10. Solution Guide to Supplementary Exercises

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Topic 11 Chemical Equilibrium

Part A Unit-based exercise

Unit 39 An introduction to chemical equilibrium

Fill in the blanks

1 Chemical reaction that take place in one direction

only are known as irreversible reactions.

2 The reaction between ethanoic acid and ethanol

does not go to completion no matter how long

the reaction mixture is heated under reflux. This

is known as a reversible reaction.

3 A dynamic equilibrium is reached when

the forward and backward reactions occur at the

same rate.

4 In an aqueous solution of potassium chromate,

the following equilibrium system is established:

2CrO42–(aq) + 2H+(aq) Cr2O7

2–(aq) + H2O(l)

a) When a little dilute sulphuric acid is added to

the system, the colour of the solution changes

from yellow to orange . This

indicates that the concentration of Cr2O72–(aq)

ions has increased while the concentration

of CrO42–(aq) ions has decreased.

b) When a little dilute sodium hydroxide solution

is added to the resulting solution, the colour

of the solution changes from orange

to yellow . This indicates that the

concentration of CrO42–(aq) ions

has increased while the concentration

of Cr2O72–(aq) ions has decreased.

5 The direction in which a net reaction will proceed

to achieve equilibrium can be predicted by

comparing reaction quotient (Qc) and equilibrium

constant (Kc).

a) When Qc < Kc, a net forward reaction

must occur until equilibrium is reached.

b) When Qc > Kc, a net backward reaction

must occur until equilibrium is reached.

True or false

Decide whether each of the following statements is true or false.

6 A dynamic equilibrium is reached when T the forward and backward reactions occur at the same rate.

7 For a system at equilibrium, the F concentrations of the reactants and the products must be the same.

8 Equilibrium can only be established in an F open system.

9 The value of Kc for a reaction can be F used to judge the rate at which equilibrium is attained.

10 If Qc is greater than Kc, the system is not T at equilibrium.

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Multiple choice questions

11 Which of the following statements is true for a reaction system at equilibrium?

A All reactions cease. B The reactions have gone to completion. C The rates of the forward and backward

reactions are equal. D The amount of products equals the amount

of reactants. C

12 Consider the following reaction:

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

Initially, SO3(g) is placed in an empty flask. How do the rates of the forward and backward reactions change as the system proceeds to equilibrium?

Forward Backward reaction rate reaction rate

A Increases increases B Increases decreases C Decreases decreases D Decreases increases D

13 Consider the following equilibrium system:

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

Which of the following graphs represents [H2(g)] after equilibrium has been established?

A B

C D

B

14 Two experiments were performed involving the following equilibrium. The temperature was the same in both experiments.

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

In experiment A, 1.00 mol dm–3 H2(g) and 1.00 mol dm–3 I2(g) were initially added to a flask and equilibrium was established. In experiment B, 2.00 mol dm–3 HI(g) were initially added to a second flask and equilibrium was established. Which of the following statements is always true about the equilibrium concentrations?

A [H2(g)] equals [HI(g)] in experiment A. B [HI(g)] equals 2[H2(g)] in experiment A. C [HI(g)] in experiment A equals [HI(g)] in

experiment B. D [HI(g)] in experiment A equals

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[I2(g)] in

experiment B. C

15 Which of the factors below is NOT a condition necessary for equilibrium?

A A closed system B A constant temperature C Equal forward and backward reaction rates D Equal concentrations of reactants and

products D

Directions: Questions 16 and 17 refer to the following information.

Consider the following reaction:

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

A 1.00 dm3 container is initially filled with NH3(g).

16 What are the changes in the rate of the forward reaction and the concentration of N2(g) as the system approaches equilibrium?

Rate of forward reaction Concentration of N2(g)

A Decreases increases B Decreases decreases C Increases increases D Increases decreases A

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17 At equilibrium, there is 0.0400 mole of N2(g) present. What is the concentration of H2(g)?

A 0.0400 mol dm–3

B 0.0600 mol dm–3

C 0.0800 mol dm–3

D 0.120 mol dm–3 D

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Which of the following chemical reactions is consistent with the above graph?

A X(g) Y(g) + 2Z(g) B X(g) Y(g) + 3Z(g) C 2X(g) Y(g) + Z(g) D 2X(g) 2Y(g) + Z(g) B

19 Consider the hypothetical reaction:

2A(g) + B(g) 3C(g)

What is the equilibrium constant, Kc?

A Kc = [C(g)]3

[A(g)]2[B(g)]

B Kc = [A(g)]2[B(g)][C(g)]3

C Kc = [C(g)]3

[A(g)]2 + [B(g)]

D Kc = [A(g)]2 + [B(g)][C(g)]3 A

20 What is the equilibrium constant, Kc, for the reaction below?

2S(s) + 3O2(g) 2SO3(g)

A Kc = 2[SO3(g)]

2[S(s)] + 3[O2(g)]

B Kc = 2[SO3(g)]3[O2(g)]

C Kc = [SO3(g)]2

[S(s)]2[O2(g)]3

D Kc = [SO3(g)]2

[O2(g)]3 D

21 CO2(g) + 3H2(g) CH3OH(g) + H2O(g)

What is the equilibrium constant, Kc, for this reaction?

A Kc = [CH3OH(g)][H2O(g)][CO2(g)][H2(g)]3

B Kc = [CO2(g)][H2(g)]3

[CH3OH(g)][H2O(g)]

C Kc = [CH3OH(g)] + [H2O(g)][CO2(g)] + 3[H2(g)]

D Kc = [CO2(g)] + 3[H2(g)][CH3OH(g)] + [H2O(g)]

A


CaCO3(s) + 2HF(g) CaF2(s) + H2O(g) + CO2(g)

Which of the following expressions represents the equilibrium concentration of CO2(g)?

A [CO2(g)] = Kc[H2O(g)][HF(g)]2

B [CO2(g)] = Kc[HF(g)]2

[H2O(g)]

C [CO2(g)] = Kc[CaCO3(s)][HF(g)]2

[H2O(g)]

D [CO2(g)] = Kc[CaCO3(s)][HF(g)]2

[CaF2(s)][H2O(g)] B

23 For which of the following systems does Kc = [O2(g)]?

A O2(l) O2(g) B 3O2(g) 2O3(g) C 2O2(g) + N2(g) N2O4(g) D 2Hg(s) + O2(g) 2HgO(s) A

24 1 mole of N2O4(g) was placed in an empty 1 dm3 container and allowed to reach equilibrium according to the following equation:

N2O4(g) 2NO2(g)

At equilibrium, x mole of N2O4(g) had dissociated. What is the value of the equilibrium constant, Kc, at the temperature of the experiment?

A 2x(1 – x)

B 2x(1 – x)2

C 4x2

(1 – x)2 D 4x2

(1 – x) D

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25 An equilibrium mixture at constant temperature and pressure,

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

was analyzed and found to contain 0.0800 mole of SO2(g), 0.200 mole of NO2(g), 0.250 mole of NO(g) and 0.480 mole of SO3(g) in a 10.0 dm3 container. What is the equilibrium constant, Kc, for this reaction?

A 7.52 B 1.14 C 0.302 D 0.133 D

26 The reaction below reaches equilibrium in a closed reaction vessel of volume 2.50 dm3.

2NO(g) + O2(g) 2NO2(g)

At equilibrium, there are 2.83 moles of NO(g), 3.00 moles of O2(g), and 18.0 moles of NO2(g). What is the equilibrium constant, Kc, for the reaction?

A 0.218 dm3 mol–1

B 1.83 dm3 mol–1

C 13.4 dm3 mol–1

D 33.7 dm3 mol–1 D

27 NO(g) and CO2(g) react according to the following equation:

NO(g) + CO2(g) NO2(g) + CO(g)

In an experiment, 4.00 moles of NO(g) and 0.900 mole of CO2(g) are placed in a 2.00 dm3 reaction vessel.

At equilibrium, 0.100 mole of CO2(g) is present. What is the equilibrium constant, Kc, for the reaction?

A 0.500 B 1.60 C 2.00 D 5.00 C


2.00 moles of each of H2(g) and I2(g) are allowed to react in a 1.00 dm3 container at a certain temperature. 3.50 moles of HI(g) are present at equilibrium.

28 What is the value of the equilibrium constant, Kc?

A 5.10 x 10–3

B 3.74 C 56.0 D 196 D

29 Which of the following graphs shows how the rates of the forward and backward reactions change when hydrogen and iodine are mixed?

A B

C D

A

30 X2(g) and Y2(g) react according to the following equation:

X2(g) + Y2(g) 2XY(g)

A mixture containing 4.00 moles each of X2(g) and Y2(g) is heated in a closed container. The system is allowed to reach equilibrium. The graph shows how the number of moles of each gas varies with time.

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What is the equilibrium constant, Kc, for the reaction?

A 0.0278 B 0.167 C 6.00 D 36.0 D


Fe3+(aq) + SCN–(aq) [Fe(SCN)]2+(aq)

50.0 cm3 of 0.100 mol dm–3 Fe3+(aq) are added to 30.0 cm3 of 0.200 mol dm–3 SCN–(aq). At equilibrium, the concentration of [Fe(SCN)]2+(aq) is found to be 0.0500 mol dm–3.


A 6.25 x 10–3 dm3 mol–1

B 0.400 dm3 mol–1

C 2.50 dm3 mol–1

D 160 dm3 mol–1 D


Equal volumes of two 1.00 mol dm–3 solutions of W and X are mixed. The reaction rapidly reaches equilibrium.

W(aq) + X(aq) Y(aq) + Z(aq)

The concentration of Z(aq) is found to be 0.300 mol dm–3.

32 What is the equilibrium concentration of W(aq)?

A 0.100 mol dm–3

B 0.200 mol dm–3

C 0.500 mol dm–3

D 0.700 mol dm–3 B

33 What is the value of Kc for the reaction?

A 9.00 B 2.25 C 0.360 D 0.184 B

34 PCl5(g) decomposes to form PCl3(g) and Cl2(g) according to the equation:

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

Four different flasks, A, B, C and D, at the same temperature, contain a mixture of PCl5(g), PCl3(g) and Cl2(g). The concentration, in mol dm–3, of these components in each of the flasks is shown below. In three of the four flasks, the mixture of gases is at equilibrium.

In which one is the mixture of gases NOT at equilibrium?

[PCl5(g)] [PCl3(g)] [Cl2(g)]

A 0.10 0.40 0.10 B 0.15 0.20 0.30 C 0.20 0.30 0.15 D 0.30 0.60 0.20 C

35 For which system does the equilibrium constant, Kc, have units of dm3 mol–1?

A H2(g) + I2(g) 2HI(g) B 2NO2(g) N2O4(g) C 2SO3(g) 2SO2(g) + O2(g) D CH3COOC2H5(l) + H2O(l) CH3COOH(l) + C2H5OH(l) B

36 For which system does the equilibrium constant, Kc, have NO units?

A C(s) + H2O(g) CO(g) + H2(g) B SO3(g) + NO(g) SO2(g) + NO2(g) C Cu2+(aq) + 4NH3(aq) [Cu(NH3)4]

2+(aq) D N2O4(g) 2NO2(g) B

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37 The equilibrium constant, Kc, for the reaction

N2O4(g) 2NO2(g)

is 6.10 x 10–3 mol dm–3 at 25 °C. What is the Kc for the following reaction?

NO2(g) 12

N2O4(g)

A 327 dm32 mol

– 12

B 164 dm32 mol

– 12

C 12.8 dm32 mol

– 12

D 3.05 x 10–3 dm32 mol

– 12 C

38 Consider the following hypothetical equilibrium systems:

A(g) B(g) Kc = 2.00 B(g) C(g) Kc = 0.0100

What is the value of Kc for the following reaction?

2C(g) 2A(g)

A 2 500 B 200 C 50.0 D 4.00 x 10–4 A

39 Starting with equal concentrations of reactants, which of the following will be closest to completion at equilibrium?

A CO(g) + Cl2(g) COCl2(g) Kc = 22 B 2H2O(g) + 2Cl2(g) 4HCl(g) + O2(g) Kc = 5.0 x 10–4

C 2HBr(g) H2(g) + Br2(g) Kc = 7.0 x 10–20

D N2(g) + O2(g) 2NO(g) Kc = 1.0 x 10–31

A

40 An equal number of moles of methane and steam are placed in a closed container at 800 °C. The following reaction occurs.

CH4(g) + H2O(g) CO(g) + 3H2(g) Kc = 1.78 x 10–3 at 800 °C

Which of the following is correct at equilibrium at this temperature?

A [CO(g)] > [CH4(g)] B [CH4(g)] > [H2(g)] C [CH4(g)] = [CO(g)] D [CO(g)] = 3[H2(g)] B

41 At a particular temperature, the equilibrium constant, Kc, for the reaction below is 65.0 dm3 mol–1.

2NO(g) + O2(g) 2NO2(g)

At equilibrium, the concentration of NO(g) is 0.600 mol dm–3 and that of O2(g) is 0.300 mol dm–3.

What is the equilibrium concentration of NO2(g)?

A 19.0 mol dm–3

B 3.44 mol dm–3

C 2.65 mol dm–3

D 0.526 mol dm–3 C

42 1.50 moles of CS2(g) and 3.00 moles of Cl2(g) are mixed and the following equilibrium is established:

CS2(g) + 3Cl2(g) S2Cl2(g) + CCl4(g)

At equilibrium, 0.300 mole of CCl4(g) is found. How much Cl2(g) is present?

A 0.900 mole B 1.80 moles C 2.10 moles D 2.70 moles C

43 At a certain temperature, the equilibrium constant, Kc, for the reaction

2NH3(g) + Cl2(g) N2H4(g) + 2HCl(g)

is 4.00.

An equilibrium mixture in a 1.00 dm3 container has 2.60 moles of NH3(g), 4.00 moles of Cl2(g) and 5.90 moles of N2H4(g).

What is the equilibrium concentration of HCl(g)?

A 4.28 mol dm–3

B 5.34 mol dm–3

C 10.7 mol dm–3

D 12.0 mol dm–3 A

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44 When 1.00 mole of phosphorus pentachloride was heated to 523 K in a closed vessel, 50.0% dissociated as shown.


How many moles of gas were present in the equilibrium mixture?

A 0.50 B 1.00 C 1.50 D 2.00 C


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

Initially, some NOCl(g) was placed in a 1.00 dm3 container. At equilibrium, there were 0.860 mole of NOCl(g), 0.0300 mole of NO(g) and 0.0150 mole of Cl2(g). How many moles of NOCl(g) were initially added to the container?

A 0.785 mole B 0.815 mole C 0.890 mole D 0.905 mole C


2O3(g) 3O2(g) Kc = 1

Which of the following correctly compares the equilibrium concentrations of the two species?

A [O3(g)] = [O2(g)]

B [O3(g)] = [O2(g)]32

C [O3(g)] = [O2(g)]23

D [O3(g)]32 = [O2(g)] B


N2O4(g) 2NO2(g) Kc = 4.50

Initially, 0.500 mole of N2O4(g) and 0.500 mole of NO2(g) are placed in a 1.00 dm3 container. Which of the following describes the changes in concentrations as the system proceeds towards equilibrium?

[N2O4(g)] [NO2(g)]

A Decreases decreases B Decreases increases C Increases decreases D Increases increases B


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

In an experiment, 0.10 mole of O2(g) and 0.10 mole of SO3(g) are added to an empty 1.0 dm3 flask and then the flask is sealed. The reaction goes backward to establish equilibrium. Which of the following must be true at equilibrium?

A [SO2(g)] = [O2(g)] = [SO3(g)] B [O2(g)] < [SO3(g)] C [O2(g)] = [SO2(g)] D [SO3(g)] < [O2(g)] D

49 Propanone can be made from propan-2-ol.

C3H8O(g) C3H6O(g) + H2(g) Kc = 0.0100

In an experiment, 6.00 moles of C3H8O(g), 0.150 mole of C3H6O(g) and 0.100 mole of H2(g) are placed in a 1.00 dm3 container and allowed to establish equilibrium.

Which of the following change(s) will occur as the system proceeds towards equilibrium?

A [C3H6O(g)] increases while [H2(g)] decreases. B [C3H6O(g)] and [H2(g)] both increase. C [C3H8O(g)] and [H2(g)] both increase. D [C3H8O(g)] and [C3H6O(g)] both decrease. B


3NO2(g) N2O5(g) + NO(g) Kc = 1.0 x 10–11 dm3 mol–1

Initially, some NO2(g), N2O5(g) and NO(g) are placed in a container and allowed to reach equilibrium. When equilibrium is established, it is found that the pressure has increased. Which of the following combinations is correct as the system proceeds towards equilibrium?

Reaction Direction quotient Qc of net reaction

A Qc > Kc net forward reaction B Qc < Kc net forward reaction C Qc > Kc net backward reaction D Qc < Kc net backward reaction C

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The equilibrium constant for the following reaction is Kc = 2.50 x 10–4 at 25 °C.

N2(g) + C2H2(g) 2HCN(g)

In an experiment, a mixture of N2(g), C2H2(g) and HCN(g), all of initial concentrations 1.00 mol dm–3, are allowed to reach equilibrium.

51 Which of the following statements is correct as the system proceeds towards equilibrium?

A A net forward reaction occurs because Qc < Kc.

B A net backward reaction occurs because Qc < Kc.

C A net forward reaction occurs because Qc > Kc.

D A net backward reaction occurs because Qc > Kc. D

52 What is the concentration of HCN(g) at equilibrium?

A 0.0158 mol dm–3

B 0.0235 mol dm–3

C 1.28 mol dm–3

D 1.53 mol dm–3 B

53 Consider the following:

(1) Constant temperature (2) Equal concentrations of reactants and

products (3) Equal rates of forward and backward

reactions

A system at equilibrium must have

A (1) and (2) only B (1) and (3) only C (2) and (3) only D (1), (2) and (3) B


2ICl(g) I2(g) + Cl2(g)

Initially, some ICl(g) is placed in an empty flask. Which of the following statements describe(s) the change(s) occurring as the system proceeds towards equilibrium?

(1) The rate of the backward reaction increases.

(2) Concentration of ICl(g) increases. (3) Concentration of Cl2(g) increases.

A (1) only B (2) only C (1) and (3) only D (2) and (3) only C


2SO2(g) + O2(g) 2SO3(g) Kc = 4.00

In an experiment, 0.600 mole of SO2(g), 0.300 mole of O2(g) and 0.600 mole of SO3(g) are placed in a 1.00 dm3 container. The system is allowed to reach equilibrium. Which of the following are correct as the system approaches equilibrium?

(1) Concentration of SO2(g) increases. (2) Concentration of O2(g) decreases. (3) Concentration of SO3(g) increases.

A (1) and (2) only B (1) and (3) only C (2) and (3) only D (1), (2) and (3) C

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Directions : Each question (Questions 56 – 60) consists of two separate statements. Decide whether each of the two statements is true or false; if both are true, then decide whether or not the second statement is a correct explanation of the first statement. Then select one option from A to D according to the following table :

A Both statements are true and the 2nd statement is a correct explanation of the 1st statement. B Both statements are true but the 2nd statement is NOT a correct explanation of the 1st statement. C The 1st statement is false but the 2nd statement is true. D Both statements are false.

1st statement 2nd statement

56 All equilibrium systems have equal concentrations At equilibrium, the rate of the forward C

of reactants and products. reaction and the rate of the backward reaction are equal.

57 When a small amount of dilute sulphuric acid When a small amount of dilute sulphuric acid A is added to an equilibrium system containing is added to the equilibrium system, the CrO4

2–(aq) ions and Cr2O72–(aq) ions, the colour concentration of CrO4

2–(aq) ions decreases while of the system changes from yellow to orange. that of Cr2O7

2–(aq) ions increases.

58 When H2(g) and I2(g) are allowed to react in a At equilibrium, both the forward and D closed container, the rate of the reaction between backward reactions stop. H2(g) and I2(g) becomes zero when equilibrium is attained.

59 A reaction with a large Kc can attain equilibrium The value of Kc can be used to predict the D rapidly. rate at which equilibrium is reached.

60 If Qc is greater than Kc, the system will undergo If Qc is greater than Kc, the system is not C a net forward reaction until equilibrium is at equilibrium. reached.

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Unit 40 Factors affecting chemical equilibrium systems

Fill in the blanks

1 When the concentration of a reactant in an

equilibrium system is increased,

a) a net forward reaction will occur;

b) the position of equilibrium will shift to the

right .

2 When the concentration of a product in an

equilibrium system is increased,

a) a net backward reaction will occur;

b) the position of equilibrium will shift to the

left .

3 An increase in pressure (i.e. a / an decrease

in volume) will bring about a net reaction

that decreases the number of moles of gas,

i.e. the position of equilibrium will shift to the

side of the equation with a fewer number

of moles of gas.

4 A decrease in pressure (i.e. a / an increase

in volume) will bring about a net reaction

that increases the number of moles of gas,

i.e. the position of equilibrium will shift to the

side of the equation with a greater number

of moles of gas.

5 When the temperature of an equilibrium

system with an exothermic forward reaction is

increased,

a) a net backward reaction will occur;

b) the position of equilibrium will shift to

the left ;

c) the value of equilibrium constant, Kc,

will decrease .

6 When the temperature of an equilibrium

system with an endothermic forward reaction is

increased,

a) a net forward reaction will occur;

b) the position of equilibrium will shift to

the right ;

c) the value of equilibrium constant, Kc,

will increase .

True or False

Decide whether each of the following statements is true or false.

7 If the position of equilibrium lies close F to the product side, the value of Kc is small.

8 When the concentration of a reactant in T an equilibrium system is increased, a net forward reaction will occur.

9 At a constant temperature, increasing F the concentration of a reactant in an equilibrium system will cause Kc to increase.

10 Removing some CaCO3(s) from an F equilibrium system of CaCO3(s), CaO(s) and CO2(g) will cause the position of equilibrium to shift to the CaCO3(s) side.

11 An increase in volume of an equilibrium T

system involving gases will bring about a net reaction that increases the number of moles of gas.

12 For an equilibrium system with an T exothermic forward reaction, decreasing the temperature will shift the position of equilibrium to the right.

13 Decreasing the temperature will cause F Kc for an endothermic reaction to increase.

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14 In the conversion of sulphur dioxide to F sulphur trioxide in the Contact process, a pressure of 200 atmospheres is usually used.

15 The optimum conditions for the Haber F process are 10 atm and 500 °C.

16 A catalyst will not affect the percentage T of the product in an equilibrium mixture.



N2H4(g) + 2O2(g) 2NO(g) + 2H2O(g)

Which of the following actions can cause the position of equilibrium to shift to the right?

A Adding N2H4(g) B Adding H2O(g) C Removing N2H4(g) D Removing O2(g) A


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

Which of the following actions would cause the concentration of H2(g) to decrease?

A Adding CO(g) B Adding H2O(g) C Removing CO2(g) D Removing H2O(g) D


2KClO3(s) 2KCl(s) + 3O2(g)

Which of the following actions will cause the position of equilibrium to shift to the left?

A Adding O2(g) B Adding KCl(s) C Removing KClO3(s) D Removing KCl(s) A


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

Some NH3(g) is added to the system. Which of the following combinations is correct?

Net change of [H2(g)] Value of Kc

A Decreases increases B Decreases remains constant C Increases increases D Increases remains constant D


Ca(OH)2(s) Ca2+(aq) + 2OH–(aq)

Adding which of the following substances will cause the equilibrium concentration of Ca2+(aq) ions to increase?

A H2O(l) B HCl(aq) C KOH(s) D Ca(OH)2(s) B

22 Consider the following system at equilibrium:

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

The position of equilibrium shifts to the right as the result of the addition of some extra H2O(g). How will this shift affect the concentrations of the other gases?

[CO(g)] [CO2(g)] [H2(g)]

A Increases decreases decreases B Increases increases decreases C Decreases decreases increases D Decreases increases increases D



2–(aq) + H2O(l)

A solution of Ba(NO3)2 is added, and a BaCrO4 precipitate forms. Which of the following combinations is correct?

Shift of position of equilibrium Value of Kc

A To the left remains constant B To the left increases C To the right remains constant D To the right decreases A

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

Which of the following graphs represents what happens when some HI(g) is removed and a new state of equilibrium is established?

A B

C D

B


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

Some H2(g) is injected into the system. Which of the combinations is correct?

Shift of position Change of [H2(g)] relative of equilibrium to previous equilibrium

A To the left increases B To the left decreases C To the right increases D To the right decreases A


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

Which of the following, when added to the system above, would result in a decrease in [H2O(g)] relative to the previous equilibrium?

A CO(g) B CO2(g) C H2(g) D H2O(g) A

27 The following equilibrium exists in aqueous bromine.

Br2(aq) + H2O(l) Br–(aq) + 2H+(aq) + OBr–(aq)yellow-brown colourless colourless

The yellow-brown colour of aqueous bromine would fade on adding a few drops of a concentrated solution of

A HCl. B KBr. C AgNO3. D NaOBr. C


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

How will the rates of the forward and backward reactions change when the volume of the reaction vessel is increased?

Rate of Rate of forward reaction backward reaction

A Increases increases B Increases decreases C Decreases increases D Decreases decreases D


2NO(g) + O2(g) 2NO2(g)

An equilibrium mixture of NO(g), O2(g) and NO2(g) is transferred from a 1 dm3 container to a 2 dm3 container.

Which of the following combinations describes what happens as the system proceeds towards a new state of equilibrium?

Shift of position Number of of equilibrium moles of NO2(g)

A To the right increases B To the right decreases C To the left increases D To the left decreases D

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At room temperature, N2O4(g) and NO2(g) exist in equilibrium as follows:

N2O4(g) 2NO2(g) pale yellow dark brown

A gas syringe is filled with a pale brown mixture of N2O4(g) and NO2(g) at equilibrium.

30 What would be observed when the plunger is quickly pushed in at time t?

A The mixture first lightens, and then becomes colourless.

B The mixture first darkens, and then becomes colourless.

C The mixture first darkens, and then lightens. D The mixture first lightens, and then darkens. C

31 Which of the following graphs represents how the concentration of NO2(g) in the mixture varies until a new state of equilibrium is established?

A B

C D

C



Which of the following combinations describes the effect of decreasing the pressure by increasing the volume?

Shift of position Number of of equilibrium moles of PCl3(g)

A To the right increases B To the right decreases C To the left increases D To the left decreases A


FeO(s) + H2(g) Fe(s) + H2O(g)

Which of the following combinations describes the effect of decreasing the volume of the system?

Shift of position of equilibrium Concentration of H2(g)

A To the right increases B To the right decreases C No change increases D No change decreases C


H2(g) + I2(g) 2HI(g) ΔH < 0 colourless purple colourless

Which of the following combinations describes the effect of increasing the pressure of the system?

Colour intensity of the mixture Value of Kc

A Decreases decreases B Decreases unchanged C Increases unchanged D Increases decreases C

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35 In which of the following systems will the position of equilibrium shift to the right when the pressure is increased at constant temperature?

A CaCO3(s) CaO(s) + CO2(g) B 2NO2(g) 2NO(g) + O2(g) C H2(g) + I2(g) 2HI(g) D PCl3(g) + Cl2(g) PCl5(g) D

36 In which of the following would the position of the equilibrium NOT be affected by a volume change at constant temperature?

A 2NF2(g) N2F4(g) B C2H6(g) C2H4(g) + H2(g) C 2NOCl(g) 2NO(g) + Cl2(g) D CO(g) + H2O(g) H2(g) + CO2(g) D

37 The value of equilibrium constant for a gaseous reaction will change when

A a catalyst is used. B the temperature changes. C the concentrations of products change. D the volume changes. B


N2(g) + 3H2(g) 2NH3(g) ΔH < 0

Which of the following sets of conditions will favour the formation of the product?

A Low pressure and low temperature B Low pressure and high temperature C High pressure and high temperature D High pressure and low temperature D


2SO2(g) + O2(g) 2SO3(g) ΔH = –197 kJ

Which of the following combinations describes the effect of increasing the temperature of the system?

Shift of position of equilibrium Value of Kc

A To the left increases B To the left decreases C To the right increases D To the right decreases B


The hydrogen used in the Haber process is made by the following reaction:

CH4(g) + H2O(g) CO(g) + 3H2(g) ΔH > 0

40 Which of the following combinations describes the effect of increasing the pressure of the system?

Equilibrium yield of H2(g) Reaction rate

A Decreases decreases B Decreases increases C Increases decreases D Increases increases B

41 Which of the following combinations describes the effect of increasing the temperature of the system?

Equilibrium yield of H2(g) Reaction rate

A Decreases decreases B Decreases increases C Increases decreases D Increases increases D

42 Consider the following graph which relates to this equilibrium system:

Fe3+(aq) + SCN–(aq) [Fe(SCN)]2+(aq) ΔH < 0

Which of the following actions would cause the concentration changes at time t?

A Addition of Fe3+(aq) ions B Removal of [Fe(SCN)]2+(aq) ions C Increase in temperature D Increase in pressure C

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Co2+(aq) + 4Cl–(aq) CoCl42–(aq)

pink blue

When the temperature is increased, the solution turns dark blue. It can be deduced that the forward reaction is

A exothermic and the value of K c has increased.

B exothermic and the value of K c has decreased.

C endothermic and the value of Kc has increased.

D endothermic and the value of Kc has decreased. C

44 Consider the following system at fixed pressure:

CH4(g) + H2O(g) CO(g) + 3H2(g) ΔH > 0

What is the effect of a small increase in temperature on the rate of the forward reaction (Rf), rate of the backward reaction (Rb), and the equilibrium constant (Kc)?

Rf Rb Kc

A Lower higher lower B Higher lower unchanged C Higher higher higher D Higher higher lower C


2H2O(g) + 2Cl2(g) 4HCl(g) + O2(g) ΔH > 0

The temperature of the system is lowered and the amount of Cl2(g) changes by 1 mole. Which of the following combinations is correct?

Amount Amount Amount of Cl2(g) of HCl(g) of O2(g)

A Increase by decrease by decrease by 1 mole 0.5 mole 2 moles B Increase by decrease by decrease by 1 mole 2 moles 0.5 mole C Decrease by increase by increase by 1 mole 0.5 mole 2 moles D Decrease by increase by increase by 1 mole 2 moles 0.5 mole B


2NO2(g) 2NO(g) + O2(g)

The value of the equilibrium constant for the reaction is 1.45 x 10–6 at 227 °C and 0.938 at 727 °C.

Which of the following combinations concerning the forward reaction is correct?

Product yield as Sign of ΔH temperature increases

A + increases B + decreases C – increases D – decreases A

47 Methanol can be produced by the reaction between carbon monoxide and hydrogen.

CO(g) + 2H2(g) CH3OH(g)

Two experiments were conducted.

Experiment 1: Some CO(g) and H2(g) were placed in a sealed vessel and the reaction allowed to proceed at constant temperature.

Experiment 2: Experiment 1 was repeated, but at a different temperature.

The graph below shows the amount of methanol produced over the course of Experiments 1 and 2.

These results show that Experiment 2 was conducted at a

A lower temperature than Experiment 1 and the forward reaction is endothermic.

B lower temperature than Experiment 1 and the forward reaction is exothermic.

C higher temperature than Experiment 1 and the forward reaction is endothermic.

D higher temperature than Experiment 1 and the forward reaction is exothermic. D

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N2(g) + 3H2(g) 2NH3(g) ΔH < 0

Which of the following actions will cause the concentration changes at time t?

A Addition of N2(g) B Removal of H2(g) C Decrease in temperature D Decrease in volume A


Consider the following equilibrium system:

CO(g) + 2H2(g) CH3OH(g) ΔH = –91 kJ

49 Which of the following actions will each cause the position of equilibrium to shift to the right?

A Increase temperature, increase volume B Increase temperature, decrease volume C Decrease temperature, decrease volume D Decrease temperature, increase volume C

50 Which of the following graphs represents the rate of the forward reaction when the temperature of the system is increased at time t?

A B

C D

B

51 In which of the following systems will the position of equilibrium shift to the left upon an increase in pressure, but to the right upon an increase in temperature?

A CO2(g) + H2(g) CO(g) + H2O(g) ΔH > 0 B C2H6(g) C2H4(g) + H2(g) ΔH > 0 C C2H4(g) + H2O(g) C2H5OH(g) ΔH < 0 D 2SO2(g) + O2(g) 2SO3(g) ΔH < 0 B

52 Phosgene (COCl2(g)) is manufactured by passing carbon monoxide and chlorine through a bed of carbon which acts as a catalyst.

CO(g) + Cl2(g) COCl2(g) ΔH < 0

Which of the following sets of conditions will favour the formation of phosgene?

A Low pressure and low temperature B Low pressure and high temperature C High pressure and low temperature D High pressure and high temperature C


N2(g) + 3H2(g) 2NH3(g) ΔH < 0

Which of the following graphs shows the effect of temperature and pressure on the percentage of NH3(g) in the equilibrium mixture?

A B

C D

D

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PCl3(g) + Cl2(g) PCl5(g) ΔH < 0

Which of the following graphs shows the relationship between concentration and time as a result of adding a catalyst at time t?

A B

C D

C



2SO2(g) + O2(g) 2SO3(g) ΔH < 0

55 How can the number of moles of O2(g) at equilibrium in the system be increased?

A Adding a catalyst B Decreasing the temperature C Decreasing the volume of the system D Adding SO3(g) to the system D

56 Which of the following graphs shows the rate of the backward reaction when a catalyst is added to the equilibrium system at time t?

A B

C D

B


PCl3(g) + Cl2(g) PCl5(g) ΔH < 0

Which of the following will NOT cause the position of equilibrium to shift to the right?

A Adding more Cl2(g) B Adding a catalyst C Increasing the pressure D Decreasing the temperature B


Ag+(aq) + Cl–(aq) AgCl(s)

Which of the following actions will increase the amount of solid silver chloride of the equilibrium system?

(1) Adding NaCl(aq) (2) Adding AgNO3(aq) (3) Removing Cl–(aq)

A (1) and (2) only B (1) and (3) only C (2) and (3) only D (1), (2) and (3) A


N2H4(g) + 2O2(g) 2NO(g) + 2H2O(g)

Some O2(g) is added to the equilibrium system and a new state of equilibrium is established. Which of the following substances have a net increase in concentration, relative to the previous equilibrium concentrations?

(1) N2H4(g) (2) O2(g) (3) NO(g) (4) H2O(g)

A (1) and (3) only B (2) and (4) only C (1), (2) and (3) only D (2), (3) and (4) only D

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60 The dissociation of a weak acid in aqueous solution is represented by the following equation:

HA(aq) H+(aq) + A–(aq)

When the above system is at equilibrium, which of the following statements are correct?

(1) Both HA(aq) and H+(aq) are present in the system.

(2) More HA(aq) dissociates upon the addition of NaOH(aq).

(3) Decreasing the pressure shifts the position of equilibrium to the right.


61 For which of the following equilibrium systems will a decrease in volume at constant temperature cause a decrease in the amounts of products?

(1) CaCO3(s) CaO(s) + CO2(g) (2) CH4(g) + H2O(g) CO(g) + 3H2(g) (3) HCl(g) + H2O(l) H3O

+(aq) + Cl–(aq)


62 The following reaction reaches equilibrium in a closed reaction vessel.

C2H4(g) + H2O(g) C2H5OH(g) ΔH < 0

Which of the following action(s) will increase the mass of C2H5OH(g) in the equilibrium mixture?

(1) Adding a catalyst (2) Decreasing the volume of the reaction

vessel (3) Increasing the temperature

A (1) only B (2) only C (1) and (3) only D (2) and (3) only B


NH4Cl(s) NH3(g) + HCl(g) ΔH > 0

Which of the following actions would favour the formation of NH3(g)?

(1) Adding a small amount of NH4Cl(s) (2) Increasing the temperature (3) Increasing the pressure



N2O4(g) 2NO2(g) ΔH > 0

Which of the following would cause the concentration changes at time t?

(1) A change in temperature (2) A change in pressure (3) A change in the concentration of NO2(g)

A (1) only B (2) only C (1) and (3) only D (2) and (3) only A

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65 Hydrogen iodide is formed when hydrogen and iodine react according to the following equation.

H2(g) + I2(g) 2HI(g) ΔH = –10 kJ

Which of the following statements about the equilibrium system is / are correct?

(1) The use of a catalyst would have no effect on the yield of HI(g).

(2) Increasing the total pressure increases the yield of HI(g).

(3) Increasing the temperature increases the yield of HI(g).


66 Which of the following statements concerning a catalyst is / are correct?

(1) It decreases the enthalpy change of the reaction.

(2) It increases the rate of formation of the products.

(3) It increases the concentration of the products.





67 At a constant temperature, increasing the Increasing the concentration of a reactant C

concentration of a reactant in an equilibrium will cause the position of equilibrium to shift system will cause Kc to increase. to the product side.

68 Adding some H2(g) to an equilibrium system of When some H2(g) is added to the equilibrium A H2(g), I2(g) and HI(g) will cause a net decrease system, a net reaction will occur to consume in the concentration of I2(g). some of the H2(g).

69 When the pressure of an equilibrium system When the pressure of an equilibrium system C involving gases is increased, the position of involving gases is increased, the rate of the equilibrium must shift to the right. forward reaction will increase.

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70 At a constant temperature, changing the The value of Kc for a reversible reaction C

pressure of an equilibrium system of N2(g), H2(g) depends only on temperature. and NH3(g) has no effect on the position of equilibrium.

71 Decreasing the pressure of an equilibrium system Decreasing the pressure will bring about a A of N2O4(g) and NO2(g) will cause more NO2(g) net reaction that increases the number of to form. moles of gas.

72 Increasing the volume of the reaction vessel of Increasing the volume of the reaction vessel D an equilibrium system of H2(g), I2(g) and HI(g) of the equilibrium system of H2(g), I2(g) and will cause the rate of the reaction between HI(g) will cause the position of equilibrium to H2(g) and I2(g) to increase. shift to the right.

73 When the temperature of an equilibrium system For an equilibrium system with an exothermic D with an exothermic forward reaction is increased forward reaction, increasing the temperature by 10 °C, the equilibrium constant Kc doubles. will shift the position of equilibrium to the right.

74 Using iron as a catalyst in the Haber process can A catalyst can increase the rate of the reaction C increase the yield of ammonia. between nitrogen and hydrogen.

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Part B

Part B Topic-based exercise


1 Phosphorus reacts with chlorine according to the following equation:

P4(s) + 6Cl2(g) 4PCl3(g)


A 4[PCl3(g)][P4(s)]6[Cl2(g)]

B 4[PCl3(g)]6[Cl2(g)]

C [PCl3(g)]4

[P4(s)][Cl2(g)]6

D [PCl3(g)]4

[Cl2(g)]6 D

2 Nitrogen dioxide decomposes on heating according to the following equation:

2NO2(g) 2NO(g) + O2(g)

When 6.80 moles of NO2(g) were put into a 1.00 dm3 container and heated, the equilibrium mixture contained 1.20 moles of O2(g).


A 0.357 mol dm–3

B 0.655 mol dm–3

C 1.53 mol dm–3

D 2.80 mol dm–3 A

3 Which of the following statements is INCORRECT for a system at equilibrium?

A The system has a constant mass. B The system acts so as to oppose

disturbances. C The forward and backward reactions proceed

at the same rate. D The reactant and product concentrations vary

with time. D



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

1.00 mole of CO2(g) and 2.00 moles of H2(g) are placed in a 2.00 dm3 container. At equilibrium, the concentration of CO(g) is 0.280 mol dm–3.

4 What is the equilibrium concentration of CO2(g)?

A 0.220 mol dm–3

B 0.360 mol dm–3

C 0.440 mol dm–3

D 0.720 mol dm–3 A

5 What is the equilibrium constant, Kc, for the reaction?

A 0.00810 B 0.495 C 2.02 D 123 B


In a reversible reaction, propanoic acid and ethanol react producing an ester and water.

propanoic acid + ethanol ester + water

In an experiment, 1.00 mole of propanoic acid, 1.00 mole of ethanol and 2.00 moles of water were mixed. At equilibrium, 2.40 moles of water were found to be present.

6 The equilibrium constant, Kc, for the reaction in this experiment has the value

A 2.67. B 1.56. C 0.643. D 0.375. A

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Part B

7 Which of the following graphs best represents how the rates of the forward and backward reactions change over time?

A B

C D

C

8 Equal volumes of 1.00 mol dm–3 solutions of W and X are mixed. The reaction below rapidly reaches equilibrium.

W(aq) + X(aq) Y(aq) + Z(aq)

At equilibrium, the concentration of Y(g) is found to be 0.400 mol dm–3. What is the value of Kc for the above reaction?

A 0.250 B 0.440 C 4.00 D 16.0 D


N2(g) + 2O2(g) 2NO2(g)

Equal number of moles of N2(g) and O2(g) are placed, under certain conditions, in a closed container. Which of the following describes the changes which occur as the system proceeds towards equilibrium?

Rate of Concentration backward reaction of NO2(g)

A Increases increases B Decreases increases C Increases decreases D Decreases decreases A


2O3(g) 3O2(g) Kc = 36.0 mol dm–3

What is the concentration of O3(g) when the equilibrium concentration of O2(g) is 5.80 x 10–2 mol dm–3?

A 2.32 x 10–3 mol dm–3

B 4.54 x 10–3 mol dm–3

C 3.87 x 10–2 mol dm–3

D 8.70 x 10–2 mol dm–3 A

11 An equal number of moles of steam and chlorine are placed in a closed container at 375 K. The following reaction occurs.

2H2O(g) + 2Cl2(g) 4HCl(g) + O2(g) Kc = 5 x 10–4 mol dm–3 at 375 K

Which of the following relates [Cl2(g)] and [HCl(aq)] at equilibrium?

A [Cl2(g)] < 2[HCl(g)] B 2[Cl2(g)] > [HCl(g)] C [Cl2(g)] = 2[HCl(g)] D 2[Cl2(g)] = [HCl(g)] B

12 Propanoic acid and ethanol react producing an ester and water.

propanoic acid + ethanol ester + water

At a certain temperature, the equilibrium constant, Kc, for the reaction is 2.45.

In an experiment, 2.00 moles of propanoic acid, 2.00 moles of ethanol and 2.40 moles of water are mixed. What is the equilibrium concentration of the ester?

A 0.900 mol dm–3

B 1.10 mol dm–3

C 1.47 mol dm–3

D 3.30 mol dm–3 A

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Part B

13 2SO2(g) + O2(g) 2SO3(g) Kc = 4.50 dm3 mol–1

In an experiment, a mixture of 4.00 moles of sulphur dioxide and 1.50 moles of oxygen is allowed to reach equilibrium.

The amount of sulphur trioxide present at equilibrium would be

A 5.50 moles. B 4.00 moles. C 3.00 moles. D less than 3 moles. D

14 Given the following equilibrium constants:

H2S(aq) H+(aq) + HS–(aq) Kc = 9.50 x 10–8 mol dm–3

HS–(aq) H+(aq) + S2–(aq) Kc = 1.00 x 10–19 mol dm–3

What is the equilibrium constant for the following reaction?

S2–(aq) + 2H+(aq) H2S(aq)

A 9.50 x 10–27

B 9.75 x 10–14

C 9.50 x 1011

D 1.05 x 1026 D


N2(g) + 3H2(g) 2NH3(g) Kc = 64.0 dm6 mol–2

A 1.00 dm3 container is filled with 0.280 mole of N2(g), 0.600 mole of H2(g) and 0.540 mole of NH3(g). The system is allowed to reach equilibrium.

Which of the following combinations is correct as the system proceeds towards equilibrium?

Direction of net reaction Pressure of system

A Forward increases B Forward decreases C Backward increases D Backward decreases B

16 Consider the following equilibrium:


A flask is filled with NOCl(g), NO(g) and Cl2(g). Initially there is a total of 5.00 moles of gases present. When equilibrium is reached, there is a total of 6.00 moles of gases present. Which of the following explains this observation?

A A net forward reaction occurs because Qc < Kc.

B A net forward reaction occurs because Qc > Kc.

C A net backward reaction occurs because Qc < Kc.

D A net backward reaction occurs because Qc > Kc. A


Fe2O3(s) + 3CO(g) 2Fe(l) + 3CO2(g)

Which of the following actions will cause the position of equilibrium to shift to the right?

A Adding Fe2O3(s) B Adding Fe(l) C Removing CO(g) D Removing CO2(g) D


CaCO3(s) CaO(s) + CO2(g) Kc = 0.100 mol dm–3

18 Which of the following changes will cause the position of equilibrium to shift to the left?

A Adding more CaO(s) B Removing CaCO3(s) C Decreasing the volume D Increasing the surface area of CaO(s) C

19 Initially, 30.0 g of CaCO3(s) were placed in a 2.00 dm3 container. What mass of CO2(g) would be present at equilibrium?

(Relative atomic masses: C = 12.0, O = 16.0)

A 0.100 g B 2.20 g C 8.80 g D 15.0 g C

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Part B


Consider the following equilibrium system:


2–(aq) + H2O(l)

A little dilute sulphuric acid is added to the above equilibrium system.

20 What would be observed?

A The mixture first becomes orange in colour, and then colourless.

B The mixture first becomes yellow in colour, and then colourless.

C The mixture becomes orange in colour. D The mixture becomes yellow in colour. C

21 Which of the following graphs represents how the concentration of CrO4

2–(aq) ions in the mixture varies until a new state of equilibrium is established?

A B

C D

B


The following equilibrium exists in aqueous bromine.

Br2(aq) + H2O(l) Br–(aq) + 2H+(aq) + OBr–(aq)

A little dilute sodium hydroxide solution is added to the above equilibrium system.


A The mixture first darkens, and then lightens. B The mixture first lightens, and then darkens. C The mixture gets lighter. D The mixture get darker. C

23 Which of the following graphs represents how the concentration of Br–(aq) ions in the mixture varies until a new state of equilibrium is established?

A B

C D

C

24 Tooth enamel, Ca5(PO4)3OH, establishes the following equilibrium:

Ca5(PO4)3OH(s) 5Ca2+(aq) + 3PO4

3–(aq) + OH–(aq)

Which of the following, when added to the above equilibrium system, will cause the position of equilibrium to shift to the right?

A H+(aq) B OH–(aq) C Ca2+(aq) D Ca5(PO4)3OH(s) A

25 Each of the following equilibrium systems is disturbed by increasing the pressure as a result of decreasing the volume.

In which of the systems will the number of moles of products increase?

A 2CO2(g) 2CO(g) + O2(g) B N2F4(g) 2NF2(g) C Si(s) + 2Cl2(g) SiCl4(g) D N2(g) + C2H2(g) 2HCN(g) C

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Part B


The diagram below shows a gas syringe containing a pale brown mixture of N2O4(g) and NO2(g) at equilibrium at room temperature:


Some N2O4(g) is added to the gas syringe at time t while the volume and the temperature of the mixture are both kept constant.


A The mixture first darkens, and then lightens. B The mixture first lightens, and then becomes

colourless. C The mixture gradually gets darker. D The mixture gradually becomes colourless. C

27 Which of the following graphs represents how the concentration of NO2(g) in the mixture varies until a new state of equilibrium is established?

A B

C D

C

28 The system below reaches equilibrium in a closed reaction vessel.

4HCl(aq) + MnO2(s) Cl2(g) + 2H2O(l) + Mn2+(aq) + 2Cl–(aq) ΔH < 0

Which of the following actions will increase the mass of Cl2(g) in the equilibrium mixture?

A Adding some MnO2(s) B Increasing the temperature C Decreasing the volume of the reaction vessel D Adding something that precipitates the

Mn2+(aq) ions D


CH3COOH(aq) + H2O(l) CH3COO–(aq) + H3O

+(aq) ΔH < 0

Which of the following actions caused the change in the concentration of H3O

+(aq) at time t?

A Addition of HCl(aq) B Decreasing the temperature C Addition of CH3COO–(aq) ions D Increasing the volume of the container C

26

Part B


The hydrogen used in the Haber process is made by the following reaction:

CH4(g) + H2O(g) CO(g) + 3H2(g) ΔHO = +206 kJ

30 Which of the following sets of conditions will favour the formation of hydrogen?

A Low pressure and low temperature B Low pressure and high temperature C High pressure and low temperature D High pressure and high temperature B

31 Equal amounts of CH4(g) and H2O(g) are placed in a reaction vessel and allowed to react. After 10 minutes, equilibrium has been reached. At that time, some H2(g) is added to the mixture and a new state of equilibrium is established.

Which of the following graphs represents the changes in the concentrations of CH4(g) and H2(g) in the reaction mixture?

A B

C D

B

32 Consider the following equilibrium system in a closed reaction vessel:

CH3CHO(g) CH4(g) + CO(g) ΔH > 0

What would happen if some CH4(g) is added to the system?

A The amounts of all substances increase, relative to their previous equilibrium amounts.

B The value of Kc increases. C The amounts of CH3CHO(g) and CO(g) both

increase, relative to their previous equilibrium amounts.

D The amounts of CH3CHO(g) and CH4(g) both increase, relative to their previous equilibrium amounts. D

33 The graph below shows the effect of temperature and pressure on the equilibrium yield of the product in a gaseous equilibrium system.

Which of the following reactions would have the relationship between the yield, temperature and pressure shown in the graph?

A H2(g) + I2(g) 2HI(g) ΔH > 0 B PCl3(g) + Cl2(g) PCl5(g) ΔH < 0 C N2O4(g) 2NO2(g) ΔH > 0 D CH3CH(OH)CH3(g) CH3COCH3(g) + H2(g) ΔH < 0 C

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Part B


Methanol is manufactured by the reaction of carbon monoxide with hydrogen in the presence of a ZnO / Cr2O3 catalyst: ZnO / Cr2O3CO(g) + 2H2(g) CH3OH(g)

34 Which of the following combinations describes the effect of increasing the pressure of the system?

Equilibrium yield Reaction rate

A Decreases increases B Decreases decreases C Increases increases D Increases decreases C

35 Which of the following combinations describes the effect of removing the catalyst from the system?

Equilibrium yield Reaction rate

A Decreases decreases B No change decreases C No change no change D Decreases no change B

36 Transport of oxygen in the body involves the complex molecules haemoglobin and oxyhaemoglobin.

haemoglobin + oxygen oxyhaemoglobin

If carbon monoxide (CO) is present in the air, poisoning can occur because

A the equilibrium constant for the reaction is reduced.

B CO reacts with oxygen to form CO2, driving the equilibrium to the left.

C the position of equilibrium shifts to the left because haemoglobin bonds strongly with CO.

D CO cata lyzes the decompos i t ion of oxyhaemoglobin into haemoglobin and oxygen. C


The equilibrium constant, Kc, for the thermal decomposition of calcium carbonate is 2.70 x 10–3 mol dm–3 at 1 000 K.

CaCO3(s) CaO(s) + CO2(g) ΔH > 0

15.0 g of CaCO3(s) are introduced into a 5.00 dm3 evacuated vessel, and the system is allowed to attain equilibrium at 1 000 K.

37 What is the percentage of decomposition of CaCO3(s) in the equilibrium system?

(Relative atomic masses: C = 12.0, O = 16.0, Ca = 40.1)

A 1.80% B 4.05% C 9.00% D 13.9% C

38 Which of the following actions will cause the percentage of decomposition of CaCO3(s) of the equilibrium system to increase?

(1) Adding some CaCO3(s) (2) Increasing the temperature (3) Removing some CaO(s)




Initially, some NOCl(g) is placed in an empty flask. Which of the following statements describe the changes which occur as the system proceeds towards equilibrium?

(1) The rate of the backward reaction increases.

(2) The concentration of NOCl(g) increases. (3) The concentration of NO(g) increases.


28

Part B


NO2Cl(g) + NO(g) NOCl(g) + NO2(g)

Adding which of the following substances will cause the equilibrium concentration of NO2(g) to increase?

(1) NOCl(g) (2) NO(g) (3) NO2Cl(g)

A (1) only B (2) only C (1) and (3) only D (2) and (3) only D


2SO2(g) + O2(g) 2SO3(g) Kc = 1.20 x 104 dm3 mol–1

Some SO2(g) is added to the equilibrium system. Which of the following statements are correct?

(1) The position of equilibrium shifts to the right.

(2) The rate of the forward reaction increases.

(3) The value of Kc increases.



N2O4(g) 2NO2(g) ΔH = +58 kJ pale yellow dark brown

Shifting the position of equilibrium to the left is accompanied by

(1) an increase in the volume of the system. (2) a release of heat by the system to the

surroundings. (3) a decrease in the average relative molecular

mass of the gas molecules in the system.



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

Which of the following statements is / are correct when the volume of the system is decreased?

(1) The position of equilibrium shifts to the right.

(2) The rate of the backward reaction is greater than that of the forward reaction.

(3) The value of Kc decreases.


44 For which of the following equilibrium systems will the amounts of reactants increase with an increase in the container volume?

(1) C(g) + CO2(g) 2CO(g) (2) N2(g) + 3H2(g) 2NH3(g) (3) 2NO(g) + O2(g) 2NO2(g)


45 Chlorine trifluoride, a colourless gas, can be decomposed into its element.

2ClF3(g) 3F2(g) + Cl2(g) ΔHO = +159 kJ

Which of the following statements is / are correct?

(1) The decomposition is a redox reaction. (2) When an equilibrium mixture is heated,

its colour fades. (3) When the volume of the container of

an equilibrium mixture is increased, more Cl2(g) will form.

A (1) only B (2) only C (1) and (3) only D (2) and (3) only C

29

Part B


2NO(g) + Br2(g) 2NOBr(g) ΔH < 0

Which of the following actions will cause the position of equilibrium to shift to the right?

(1) Adding some NO(g) (2) Increasing the volume (3) Decreasing the temperature


47 Consider the following equilibrium system in a closed reaction vessel:

CO(g) + 2H2(g) CH3OH(g) ΔH < 0

Which of the following action(s) can cause an increase in the value of Kc?

(1) Adding some CO(g) (2) Decreasing the temperature (3) Transferring the reaction mixture to a vessel

of larger volume



Cu2+(aq) + 4Br–(aq) CuBr42–(aq)

blue colourless green

Cooling the system changes its colour from green to blue.

Which of the following statements is / are correct?

(1) The forward reaction is exothermic. (2) The value of Kc decreases when the system

is cooled. (3) A net backward reaction occurs when the

system is cooled.



CO(g) + H2O(g) CO2(g) + H2(g) ΔH = –41 kJ mol–1

Which of the following will cause a shift in the position of equilibrium?

(1) Adding a catalyst (2) Changing the temperature (3) Changing the volume


50 Which of the following statements about enzymes are correct?

(1) Enzymes are proteins. (2) Enzymes increase the rate of biochemical

reactions. (3) Enzymes increase the equilibrium constant

of biochemical reactions.


30

Part B




51 All chemical equilibrium systems have equal At equilibrium, both the forward and backward D concentrations of reactants and products. reactions stop.

52 Adding some O2(g) to an equilibrium system of When some O2(g) is added to the equilibrium A SO2(g), O2(g) and SO3(g) will cause a net increase system, a net reaction will occur to consume in the concentration of SO3(g). some of the O2(g).

53 Adding some CaO(s) to an equilibrium system Adding some CaO(s) to the equilibrium system D of CaCO3(s), CaO(s) and CO2(g) will cause the will cause the value of Kc to increase. position of equilibrium to shift to the left.

54 Keeping the volume constant, adding some The new equilibrium concentration of NO2(g) C NO2(g) to an equilibrium system of N2O4(g) and will increase relative to its previous equilibrium NO2(g) will cause the mixture to get darker concentration. gradually.

55 Decreasing the volume of the container of an Decreasing the volume of the container of A equilibrium system of H2(g), I2(g) and HI(g) will the equilibrium system will cause the cause the purple colour of the system to become concentration of I2(g) to increase. deeper.

56 When the volume of the gas syringe containing Increasing the volume of the gas syringe C an equilibrium system of N2O4(g) and NO2(g) is containing the equilibrium system will cause increased, the mixture first darkens and then more NO2(g) to form. lightens.

57 For an equilibrium system with an endothermic For an equilibrium system with an endothermic C forward reaction, increasing the temperature forward reaction, increasing the temperature will cause the position of equilibrium to shift will cause the value Kc to increase. to the left.

58 A temperature of 1 000 °C is usually used in Increasing the temperature will increase the D the Haber process. yield of ammonia in the Haber process.

31

Part B

59 The pressure used for converting sulphur dioxide Increasing the pressure will increase the yield C to sulphur trioxide in the Contact process is of sulphur trioxide. usually 200 atmospheres.

60 Catalysts are used in many industrial processes. Catalysts will not affect the percentage of the B product in the equilibrium mixture.

Short questions

61 For each of the following equilibrium system,

(i) write an expression for the equilibrium constant, Kc;

(ii) State the units of Kc.

a) Fe2O3(s) + 3CO(g) 2Fe(s) + 3CO2(g) (2 marks)

Kc =

[CO2(g)]3

[CO(g)]3 (1)

Kc has no units. (1)

b) MgCO3(s) MgO(s) + CO2(g) (2 marks)

Kc = [CO2(g)] (1)

The units of Kc are mol dm–3. (1)

c) 3O2(g) 2O3(g) (2 marks)

Kc =

[O3(g)]2

[O2(g)]3 (1)

The units of Kc are dm3 mol–1. (1)

d) CO(g) + 2H2(g) CH3OH(g) (2 marks)

Kc =

[CH3OH(g)][CO(g)][H2(g)]2 (1)

The units of Kc are dm6 mol–2. (1)

62 Given the following reactions and their corresponding equilibrium constants at 1 292 °C:

2H2O(g) 2H2(g) + O2(g) Kc1 = 1.60 x 10–11 mol dm–3

2CO2(g) 2CO(g) + O2(g) Kc2 = 1.30 x 10–10 mol dm–3

Calculate the equilibrium constant, Kc3, for the following reaction

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

at the same temperature. (3 marks)

Kc1 =

[H2(g)]2[O2(g)][H2O(g)]2

Kc2 =

[CO(g)]2[O2(g)][CO2(g)]2

32

Part B

Kc3 =

[H2O(g)][CO(g)][CO2(g)][H2(g)] (1)

= 1Kc1

x Kc2

= 1

1.60 x 10–11 mol dm–3 x 1.30 x 10–10 mol dm–3

(1)

= 2.85 (1)

63 When bromine is dissolved in water, the following equilibrium system is established:

Br2(l) + H2O(l) H+(aq) + Br–(aq) + HOBr(aq) yellow-brown colourless

In an experiment, dilute sodium hydroxide solution and then dilute hydrochloric acid are added to the system. The observations made are listed below.

Step Procedure Observation

I Adding dilute sodium hydroxide solution A colourless solution is formed

II Adding dilute hydrochloric acid to the resulting solution A yellow-brown solution is formed

Explain the observations using Le Chatelier’s principle. (6 marks)

When dilute sodium hydroxide solution is added, the hydroxide ions react with the hydrogen ions to form water. Thus, the

concentration of hydrogen ions decreases. (1)

The system responds by reducing this change. (1)

A net forward reaction occurs to produce more hydrogen ions. (1)

A colourless solution is observed as the concentration of Br2(aq) decreases. (1)

When hydrogen ions are added, the system responds by reducing this change.

A net backward reaction occurs to consume some of the hydrogen ions. (1)

A yellow-brown solution is observed as the concentration of Br2(aq) increases. (1)

33

Part B

64 Each of the following equilibrium systems is disturbed by increasing the pressure as a result of decreasing the volume of the reaction vessel. Decide whether the number of moles of reaction product(s) will increase, decrease, or remain the same. Explain your answer in each case.

a) 2CO2(g) 2CO(g) + O2(g) (3 marks)

An increase in pressure will bring about a net reaction that decreases the number of moles of gas. This helps to reduce

the pressure. (1)

A net backward reaction occurs. (1)

Thus, the number of moles of reaction products will decrease. (1)

b) CO(g) + H2O(g) CO2(g) + H2(g) (3 marks)

The number of moles of gas is the same on both sides of the equation, (1)

so changing the pressure has no effect on the position of equilibrium. (1)

Thus, the number of moles of reaction products will remain the same. (1)

c) Si(s) + 2Cl2(g) SiCl4(g) (3 marks)

An increase in pressure will bring about a net reaction that decrease the number of moles of gas. This helps to reduce

the pressure. (1)

A net forward reaction occurs. (1)

Thus, the number of moles of reaction product will increase. (1)

34

Part B

65 Consider the reaction between copper(II) ions and chloride ions:

Cu2+(aq) + 4Cl–(aq) CuCl42–(aq)

blue yellow

The Cu2+(aq) ion is blue while the CuCl42–(aq) ion is yellow, so a mixture of these two appears green.

The table below lists the colour of two samples of equilibrium mixture of the same composition kept at different temperatures.

Temperature Colour of equilibrium mixture

10 °C blue

90 °C green

Deduce and explain whether the forward reaction is exothermic or endothermic. (3 marks)

When the system is cooled, it appears blue because the concentration of Cu2+(aq) ions increases. (1)

It can be deduced that when the temperature is decreased, the system will undergo a net backward reaction so as to raise

the temperature. (1)

Thus, the backward reaction should be an exothermic reaction. / The forward reaction should be an endothermic reaction. (1)

66 Consider the following equilibrium system in a closed container:

CaCO3(s) CaO(s) + CO2(g) ΔH > 0

Complete the table to describe the effect of various actions on the position of equilibrium of the system. (5 marks)

Action Position of equilibrium

(a) The volume of the reaction vessel is increased. shifts to the right (1)

(b) Some CaCO3(s) is removed. no effect (1)

(c) Some CO2(g) is added. shifts to the left (1)

(d) A few drops of NaOH(aq) are added. shifts to the right (1)

(e) The temperature is increased. shifts to the right (1)

35

Part B

67 In the manufacture of nitric acid, ammonia is oxidized to nitrogen monoxide by the following reaction:

4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g) ΔHO = –906 kJ

In an experiment, a mixture of NH3(g), O2(g), NO(g) and H2O(g) is allowed to reach equilibrium in a cylinder fitted with a movable piston. Give the effect of each of the following changes on items listed in the table below:

a) Adding a catalyst (2 marks)

b) Increasing the temperature (2 marks)

c) Increasing the pressure by decreasing the volume (2 marks)

Effect of change on (a) Adding a catalyst(b) Increasing the temperature

(c) Increasing the pressure

the rate of the forward reaction

increases (0.5) increases (0.5) increases (0.5)

the rate of the backward reaction

increases (0.5) increases (0.5) increases (0.5)

the position of equilibrium

no effect (0.5) shifts to the left (0.5) shifts to the left (0.5)

the equilibrium constant, Kc

no effect (0.5) decreases (0.5) no effect (0.5)

68 A particular industrial process involves the following steps.

B(g) and C(g) recycling

Reaction 1

A(g) + B(g) 2C(g)

ΔH = +100 kJ

Reaction 2

2C(g) + B(g) 2D(g)

ΔH = –150 kJtemperature 300 °C pressure 10 atm

A(g) B(g)B(g)

separation chamber product D(g)

36

Part B

a) It is possible to alter the temperature and pressure at which Reaction 2 occurs.

In the table below, indicate what effect the following changes would have on the rate, equilibrium yield and value of the equilibrium constant, Kc, for Reaction 2. (3 marks)

Would the rate of Reaction 2 become

higher, lower or remain unchanged?

Would the equilibrium yield of Reaction 2

become higher, lower or remain unchanged?

Would the value of Kc of Reaction 2 become

higher, lower or remain unchanged?

The temperature of Reaction 2 is increased to 600 °C.

higher (0.5) lower (0.5) lower (0.5)

The pressure of Reaction 2 is decreased to 5 atm at constant temperature.

lower (0.5) lower (0.5) remain unchanged (0.5)

b) Heat is released in Reaction 2. Describe how the heat can be used within this industrial process. (1 mark)

Provide heat for Reaction 1. / Increase the rate of Reaction 1. / Heat the incoming reactants. / Generate electricity. (1)

Structured questions

69 When nitrogen monoxide reacts with oxygen, a dynamic equilibrium is established.

2NO(g) + O2(g) 2NO2(g) ΔHO = –115 kJ

a) State TWO features of a system that is in dynamic equilibrium. (2 marks)

Rate of forward reaction = rate of backward reaction. (1)

Concentrations of the reactants and products remain constant. (1)

b) At a certain temperature, the equilibrium constant, Kc, for the reaction is 65.0 dm3 mol–1.

The concentrations of NO(g) and O2(g) in an equilibrium mixture are 0.600 mol dm–3 and 0.300 mol dm–3 respectively. What is the equilibrium concentration of NO2(g)? (2 marks)

Kc = [NO2(g)]2

[NO(g)]2[O2(g)]

65.0 dm3 mol–1 = [NO2(g)]2

(0.600 mol dm–3)2(0.300 mol dm–3)

(1)

[NO2(g)] = 2.65 mol dm–3 (1)

37

Part B

c) In each of the following cases, sketch on the given graph to show the expected variation in the concentration of NO2(g) in the equilibrium mixture in (b) until the attainment of a new state of equilibrium. Explain your answer in each case.

i) Some O2(g) is introduced into the equilibrium mixture at time t1, while the volume and the temperature are both kept constant. (2 marks)

An increase in the concentration of O2(g) will shift the position of equilibrium to the right. (1)

Thus, the concentration of NO2(g) will increase.

ii) The temperature of the equilibrium mixture is increased at time t2. (3 marks)

When the temperature is increased, the system will respond by reducing the temperature. (1)

As the backward reaction is endothermic, the system will undergo a net backward reaction. Thus, the concentration

of NO2(g) will decrease. (1)

d) State and explain how the pressure should be changed to give a higher yield of NO2(g). (2 marks)

Increase the pressure. (1)


the pressure. (1)

A net forward reaction will occur.

Thus, the yield of NO2(g) will increase.

38

Part B

70 The ester ethyl ethanoate is hydrolyzed when it is heated with water in the presence of an acid catalyst. An equilibrium is established.

CH3COOCH2CH3(l) + H2O(l) CH3COOH(l) + CH3CH2OH(l)

a) A 0.600 mole sample of ethyl ethanoate was heated with 4.00 mole of water. At equilibrium, 68.0% of the ester was hydrolyzed. Calculate the equilibrium constant, Kc, for this reaction. (4 marks)

Number of moles of ester reacted = 0.600 mol x 68.0% = 0.408 mol

Number of moles of ester in equilibrium mixture = (0.600 – 0.408) mol = 0.192 mol (0.5)

Number of moles of water in equilibrium mixture = (4.00 – 0.408) mol = 3.59 mol (0.5)

Number of moles of ethanoic acid in equilibrium mixture = 0.408 mol (0.5)

Number of moles of ethanol in equilibrium mixture = 0.408 mol (0.5)

Let V dm3 be the total volume of the equilibrium mixture.

Kc =

[CH3COOH(l)][CH3CH2OH(l)][CH3COOCH2CH3(l)][H2O(l)]

= ( 0.408

V ) ( 0.408V )

( 0.192V ) ( 3.59

V ) (1)

= 0.242 (1)

b) A student repeated the experiment using the same initial quantities. During the experiment she noticed that the water in the condenser had stopped flowing. There was a sweetish smell coming from the top of the condenser.

What effect, if any, will this problem have on

i) the concentrations of the products in the flask? (2 marks)

Ester was lost. (1)

The concentrations of the products would decrease. (1)

ii) the value of the equilibrium constant, Kc?

Explain your answer in each case. (2 marks)

The value of Kc would remain constant. (1)

The value of Kc does not change with concentration. / The value of Kc only changes with temperature. (1)

39

Part B

71 Hydrogen gas can be made from carbon monoxide and steam as shown by the following equation.

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

The diagram below shows how the concentrations of H2O(g) and H2(g) change with time as equilibrium is established.

a) On the time axis mark with an X the time at which equilibrium is first established. (1 mark)

b) State Le Chatelier’s principle. (2 marks)

Le Chatelier’s principle states that if the condition of a system in equilibrium is changed, the position of equilibrium

will shift (1)

so as to reduce that change. (1)

c) The volume and the temperature of the equilibrium mixture are both kept constant. Predict the effect of each of the following changes on the position of equilibrium:

i) adding carbon monoxide to the equilibrium mixture; and (2 marks)

The system responds by reducing the change. A net forward reaction occurs to use up some of the extra carbon

monoxide. (1)

The position of equilibrium shifts to the right. (1)

ii) removing steam from the equilibrium mixture. (2 marks)

The system responds by reducing the change. A net backward reaction occurs to produce more steam. (1)

The position of equilibrium shifts to the left. (1)

40

Part B

d) This reaction is usually carried out at 450 °C.

i) State and explain the effect on the rate of production of hydrogen if this reaction is carried out at a temperature above 450 °C. (3 marks)

The rate of production of hydrogen will increase. (1)

The reactant particles have more energy and collide more often. (1)

A larger portion of the reactant particles have energy equal to or greater than the activation energy. (1)

ii) State and explain the effect on the yield of hydrogen if the reaction is carried out in the presence of a catalyst. (2 marks)

No effect on the yield of hydrogen. (1)

A catalyst increases the rates of both the forward reaction and the backward reaction to the same extent. (1)

72 Methanol can be produced by using a reversible reaction between carbon monoxide and hydrogen.

2H2(g) + CO(g) CH3OH(g)

When 2.00 mole of hydrogen and 1.00 mole of carbon monoxide are mixed and heated to a high temperature in a container of volume 1.50 dm3, the equilibrium yield of methanol is 0.800 mole.

a) Calculate a value for the equilibrium constant, Kc, for this reaction at this temperature and give its units. (4 marks)


According to the equation, 2 moles of H2(g) react with 1 mole of CO(g) to give 1 mole of CH3OH(g).

As 0.800 mole of CH3OH(g) is produced at equilibrium, the amount of H2(g) would decrease by 2 x 0.800 mole while

that of CO(g) would decrease by 0.800 mole.


Equilibrium concentration ( 2.00 – 2 x 0.800

1.50 ) mol dm–3 ( 1.00 – 0.8001.50 ) mol dm–3

0.8001.50

mol dm–3

= 0.267 mol dm–3 = 0.133 mol dm–3 = 0.533 mol dm–3 (1)

Kc =

0.533 mol dm–3

(0.267 mol dm–3)2(0.133 mol dm–3) (1)

= 56.3 (1) dm6 mol–2 (1)

41

Part B

b) The pressure of the equilibrium mixture is increased whilst keeping the temperature constant. The position of equilibrium shifts to the right.

i) Explain why the position of equilibrium shifts to the right. (1 mark)

An increase in pressure will bring about a net reaction that decreases the number of moles of gas. (1)

ii) What is the effect, if any, on the value of Kc? (1 mark)

No effect. (1)

c) The temperature of the equilibrium mixture is increased whilst keeping the pressure constant. The value of Kc decreases.

i) Explain what will happen to the position of equilibrium. (2 marks)

As the value of Kc decreases, it can be deduced that the concentration of the product decreases while the

concentrations of the reactants increase, (1)

i.e. the position of equilibrium shifts to the left. (1)

ii) Deduce the sign of the enthalpy change for the reaction between carbon monoxide and hydrogen. (3 marks)

It can be deduced that when the temperature is increased, the mixture will undergo a net backward reaction so as

to lower the temperature. (1)

Hence the backward reaction should be endothermic reaction / the reaction between carbon monoxide and

hydrogen should be an exothermic reaction. (1)

The sign of the enthalpy change for the reaction between carbon monoxide and hydrogen should be negative. (1)

42

Part B

d) Describe and explain what will happen to the rate of the reaction between carbon monoxide and hydrogen when the pressure is increased by decreasing the volume of the reaction vessel. (3 marks)

Decreasing the volume of the reaction vessel will cause the concentrations of the reactants to increase. (1)

The chance of collision between reactant particles increases and the frequency of effective collisions increases. (1)

Hence the rate of the reaction will increase. (1)

73 The equilibrium constant, Kc, for the following reaction is 5.00 mol dm–3 at 873 K.

COCl2(g) CO(g) + Cl2(g)

2.00 moles of COCl2(g) are introduced into an evacuated vessel of 4.00 dm3 kept at 873 K and allowed to achieve equilibrium.

a) Calculate the percentage dissociation of COCl2(g) when equilibrium is achieved. (4 marks)

Let y% be the percentage dissociation of COCl2(g).

COCl2(g) CO(g) + Cl2(g)

Initial concentration

2.004.00

mol dm–3 0 mol dm–3 0 mol dm–3

Equilibrium concentration 2.00 – 2.00 x

y100

4.00 mol dm–3

2.00 x y

1004.00

mol dm–3 2.00 x

y100

4.00 mol dm–3

(1)

Kc = 5.00 mol dm–3 =

( 2.00 x y

1004.00

mol dm–3) ( 2.00 x y

1004.00

mol dm–3)2.00 – 2.00 x

y100

4.00 mol dm–3

(1)

Rearranging the equation gives

5.00 ( 200 – 2y

400 ) = ( 2y400 )2

y2 + 1 000 y – 100 000 = 0

Solving the quadratic equation gives two solutions:

y = 91.6 or – 1 092 (rejected) (2)

∴ the percentage dissociation of COCl2(g) is 91.6%.

43

Part B

b) While keeping the temperature of the system at 873 K, the volume of the reaction vessel is decreased. State and explain the effect on

i) the position of equilibrium; (3 marks)

The pressure of the system will increase when the volume of the reaction vessel decreases. (1)

This will bring about a net reaction that decreases the number of moles of gas. This helps to reduce the pressure. (1)

The position of equilibrium will shift to the left. (1)

ii) the value of Kc. (2 marks)

No effect. (1)

The value of Kc only changes with temperature. (1)

74 When heated, phosphorus pentachloride dissociates.

PCl5(g) PCl3(g) + Cl2(g) ΔHO = +93 kJ mol–1

At 227 °C, an equilibrium mixture in a 1.00 dm3 vessel contains 0.400 mole of PCl5(g), 0.108 mole of PCl3(g) and 0.900 mole of Cl2(g).

a) Calculate the equilibrium constant, Kc, for the reaction at 227 °C. (3 marks)

Kc =

[PCl3(g)][Cl2(g)][PCl5(g)]

= ( 0.108

1.00 mol dm–3) ( 0.900

1.00 mol dm–3)

0.4001.00

mol dm–3

(1)

= 0.243(1) mol dm–3(1)

b) i) State the effect, if any, on the value of Kc of adding more PCl5(g) at a constant temperature. Explain your answer. (2 marks)

No effect. (1)

The value of Kc does not change with concentration. / The value of Kc only changes with temperature. (1)

44

Part B

ii) State the effect, if any, on the value of Kc of increasing the temperature of the reaction vessel. Explain your answer. (3 marks)

The value of Kc increases. (1)

When the temperature is increased, the system will undergo a net reaction so as to lower the temperature. (1)

As the forward reaction is endothermic, the system will undergo a net forward reaction. More PCl3(g) and Cl2(g)

are produced. (1)

So, the value of Kc increases.

c) In another experiment, a 1.00 dm3 reaction vessel is filled with 0.100 mole of PCl5(g), 0.100 mole of PCl3(g) and 0.100 mole of Cl2(g) at 227 °C.

i) Calculate Qc of the system and decide whether the system is at equilibrium. (2 marks)

Qc = ( 0.100

1.00 mol dm–3) ( 0.100

1.00 mol dm–3)

0.1001.00

mol dm–3

= 0.100 mol dm–3 (1)

Qc ≠ Kc, thus the system is not at equilibrium. (1)

ii) If the system is not at equilibrium, in which direction will the net reaction proceed? Explain your answer. (2 marks)

When Qc < Kc, the concentration of PCl3(g) and Cl2(g) will increase while that of PCl5(g) will decrease until

Qc = Kc. (1)

Thus a net forward reaction will occur. (1)

45

Part B

75 X(g) reacts with Y(g) reversibly to give Z(g). A mixture of X(g) and Y(g) is allowed to react in a closed container of volume 1.0 dm3 kept at a constant temperature. The graph below shows the changes in concentrations of X(g), Y(g) and Z(g) in the container with time.

(X, Y and Z do not represent symbols of elements.)

a) With reference to the above graph,

i) deduce the chemical equation for the reversible reaction between X(g), Y(g) and Z(g). (1 mark)

From the curve, it can be deduced that 3 moles of X(g) react with 1 mole of Y(g) to give 2 moles of Z(g).

The equation for the reaction is 3X(g) + Y(g) 2Z(g). (1)

ii) write an expression for the equilibrium constant, Kc, for the reaction. (1 mark)

Kc =

[Z(g)]2

[X(g)]3[Y(g)] (1)

b) Compare the rate of the forward reaction and that of the backward reaction

i) at the 10th minute after X(g) and Y(g) are mixed. (1 mark)

At the 10th minute, the rate of forward reaction is greater than the rate of backward reaction. (1)

46

Part B

ii) at the 60th minute after X(g) and Y(g) are mixed. (1 mark)

(You are not required to perform any calculation.)

At the 60th minute, the rate of forward reaction is equal to the rate of backward reaction. (1)

c) If the mixture X(g) and Y(g) is allowed to react at the same temperature but in a closed container of volume 0.5 dm3 instead,

i) will the rate of attainment of equilibrium remain the same? Explain. (3 marks)

The rate of attainment of equilibrium will increase. (1)

Decreasing the volume of the container will cause the concentration of the reactants to increase. (1)


ii) will the yield of Z(g) be the same? Explain. (3 marks)

The yield of Z(g) will increase. (1)

The pressure of the system will increase when the volume of the container decreases. (1)

The position of equilibrium will shift to the side with a fewer number of moles of gas. (1)

More Z(g) will form.

76 Methoxymethane, CH3OCH3, is used as an environmentally friendly propellant in spray cans. It can be made from methanol according to the following equation:

2CH3OH(g) CH3OCH3(g) + H2O(g) ΔH = –24 kJ

The equilibrium constant, Kc, for this reaction at 350 °C is 5.74.

a) Write an expression for Kc for this reaction. (1 mark)

Kc =

[CH3OCH3(g)][H2O(g)][CH3OH(g)]2 (1)

47

Part B

b) Calculate the value of Kc’, at 350 °C for the following reaction. (1 mark)

CH3OCH3(g) + H2O(g) 2CH3OH(g)

Kc’ =

[CH3OH(g)]2

[CH3OCH3(g)][H2O(g)]

=

1Kc

=

15.74

= 0.174 (1)

c) A 1.00 dm3 vessel at 350 °C contains 0.500 mole of CH3OH(g), 0.100 mole of CH3OCH3(g) and 0.700 mole of H2O(g).

i) Calculate Qc of the system and decide whether the system is at equilibrium. (2 marks)

Qc =

[CH3OCH3(g)][H2O(g)][CH3OH(g)]2

= ( 0.100 mol

1.00 dm3 ) ( 0.700 mol1.00 dm3 )

( 0.500 mol1.00 dm 3 )2

= 0.280 (1)

Qc ≠ Kc, thus the system is not at equilibrium. (1)

ii) If the system is not at equilibrium, in which direction will a net reaction proceed? Explain your answer. (2 marks)

When Qc < Kc, the concentrations of CH3OCH3(g) and H2O(g) will increase while that of CH3OH(g) will decrease until

Qc = Kc. (1)

Thus, a net forward reaction will occur. (1)

d) State and explain the effect of an increase in temperature on the position of equilibrium of the above system. (3 marks)


As the backward reaction is endothermic, the system will undergo a net backward reaction, (1)

i.e. the position of equilibrium will shift to the left. (1)

48

Part B

e) State and explain the effect of an increase in pressure on the position of equilibrium of the above system. (2 marks)

The number of moles of gas is the same on both sides of the equation, (1)

so changing the pressure has no effect on the position of equilibrium. (1)

f) In an experiment, methanol is pumped into an empty 20.0 dm3 reactor vessel. At equilibrium the vessel contains 0.340 mole of methanol at 350 °C.

i) Calculate the concentration, in mol dm–3, of methanol at equilibrium. (1 mark)

Concentration of methanol =

0.340 mol20.0 dm3

= 0.0170 mol dm–3 (1)

ii) Calculate the number of moles of methoxymethane present at equilibrium. (2 marks)

Let x mol dm–3 be the equilibrium concentration of CH3OCH3(g).

2CH3OH(g) CH3OCH3(g) + H2O(g)

Equilibrium concentration 0.0170 mol dm–3 x mol dm–3 x mol dm–3

Kc = 5.74 =

(x mol dm–3)2

(0.0170 mol dm–3)2

x = 0.0407 (1)

Number of moles of CH3OCH3(g) = 0.0407 mol dm–3 x 20.0 dm3

= 0.814 mol (1)

iii) Calculate the number of moles of methanol initially pumped into the reaction vessel. (2 marks)

Number of moles of CH3OH(g) reacted = 2 x 0.814 mol

= 1.63 mol (1)

Number of moles of CH3OH(g) at equilibrium = 0.340 mol

Number of moles of CH3OH(g) initially = (1.63 + 0.340) mol

= 1.97 mol (1)

49

Part B

77 When hydrogen and iodine gases are allowed to react, an equilibrium is established according to the following equation.

H2(g) + I2(g) 2HI(g) ΔHO = –10 kJ colourless purple colourless

A 2.00 dm3 container is filled with 0.0700 mole of H2(g) and 0.0600 mole of I2(g). Equilibrium is established after 15.0 minutes when there is 0.0600 mole of HI(g) present.

a) Sketch and label the curves for changes in the concentrations of H2(g), I2(g) and HI(g) for the time interval of 0 to 30.0 minutes. (3 marks)

b) Calculate the equilibrium constant, Kc, for the reaction. (2 marks)

Kc =

[HI(g)]2

[H2(g)][I2(g)]

= (0.0300 mol dm–3)2

(0.0200 mol dm–3)(0.0150 mol dm–3) (1)

= 3.00 (1)

c) Calculate the equilibrium constant, Kc’, for the following reaction. (1 mark)

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

Kc’ for this reaction =

[H2(g)][I2(g)]

[HI(g)]2

=

13.00

= 0.333 (1)

50

Part B

d) Explain the following observations when changes are made to an equilibrium mixture of H2(g), I2(g) and HI(g).

Include reference to the equilibrium position and any other factors.

i) When the temperature is decreased, the purple colour becomes paler. (2 marks)

The purple colour becomes paler because the position of equilibrium shifts to the right. (1)

The forward reaction is exothermic.

When the temperature is decreased, the system will undergo a net forward reaction so as to raise the

temperature. (1)

ii) When the pressure is increased by decreasing the volume, the purple colour becomes deeper. (3 marks)

The molecules are pushed closer together. / The concentration of I2(g) increases. (1)

Increasing the pressure has no effect on the position of equilibrium, (1)

because the number of moles of gas is the same on both sides of the equation. (1)

78 At high temperatures, nitrogen is oxidized by oxygen to form nitrogen monoxide in a reversible reaction as shown in the equation below.

N2(g) + O2(g) 2NO(g) ΔHO = +180 kJ

a) The equilibrium constant, Kc, for the above reaction is 5.00 x 10–3 at 3 000 °C.

What is the equilibrium constant, Kc’, for the following reaction at 3 000 °C?

4NO(g) 2N2(g) + 2O2(g) (2 marks)

Kc = 5.00 x 10–3

=

[NO(g)]2

[N2(g)][O2(g)]

Kc’ =

[N2(g)]2[O2(g)]2

[NO(g)]4 (1)

=

( 15.00 x 10–3 )

2

= 4.00 x 104 (1)

51

Part B

b) The rate at which equilibrium is reached could be increased by increasing the pressure or temperature.

In each case explain why the rate increases.

i) Increasing the pressure (2 marks)

Increasing the pressure will cause the concentrations of the reactants to increase. (1)


ii) Increasing the temperature (2 marks)

The reactant particles have more energy and collide more often. (1)

A larger portion of the reactant have energy equal to or greater than the activation energy. (1)

c) State and explain the effect of an increase in pressure and the effect of an increase in temperature, on the yield of nitrogen monoxide.

i) Effect of an increase in pressure. (3 marks)

The yield of nitrogen monoxide does not change. (1)

The number of moles of gas is the same on both sides of the equation. (1)

So, changing the pressure has no effect on the position of equilibrium. (1)

ii) Effect of an increase in temperature. (3 marks)

The yield of nitrogen monoxide increases. (1)


As the forward reaction is endothermic, the system will undergo a net forward reaction. (1)

52

Part B

d) N2(g), O2(g) and NO(g) were mixed together and allowed to reach equilibrium. The concentrations of the gases were then measured at various times and the results plotted. At time t, a change was made to the composition of the mixture.

i) What change was made to the mixture at time t? (1 mark)

Nitrogen was added. (1)

ii) Explain the changes that happen to the mixture after time t. (2 marks)

The concentration of NO(g) increased while those of N2(g) and O2(g) decreased.

It can be deduced that the position of equilibrium shifted to the right. (1)

A new state of equilibrium was attained (1)

when the concentrations of the substances become constant.

79 Chemists were investigating the production of a chemical, XY2(g), that can be formed from X2(g) and Y2(g) as shown in the reaction below.

X2(g) + 2Y2(g) 2XY2(g)

a) Write an expression for the equilibrium constant, Kc, for the reaction, and deduce its units. (2 marks)

Kc =

[XY2(g)]2

[X2(g)][Y2(g)]2 (1)

Units of Kc is dm3 mol–1. (1)

b) State and explain the effect of a decrease in pressure on

i) the position of equilibrium; (2 marks)

A decrease in pressure will bring about a net reaction that increases the number of moles of gas. This helps to

increase the pressure. (1)


53

Part B

ii) the rate of reaction. (3 marks)

Decreasing the pressure will cause the concentrations of the reactants to decrease. (1)

The chance of collision between reactant particles decreases and the frequency of effective collisions decreases. (1)

Hence the rate of reaction will decrease. (1)

c) The chemists measured the percentage conversion of X2(g) at various temperatures. The results are shown in the graph below.

i) Use the graph to predict the percentage conversion of X2(g) at 350 °C. (1 mark)

13.5% (1)

ii) Deduce from the graph whether the production of XY2(g) from X2(g) and Y2(g) is an exothermic or an endothermic reaction. Explain your answer. (3 marks)

The production of XY2(g) from X2(g) and Y2(g) is an endothermic reaction. (1)

The percentage conversion of X2(g) increases with increasing temperature. (1)

It can be deduced that when the temperature is increased, the system will undergo a net forward reaction so as to

lower the temperature. (1)

Hence the production of XY2(g) (the forward reaction) should be an endothermic reaction.

54

Part B

d) The chemists decided to use a catalyst in the process. State the effect of using a catalyst on:

i) the rate of conversion of X2(g) and Y2(g) into XY2(g); (1 mark)

The rate of conversion increases. (1)

ii) the percentage conversion of X2(g) and Y2(g) into XY2(g) at equilibrium. (1 mark)

No change. (1)

80 Fe3+(aq) ions and SCN–(aq) ions react in solution to give a deep red [Fe(SCN)]2+(aq) ions.

Fe3+(aq) + SCN–(aq) [Fe(SCN)]2+(aq) yellow-brown colourless deep red

a) A student makes changes to separate samples of the equilibrium mixture. The changes are listed in the table below.

Complete the table by placing ticks in the appropriate boxes to indicate the effect of each change on

(1) the intensity of the red colour of the mixture; and

(2) the concentration of Fe3+(aq) ions

once a new state of equilibrium has been attained. (3 marks)

Change to the equilibrium mixture

(1) Intensity of the red colour of mixture compared with initial equilibrium

(2) Concentration of Fe3+(aq) ions compared with initial equilibrium

less intense more intense decreases increases

(i) Adding 1 drop of FeCl3(aq) ✔ (0.5) ✔ (0.5)

(ii) Adding 1 drop of AgNO3(aq) (Ag+(aq) ions will form a AgSCN precipitate)

✔ (0.5) ✔ (0.5)

(iii) Adding NH4Cl(s) (NH4Cl(s) reacts with Fe3+(aq) ions to form complex ions)

✔ (0.5) ✔ (0.5)

55

Part B

b) The forward reaction is exothermic. The graph below shows the initial equilibrium concentration of the ions. The temperature of the system is increased at t1.

i) Sketch on the graph to show the expected variation in the concentrations until the attainment of a new state of equilibrium at t2. (2 marks)

ii) Explain your answer. (2 marks)

When the temperature is increased, the system will respond by lowering the temperature. (1)

As the forward reaction is exothermic, the system will undergo a net backward reaction. (1)

Thus, the concentrations of Fe3+(aq) ions and SCN–(aq) ions increase while that of [Fe(SCN)]2+(aq) ions decreases.

c) The following table lists the compositions of two equilibrium mixtures.

Fe3+(aq) SCN–(aq) [Fe(SCN)]2+(aq)

Concentration in equilibrium mixture 1 (mol dm–3) 3.91 x 10–2 8.02 x 10–5 9.22 x 10–4

Concentration in equilibrium mixture 2 (mol dm–3) 6.27 x 10–3 3.65 x 10–4 ?

Calculate the concentration of [Fe(SCN)]2+(aq) ions in equilibrium mixture 2. (2 marks)

Kc =

9.22 x 10–4 mol dm–3

(3.91 x 10–2 mol dm–3)(8.02 x 10–5 mol dm–3)

= [[Fe(SCN)]2+(aq)]

(6.27 x 10–3 mol dm–3)(3.65 x 10–4 mol dm–3) (1)

[[Fe(SCN)]2+(aq)] = 6.73 x 10–4 mol dm–3 (1)

56

Part B

81 Under suitable conditions, the equilibrium represented below was established.

2CH4(g) 3H2(g) + C2H2(g) ΔHO = +377 kJ

The following graph shows the concentrations of the substances in the equilibrium mixture at a certain temperature and pressure.

a) i) Write an expression for the equilibrium constant, Kc, for the reaction. (1 mark)

Kc =

[H2(g)]3[C2H2(g)][CH4(g)]2 (1)

ii) With reference to the above graph, calculate Kc under the experimental conditions. (2 marks)

Kc =

(1.20 mol dm–3)3(0.400 mol dm–3)(1.60 mol dm–3)2

= 0.270 (1) mol2 dm–6 (1)

57

Part B

b) At time t1, the concentration of CH4(g) was increased to 2.00 mol dm–3 and a new state of equilibrium was established at t2.

i) Calculate the equilibrium concentration of each gas at time t2. (6 marks)

Let x mol dm–3 be the increase in concentration of C2H2(g) when a new state of equilibrium was established.

2CH4(g) 3H2(g) + C2H2(g)

Initial concentration 2.00 mol dm–3 1.20 mol dm–3 0.400 mol dm–3

New equilibrium concentration (2.00 – 2x) mol dm–3 (1.20 + 3x) mol dm–3 (0.400 + x) mol dm–3

Kc = 0.270 mol2 dm–6 =

((1.20 + 3x) mol dm–3)3((0.400 + x) mol dm–3)

((2.00 – 2x) mol dm–3)2 (1)

0.270 = 27(0.400 + x)4

4(1.00 – x)2

0.0400 = (0.400 + x)4

(1.00 – x)2


0.200 (1.00 – x) = (0.400 + x)2

x2 + x – 0.0400 = 0


x = 0.0385 or –1.04 (rejected) (2)

∴ the equilibrium concentrations at time t2 are as follows:

[CH4(g)] = (2.00 – 2 x 0.0385) mol dm–3

= 1.92 mol dm–3 (1)

[H2(g)] = (1.20 + 3 x 0.0385) mol dm–3

= 1.32 mol dm–3 (1)

[C2H2(g)] = (0.400 + 0.0385) mol dm–3

= 0.439 mol dm–3 (1)

ii) Sketch on the graph to show the changes in concentration of CH4(g) and H2(g). (2 marks)

58

Part B

c) Describe the conditions that would produce a high yield of hydrogen at equilibrium. Explain your answers.

i) Temperature (3 marks)

A high temperature is required. (1)

When a high temperature is used, the system will respond by reducing the temperature. (1)

As the production of hydrogen is endothermic, the system will undergo a net forward reaction. (1)

Thus, the yield of hydrogen will increase.

ii) Pressure (2 marks)

A low pressure is required. (1)

A decrease in pressure will bring about a net reaction that increases the number of moles of gas. This helps to

increase the pressure. (1)



d) A different equilibrium mixture was produced, starting from CH4(g) alone.

When 1.00 moles of CH4(g) were charged into a container of volume 0.250 dm3, it was found that the equilibrium mixture formed contained 0.730 moles of CH4(g). Calculate the number of moles of H2(g) and C2H2(g) present in this equilibrium mixture. (3 marks)

Number of moles of CH4(g) reacted = (1.00 – 0.730) mol

= 0.270 mol (1)

Number of moles of H2(g) in the equilibrium mixture =

32

x 0.270 mol

= 0.405 mol (1)

Number of moles of C2H2(g) in the equilibrium mixture =

0.2702

mol

= 0.135 mol (1)

59

Part B

82 The hydrogen used in the Haber process is made by the following reaction:


a) Initially, 0.0600 mole of CH4(g), 0.0800 mole of H2O(g), 0.280 mole of CO(g) and 0.740 mole of H2(g) are placed in a 4.00 dm3 container. At equilibrium, the concentration of H2(g) is 0.200 mol dm–3. Calculate the equilibrium constant, Kc, for the reaction. (5 marks)

CH4(g) + H2O(g) CO(g) + 3H2(g)


0.06004.00

mol dm–3 0.08004.00

mol dm–3 0.2804.00

mol dm–3 0.7404.00

mol dm–3

= 0.0150 mol dm–3 = 0.0200 mol dm–3 = 0.0700 mol dm–3 = 0.185 mol dm–3

As the concentration of H2(g) increased by 0.0150 mol dm–3 when equilibrium is reached, that of CO(g) would increase

by 5.00 x 10–3 mol dm–3 while that of both CH4(g) and H2O(g) would decrease by 5.00 x 10–3 mol dm–3.

Species

CH4(g) H2O(g) CO(g) H2(g)

Initial concentration (mol dm–3) 0.0150 0.0200 0.0700 0.185

Change in concentration (mol dm–3) –0.00500 –0.00500 +0.00500 +0.0150 (2)

Equilibrium concentration (mol dm–3) 0.0100 0.0150 0.0750 0.200

Kc =

[CO(g)][H2(g)]3

[CH4(g)][H2O(g)]

=

(0.0750 mol dm–3)(0.200 mol dm–3)3

(0.0100 mol dm–3)(0.0150 mol dm–3) (1)

= 4.00 (1) mol2 dm–6 (1)

b) In another experiment, some methane and steam are placed in a closed container and allowed to react at a fixed temperature. The following graph shows the changes in concentrations of methane and carbon monoxide as the reaction proceeds.

60

Part B

i) On the graph above, draw a line to show the change in concentration of hydrogen as the reaction proceeds. Label this line. (1 mark)

ii) On the graph above, draw a line to show how the formation of carbon monoxide would differ over time in the presence of a catalyst. Label this line. (1 mark)

c) A high temperature of over 1 000 °C is used in the production of hydrogen by this reaction.

i) Explain why a high temperature is needed to produce a high yield of hydrogen. (2 marks)

When a high temperature is used, the system will respond by reducing the temperature. (1)

As the production of hydrogen is endothermic, the system will undergo a net forward reaction. (1)


ii) Give one disadvantage of using temperatures much higher than 1 000 °C. (1 mark)

The cost of energy is high. / The amount of energy used is high. (1)

d) State and explain how the overall pressure must be changed to increase the yield of hydrogen. (2 marks)

Decrease the pressure. (1)

A decrease in pressure will bring about a net reaction that increases the number of moles of gas. This helps to increase

the pressure. (1)



83 Dinitrogen tetroxide is a pale yellow gas. It exists in equilibrium with nitrogen dioxide, a dark brown gas.


a) 0.400 mole of N2O4(g) is heated in a closed reaction vessel of volume 16.0 dm3. A state of equilibrium is established at a constant temperature. The equilibrium mixture is found to contain 0.180 mole of N2O4(g).

i) What is the number of moles of NO2(g) in the equilibrium mixture? (2 marks)

Number of moles of N2O4(g) reacted = (0.400 – 0.180) mol

= 0.220 mol (1)

Number of moles of NO2(g) in the equilibrium mixture = 2 x 0.220 mol

= 0.440 mol (1)

61

Part B

ii) Calculate the equilibrium constant, Kc, for the reaction. (3 marks)

Kc =

[NO2(g)]2

[N2O4(g)]

= ( 0.440

16.0 mol dm–3)2

0.18016.0

mol dm–3

(1)

= 0.0672 (1) mol dm–3 (1)

b) The diagram below shows a gas syringe containing a pale brown mixture of N2O4(g) and NO2(g) at equilibrium at 25 °C.

i) Keeping the volume constant, the temperature is then raised to 35 °C. The brown colour becomes more intense.

Is the forward reaction below exothermic or endothermic? Explain your answer.

N2O4(g) 2NO2(g) (3 marks)

The forward reaction is endothermic. (1)

When the temperature is increased, the brown colour becomes more intense because more NO2(g) is produced. (1)

It can be deduced that when the temperature is increased, the system will undergo a net forward reaction so as to

lower the temperature. (1)

Hence the forward reaction should be an endothermic reaction.

ii) Keeping the temperature at 35 °C, the plunger of the syringe is then pushed in so as to halve the volume at time t.

(1) State and explain your expected observations. (5 marks)

The brown colour of the mixture becomes more intense for a moment (1)

and then gets paler gradually. (1)

When the volume of the syringe is halved, the concentration of NO2(g) doubles. (1)

A decrease in volume will bring about a net reaction that decreases the number of moles of gas.


62

Part B

The concentration of NO2(g) decreases, but not back to its concentration before time t. (1)

(2) Sketch on the given graph to show the expected variation in the concentration of NO2(g) in the mixture until the attainment of a new state of equilibrium. (1 mark)

84 The dissociation of NH4HS(s) can be represented by the following equation:

NH4HS(s) NH3(g) + H2S(g) ΔH > 0

0.700 mole of NH4HS(s) is introduced into an evacuated vessel of 2.00 dm3 kept at 310 K. When equilibrium is attained, 1.00% of the NH4HS(s) is found to have dissociated.

a) Write an expression for the equilibrium constant, Kc, for the dissociation of NH4HS(s). (1 mark)

Kc = [NH3(g)][H2S(g)] (1)

b) Calculate the equilibrium constant, Kc, for the dissociation at 310 K. (3 marks)

Number of moles of NH4HS(s) dissociated = 0.700 mol x 1.00%

= 7.00 x 10–3 mol

According to the equation, 1 mole of NH4HS(s) dissociates to give 1 mole of NH3(g) and 1 mole of H2S(g),

7.00 x 10–3 mole of NH4HS(s) has dissociated.

∴ number of mole of NH3(g) formed = number of moles of H2S(g) formed

= 7.00 x 10–3 mol (1)

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

=

( 7.00 x 10–3

2.00 mol dm–3)( 7.00 x 10–3

2.00 mol dm–3)

= 1.23 x 10–5 (1) mol2 dm–6 (1)

63

Part B

c) If more NH4HS(s) is added to the equilibrium system at 310 K, will the equilibrium concentration of NH3(g) change? Explain your answer. (2 marks)

No effect. (1)

The equilibrium constant depends only on temperature. Adding NH4HS(s) will not affect the equilibrium concentration of

NH3(g). (1)

d) What will happen to the percentage dissociation of NH4HS(s) if the temperature of the above equilibrium system is increased? Explain your answer. (3 marks)

When the temperature is increased, the system will respond by lowering the temperature. (1)

As the dissociation of NH4HS(s) is endothermic, the system will undergo a net forward reaction. (1)

Thus, the percentage dissociation of NH4HS(s) will increase. (1)

85 Part of the process by which coal can be converted into a combustible mixture of gases involves passing steam over white hot coke.

H2O(g) + C(s) H2(g) + CO(g) ΔH = +131 kJ

a) Write an expression for the equilibrium constant, Kc, for the reaction. State the units of Kc. (2 marks)

Kc =

[H2(g)][CO(g)][H2O(g)]

(1)

Units of Kc : mol dm–3 (1)

b) At 1 000 K, the value of Kc for the reaction is 3.00 x 10–2.

In an experiment, 6.00 moles of steam and an excess of solid carbon are heated in a 5.00 dm3 container. Calculate the equilibrium concentration of H2O(g), H2(g) and CO(g). (5 marks)

Let x mol dm–3 be the decrease in the concentration of H2O(g) when equilibrium is reached.

H2O(g) + C(s) H2(g) + CO(g)


6.005.00

mol dm–3 0 mol dm–3 0 mol dm–3

= 1.20 mol dm–3

Equilibrium concentration (1.20 – x) mol dm–3 x mol dm–3 x mol dm–3

Kc = 3.00 x 10–2 mol dm–3 =

(x mol dm–3)2

(1.20 – x) mol dm–3 (1)

64

Part B


3.00 x 10–2 (1.20 – x) = x2

x2 + 0.0300 x – 0.0360 = 0


x = 0.175 or –2.05 (rejected) (2)

i.e. the decrease in the concentration of H2O(g) is 0.175 mol dm–3.

The equilibrium concentrations are as follows:

[H2O(g)] = (1.20 – 0.175) mol dm–3

= 1.03 mol dm–3 (1)

[H2(g)] = [CO(g)]

= 0.175 mol dm–3 (1)

c) State and explain how the amount of steam in the equilibrium mixture would change if there is an increase in

i) the pressure; (3 marks)

The amount of steam will increase. (1)

An increase in pressure will bring about a net reaction that decrease the number of moles of gas. This helps to

reduce the pressure. (1)


ii) the temperature. (3 marks)

The amount of steam will decrease. (1)


As the forward reaction is endothermic, the system will undergo a net forward reaction. (1)

65

Part B

86 A crucial reaction in the manufacture of sulphuric acid involves the oxidation of sulphur dioxide to sulphur trioxide by oxygen in the air.

2SO2(g) + O2(g) 2SO3(g) ΔH = –197 kJ

A catalyst is used in the process.

a) Name the catalyst used in the process. (1 mark)

Vanadium(V) oxide (1)

b) The high yield is only achieved under certain conditions. After each condition, explain why it leads to an increased yield of sulphur trioxide.

i) There needs to be an excess of air in the reacting gas mixture. (2 marks)

An excess of air is needed to provide excess oxygen (1)

to drive the position of equilibrium to the right (1)

so that a high yield of sulphur trioxide can be obtained.

ii) The catalyst needs to be cooled. (3 marks)

When the temperature is decreased, the system will respond by raising the temperature. (1)

As the forward reaction is exothermic, the system will undergo a net forward reaction. (1)

Thus, a high yield of sulphur trioxide can be obtained.

Also the catalyst gets heated up and a high temperature may denature the catalyst. (1)

c) In an experiment, a mixture of 1.59 moles of sulphur dioxide and 0.855 mole of oxygen is introduced to a 30.0 dm3 container at 800 K. When equilibrium is reached, it is found that 94.3% of sulphur dioxide is converted to sulphur trioxide.

Calculate the equilibrium constant, Kc, at 800 K. (4 marks)

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

According to the equation, 2 moles of SO2(g) react with 1 mole of O2(g) to give 2 moles of SO3(g). As 94.3% of SO2(g)

is converted to SO3(g), the amount of SO2(g) would decrease by (1.59 x 0.943) mole and that of O2(g) would decrease

by ( 1.59 x 0.9432 ) mole; the amount of SO3(g) would increase by (1.59 x 0.943) mole.

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

Equilibrium

concentration ( 1.59 – 1.59 x 0.943

30.0 ) mol dm–3 ( 0.855 –

1.59 x 0.9432

30.0 ) mol dm–3 ( 1.59 x 0.94330.0 ) mol dm–3

= 3.02 x 10–3 mol dm–3 = 3.51 x 10–3 mol dm–3 = 0.0500 mol dm–3 (1)

66

Part B

Kc =

(0.0500 mol dm–3)2

(3.02 x 10–3 mol dm–3)2(3.51 x 10–3 mol dm–3) (1)

= 7.81 x 104 (1) dm3 mol–1 (1)

87 Consider the following equations which show reversible reactions.

Reaction 1 4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g) ΔHO = –900 kJ

Reaction 2 CO(g) + 2H2(g) CH3OH(g) ΔHO = –91 kJ

a) In industry these reactions are carried out in the presence of catalysts. A platinum catalyst is used in Reaction 1 and copper catalyst is used in Reaction 2.

State and explain the effect on the yield of a reaction when a catalyst is used. (2 marks)

No effect. (1)


b) State and explain which of the above reactions will give an increase in the yield of product(s) when the pressure is increased at a constant temperature. (3 marks)

Reaction 2 (1)


the pressure. (1)

The position of equilibrium will shift to the side of the equation with a fewer number of moles of gas, i.e. the product

side of Reaction 2. (1)

Thus, the yield of the product in Reaction 2 will increase.

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Part B

c) State and explain the effect on the yield of NO(g) when the temperature is increased in Reaction 1 at a constant pressure. (3 marks)


As the backward reaction is endothermic, the system will undergo a net backward reaction. (1)

Thus, the yield of NO(g) will decrease. (1)

d) i) Write an expression for the equilibrium constant, Kc, for Reaction 2. State the units of Kc. (2 marks)

Kc =

[CH3OH(g)][CO(g)][H2(g)]2

(1)

Units of Kc : dm6 mol–2 (1)

ii) The numerical value for the equilibrium constant, Kc, of Reaction 2 is 14.5 at 500 K. At equilibrium the concentrations of CO(g) and H2(g) are 0.120 and 0.0953 mol dm–3 respectively. Calculate the equilibrium concentration of CH3OH(g). (2 marks)

Kc = 14.5 dm6 mol–2 =

[CH3OH(g)]

(0.120 mol dm–3)(0.0953 mol dm–3)2 (1)

[CH3OH(g)] = 0.0158 mol dm–3 (1)

iii) What does the size of the equilibrium constant, Kc, tell you about the position of equilibrium for Reaction 2? (1 mark)

The position of equilibrium lies to the product side. (1)

88 In the Haber process, ammonia is synthesized by the exothermic reaction of nitrogen and hydrogen.

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

The table shows the percentage yield of ammonia, under different conditions of pressure and temperature, when the reaction has reached equilibrium.

Temperature (K) 600 800 1 000

% yield of ammonia at 100 atm 50 10 2



a) i) Explain why, at a given temperature, the percentage yield of ammonia increases with an increase in pressure. (2 marks)

An increase in pressure will bring about a net reaction that decreases the number of moles of gas. This helps to



Thus, the percentage yield of ammonia increases.

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Part B

ii) Give a reason why a high pressure of 500 atmospheres is NOT normally used in the Haber process. (1 mark)

Too expensive to generate. / The cost of building and running the plant is too high. (1)

b) Many industrial ammonia plants operate at a compromise temperature of about 723 K.

i) State one advantage, other than cost, of using a temperature lower than 723 K. Explain your answer. (3 marks)

The yield of ammonia increases. (1)



ii) State the major advanatge of using a temperature higher than 723 K. (1 mark)

The rate of production of ammonia is higher. (1)

iii) Hence explain why 723 K is referred to as a compromise temperature. (1 mark)

723 K is a balance between rate and yield. (1)

c) In a simulation of the process, a mixture of nitrogen and hydrogen is placed in a closed container. The initial concentrations of nitrogen and hydrogen are 0.500 mol dm–3 and 1.50 mol dm–3 respectively. When the equilibrium is attained at 723 K, 25.0% of the original nitrogen are consumed.

i) Calculate the respective concentrations of nitrogen, hydrogen and ammonia in the equilibrium mixture. (3 marks)

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

According to the equation, 1 mole of N2(g) reacts with 3 moles of H2(g) to give 2 moles of NH3(g).

As 25.0% of N2(g) are consumed, therefore the concentration of N2(g) decreases by (0.500 x 0.250) mol dm–3, that

of H2(g) decreases by (3 x 0.500 x 0.250) mol dm–3 and that of NH3(g) increases by (2 x 0.500 x 0.250) mol dm–3.

Equilibrium concentration of N2(g) = (0.500 – 0.500 x 0.250) mol dm–3

= 0.375 mol dm–3 (1)

Equilibrium concentration of H2(g) = (1.50 – 3 x 0.500 x 0.250) mol dm–3

= 1.13 mol dm–3 (1)

Equilibrium concentration of NH3(g) = (2 x 0.500 x 0.250) mol dm–3

= 0.250 mol dm–3 (1)

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Part B

ii) Calculate Kc for the reaction at 723 K. (3 marks)

Kc =

(0.250 mol dm–3)2

(0.375 mol dm–3)(1.13 mol dm–3)3 (1)

= 0.116 (1) dm6 mol–2 (1)

89 In the Contact process, sulphur dioxide is catalytically oxidized to sulphur trioxide according to the equation:

2SO2(g) + O2(g) 2SO3(g) ΔH < 0

The following table summarizes several possible conditions for the preparation of sulphur trioxide:

CaseConditions

Reactants Pressure (atm) Temperature (°C)

A SO2(g) + excess O2(g) 500 400

B excess SO2(g) + O2(g) 500 1 000

C excess SO2(g) + air 1 1 000

D SO2(g) + excess air 1 400

Which of the cases would represent

a) the theoretical conditions for obtaining the maximum yield of sulphur trioxide? Explain your answer. (5 marks)

Case A (1)


the pressure. (1)


Thus, a high pressure will increase the yield of SO3(g).



Thus, a low temperature will increase the yield of SO3(g).

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Part B

b) the most economical conditions for the industrial preparation of sulphur trioxide? Explain your answer. (4 marks)

Case D (1)

Air is readily available and cheap. (1)

At lower pressure, the cost of building and running the plant is lower. (1)

At lower temperature, the cost of energy is lower / less fuel is required. (1)

90 Ethanol is manufactured by catalytic hydration of ethene:

CH2=CH2(g) + H2O(g) CH3CH2OH(g)

a) The reaction represented by the above equation can reach a position of dynamic equilibrium. State TWO features of a system that is in dynamic equilibrium. (2 marks)

Rate of forward reaction = rate of backward reaction (1)

Concentration of the reactants and product remain constant. (1)

b) In a simulation study of the manufacturing process, 0.900 mole of ethene and 0.600 mole of steam are mixed in a closed container of volume 1.20 dm3. When equilibrium is attained, 5.00% of the ethene are converted into ethanol.

Calculate the equilibrium constant, Kc, under the above conditions. (4 marks)


According to the equation, 1 mole of CH2=CH2(g) reacts with 1 mole of H2O(g) to give 1 mole of CH3CH2OH(g).

As 5.00% of CH2=CH2(g) are converted to CH3CH2OH(g), the amounts of both CH2=CH2(g) and H2O(g) would decrease

by (0.900 x 0.0500) mole while that of CH3CH2OH(g) would increase by (0.900 x 0.0500) mole.


Equilibrium

concentration ( 0.900 – 0.900 x 0.0500

1.20 ) mol dm–3 ( 0.600 – 0.900 x 0.05001.20 ) mol dm–3 ( 0.900 x 0.0500

1.20 ) mol dm–3

= 0.713 mol dm–3 = 0.463 mol dm–3 = 0.0375 mol dm–3 (1)

Kc =

(0.0375 mol dm–3)(0.713 mol dm–3)(0.463 mol dm–3)

(1)

= 0.114 (1) dm3 mol–1 (1)

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Part B

c) The following table lists the percentage conversion of ethene using excess steam at various reaction conditions used in industry.

Pressure (atm) Temperature (°C) Percentage conversion (%)

50 200 45

50 320 30

80 200 60

80 320 45

i) State and explain the effect of increasing the pressure on the percentage conversion. (3 marks)

The percentage conversion increases with pressure. (1)

An increase in pressure will bring about a net reaction that decreases the number of moles of gas. This helps to


The position of equilibrium will shift to the side of the equation with a fewer number of moles of gas, i.e. shift to

the right. (1)

ii) Deduce the sign of the enthalpy change for the forward reaction. Explain your answer. (3 marks)

When the temperature is increased, the percentage conversion decreases. (1)

It can be deduced that when the temperature is increased, the system will undergo a net backward reaction so as

to lower the temperature. (1)

Hence the backward reaction should be an endothermic reaction / the forward reaction should be an exothermic

reaction.

The sign of the enthalpy change for the forward reaction should be negative. (1)

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Part B

iii) The equation for the formation of ethanol shows that equal numbers of moles of ethene and steam are required. In industry however excess steam is used.

Suggest why excess steam in used. (1 mark)

Using excess steam will shift the position of equilibrium to the right. (1)

91 Read the following passage and answer the questions that follow.

Chlorine for disinfectionChlorine is used in water treatment for disinfection.

When chlorine is added to pure water, hypochlorous acid (HOCl) and hydrochloric acid (HCl) are formed.

Cl2(aq) + H2O(l) HOCl(aq) + H+(aq) + Cl–(aq)

The principal disinfecting action of aqueous chlorine is due to the hypochlorous acid formed.

Hypochlorous acid dissociated into hydrogen ions and hypochlorite ions.

HOCl(aq) H+(aq) + OCl–(aq)

The concentrations of hypochlorous acid and hypochlorite ions in chlorinated water depend on the pH of the water.

Instead of using chlorine gas, some plants apply sodium hypochlorite or calcium hypochlorite to water. Sodium hypochlorite completely dissociates in water to form sodium ions and hypochlorite ions. In solution, the hypochlorite ions hydrolyze to form the disinfectant hypochlorous acid according to the following equation:

OCl–(aq) + H2O(l) HOCl(aq) + OH–(aq)

a) State Le Chatelier’s principle. (2 marks)

Le Chatelier’s principle states that if the conditions of a system in equilibrium is changed, the position of equilibrium

will shift (1)

so as to reduce that change. (1)

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Part B

b) Use Le Chatelier’s principle to explain how the pH of the chlorinated water will affect the concentrations of hypochlorous acid and hypochlorite ions in the water. (4 marks)

Decreasing the pH of the chlorinated water means increasing the concentration of hydrogen ions. (1)

The system will respond by reducing this change. (1)


More hypochlorous acid will form. (1)

OR

Increasing the pH of the chlorinated water means decreasing the concentration of hydrogen ions. (1)

The system will respond by reducing this change. (1)

The position of equilibrium will shift to the right. (1)

More hypochlorite ions will form. (1)

c) Suppose the equilibrium constant for the following reaction is K1.

OCl–(aq) + H2O(l) HOCl(aq) + OH–(aq)

What is the equilibrium constant K2 for the following reaction?

2HOCl(aq) + 2OH–(aq) 2OCl–(aq) + 2H2O(l) (2 marks)

K1 = [HOCl(aq)][OH–(aq)]

[OCl–(aq)][H2O(l)] (1)

K2 = [OCl–(aq)]2[H2O(l)]2

[HOCl(aq)]2[OH–(aq)]2

= 1

K12 (1)

d) The hypochlorous acid produced in a solution of sodium hypochlorite can react further to produce small amount of chlorine according to the following equation:

HOCl(aq) + H+(aq) + Cl–(aq) Cl2(g) + H2O(l)

What will happen to the concentration of chlorine if a little sodium hydroxide solution is added to a sodium hypochlorite solution? Explain your answer. (4 marks)

When sodium hydroxide solution is added, the hydroxide ions react with the hydrogen ions to form water. (1)

Thus, the concentration of hydrogen ions decreases. (1)

The system responds by reducing this change.

A net backward reaction occurs to produce more hydrogen ions. (1)

Thus, the concentration of chlorine will decrease. (1)

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Part B

92 Read the following passage and answer the questions that follow.

Carbon dioxide and formation of stalactites and stalagmites in limestone cavesIn addition to being a component of the atmosphere, carbon dioxide also dissolves in the water of the oceans. The dissolving process can be described by the following equations:

CO2(g) CO2(aq) ...... equation (1)

CO2(aq) + H2O(l) H+(aq) + HCO3–(aq) ...... equation (2)

In nature, surface water often becomes acidic because atmospheric carbon dioxide dissolves in it. This acidic solution can dissolve limestone:

CaCO3(s) + CO2(aq) + H2O(l) Ca2+(aq) + 2HCO3–(aq) ...... equation (3)

Openings formed in the limestone as the calcium carbonate dissolves.

Slight cooling of the water saturated with carbon dioxide can reduce the solubility of carbon dioxide. The position of equilibrium shifts, resulting in the precipitation of calcium carbonate. This precipitate, the stalactite, is formed immediately when the seeping water comes into contact with air currents in a cave. Stalagmites are formed on the floors of caves in the same way.

a) Suggest why the balance between CO2(g) in the atmosphere and CO2(aq) in the oceans CANNOT be regarded as a true dynamic equilibrium. (1 mark)

The system is not closed. (1)

b) Based on equations (1) and (2), explain the likely effect of the increasing concentration of atmospheric carbon dioxide on the pH of water at the ocean surface. (5 marks)

When the concentration of CO2(g) increases, the system responds by reducing the change. (1)

The position of equilibrium of the first system will shift to the right. (1)

The concentration of CO2(aq) will increase. (1)

When the concentration of CO2(aq) increases, the position of equilibrium of the second system will shift to the right.

The concentration of hydrogen ions will increase. (1)

Thus, the pH of the water at the ocean surface will decrease. (1)

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Part B

c) Use Le Chatelier’s principle to explain why slight cooling of the water saturated with carbon dioxide will result in the precipitation of calcium carbonate. (3 marks)

Slight cooling of the water saturated with carbon dioxide can reduce the solubility of carbon dioxide.

Some carbon dioxide bubbles out of the solution and the concentration of carbon dioxide decreases. (1)

The system responds by reducing the change. (1)

The position of equilibrium of the system represented by equation (3) shifts to the left, (1)

resulting in the precipitation of calcium carbonate.

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Part B

93 Ethanoic acid and ethanol react to form ethyl ethanoate and water in the presence of concentrated sulphuric acid.

CH3COOH(l) + CH3CH2OH(l) CH3COOCH2CH3(l) + H2O(l)

Describe briefly how you can carry out an experiment to determine the equilibrium constant, Kc, for the esterification reation. Suggest also how you can calculate Kc from the experimental results obtained. (No actual calculation is required.)

(For this question, you are required to give answers in paragraph form.) (9 marks)

Mix equal number of moles of ethanoic acid and ethanol in a pear-shaped flask.

Withdraw 1.00 cm3 of this mixture and add to a conical flask containing 25 cm3 of distilled water. (0.5)

Titrate the contents of the conical flask against standard sodium hydroxide solution. V1 cm3 of the alkali are required to

reach the end point. (0.5)

Add a few drops of concentrated sulphuric acid to the acid-alcohol mixture in the pear-shaped flask. (0.5)

Titrate 1.00 cm3 of the resulting mixture against standard sodium hydroxide solution. V2 cm3 of the alkali are required to

reach the end point. (0.5)

Heat the mixture in the pear-shaped flask under reflux for 2 hours. After cooling, withdraw 1.00 cm3 of the mixture

and (0.5)

titrate against standard sodium hydroxide solution. Vf cm3 of the alkali are required to reach the end point. (0.5)

The concentration of CH3COOH(l) in the original mixture (Xi) can be obtained from V1 and the concentration of

NaOH(aq). (0.5)

The concentration of CH3COOH(l) in the mixture after reflux (Xf) can be obtained from [Vf – (V2 – V1)] and the concentration

of NaOH(aq). (0.5)

Change in concentration of CH3COOH(l) = Xf – Xi

According to the equation, 1 mole of CH3COOH(l) reacts with 1 mole of CH3CH2OH(l) to give 1 mole of CH3COOCH2CH3(l)

and H2O(l).

The concentrations of substances in the mixture after reflux are as follows:

[CH3CH2OH(l)] = Xf

[CH3COOCH2CH3(l)] = [H2O(l)] = Xf – Xi

Equilibrium constant (Kc) = [CH3COOCH2CH3(l)][H2O(l)][CH3COOH(l)][CH3CH2OH(l)] (1)

= (Xf – Xi)

2

Xf2

(1)

(3 marks for organization and presentation)

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Part B

94 The hydrogen used in the Haber process is made by the following reaction:


Discuss how the yield of hydrogen in the process is affected by changing the pressure, changing the temperature and using a catalyst.

(For this question, you are required to give answers in paragraph form.) (9 marks)

• An increase in pressure will bring about a net reaction that decreases the number of moles of gas. This helps to reduce

the pressure. (1)

A net backward reaction will occur. Thus, the yield of hydrogen will decrease. (1)

OR

A decrease in pressure will bring about a net reaction that increases the number of moles of gas. This helps to increase

the pressure. (1)

A net forward reaction will occur. Thus, the yield of hydrogen will increase. (1)

• When the temperature is increased, the system will respond by reducing the temperature. (1)

As the production of hydrogen is endothermic, the system will undergo a net forward reaction. Thus, the yield of

hydrogen will increase. (1)

OR


As the production of hydrogen is endothermic, the system will undergo a net backward reaction. Thus, the yield of

hydrogen will decrease. (1)

• The yield of hydrogen is not affected by a catalyst. (1)


(3 marks for organization and presentation)

10. Solution Guide to Supplementary Exercises

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