(aq)][Ag(aq)][Fe

(aq)][Fe

[1]

(ii) After mixing the two solutions, the total volume of solution is 50 cm3.

The concentration of Ag+(aq) after mixing = 0.10 mol dm3 × 50

25 = 0.05 mol dm3 [1]

The concentration of Fe2+(aq) after mixing = 0.12 mol dm3 × 50

25 = 0.06 mol dm3 [1]

(iii) (1) Ag+(aq) + SCN(aq) AgSCN(s) [1]

(2) Number of moles of KSCN used = 0.02 mol dm3 × 1000

14.2dm3 = 0.000284 mol [1]

From the equation, mole ratio of Ag+ : SCN = 1 : 1

Number of moles of Ag+(aq) in the 25.0 cm3 resultant solution = 0.000284 mol

Concentration of Ag+(aq) at equilibrium = 3dm

1000

25mol 0.000284

= 0.01136 mol dm3 [1]

(d)

Concentration

(mol dm3) Ag+(aq) + Fe2+(aq) Ag(s) + Fe3+ (aq)

Initial 0.05 0.06 / 0

Change

0.01136

0.05

= 0.03864

0.03864 / +0.03864

Equilibrium 0.01136 0.02136 / 0.03864

[2]

Kc = 0.02136)(0.01136)(

0.03864 [1]

= 159.24 mol1 dm3 [1]

48 ##

(a) (i) The red substance formed is thiocyanatoiron(III) ion/FeSCN2+(aq) [1]

(ii) Fe3+ (aq) + SCN(aq) FeSCN2+(aq) [1]

(b) (i) Addition of Fe(NO3)3(aq) increases the concentration of Fe3+(aq) in the mixture. This shifts

the equilibrium position of the system to the right. [1] Hence, the colour of the solution turns

dark red. [1]

(ii) Addition of KSCN(aq) increases the concentration of SCN(aq) in the mixture. This shifts the

equilibrium position of the system to the right. [1] Hence, the colour of the solution turns dark

red. [1]

(iii) Addition of KOH(aq) removes the Fe3+(aq) in the mixture. Fe3+(aq) react with OH(aq) to

give a precipitate Fe(OH)3. [1] This shifts the equilibrium position of the mixture to the left.

[1] Hence, the colour of solution turns yellow. [1]

(iv) Addition of Na2HPO4 crystal removes the Fe3+(aq) in the mixture. Fe3+(aq) react with

HPO42(aq) to give a colourless complex FeHPO4

+. [1] This shifts the equilibrium position of

the mixture to the left. [1] Hence, the colour of solution becomes paler. [1]

49 (a) More CaCO3(s) will be precipitated back and deposit at the bottom of the container. [1]

(b) More CaCO3(s) will be consumed and some gas bubbles are evolved from the surface of CaCO3

solid. [1]

(c) More CaCO3(s) will be consumed and some gas bubbles are evolved from the surface of CaCO3

solid. [1]

(d) More CaCO3(s) will be precipitated back and deposit at the bottom of the container. [1]

50 (a) When concentrated hydrochloric acid is added, this increases the concentration of H+(aq), thus the

equilibrium position of the system shifts to the left. [1] The colour of bromine water becomes

darker. [1]

(b) When concentrated potassium hydroxide is added, the OH(aq) removes H+(aq) from bromine

water, thus the equilibrium position of the system shifts to the right. [1] The colour of bromine

water becomes paler. [1]

(c) When silver nitrate solution is added, Ag+(aq) removes Br(aq) from bromine water, the

equilibrium position of the system shifts to the right. [1] The colour of bromine water becomes

paler. [1]

51 (a) When concentrated hydrochloric acid is added, the concentrations of H+(aq) and Cl(aq) increase,

thus, the equilibrium position of the system shifts to the left. [1] The white precipitate dissolves to

give a clear solution. [1]

(b) When a large amount of water is added, the concentrations of H+(aq) and Cl(aq) decrease, thus the

equilibrium position of the system shifts to the right.[1] More white precipitate is formed. [1]

(c) When saturated sodium chloride solution is added, the concentration of Cl(aq) increases, thus the

equilibrium position of the system shifts to the left. [1] The white precipitate dissolves to give a

clear solution. [1]

52 (a) When water is added, the equilibrium position of the system shifts to the left, [1] more

tetrachlorocobaltate(II) becomes cobalt(II) hexahydrate. Hence, the colour of cobalt chloride paper

changes from blue to pink. [1]

(b) As the forward reaction is endothermic, [1] heating the pellets shifts the equilibrium position to the

right. [1] Hence, the blue colour of pellets can be restored.

53 (a) Kc =

eqm2eqm2

eqmeqm2

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

[CO(g)]O(g)][H [1]

(b) Qc = (g)](g)][CO[H

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

22

2

= 33

33

dm mol 0.30×dm mol 0.20

dm mol 0.55×dm mol 0.55

[1]

= 5.042 [1]

As Qc is not equal to Kc at 1800 K, therefore the system has not reached the state of equilibrium.

[1]

(c) As Qc (= 5.042) is smaller than Kc (= 6.0) at 1800 K, the equilibrium position will shift to the

right. [1], more products should be formed and less reactants should be remained, until the value of

Qc equals Kc. [1]

(d) Let x mol dm3 be the change in the concentration of H2O(g).

Concentration

(mol dm3) H2(g) + CO2(g) ⇌ H2O(g) + CO(g)

Initial 0.20 0.30 0.55 0.55

Change x x +x +x

Equilibrium 0.20 x 0.30 x 0.55 + x 0.55 + x

[2]

Kc = eqm2eqm2

eqmeqm2

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

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

6.0 = ) )(0.30 (0.20

)+(0.55 2

xx

x

[1]

6.0 = 2

2

0.5 0.06

+1.1+0.3025

xx

xx

0.36 3x + 6x2 = 0.3025 + 1.1x + x2

5x2 4.1x + 0.0575 = 0

x = 0.01427 or 0.8057 (rejected) [1]

[H2(g)]eqm = 0.20 0.01427 = 0.186 mol dm3 [1]

[CO2(g)]eqm = 0.30 0.01427 = 0.286 mol dm3 [1]

[H2O(g)]eqm = 0.55 + 0.01427 = 0.564 mol dm3 [1]

[CO(g)]eqm = 0.55 + 0.01427 = 0.564 mol dm3 [1]

54 (a) The equilibrium position will shift to the left. [1] Since the forward reaction is exothermic, [1]

increasing the temperature shifts the equilibrium position to the left.

(b) The equilibrium position will shift to the left. [1] According to Le Châtelier’s Principle, the

equilibrium position will shift to the left producing more SO2(g). [1]

(c) The equilibrium position will shift to the right. [1] According to Le Châtelier’s Principle, the

equilibrium position will shift to the right producing more SO3(g). [1]

55

Addition of

H2O(g) Addition of H2(g)

Change in

temperature

Effect on [CO2(g)] Increases [1] Decreases [1] Decreases [1]

Effect on [CO(g)] Decreases [1] Increases [1] Increases [1]

Effect on equilibrium

position

Shifts to the right

[1]

Shifts to the left

[1]

Shifts to the left

[1]

Effect on equilibrium

constant No change [1] No change [1] Decreases [1]

56 (a) The equilibrium position shifts to the left. [1]

Since an increase in volume can decrease the pressure, the equilibrium position shifts to the left,

raising the number of gas molecules and bringing the pressure back up. [1]

(b) The equilibrium position shifts to the right. [1]

Since the concentration of hydrogen decreases, the equilibrium position shifts to the right to

produce more hydrogen. [1]

(c) The equilibrium position shifts to the right. [1]

Since the pressure is decreased, the equilibrium position shifts to the right to produce more gases.

[1]

(d) There is no change in equilibrium position. [1]

It is because the numbers of moles of gas molecules are equal on both sides of chemical equation.

[1]

(e) The equilibrium position shifts to the right. [1]

Since sodium hydroxide removes HCl once it is formed, the concentration of HCl decreases. Thus

the equilibrium position shifts to the right to produce more HCl. [1]

57 (a) When the volume of the vessel is decreased, the pressure is increased. [1]

The equilibrium position shifts to the left to lower the number of gas molecules and bring the

pressure back down. [1]

(b) The forward reaction is exothermic [1] because a decrease in temperature favours an exothermic

change of an equilibrium system. [1]

(c) His statement is incorrect. [1]

Adding a catalyst does not affect the equilibrium position [1] because a catalyst increases the rates

of forward and backward reactions to the same extent. [1]

58 (a) An increase in temperature will shift the equilibrium position to the left. [1]

Since the backward reaction is endothermic, an increase in temperature favours the endothermic

change of an equilibrium system. [1]

(b) (i) The reaction quotient, Qc = 32

24

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

)]g(OH)][g(CH[ [1]

= 333

33

)dmmol8)(dmmol4(

)dmmol6)(dmmol6(

= 0.0176 mol2 dm6 [1]

(ii) Since Qc < Kc, more CH4(g) and H2O(g) will be formed and more CO(g) and H2(g) will be

consumed until Qc = Kc. [1] Hence, the equilibrium position will shift to the right. [1]

59 (a) The equilibrium constant, Kc =

eqm2eqm2

eqm2eqm

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

)]g(OH[)]g(CO[ [1]

= )dmmol0717.0)(dmmol0517.0(

)dmmol0483.0)(dmmol0483.0(33

33

= 0.629 [1]

Let x mol dm3 be the change in concentration of CO2(g).

Concentration

(mol dm3) CO2(g) + H2(g) CO(g) + H2O(g)

Initial 0.0517 + 0.02

= 0.0717 0.0717 0.0483 0.0483

Change x x +x +x

Equilibrium 0.0717 x 0.0717 x 0.0483 + x 0.0483 + x

0.629 = )dmmol0717.0)(dmmol0717.0(

)dmmol0483.0)(dmmol0483.0(33

33

xx

xx

Solving for x,

x = 4.78 × 103 or x = 0.508 (rejected)

new equilibrium concentrations

[CO2(g)]eqm = 0.0717 4.78 × 103 mol dm3 = 0.0669 mol dm3 [1]

[H2(g)]eqm = 0.0717 4.78 × 103 mol dm3 = 0.0669 mol dm3 [1]

[CO(g)]eqm = 0.0483 + 4.78 × 103 mol dm3 = 0.0531 mol dm3 [1]

[H2O(g)]eqm = 0.0483 + 4.78 × 103 mol dm3 = 0.0531 mol dm3 [1]

(b) The equilibrium position will remain unchanged. [1] Since the numbers of moles of gas molecules

are equal on both sides of reaction, [1] the change in volume does not affect the equilibrium

position. [1]

60 (a) Let x mol dm3 be the change in concentration of Ag+(aq).

Concentration

(mol dm3) Ag+(aq) + Fe2+(aq) Ag(s) + Fe3+(aq)

Initial 0.200 0.100 / 0.300

Change +x +x / x

Equilibrium 0.200 + x 0.100 + x / 0.300 x

Kc = eqm

2eqm

eqm3

)]aq(Fe[)]aq(Ag[

)]aq(Fe[

[1]

2.98 = )100.0)(200.0(

300.0

xx

x

Solving for x,

x = 0.108 or x = 0.744 (rejected)

equilibrium concentrations

[Ag+(aq)]eqm = 0.200 + 0.108 mol dm3 = 0.308 mol dm3 [1]

[Fe2+(aq)]eqm = 0.100 + 0.108 mol dm3 = 0.208 mol dm3 [1]

[Fe3+(aq)]eqm = 0.300 0.108 mol dm3 = 0.192 mol dm3 [1]

(b) The equilibrium position will shift to the left. [1]

Since Cl(aq) reacts with Ag+(aq) to form AgCl(s), a white precipitate, the concentration of

Ag+(aq) decreases. [1] Thus, the equilibrium shifts to the left, producing more Ag+(aq) and

Fe2+(aq).

61 (a) Kc =

eqm2eqm42

eqm52

])g(O[H])g(H[C

])g(OHH[C [1]

(b) (i) A high pressure should be used. [1]

Since the number of moles of gas molecules on the reactant side is greater than that on the

product side, a high pressure favours the forward reaction and gives more products. [1]

(ii) Advantage: The equilibrium position can be shifted to the product side, giving more products.

[1]

Disadvantage: The manufacture cost is higher. [1]/The danger of production is enhanced. [1]

(c) (i) Fermentation [1]

(ii) C6H12O6 2CH3CH2OH + 2CO2 [1]

62 (a) Vanadium(V) oxide [1]

(b) His statement is incorrect. [1] It is because the value of equilibrium constant depends on

temperature only. [1]

(c) Determine the equilibrium constants of the forward reaction at various temperatures. [1] If the

equilibrium constants decrease with the increase in temperature, then the forward reaction is

exothermic. [1] It is because an increase in temperature favours an endothermic reaction only. [1]

63

Change

Effect on the

equilibrium position

Indicated

quantity Effect

(a) Adding CO(g) Shift to the right Ni(s) Decreases

(b) Removing CO(g) Shift to the left Kc No change

(c) Removing

Ni(CO)4(g)

Shift to the right CO(g) Decreases

(d) Increasing the

temperature at

constant pressure

Shift to the left Kc Decreases

(e) Decreasing the

temperature at

constant pressure

Shift to the right CO(g) Decreases

(f) Decreasing the

volume at constant

temperature

Shift to the right Ni(CO)4(g) Increases

Each correct answer [1]

64 (a) (i) The equilibrium position does not change. [1]

(ii) The equilibrium position shifts to the left. [1]

(iii) The equilibrium position shifts to the right. [1]

(iv) The equilibrium position shifts to the left. [1]

(v) The equilibrium position shifts to the left. [1]

(b) (i)

Each correct answer [2

1]

(ii)

I2(g)

CO2(g)

Con

cent

rati

on (

mol

dm

3)

CO(g)

Time (min)10


1]

(iii)


1]

65 (a) The equilibrium position will shift to the right. [1]

According to Le Châtelier’s Principle, the equilibrium position will shift to the right, consuming

some of added HA(aq) and decreasing its concentration. [1]

(b) The equilibrium position will shift to the left. [1]

If the pH of the reaction mixture is decreased, the concentration of H3O+(aq) is increased. [1]

According to Le Châtelier’s Principle, the equilibrium position will shift to the left, consuming

some of added H3O+(aq) and lowering its concentration. [1]

(c) The equilibrium position will shift to the right. [1]

The forward reaction is an endothermic reaction. [1]

According to Le Châtelier’s Principle, the equilibrium position will shift to the right to remove the

heat added. [1]

66 (a) Since the forward reaction is an exothermic reaction, so a lower temperature favours the formation

of SO3. [1]

(b) Although the percentage of conversion is higher at lower temperature, the rate of reaction is slow.

[1]

I2(g)

CO2(g)

Con

cent

rati

on (

mol

dm

3)

CO(g)

Time (min)10

I2(g)

CO2(g)

Con

cent

rati

on (

mol

dm

3)

CO(g)

Time (min)10

(c) (i) When the volume of container is reduced, the pressure of the system increases. [1] The

equilibrium position shifts to the right. [1]

(ii) When sulphur trioxide is removed from the system, the equilibrium position shifts to the

right. [1]

67 (a) Kc =

eqm2

eqm2eqm2

[NOCl(g)]

(g)][Cl[NO(g)] [1]

(b) Initial concentration of NOCl(g):

= 3dm 5

mol 1 = 0.2 mol dm3

Concentration

(mol dm3) 2NOCl(g) 2NO(g) + Cl2(g)

Initial 0.2 0 0

Change 0.2 × (0.2)

= 0.04

+0.2 × (0.2)

= +0.04

+0.2 × (0.1)

= +0.02

Equilibrium 0.2 0.04

= 0.16 0.04 0.02

[2]

Kc = 2

2

(0.16)

(0.02)(0.04) [1]

= 0.00125 mol dm3 [1]

(c) Since the forward reaction is an endothermic reaction, [1] an increase in temperature shifts the

equilibrium position of the system to the right. Hence, the degree of dissociation increases. [1]

(d) The numbers of moles of gases on the left-hand side of the balanced equation is less than that on

the right-hand side.[1] An increase in pressure shifts the equilibrium position of the system to the

left, hence the degree of dissociation decreases. [1]

68 (a) [SO3(g)] increases. [1]

(b) [SO3(g)] decreases. [1]

(c) [SO3(g)] increases. [1]

(d) [SO3(g)] increases. [1]

69

(a) Kc = eqm2eqm4

eqm3

2eqm

O(g)][H(g)][CH

(g)][H[CO(g)] [1]

(b) (i) The equilibrium position of the system will shift to the left [1] because the number of moles

of gases on the left-hand side of the equation is less than that on the right-hand side. [1]

(ii) The equilibrium position of the system will shift to the right [1] because the forward reaction

is endothermic. [1]

(iii) The equilibrium position of the system will not be affected [1] because the catalyst can only

affect the time to attain equilibrium but not the equilibrium position. [1]

(c) (i) The value of Kc remains unchanged. [1]

(ii) The value of Kc increases. [1]

(iii) The value of Kc remains unchanged. [1]

70 (a) Kc = [H2S(g)]eqm[NH3(g)]eqm [1]

(b)

Concentration

(mol dm3) NH4HS(s) H2S(g) + NH3(g)

Initial 0 0.60

Change +0.27 0.87 0.60

= +0.27

Equilibrium 0.27 0.87

Kc = [H2S(g)]eqm[NH3(g)]eqm

= (0.27 mol dm3)(0.87 mol dm3) [1]

= 0.2349 mol2 dm6 [1]

(c) Let V be the volume of the sealed flask.

Doubling the volume of the sealed flask, the new volume = 2V

The initial number of moles of ammonia gas = molarity × volume = 0.60 × V

The initial concentration of ammonia gas = volumenew

gas ammonia of moles ofnumber

= V

V

2

60.0 = 0.30 mol dm3 [1]

Let x mol dm3 be the change in the concentration of H2S(g).

Concentration

(mol dm3) NH4HS(s) H2S(g) + NH3(g)

Initial 0 0.30

Change +x +x

Equilibrium x 0.30 + x

0.2349 = [H2S(g)][NH3(g)]

0.2349 = (x)(0.30 + x) [1]

0.2349 = 0.3x + x2

x = 0.357 or x = 0.657 (rejected) [1]

∴ [NH3(g)]eqm = 0.30 + 0.357 = 0.657 mol dm3 [1]

71 (a) Kc =

eqm3

2eqm2

eqm2

3

(g)][F(g)][Cl

(g)][ClF [1]

(b) (1) Decreasing the temperature of the system. [1]

(2) Adding either Cl2(g) or F2(g) to the system. [1]

(3) Increasing the pressure by decreasing the volume of the system. [1]

[1]

(c) It was wrong to predict the amount of reactants and products in the mixture by referring to the

enthalpy change of the reaction [1] because enthalpy change only implies that the products are

more energetically stable than that of the reactants, but it gives no information on the amount of

reactants and products in the mixture. [1] Only Kc can tell the ratio of the concentrations of

reactants and products. [1]

72 (a) As the forward reaction is exothermic, [1] an increase in the temperature of the reaction vessel

causes the equilibrium position of the system to shift to the left and the value of Kc will decrease.

[1]

(b) As the numbers of moles of gases on the left-hand side of the balanced equation are greater than

that on the right-hand side, [1] an increase in the volume of the reaction vessel causes the

equilibrium position of the system to shift to the left. But the value of Kc will remain unchanged.

[1]

(c) Adding C(graphite) to the reaction vessel does not affect the concentration of C(graphite) in the

system. [1] Hence, there is no effect on both the equilibrium position and the value of Kc. [1]

73

(a) Kc = 3eqm2eqm2

2eqm3

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

)]g(NH[ [1]

(b) (i) The reaction quotient, Qc, has the same form as the equilibrium constant, Kc. [1] It can be

calculated from concentrations at any particular moment, not necessarily at equilibrium. [1]

(ii) Once an equilibrium system is disturbed, more products will be formed if Qc Kc and more

reactants will be formed if Qc Kc. The equilibrium is established if Qc = Kc. [3]

sed. [1]

(c) (i) If the volume of the system is decreased, the pressure of the system is increased. [1] Since the

number of moles of gas molecules on the product side is less than that on the reactant side, [1]

an increase in pressure shifts the equilibrium position to the right, [1] producing more

NH3(g). Thus, the yield of NH3(g) is increa

(ii) Since the backward reaction is an endothermic change, an increase in temperature favours an

endothermic reaction. Thus, the equilibrium position shifts to the left, [1] removing some of

added heat and consuming some NH3(g). Therefore, the yield of NH3(g) is decreased. [1]

74

(a) Kc = 3eqm2eqm2

2eqm3

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

)]g(NH[ [1]

(b) (i) The equilibrium position will shift to the right. [1]

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

exothermic reaction. [1]

(ii) The equilibrium position will shift to the left. [1]

When the volume of the system is increased, the pressure of the system is decreased. [1]

Since the number of moles of gas molecules on the reactant side is greater than that on the

product side, [1] the system shifts to the left, increasing the number of gas molecules and

bringing the pressure back up. [1]

equlibrium sq ans

Documents

rate of backward reaction

reversible reaction

rate of forward reaction

reaction system

reaction mixture

backward reactions

b nh3g

c ethanoic acid