Edexcel A-Level Chemistry Paper 1, June 2017: Question 10
12 marks · Medium difficulty · Calculations
Analyze equilibrium properties including Kp expression, Le Chatelier's principle, Kp calculation, and reaction quotient for reversible gas-phase reactions involving hydrogen production and carbon monoxide.
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Question text
10 Hydrogen is produced on a large scale by several different processes.
(a) One process for producing hydrogen involves reacting white-hot carbon with steam.
C(s) + H O(g) U H (g) + CO(g) ¨H = +131 kJ mol–1
The expression for the equilibrium constant, Kp, is
p(H2) p(CO)
Kp = —
p(H2O)
(i) Give a reason why the partial pressure of carbon is not included in the expression.
(1)
(ii) Explain the effect of an increase in pressure on the equilibrium position of this reaction.
(2)
(iii) Explain, by reference to any change in the value of Kp, the effect of an increase
in temperature on the equilibrium position of this reaction.
(2)
(iv) At 1000 K and a total pressure of 2.0 atm, 1.00 mol of steam reacted with excess carbon.
At equilibrium, 0.81 mol of hydrogen was present.
Calculate the value of Kp at 1000 K, stating any units.
(4)
(b) Carbon monoxide reacts with steam.
*P48058A02428*CO(g) + H2O(g)UCO2(g) + H2(g)
At 1100 K, Kc = 1.00
In an experiment, 1 mol of carbon monoxide was mixed with 1 mol of steam,
2 mol of carbon dioxide and 2 mol of hydrogen.
Deduce, with reasons, the direction in which the reaction will shift to reach equilibrium.
(3)
(Total for Question 10 = 12 marks)
Mark scheme
Show the mark scheme
Question
Answer Additional Guidance Mark
Number
10(a)(i) An answer that makes reference to the following point: (1)
carbon / solid has no (vapour / partial) pressure Allow the reaction is heterogeneous and
or (partial) pressure of a pure solid is not
(partial) pressure of carbon / solid is constant included (in Kp expression)
or
carbon does not contribute to the overall pressure (of Do not allow just ‘because carbon is a
the system) solid’ or ‘carbon is not a gas’
Question
Answer Additional Guidance Mark
Number
10(a)(ii) An explanation that makes reference to the following (2)
points:
there are fewer moles / molecules / particles of gas on Allow 2 moles / molecules of gas on
the left / reactant side (1) right and 1 mole / molecule on left
so equilibrium position/ it moves / shifts to the left / M2 is conditional on M1 or the idea of
reactant side (1) fewer particles on the left / decreasing
the value of the quotient / Q
Do not allow any indication of Kp
changing
Question
Answer Additional Guidance Mark
Number
10(a)(iii) An explanation that makes reference to the following points: (2)
(forward) reaction is endothermic
and Ignore references to ∆G and ∆S
so equilibrium constant / Kp increases as temperature
increases (1)
so equilibrium position / it moves / shifts to the right / M2 is conditional on M1 or
product side (1) endothermic or equilibrium constant
increases
Question
Answer Additional Guidance Mark
Number
10(a)(iv) Example of calculation: (4)
H2O(g) H2(g) CO(g)
Initial 1.00 0 0
moles
calculation of moles of each substance at Eqm 1.00 – 0.81 0.81 0.81
equilibrium (1) moles = 0.19
Total moles = 0.19 + 0.81 + 0.81 = 1.81
Mole 0.19/1.81 0.81/1.81 0.81/1.81
fraction = 0.10497 = 0.4475 = 0.4475
Partial 0.10497 x 0.4475 x 0.4475 x
pressure 2.0 2.0 2.0
calculation of partial pressure of each /atm =0.20994 = 0.895 =0.895
substance (1)
Kp = 0.895 x 0.895
0.20994
calculation of Kp (1) = 3.815 / 3.82 / 3.8 atm
units (stand alone mark) (1) 3.8144 / 3.814 / 3.81 / 3.8 atm from 0.105
and 0.210
Correct answer with units but no working
scores (4)
Allow TE for M2 and M3
Ignore SF except 1 SF
Question
Answer Additional Guidance Mark
Number
10(b) Allow amounts / moles / (partial) (3)
pressures for concentrations
the quotient / Q: Allow calculated Kc / the quotient / Q
[CO2][H2] = 2 x 2 = 4, which is larger than Kc will be greater than 1
[CO][H2O] 1 x 1
or
(since Kc = 1) the concentrations of the products must be
equal to the concentrations of the reactants at
equilibrium (1)
the concentrations of CO and H / products need to Allow shift so that there is 1.5 mol of
decrease each substance
and
those of CO and H2O / reactants need to increase (1)
so reaction shifts to the left (1) M3 conditional on some explanation
(Total for Question 10 = 12 marks)
TOTAL FOR PAPER = 90 MARKS
How to answer it
Industrial Production of Hydrogen & Equilibria Study Guide
What this question tests
This question assesses your mastery of chemical equilibria, specifically heterogeneous systems, Le Chatelier's Principle regarding pressure and temperature changes, calculating equilibrium constants (Kp) using ice tables and mole fractions, and predicting the direction of equilibrium using the reaction quotient (Q or Kc comparison).
Excluding Carbon from Kp
✅ Correct Answer
Carbon is a solid, therefore it has no vapor/partial pressure, its partial pressure is constant, or it does not contribute to the overall pressure of the system.
❌ Common Errors
Writing simplistic statements like "because carbon is a solid" or "carbon is not a gas" without explaining the implication on partial pressure or system composition. Examiners require reference to pressure or constant concentration/activity.
Effect of Pressure Increase on Equilibrium Position
✅ Correct Answer
- There are fewer moles of gas on the right (2 moles) compared to the left (1 mole).
- Therefore, the equilibrium shifts to the left (reactant side) to oppose the increase in pressure.
🧠 Exam Technique
Always structure Le Chatelier explanations in two steps: (1) Compare gas moles on each side, and (2) State the direction of shift and how it opposes the change. Note that M2 is conditional on M1.
Effect of Temperature Increase on Equilibrium
✅ Correct Answer
- The forward reaction is endothermic (ΔH is positive), so increasing temperature increases the value of Kp.
- The equilibrium position shifts to the right (product side) to absorb the added heat.
💡 Key Knowledge
Temperature is the only factor that changes the value of the equilibrium constant (Kp or Kc). Pressure changes alter equilibrium position via the quotient Q, but Kp remains constant at a fixed temperature.
Calculating Kp and Units
📐 Step-by-Step Calculation
Step 1: Construct an ICE Table (Moles)
• Steam (H₂O): Initial = 1.00 mol, Eqm = 1.00 - 0.81 = 0.19 mol
• Hydrogen (H₂): Initial = 0 mol, Eqm = 0.81 mol
• Carbon Monoxide (CO): Initial = 0 mol, Eqm = 0.81 mol
• Total moles at equilibrium = 0.19 + 0.81 + 0.81 = 1.81 mol
Step 2: Calculate Mole Fractions
• H₂O = 0.19 / 1.81 = 0.10497
• H₂ = 0.81 / 1.81 = 0.4475
• CO = 0.81 / 1.81 = 0.4475
Step 3: Calculate Partial Pressures (Total P = 2.0 atm)
• p(H₂O) = 0.10497 × 2.0 = 0.2099 atm
• p(H₂) = 0.4475 × 2.0 = 0.895 atm
• p(CO) = 0.4475 × 2.0 = 0.895 atm
Step 4: Expression, Kp Value, and Units
• Kp = (p(H₂) × p(CO)) / p(H₂O)
• Kp = (0.895 × 0.895) / 0.2099 = 3.82 atm (Accept 3.8 to 3.82)
❌ Common Calculation Traps
- Unit Errors: For this expression, units are atm . Missing or incorrect units lose the final standalone mark.
- ICE Math: Forgetting to subtract the reacted moles from initial steam to find equilibrium moles.
Deducing Equilibrium Shift using Reaction Quotient (Q)
✅ Correct Answer
- Calculate the reaction quotient Q:
Q = ([CO₂][H₂]) / ([CO][H₂O]) = (2 × 2) / (1 × 1) = 4 - Since Q (4) is greater than Kc (1.00), the concentrations of products are too high relative to reactants.
- Therefore, the reaction shifts to the left (reactant side) to reach equilibrium.
🧠 Exam Technique & Examiner Commentary
Top-level responses immediately calculate Q (or compare the actual ratio to Kc) and explicitly state whether product or reactant concentrations need to decrease. Make sure to clearly link numerical evidence to the direction of the shift.
Topics
Physical Chemistry · Topic 10: Equilibrium I · Topic 11: Equilibrium II
Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 1, June 2017. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.