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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.

Verified by Dennis M.
Verified by Dennis M.
Paper
Difficulty
Status
Level
Question 1
SL • Paper 1A
Easy
Calculator Permitted

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
Easy
Calculator Permitted

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
Easy
Calculator Permitted

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
Easy
Calculator Permitted

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.

2.5×1042.5\times10^{-4}

B.

1.6×1071.6\times10^7

C.

4.0×1034.0\times10^3

D.

4.0×1034.0\times10^{-3}

Question 5
SL • Paper 1A
Easy
Calculator Permitted

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
Easy
Calculator Permitted

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 sealed flask containing a small layer of liquid at the bottom and coloured vapour above it. The diagram should show particles moving between liquid and vapour with arrows in both directions, but should not label the rates or state that equilibrium has been reached.
A

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

[1]
Write your answer here...
B

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

[2]
Write your answer here...

0

Question 7
SL • Paper 2
Easy
Calculator Permitted

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

Deduce the equilibrium constant expression for the reaction as written.

[1]
Write your answer here...
B

At a fixed temperature the value of KK for the reaction as written is 3.20×1033.20 \times 10^{-3}. Calculate the value of KK for the reverse reaction at the same temperature and state which side is favoured for the reaction as written.

[2]
Write your answer here...

0

Question 8
SL • Paper 2
Easy
Calculator Permitted

Two weak monoprotic acids have the following acid dissociation constants at the same temperature.

Ka(HA)=1.8×105K_a(HA) = 1.8 \times 10^{-5}

Ka(HB)=6.3×1010K_a(HB) = 6.3 \times 10^{-10}

A

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

[2]
Write your answer here...
B

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

[1]
Write your answer here...

0

Question 9
SL • Paper 1A
Medium
Calculator Permitted

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
Medium
Calculator Permitted

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 dm3[H_2]=0.20\ \text{mol dm}^{-3}, [I2]=0.30 mol dm3[I_2]=0.30\ \text{mol dm}^{-3} and [HI]=0.60 mol dm3[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
Medium
Calculator Permitted

For the homogeneous equilibrium

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

1.00 mol dm31.00\ \text{mol dm}^{-3} of AA is placed in a sealed container. At equilibrium, [A]=0.64 mol dm3[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
Medium
Calculator Permitted

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 dm3[A]=0.20\ \text{mol dm}^{-3} and [B]=0.10 mol dm3[B]=0.10\ \text{mol dm}^{-3}.

What is [C][C] at equilibrium?

A.

0.020 mol dm30.020\ \text{mol dm}^{-3}

B.

500 mol dm3500\ \text{mol dm}^{-3}

C.

2.0 mol dm32.0\ \text{mol dm}^{-3}

D.

0.20 mol dm30.20\ \text{mol dm}^{-3}

Question 13
HL • Paper 1A
Medium
Calculator Permitted

For a reaction at 298 K298\ \text{K}, K=1.0×102K=1.0\times10^{-2}.

Using ΔG=RTlnK\Delta G^\circ=-RT\ln K, what is ΔG\Delta G^\circ?

A.

11.4 kJ mol1-11.4\ \text{kJ mol}^{-1}

B.

+11.4 kJ mol1+11.4\ \text{kJ mol}^{-1}

C.

5.70 kJ mol1-5.70\ \text{kJ mol}^{-1}

D.

+5.70 kJ mol1+5.70\ \text{kJ mol}^{-1}

Question 14
SL • Paper 2
Medium
Calculator Permitted

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 CrO42CrO_4^{2-}, orange Cr2O72Cr_2O_7^{2-}

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

A

Deduce the equilibrium constant expression for this reaction, omitting water.

[1]
Write your answer here...
B

Predict and explain the colour change after adding sodium hydroxide.

[2]
Write your answer here...
C

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

[1]
Write your answer here...

0

Question 15
SL • Paper 2
Medium
Calculator Permitted

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

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

[1]
Write your answer here...
B

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

[2]
Write your answer here...

0

Question 16
SL • Paper 2
Medium
Calculator Permitted

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

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

[3]
Write your answer here...
B

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

[1]
Write your answer here...

0

Question 17
HL • Paper 2
Medium
Calculator Permitted

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 dm3[H_2] = 0.0200\ \text{mol dm}^{-3}, [I2]=0.0100 mol dm3[I_2] = 0.0100\ \text{mol dm}^{-3} and [HI]=0.0600 mol dm3[HI] = 0.0600\ \text{mol dm}^{-3}.

A

Calculate the reaction quotient, QQ.

[2]
Write your answer here...
B

Determine the direction in which the reaction proceeds to reach equilibrium.

[1]
Write your answer here...

0

Question 18
HL • Paper 2
Medium
Calculator Permitted

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 dm3[A] = 0.500\ \text{mol dm}^{-3} and [B]=0.500 mol dm3[B] = 0.500\ \text{mol dm}^{-3}, with no CC present. At equilibrium, [C]=0.300 mol dm3[C] = 0.300\ \text{mol dm}^{-3}.

A

Determine the equilibrium concentrations of AA and BB.

[1]
Write your answer here...
B

Calculate the value of KK at this temperature.

[2]
Write your answer here...

0

Question 19
SL • Paper 1B
Medium
Calculator Permitted

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.

Rates of evaporation and condensation of bromine in a sealed vessel over time.
A

Identify the time at which dynamic equilibrium is first reached.

[1]
Write your answer here...
B

Describe two characteristics of the system after this time.

[2]
Write your answer here...
C

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

[1]
Write your answer here...

0

Question 20
HL • Paper 1A
Medium
Calculator Permitted

A weak acid dissociates according to

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

For 0.100 mol dm30.100\ \text{mol dm}^{-3} HAHA, Ka=1.6×105K_a=1.6\times10^{-5}. Using the small-KK approximation, what is the equilibrium concentration of H+(aq)H^+(aq)?

A.

1.3×103 mol dm31.3\times10^{-3}\ \text{mol dm}^{-3}

B.

4.0×102 mol dm34.0\times10^{-2}\ \text{mol dm}^{-3}

C.

1.6×106 mol dm31.6\times10^{-6}\ \text{mol dm}^{-3}

D.

1.6×104 mol dm31.6\times10^{-4}\ \text{mol dm}^{-3}

Question 21
HL • Paper 1A
Medium
Calculator Permitted

For a reaction at 298 K298\ \text{K}, ΔG=17.1 kJ mol1\Delta G^\circ=-17.1\ \text{kJ mol}^{-1}.

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

A.

1.0×1031.0\times10^{-3}

B.

1.0×1011.0\times10^{1}

C.

1.0×1011.0\times10^{-1}

D.

1.0×1031.0\times10^{3}

Question 22
HL • Paper 2
Medium
Calculator Permitted

A 0.200 mol dm30.200\ \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=1.6×105K_a = 1.6 \times 10^{-5}

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

A

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

[2]
Write your answer here...
B

Calculate the pH of the solution.

[1]
Write your answer here...
C

State why the approximation is reasonable in this calculation.

[1]
Write your answer here...

0

Question 23
HL • Paper 2
Medium
Calculator Permitted

For a reaction at 298 K298\ \text{K}, the standard Gibbs energy change is +12.4 kJ mol1+12.4\ \text{kJ mol}^{-1}.

Use ΔG=RTlnK\Delta G^\circ = -RT\ln K and R=8.31 J K1 mol1R = 8.31\ \text{J K}^{-1}\ \text{mol}^{-1}.

A

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

[2]
Write your answer here...
B

State whether reactants or products are favoured under standard conditions.

[1]
Write your answer here...

0

Question 24
HL • Paper 2
Medium
Calculator Permitted

At 350 K350\ \text{K}, a homogeneous equilibrium has K=2.50×104K = 2.50 \times 10^4.

Use ΔG=RTlnK\Delta G^\circ = -RT\ln K and R=8.31 J K1 mol1R = 8.31\ \text{J K}^{-1}\ \text{mol}^{-1}.

A

Calculate ΔG\Delta G^\circ in kJ mol1\text{kJ mol}^{-1}.

[2]
Write your answer here...
B

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

[1]
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0

Question 25
SL • Paper 1B
Medium
Calculator Permitted

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.

SpeciesEquilibrium concentration / mol dm^-3
NOCl0.100
NO0.300
Cl20.150
A

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

[2]
Write your answer here...
B

Calculate KK using the equilibrium concentrations in the table.

[1]
Write your answer here...
C

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

[1]
Write your answer here...

0

Question 26
SL • Paper 1B
Medium
Calculator Permitted

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)

Equilibrium constant for methanol formation vs temperature.
A

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

[1]
Write your answer here...
B

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

[1]
Write your answer here...
C

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

[2]
Write your answer here...

0

Question 27
SL • Paper 1B
Medium
Calculator Permitted

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.

Absorbance changes over time after two additions.
A

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

[2]
Write your answer here...
B

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

[1]
Write your answer here...
C

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

[1]
Write your answer here...

0

Question 28
HL • Paper 1B
Medium
Calculator Permitted

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] / mol dm^-3[I2] / mol dm^-3[HI] / mol dm^-3
A0.2000.2000.400
B0.05000.1000.500
A

Calculate the reaction quotient, QQ, for mixture A.

[1]
Write your answer here...
B

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

[2]
Write your answer here...
C

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

[1]
Write your answer here...

0

Question 29
HL • Paper 2
Medium
Calculator Permitted

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 dm3[A] = 0.400\ \text{mol dm}^{-3} and [B]=0.0200 mol dm3[B] = 0.0200\ \text{mol dm}^{-3}.

A

Calculate QQ for the mixture at this instant.

[2]
Write your answer here...
B

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

[2]
Write your answer here...

0

Question 30
SL • Paper 1B
Hard
Calculator Permitted

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.

An annotated stimulus showing a sealed gas syringe containing carbon dioxide gas above water. The plunger is shown before and after being pushed in, with constant temperature indicated. A small inset graph shows dissolved carbon dioxide concentration increasing after the pressure increase and then becoming constant at a new equilibrium.
A

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

[1]
Write your answer here...
B

Explain the change using Le Châtelier's principle.

[2]
Write your answer here...
C

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

[1]
Write your answer here...

0

Question 31
SL • Paper 1B
Hard
Calculator Permitted

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 mol1\Delta H_r^\circ=+57\ \text{kJ mol}^{-1}. The diagrams show the colour intensity of sealed tubes under different conditions.

A set of annotated sealed tubes for the nitrogen dioxide dinitrogen tetroxide equilibrium. Tubes are labelled low temperature, high temperature, no catalyst and catalyst. Brown colour intensity is shown qualitatively, with high temperature darker than low temperature. A small concentration-time inset compares catalysed and uncatalysed samples reaching the same final equilibrium composition at different times.
A

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

[2]
Write your answer here...
B

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

[2]
Write your answer here...
C

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

[1]
Write your answer here...

0

Question 32
HL • Paper 1B
Hard
Calculator Permitted

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.

SpeciesInitial concentration / mol dm^-3Equilibrium concentration / mol dm^-3
A(g)0.500
B(g)0.500
C(g)0.0000.250
A

Complete the equilibrium concentrations of AA, BB and CC.

[2]
Write your answer here...
B

Calculate KK for the equilibrium.

[1]
Write your answer here...
C

State whether products or reactants are favoured by this equilibrium constant.

[1]
Write your answer here...

0

Question 33
HL • Paper 1B
Hard
Calculator Permitted

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=1.8×105K_a=1.8\times10^{-5} for HA. The initial concentration of HA is 0.100 mol dm30.100\ \text{mol dm}^{-3}.

StateHA / mol dm^-3H+ / mol dm^-3A- / mol dm^-3
Initial0.10000
Change-x+x+x
Equilibrium0.100 - xxx
A

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

[2]
Write your answer here...
B

Calculate the pH of the solution.

[1]
Write your answer here...
C

Justify the use of the approximation for this solution.

[1]
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0

Question 34
HL • Paper 1B
Hard
Calculator Permitted

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=RTlnK\Delta G^\circ=-RT\ln K

QuantityValueUnits
Equilibrium constant, K3.2 × 10^4dimensionless
Temperature, T298K
Gas constant, R8.31J mol^-1 K^-1
A

Calculate ΔG\Delta G^\circ in kJ mol1\text{kJ mol}^{-1} at 298 K298\ \text{K}.

[2]
Write your answer here...
B

State which side of the equilibrium is favoured under standard conditions.

[1]
Write your answer here...
C

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

[1]
Write your answer here...

0

Question 35
SL • Paper 2
Hard
Calculator Permitted

A small amount of liquid bromine is placed in a sealed flask. After some time the intensity of the brown vapour becomes constant.

A sealed flask containing liquid bromine at the bottom and brown bromine vapour above it, with a time sequence indicating increasing then constant vapour colour intensity without showing particle-rate labels.
A

The bromine system reaches a physical equilibrium.

I.

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

[2]
Write your answer here...
II.

State one macroscopic observation that indicates equilibrium has been reached.

[1]
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B

Compare and contrast this physical equilibrium with a chemical equilibrium.

[3]
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0

Question 36
SL • Paper 2
Hard
Calculator Permitted

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.

Deduce the equilibrium constant expression for the reaction as written.

[1]
Write your answer here...
II.

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

[2]
Write your answer here...
B

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

[3]
Write your answer here...

0

Question 37
SL • Paper 2
Hard
Calculator Permitted

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.

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

[2]
Write your answer here...
II.

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

[2]
Write your answer here...
B

Discuss the effect of increasing temperature on the Haber equilibrium.

[2]
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0

Question 38
HL • Paper 1B
Hard
Calculator Permitted

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.

Reaction quotient Q vs time for two starting conditions; K = 12.
A

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

[1]
Write your answer here...
B

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

[2]
Write your answer here...
C

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

[2]
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0

Question 39
HL • Paper 1B
Hard
Calculator Permitted

For the equilibrium below at 298 K298\ \text{K}, the standard Gibbs energy change is +4.80 kJ mol1+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.

ItemValueUnits
ReactionN2O4(g) ⇌ 2NO2(g)
T298K
ΔG°+4.80kJ mol^-1
Q (prepared mixture)0.050dimensionless
A

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

[2]
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B

Determine KK for the reverse reaction at the same temperature.

[1]
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C

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

[1]
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Question 40
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.

Predict and explain the colour change when concentrated hydrochloric acid is added.

[2]
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II.

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

[2]
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B

Discuss 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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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 sealed gas syringe containing a brown gas mixture before and after the plunger is pushed in, showing reduced volume but not indicating the final colour or equilibrium direction.
A

The plunger is pushed in at constant temperature.

I.

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

[2]
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II.

State the effect of compression on the value of KK.

[1]
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B

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

[3]
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Question 42
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 closed vessel with gas $X$ above an aqueous solution containing dissolved $X$, with labels for gas phase, aqueous phase, and a piston to change pressure.
A

Pressure and temperature are changed independently.

I.

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

[2]
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II.

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

[2]
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B

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

[3]
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Question 43
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 dm3[CO]=0.50\ \text{mol dm}^{-3}, [H2O]=0.40 mol dm3[H_2O]=0.40\ \text{mol dm}^{-3}, [CO2]=0.60 mol dm3[CO_2]=0.60\ \text{mol dm}^{-3} and [H2]=0.20 mol dm3[H_2]=0.20\ \text{mol dm}^{-3}.

A

The reaction quotient is used before equilibrium is reached.

I.

Calculate QQ for the mixture at this instant.

[2]
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II.

Deduce the direction in which the reaction proceeds to reach equilibrium.

[2]
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B

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

[2]
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Question 44
HL • Paper 2
Hard
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Methanoic acid is a weak acid. At 298 K298\ \text{K}, Ka=1.8×104K_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 dm30.100\ \text{mol dm}^{-3} methanoic acid.

A

The small-KK approximation may be used.

I.

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

[2]
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II.

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

[3]
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B

Discuss 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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At 298 K298\ \text{K}, the equilibrium constant for a reaction is K=3.2×105K=3.2\times10^{-5}. The standard Gibbs energy change is related to KK by

ΔG=RTlnK\Delta G^\circ=-RT\ln K

A

The thermodynamic position of equilibrium is considered.

I.

Calculate ΔG\Delta G^\circ in kJ mol1\text{kJ mol}^{-1} using R=8.31 J K1mol1R=8.31\ \text{J K}^{-1}\text{mol}^{-1}.

[3]
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II.

Deduce which side of the equilibrium is favoured under standard conditions.

[1]
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B

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

[2]
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Question 46
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 dm30.600\ \text{mol dm}^{-3} of AA, 0.600 mol dm30.600\ \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]
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II.

Calculate the equilibrium concentration of CC.

[3]
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B

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

[2]
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Question 47
HL • Paper 2
Hard
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At 350 K350\ \text{K}, a reaction has ΔG=18.0 kJ mol1\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.

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

[3]
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II.

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

[1]
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B

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

[3]
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Question 48
HL • Paper 2
Hard
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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 dm3[X]=0.300\ \text{mol dm}^{-3}, [Y]=0.150 mol dm3[Y]=0.150\ \text{mol dm}^{-3} and [Z]=0.150 mol dm3[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]
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II.

Determine the initial concentrations of XX and YY.

[4]
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B

Evaluate 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