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R2.3 How far? The extent of chemical change

Practice exam-style IB Chemistry questions for How far? The extent of chemical change, aligned with the syllabus and grouped by topic.

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Question 1
SL • Paper 1A
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SL • Paper 1A
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A sealed flask contains liquid propanone and propanone vapour at constant temperature. The system has reached dynamic equilibrium.

What is true for this system?

A.

Condensation has stopped and evaporation continues.

B.

The rates of evaporation and condensation are equal.

C.

Evaporation has stopped and condensation continues.

D.

The amounts of liquid and vapour must be equal.

Question 2
SL • Paper 1A
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SL • Paper 1A
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A catalyst is added to a reversible reaction mixture at constant temperature before equilibrium has been reached.

What is the effect of the catalyst?

A.

The time taken to reach equilibrium decreases and KK increases.

B.

The equilibrium composition changes and KK is unchanged.

C.

The equilibrium composition changes and KK increases.

D.

The time taken to reach equilibrium decreases and KK is unchanged.

Question 3
SL • Paper 1A
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SL • Paper 1A
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For the homogeneous equilibrium

2NO2(g)⇌N2O4(g)2NO_2(g) \rightleftharpoons N_2O_4(g)

what is the equilibrium constant expression?

A.

K=[N2O4]2[NO2]K=\dfrac{[N_2O_4]}{2[NO_2]}

B.

K=2[NO2][N2O4]K=\dfrac{2[NO_2]}{[N_2O_4]}

C.

K=[NO2]2[N2O4]K=\dfrac{[NO_2]^2}{[N_2O_4]}

D.

K=[N2O4][NO2]2K=\dfrac{[N_2O_4]}{[NO_2]^2}

Question 4
SL • Paper 1A
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SL • Paper 1A
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At a fixed temperature, the equilibrium constant for

A(g)+2B(g)⇌C(g)A(g)+2B(g) \rightleftharpoons C(g)

is 4.0×1034.0\times10^3.

What is the equilibrium constant for the reverse reaction at the same temperature?

A.

2.5×10−42.5\times10^{-4}

B.

1.6×1071.6\times10^7

C.

4.0×1034.0\times10^3

D.

4.0×10−34.0\times10^{-3}

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Question 5
SL • Paper 1A
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SL • Paper 1A
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Carbon dioxide is in equilibrium between air and water in a closed container.

CO2(g)⇌CO2(aq)CO_2(g) \rightleftharpoons CO_2(aq)

The pressure is increased at constant temperature.

What is the effect on the equilibrium position and on KK?

A.

Equilibrium shifts to CO2(aq)CO_2(aq) and KK is unchanged.

B.

Equilibrium shifts to CO2(aq)CO_2(aq) and KK increases.

C.

Equilibrium shifts to CO2(g)CO_2(g) and KK is unchanged.

D.

Equilibrium shifts to CO2(g)CO_2(g) and KK decreases.

Question 6
SL • Paper 2
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SL • Paper 2
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A small amount of liquid bromine is sealed in a flask at constant temperature. After some time the intensity of the orange-brown vapour remains constant.

A

why the flask must be sealed for equilibrium to be established.

[1]
B

two characteristics of the system when dynamic equilibrium has been reached.

[2]
Question 7
SL • Paper 2
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SL • Paper 2
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Nitrosyl chloride decomposes according to the following homogeneous equilibrium.

2NOCl(g)⇌2NO(g)+Cl2(g)2\text{NOCl}(g) \rightleftharpoons 2\text{NO}(g) + Cl_2(g)

A

the equilibrium constant expression for the reaction as written.

[1]
B

At a fixed temperature the value of KK for the reaction as written is 3.20×10−33.20 \times 10^{-3}. the value of KK for the reverse reaction at the same temperature and which side is favoured for the reaction as written.

[2]
Question 8
SL • Paper 2
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SL • Paper 2
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Two weak monoprotic acids have the following acid dissociation constants at the same temperature.

Ka(HA)=1.8×10−5K_a(HA) = 1.8 \times 10^{-5}

Ka(HB)=6.3×10−10K_a(HB) = 6.3 \times 10^{-10}

A

which acid is stronger and your answer in terms of extent of ionization.

[2]
B

the equilibrium constant for the reverse of the dissociation of HAHA at the same temperature.

[1]

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Question 9
SL • Paper 1A
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SL • Paper 1A
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For the equilibrium

N2(g)+3H2(g)⇌2NH3(g)ΔH<0N_2(g)+3H_2(g) \rightleftharpoons 2NH_3(g)\qquad \Delta H<0

what happens when the temperature is increased?

A.

The equilibrium shifts right and KK decreases.

B.

The equilibrium shifts left and KK decreases.

C.

The equilibrium shifts left and KK increases.

D.

The equilibrium shifts right and KK increases.

Question 10
HL • Paper 1A
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HL • Paper 1A
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For the equilibrium

H2(g)+I2(g)⇌2HI(g)H_2(g)+I_2(g) \rightleftharpoons 2HI(g)

K=50K=50 at a particular temperature. At one instant, [H2]=0.20 mol dm−3[H_2]=0.20\ \text{mol dm}^{-3}, [I2]=0.30 mol dm−3[I_2]=0.30\ \text{mol dm}^{-3} and [HI]=0.60 mol dm−3[HI]=0.60\ \text{mol dm}^{-3}.

What is the value of QQ and the direction favoured?

A.

Q=6.0Q=6.0; reverse reaction favoured

B.

Q=50Q=50; the mixture is at equilibrium

C.

Q=6.0Q=6.0; forward reaction favoured

D.

Q=0.17Q=0.17; forward reaction favoured

Question 11
HL • Paper 1A
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HL • Paper 1A
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For the homogeneous equilibrium

A(g)⇌2B(g)A(g) \rightleftharpoons 2B(g)

1.00 mol dm−31.00\ \text{mol dm}^{-3} of AA is placed in a sealed container. At equilibrium, [A]=0.64 mol dm−3[A]=0.64\ \text{mol dm}^{-3}.

What is the value of KK?

A.

0.200.20

B.

0.810.81

C.

0.410.41

D.

1.131.13

Question 12
HL • Paper 1A
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HL • Paper 1A
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For the equilibrium

A(g)+B(g)⇌C(g)A(g)+B(g) \rightleftharpoons C(g)

K=10K=10 at a fixed temperature. At equilibrium, [A]=0.20 mol dm−3[A]=0.20\ \text{mol dm}^{-3} and [B]=0.10 mol dm−3[B]=0.10\ \text{mol dm}^{-3}.

What is [C][C] at equilibrium?

A.

0.020 mol dm−30.020\ \text{mol dm}^{-3}

B.

500 mol dm−3500\ \text{mol dm}^{-3}

C.

2.0 mol dm−32.0\ \text{mol dm}^{-3}

D.

0.20 mol dm−30.20\ \text{mol dm}^{-3}

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Question 13
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HL • Paper 1A
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For a reaction at 298 K298\ \text{K}, K=1.0×10−2K=1.0\times10^{-2}.

Using ΔG∘=−RTln⁡K\Delta G^\circ=-RT\ln K, what is ΔG∘\Delta G^\circ?

A.

−11.4 kJ mol−1-11.4\ \text{kJ mol}^{-1}

B.

+11.4 kJ mol−1+11.4\ \text{kJ mol}^{-1}

C.

−5.70 kJ mol−1-5.70\ \text{kJ mol}^{-1}

D.

+5.70 kJ mol−1+5.70\ \text{kJ mol}^{-1}

Question 14
SL • Paper 2
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SL • Paper 2
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The chromate-dichromate equilibrium is shown.

2CrO42−(aq)+2H+(aq)⇌Cr2O72−(aq)+H2O(l)2CrO_4^{2-}(aq) + 2H^+(aq) \rightleftharpoons Cr_2O_7^{2-}(aq) + H_2O(l)

yellow CrO42−CrO_4^{2-}, orange Cr2O72−Cr_2O_7^{2-}

A few drops of aqueous sodium hydroxide are added to an equilibrium mixture at constant temperature.

A

the equilibrium constant expression for this reaction, omitting water.

[1]
B

and the colour change after adding sodium hydroxide.

[2]
C

the effect of adding sodium hydroxide on the value of KK.

[1]
Question 15
SL • Paper 2
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SL • Paper 2
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In a sealed bottle of carbonated water, carbon dioxide is involved in the heterogeneous equilibrium:

CO2(g)⇌CO2(aq)CO_2(g) \rightleftharpoons CO_2(aq)

The bottle is opened at constant temperature.

A

the change in pressure of CO2(g)CO_2(g) when the bottle is opened.

[1]
B

the effect on the equilibrium position and the value of KK.

[2]
Question 16
SL • Paper 2
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SL • Paper 2
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Dinitrogen tetroxide and nitrogen dioxide form the following equilibrium in a sealed syringe.

N2O4(g)⇌2NO2(g)N_2O_4(g) \rightleftharpoons 2NO_2(g)

The forward reaction is endothermic. N2O4N_2O_4 is colourless and NO2NO_2 is brown.

A

and the effect of increasing the temperature on the colour intensity and on KK.

[3]
B

the effect of decreasing the volume of the syringe at constant temperature on the equilibrium position.

[1]

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Question 17
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At a particular temperature, K=49.0K = 49.0 for the equilibrium:

H2(g)+I2(g)⇌2HI(g)H_2(g) + I_2(g) \rightleftharpoons 2HI(g)

At one instant, the concentrations are:

[H2]=0.0200 mol dm−3[H_2] = 0.0200\ \text{mol dm}^{-3}, [I2]=0.0100 mol dm−3[I_2] = 0.0100\ \text{mol dm}^{-3} and [HI]=0.0600 mol dm−3[HI] = 0.0600\ \text{mol dm}^{-3}.

A

the reaction quotient, QQ.

[2]
B

the direction in which the reaction proceeds to reach equilibrium.

[1]
Question 18
HL • Paper 2
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HL • Paper 2
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A homogeneous equilibrium is established at constant temperature.

A(g)+B(g)⇌C(g)A(g) + B(g) \rightleftharpoons C(g)

Initially, [A]=0.500 mol dm−3[A] = 0.500\ \text{mol dm}^{-3} and [B]=0.500 mol dm−3[B] = 0.500\ \text{mol dm}^{-3}, with no CC present. At equilibrium, [C]=0.300 mol dm−3[C] = 0.300\ \text{mol dm}^{-3}.

A

the equilibrium concentrations of AA and BB.

[1]
B

the value of KK at this temperature.

[2]
Question 19
SL • Paper 1B
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SL • Paper 1B
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A small amount of liquid bromine is placed in a sealed glass vessel at constant temperature. The graph shows the rates of evaporation and condensation of bromine over time.

A

Identify the time at which dynamic equilibrium is first reached.

[1]
B

two characteristics of the system after this time.

[2]
C

why the same equilibrium would not be established if the vessel were open.

[1]
Question 20
HL • Paper 1A
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HL • Paper 1A
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A weak acid dissociates according to

HA(aq)⇌H+(aq)+A−(aq)HA(aq) \rightleftharpoons H^+(aq)+A^-(aq)

For 0.100 mol dm−30.100\ \text{mol dm}^{-3} HAHA, Ka=1.6×10−5K_a=1.6\times10^{-5}. Using the small-KK approximation, what is the equilibrium concentration of H+(aq)H^+(aq)?

A.

1.3×10−3 mol dm−31.3\times10^{-3}\ \text{mol dm}^{-3}

B.

4.0×10−2 mol dm−34.0\times10^{-2}\ \text{mol dm}^{-3}

C.

1.6×10−6 mol dm−31.6\times10^{-6}\ \text{mol dm}^{-3}

D.

1.6×10−4 mol dm−31.6\times10^{-4}\ \text{mol dm}^{-3}

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Question 21
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For a reaction at 298 K298\ \text{K}, ΔG∘=−17.1 kJ mol−1\Delta G^\circ=-17.1\ \text{kJ mol}^{-1}.

Using ΔG∘=−RTln⁡K\Delta G^\circ=-RT\ln K, what is the approximate value of KK?

A.

1.0×10−31.0\times10^{-3}

B.

1.0×1011.0\times10^{1}

C.

1.0×10−11.0\times10^{-1}

D.

1.0×1031.0\times10^{3}

Question 22
HL • Paper 2
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HL • Paper 2
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A 0.200 mol dm−30.200\ \text{mol dm}^{-3} solution of a weak acid, HAHA, is prepared at 298 K298\ \text{K}.

HA(aq)⇌H+(aq)+A−(aq)HA(aq) \rightleftharpoons H^+(aq) + A^-(aq)

Ka=1.6×10−5K_a = 1.6 \times 10^{-5}

Use the approximation [HA]initial≈[HA]eqm[HA]_{\text{initial}} \approx [HA]_{\text{eqm}}.

A

[H+][H^+] in the solution.

[2]
B

the pH of the solution.

[1]
C

why the approximation is reasonable in this calculation.

[1]
Question 23
HL • Paper 2
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HL • Paper 2
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For a reaction at 298 K298\ \text{K}, the standard Gibbs energy change is +12.4 kJ mol−1+12.4\ \text{kJ mol}^{-1}.

Use ΔG∘=−RTln⁡K\Delta G^\circ = -RT\ln K and R=8.31 J K−1 mol−1R = 8.31\ \text{J K}^{-1}\ \text{mol}^{-1}.

A

the equilibrium constant, KK, at 298 K298\ \text{K}.

[2]
B

whether reactants or products are favoured under standard conditions.

[1]
Question 24
HL • Paper 2
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HL • Paper 2
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At 350 K350\ \text{K}, a homogeneous equilibrium has K=2.50×104K = 2.50 \times 10^4.

Use ΔG∘=−RTln⁡K\Delta G^\circ = -RT\ln K and R=8.31 J K−1 mol−1R = 8.31\ \text{J K}^{-1}\ \text{mol}^{-1}.

A

ΔG∘\Delta G^\circ in kJ mol−1\text{kJ mol}^{-1}.

[2]
B

what the sign of ΔG∘\Delta G^\circ indicates about the equilibrium position.

[1]

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Question 25
SL • Paper 1B
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SL • Paper 1B
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Nitrosyl chloride decomposes in a sealed vessel according to the homogeneous equilibrium:

2NOCl(g)⇌2NO(g)+Cl2(g)2\text{NOCl}(g) \rightleftharpoons 2\text{NO}(g) + \text{Cl}_2(g)

The table gives equilibrium concentrations at one temperature.

Species

Equilibrium concentration / mol dm−3^{-3}

NOCl

0.100

NO

0.300

Cl2

0.150

A

the equilibrium constant expression, KK, for the reaction as written.

[2]
B

KK using the equilibrium concentrations in the table.

[1]
C

what the magnitude of KK suggests about the extent of the reaction as written.

[1]
Question 26
SL • Paper 1B
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SL • Paper 1B
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The equilibrium constant for the formation of methanol is shown at different temperatures.

CO(g)+2H2(g)⇌CH3OH(g)\text{CO}(g)+2\text{H}_2(g) \rightleftharpoons \text{CH}_3\text{OH}(g)

A

Identify the temperature at which methanol is most favoured at equilibrium.

[1]
B

At 500 K500\ \text{K}, K=5.0K=5.0 for the reaction as written. KK for the reverse reaction at this temperature.

[1]
C

Use the data to whether the forward reaction is exothermic or endothermic.

[2]
Question 27
SL • Paper 1B
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SL • Paper 1B
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The equilibrium below is established in aqueous solution. The product ion is red.

Fe3+(aq)+SCN−(aq)⇌FeSCN2+(aq)\text{Fe}^{3+}(aq)+\text{SCN}^-(aq) \rightleftharpoons \text{FeSCN}^{2+}(aq)

Absorbance is used to monitor the concentration of FeSCN2+\text{FeSCN}^{2+} after different additions.

A

and the effect of adding Fe3+(aq)\text{Fe}^{3+}(aq) on the equilibrium mixture.

[2]
B

Silver ions react with SCN−(aq)\text{SCN}^-(aq). the change after adding silver ions.

[1]
C

the effect of these concentration changes on the value of KK at constant temperature.

[1]
Question 28
HL • Paper 1B
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HL • Paper 1B
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At a fixed temperature, K=50.0K=50.0 for the equilibrium:

H2(g)+I2(g)⇌2HI(g)\text{H}_2(g)+\text{I}_2(g) \rightleftharpoons 2\text{HI}(g)

The table gives concentrations in two reaction mixtures.

Mixture

[H2][H_2] / mol dm−3^{-3}

[I2][I_2] / mol dm−3^{-3}

[HI][HI] / mol dm−3^{-3}

A

0.200

0.200

0.400

B

0.0500

0.100

0.500

A

the reaction quotient, QQ, for mixture A.

[1]
B

the direction in which mixture A will proceed to reach equilibrium.

[2]
C

Mixture B has Q=50.0Q=50.0. what this indicates about mixture B.

[1]

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Question 29
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The equilibrium constant for the reaction below is K=0.250K = 0.250 at a particular temperature.

2A(g)⇌B(g)2A(g) \rightleftharpoons B(g)

At one instant, [A]=0.400 mol dm−3[A] = 0.400\ \text{mol dm}^{-3} and [B]=0.0200 mol dm−3[B] = 0.0200\ \text{mol dm}^{-3}.

A

QQ for the mixture at this instant.

[2]
B

the direction in which the reaction is favoured and the sign of ΔG\Delta G under these conditions.

[2]
Question 30
SL • Paper 1B
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SL • Paper 1B
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Carbon dioxide is allowed to reach equilibrium between the gas phase and water in a sealed syringe.

CO2(g)⇌CO2(aq)\text{CO}_2(g) \rightleftharpoons \text{CO}_2(aq)

The plunger is then pushed in at constant temperature.

A

the change in the concentration of dissolved CO2\text{CO}_2 after the plunger is pushed in.

[1]
B

the change using Le Châtelier's principle.

[2]
C

the effect of the pressure change on KK for this equilibrium.

[1]
Question 31
SL • Paper 1B
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SL • Paper 1B
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Dinitrogen tetroxide and nitrogen dioxide form the equilibrium below. Nitrogen dioxide is brown and dinitrogen tetroxide is colourless.

N2O4(g)⇌2NO2(g)\text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g)

For the reaction as written, ΔHr∘=+57 kJ mol−1\Delta H_r^\circ=+57\ \text{kJ mol}^{-1}. The diagrams show the colour intensity of sealed tubes under different conditions.

A

and the effect of increasing temperature on the colour of the mixture.

[2]
B

the effect of adding a catalyst with the effect of increasing temperature.

[2]
C

the effect of cooling on the value of KK for the reaction as written.

[1]
Question 32
HL • Paper 1B
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HL • Paper 1B
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A homogeneous equilibrium is established at constant temperature.

A(g)+B(g)⇌C(g)A(g)+B(g) \rightleftharpoons C(g)

The table shows initial concentrations and one measured equilibrium concentration.

Species

Initial concentration / mol dm−3^{-3}

Equilibrium concentration / mol dm−3^{-3}

A(g)

0.500

B(g)

0.500

C(g)

0.000

0.250

A

Complete the equilibrium concentrations of A and B.

[2]
B

KK for the equilibrium.

[1]
C

whether products or reactants are favoured by this equilibrium constant.

[1]

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Question 33
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A weak monoprotic acid, HA, is dissolved in water.

HA(aq)⇌H+(aq)+A−(aq)\text{HA}(aq) \rightleftharpoons \text{H}^+(aq)+\text{A}^-(aq)

At 298 K298\ \text{K}, Ka=1.8×10−5K_a=1.8\times10^{-5} for HA. The initial concentration of HA is 0.100 mol dm−30.100\ \text{mol dm}^{-3}.

State

HA / mol dm−3^{-3}

H+^+ / mol dm−3^{-3}

A−^- / mol dm−3^{-3}

Initial

0.100

0

0

Change

−x-x

+x+x

+x+x

Equilibrium

0.100−x0.100 - x

xx

xx

A

Using the small-KaK_a approximation, [H+][\text{H}^+] at equilibrium.

[2]
B

the pH of the solution.

[1]
C

Justify the use of the approximation for this solution.

[1]
Question 34
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HL • Paper 1B
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The equilibrium constant for a reaction at 298 K298\ \text{K} is measured as K=3.2×104K=3.2\times10^4.

The relationship between standard Gibbs energy change and KK is:

ΔG∘=−RTln⁡K\Delta G^\circ=-RT\ln K

Quantity

Value

Units

Equilibrium constant, KK

3.2×1043.2 \times 10^{4}

dimensionless

Temperature, TT

298

KK

Gas constant, RR

8.31

J mol−1 K−1\text{J mol}^{-1}\ \text{K}^{-1}

A

ΔG∘\Delta G^\circ in kJ mol−1\text{kJ mol}^{-1} at 298 K298\ \text{K}.

[2]
B

which side of the equilibrium is favoured under standard conditions.

[1]
C

the value of ΔG\Delta G when the reaction mixture has reached equilibrium.

[1]
Question 35
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SL • Paper 2
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A small amount of liquid bromine is placed in a sealed flask. After some time the intensity of the brown vapour becomes constant.

A

The bromine system reaches a physical equilibrium.

I.

why the flask must be sealed for equilibrium to be established.

[2]
II.

one macroscopic observation that indicates equilibrium has been reached.

[1]
B

and this physical equilibrium with a chemical equilibrium.

[3]
Question 36
SL • Paper 2
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SL • Paper 2
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Methanol can be manufactured from carbon monoxide and hydrogen in a reversible reaction.

CO(g)+2H2(g)⇌CH3OH(g)CO(g)+2H_2(g)\rightleftharpoons CH_3OH(g)

A

The equilibrium law can be applied to this homogeneous equilibrium.

I.

the equilibrium constant expression for the reaction as written.

[1]
II.

how the stoichiometry of the equation is represented in the equilibrium expression.

[2]
B

At a certain temperature K=12K=12 for the reaction as written. what this value of KK indicates about the equilibrium, KK for the reverse reaction, and what happens to KK if more H2H_2 is added at constant temperature.

[3]

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Question 37
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SL • Paper 2
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Ammonia is produced in the Haber process.

N2(g)+3H2(g)⇌2NH3(g)ΔH<0N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)\qquad \Delta H<0

A

Two changes are considered for an equilibrium mixture at constant temperature.

I.

the effect of increasing pressure on the equilibrium yield of ammonia.

[2]
II.

the effect of adding an iron catalyst on the equilibrium composition and on KK.

[2]
B

the effect of increasing temperature on the Haber equilibrium.

[2]
Question 38
HL • Paper 1B
Hard
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HL • Paper 1B
Hard
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For a reversible reaction at constant temperature, the reaction quotient QQ was calculated at different times after mixing the reactants. The equilibrium constant is K=12K=12 at this temperature.

A

For the curve that starts below KK, identify the direction favoured at the first labelled time.

[1]
B

For the curve that starts above KK, why the reverse reaction is favoured initially.

[2]
C

the relationship between QQ, KK and Gibbs energy when equilibrium is reached.

[2]
Question 39
HL • Paper 1B
Hard
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HL • Paper 1B
Hard
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For the equilibrium below at 298 K298\ \text{K}, the standard Gibbs energy change is +4.80 kJ mol−1+4.80\ \text{kJ mol}^{-1}.

N2O4(g)⇌2NO2(g)\text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g)

A mixture is then prepared with non-equilibrium concentrations of the two gases.

Item

Value

Units

Reaction

N2O4(g)⇌2NO2(g)\text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g)

TT

298

K

ΔG∘^\circ

+4.80

kJ mol−1\text{kJ mol}^{-1}

QQ (prepared mixture)

0.050

dimensionless

A

KK for the reaction as written at 298 K298\ \text{K}.

[2]
B

KK for the reverse reaction at the same temperature.

[1]
C

For the prepared mixture, Q=0.050Q=0.050. the direction in which the reaction will proceed.

[1]
Question 40
SL • Paper 2
Hard
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SL • Paper 2
Hard
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An aqueous equilibrium involving cobalt ions is shown.

[Co(H2O)6]2+(aq)+4Cl−(aq)⇌[CoCl4]2−(aq)+6H2O(l)[\text{Co(H}_2\text{O)}_6]^{2+}(aq)+4Cl^-(aq)\rightleftharpoons [\text{CoCl}_4]^{2-}(aq)+6H_2O(l)

The forward reaction is endothermic. The octahedral complex is pink and the tetrahedral complex is blue.

A

Changes are made to a solution at equilibrium.

I.

and the colour change when concentrated hydrochloric acid is added.

[2]
II.

and the effect of heating the equilibrium mixture on the value of KK.

[2]
B

whether adding water as solvent should be included in the equilibrium expression for this reaction.

[2]

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Question 41
SL • Paper 2
Hard
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SL • Paper 2
Hard
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Nitrogen dioxide and dinitrogen tetroxide form an equilibrium in a sealed gas syringe.

N2O4(g)⇌2NO2(g)N_2O_4(g)\rightleftharpoons 2NO_2(g)

N2O4N_2O_4 is colourless and NO2NO_2 is brown.

A

The plunger is pushed in at constant temperature.

I.

the direction of the equilibrium shift after compression and your answer.

[2]
II.

the effect of compression on the value of KK.

[1]
B

the statement: "The mixture becomes less brown after compression because compression removes NO2NO_2 molecules from the syringe."

[3]
Question 42
SL • Paper 2
Hard
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SL • Paper 2
Hard
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A gas XX is used in an industrial absorption process represented by the heterogeneous equilibrium:

X(g)⇌X(aq)X(g)\rightleftharpoons X(aq)

The dissolution of XX is exothermic.

A

Pressure and temperature are changed independently.

I.

the effect of increasing pressure on the amount of X(aq)X(aq) at equilibrium.

[2]
II.

the effect of decreasing temperature on the value of KK for the equilibrium as written.

[2]
B

the proposal that the absorption process should use high pressure, low temperature and a catalyst to maximize the equilibrium amount of dissolved XX.

[3]
Question 43
HL • Paper 2
Hard
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HL • Paper 2
Hard
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At 700 K700\ \text{K}, the equilibrium constant for the reaction below is K=4.0K=4.0.

CO(g)+H2O(g)⇌CO2(g)+H2(g)CO(g)+H_2O(g)\rightleftharpoons CO_2(g)+H_2(g)

At one instant, [CO]=0.50 mol dm−3[CO]=0.50\ \text{mol dm}^{-3}, [H2O]=0.40 mol dm−3[H_2O]=0.40\ \text{mol dm}^{-3}, [CO2]=0.60 mol dm−3[CO_2]=0.60\ \text{mol dm}^{-3} and [H2]=0.20 mol dm−3[H_2]=0.20\ \text{mol dm}^{-3}.

A

The reaction quotient is used before equilibrium is reached.

I.

QQ for the mixture at this instant.

[2]
II.

the direction in which the reaction proceeds to reach equilibrium.

[2]
B

why comparing QQ with a value of KK measured at a different temperature would be invalid.

[2]
Question 44
HL • Paper 2
Hard
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HL • Paper 2
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Methanoic acid is a weak acid. At 298 K298\ \text{K}, Ka=1.8×10−4K_a=1.8\times10^{-4} for:

HCOOH(aq)⇌H+(aq)+HCOO−(aq)HCOOH(aq)\rightleftharpoons H^+(aq)+HCOO^-(aq)

A solution initially contains 0.100 mol dm−30.100\ \text{mol dm}^{-3} methanoic acid.

A

The small-KK approximation may be used.

I.

why [HCOOH]initial≈[HCOOH]eqm[HCOOH]_{initial}\approx[HCOOH]_{eqm} is reasonable.

[2]
II.

[H+][H^+] and the pH of the solution using this approximation.

[3]
B

how the magnitude of KaK_a relates to acid strength.

[2]

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Question 45
HL • Paper 2
Hard
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HL • Paper 2
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At 298 K298\ \text{K}, the equilibrium constant for a reaction is K=3.2×10−5K=3.2\times10^{-5}. The standard Gibbs energy change is related to KK by

ΔG∘=−RTln⁡K\Delta G^\circ=-RT\ln K

A

The thermodynamic position of equilibrium is considered.

I.

ΔG∘\Delta G^\circ in kJ mol−1\text{kJ mol}^{-1} using R=8.31 J K−1mol−1R=8.31\ \text{J K}^{-1}\text{mol}^{-1}.

[3]
II.

which side of the equilibrium is favoured under standard conditions.

[1]
B

a student's claim that a positive ΔG∘\Delta G^\circ means the reaction can never form products.

[2]
Question 46
HL • Paper 2
Hard
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HL • Paper 2
Hard
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At a particular temperature, K=1.80K=1.80 for the equilibrium:

A(g)+B(g)⇌2C(g)A(g)+B(g)\rightleftharpoons 2C(g)

A mixture initially contains 0.600 mol dm−30.600\ \text{mol dm}^{-3} of AA, 0.600 mol dm−30.600\ \text{mol dm}^{-3} of BB and no CC.

A

The mixture is allowed to reach equilibrium.

I.

Set up the equilibrium concentrations in terms of xx, where xx is the decrease in [A][A].

[2]
II.

the equilibrium concentration of CC.

[3]
B

whether the equilibrium mixture is product-favoured based on the value of KK and the calculated composition.

[2]
Question 47
HL • Paper 2
Hard
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HL • Paper 2
Hard
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At 350 K350\ \text{K}, a reaction has ΔG∘=−18.0 kJ mol−1\Delta G^\circ=-18.0\ \text{kJ mol}^{-1} for the equation:

D(g)⇌E(g)D(g)\rightleftharpoons E(g)

A

The equilibrium constant is calculated from the Gibbs energy change.

I.

KK for the reaction at 350 K350\ \text{K}.

[3]
II.

KK for E(g)⇌D(g)E(g)\rightleftharpoons D(g) at the same temperature.

[1]
B

how the signs and magnitudes of ΔG∘\Delta G^\circ and KK the position of equilibrium.

[3]
Question 48
HL • Paper 2
Hard
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HL • Paper 2
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At a fixed temperature, K=0.250K=0.250 for the homogeneous equilibrium:

2X(g)⇌Y(g)+Z(g)2X(g)\rightleftharpoons Y(g)+Z(g)

At equilibrium, [X]=0.300 mol dm−3[X]=0.300\ \text{mol dm}^{-3}, [Y]=0.150 mol dm−3[Y]=0.150\ \text{mol dm}^{-3} and [Z]=0.150 mol dm−3[Z]=0.150\ \text{mol dm}^{-3}. The mixture was prepared initially with only XX and YY present.

A

The equilibrium data are checked and then used to infer the initial mixture.

I.

Verify that the stated equilibrium concentrations are consistent with K=0.250K=0.250.

[2]
II.

the initial concentrations of XX and YY.

[4]
B

the conclusion that because K<1K<1, very little YY and ZZ can be present at equilibrium.

[2]

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R2.2 How fast? The rate of chemical change

R3.1 Proton transfer reactions