Equilibrium constant KpAQA A-Level Chemistry: Topic test

20 questions, 54 marks

AQA A-Level Chemistry

Equilibrium constant Kp topic test

Total 54 marks

Name

Class

Date

1
Hydrogen is produced industrially by the water–gas shift reaction: CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g), ΔH = −41 kJ mol⁻¹. The reaction is carried out in a sealed reactor with an iron-based catalyst.
(a)
Which expression is Kp for this reaction?
[1 mark]
AKp = (pCO × pH₂O) / (pCO₂ × pH₂)
BKp = (pCO₂ + pH₂) / (pCO + pH₂O)
CKp = (pCO₂ × pH₂) / (pCO × pH₂O)
DKp = ([CO₂][H₂]) / ([CO][H₂O])
(b)
The total pressure in the reactor is increased at constant temperature. What happens?
[1 mark]
AThe position of equilibrium does not change because there are equal numbers of moles of gas on each side
BThe equilibrium shifts to the right because the forward reaction is exothermic
CThe equilibrium shifts to the left because the pressure has increased
DThe value of Kp increases because the partial pressures increase
(c)
Explain the effect of increasing the temperature on the value of Kp and on the equilibrium yield of hydrogen.
[2 marks]
Total for question 1: 4 marks
2
Methane reacts with steam in a reformer: CH₄(g) + H₂O(g) ⇌ CO(g) + 3H₂(g). In an experiment at constant temperature and a total pressure of 200 kPa, the equilibrium mixture contained 0.20 mol CH₄, 0.30 mol H₂O, 0.10 mol CO and 0.30 mol H₂.
(a)
What is the partial pressure of CH₄ in the equilibrium mixture?
[1 mark]
A22 kPa
B40 kPa
C67 kPa
D44 kPa
(b)
What are the units of Kp for this reaction?
[1 mark]
AkPa⁻²
BkPa²
CkPa⁴
DNo units
(c)
The total pressure of the reformer is increased at constant temperature. Predict and explain the effect on the equilibrium yield of hydrogen and on the value of Kp.
[2 marks]
Total for question 2: 4 marks
3
Ethanol is made by the hydration of ethene: C₂H₄(g) + H₂O(g) ⇌ C₂H₅OH(g). A mixture of 1.00 mol of ethene and 1.00 mol of steam is heated in a sealed vessel at constant temperature with an acid catalyst. At equilibrium the vessel contains 0.250 mol of ethanol and the total pressure is 6.50 MPa.
(a)
Calculate the amounts of each gas at equilibrium and the partial pressure of each gas in MPa.
[3 marks]
(b)
Write the expression for Kp and calculate its value at this temperature, including units.
[4 marks]
Total for question 3: 7 marks
4
Methanol is made from synthesis gas: CO(g) + 2H₂(g) ⇌ CH₃OH(g), ΔH = −91 kJ mol⁻¹. A mixture of 1.00 mol of carbon monoxide and 2.00 mol of hydrogen is allowed to reach equilibrium over a copper-based catalyst at constant temperature and a total pressure of 5000 kPa. At equilibrium 0.600 mol of methanol is present.
(a)
Calculate the value of Kp at this temperature, including units. Show how you obtain the equilibrium partial pressures.
[6 marks]
(b)
Explain how changing (i) the temperature, (ii) the pressure and (iii) the use of a catalyst affects the rate of reaction, the equilibrium yield of methanol and the value of Kp. Suggest why a compromise temperature is used industrially.
[6 marks]
Total for question 4: 12 marks
5
Phosgene dissociates on heating: COCl₂(g) ⇌ CO(g) + Cl₂(g). At a certain temperature the equilibrium mixture, at a total pressure of 150 kPa, has mole fractions of 0.20 for COCl₂, 0.40 for CO and 0.40 for Cl₂.
(a)
What is the partial pressure of CO in the equilibrium mixture?
[1 mark]
A30 kPa
B60 kPa
C40 kPa
D150 kPa
(b)
What is the value of Kp for this reaction at this temperature?
[1 mark]
A0.0083 kPa⁻¹
B0.80
C3600 kPa²
D120 kPa
(c)
When the temperature is raised, the value of Kp increases. State whether the forward reaction is exothermic or endothermic and explain your answer.
[2 marks]
Total for question 5: 4 marks
6
In a car engine at about 2000 K, nitrogen and oxygen from the air react: N₂(g) + O₂(g) ⇌ 2NO(g), ΔH = +180 kJ mol⁻¹. In one equilibrium mixture at 2000 K the partial pressures are N₂ 78.0 kPa, O₂ 20.0 kPa and NO 1.20 kPa.
(a)
What are the units of Kp for this reaction?
[1 mark]
AKp has no units, because the number of moles of gas is the same on each side
BkPa
CkPa⁻¹
DkPa²
(b)
The temperature of the engine is increased. What happens to the value of Kp?
[1 mark]
AIt decreases, because the forward reaction is endothermic
BIt stays the same, because the pressure has not changed
CIt increases, because the forward reaction is endothermic
DIt stays the same, because there are equal moles of gas on each side
(c)
Calculate the value of Kp for this equilibrium at 2000 K. Give your answer to two significant figures.
[2 marks]
Total for question 6: 4 marks
7
Nitrogen monoxide is oxidised in air: 2NO(g) + O₂(g) ⇌ 2NO₂(g). A mixture of 2.00 mol of NO and 1.00 mol of O₂ is sealed in a vessel and reaches equilibrium at constant temperature and a total pressure of 300 kPa. At equilibrium 1.20 mol of NO₂ is present.
(a)
Calculate the amounts of each gas at equilibrium and the partial pressure of each gas in kPa.
[3 marks]
(b)
Write the expression for Kp and calculate its value, including units.
[4 marks]
Total for question 7: 7 marks
8
Ethene is hydrogenated over a nickel catalyst: C₂H₄(g) + H₂(g) ⇌ C₂H₆(g), ΔH = −137 kJ mol⁻¹. At one temperature, the equilibrium partial pressures of ethene and of hydrogen are each 50.0 kPa. At this temperature Kp = 0.0300, with units to be deduced.
(a)
Deduce the units of Kp. Calculate the partial pressure of ethane and the mole fraction of ethane in the equilibrium mixture.
[6 marks]
(b)
The equilibrium mixture described in part (a) is compressed at constant temperature so that its total pressure doubles instantly. Use calculations to explain how the composition of the mixture changes afterwards, and state what happens to the value of Kp.
[6 marks]
Total for question 8: 12 marks

End of questions

Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).