Edexcel A-Level Chemistry Paper 3, June 2023: Question 10
15 marks · Hard difficulty · Calculations
Calculate entropy changes, Gibbs free energy sign, and Kp for reactions involved in the Mond Process of nickel purification, and explain entropy changes qualitatively.
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Question text
10 The Mond Process is an industrial method of purifying nickel.
(a) The first step involves the reaction of nickel oxide with hydrogen gas at 473K.
NiO(s) + H2(g) → Ni(s) + H2O(g)
The nickel is not pure because the impurities also react with the hydrogen gas.
Complete the electronic configuration of the Ni2+ ion.
(1)
1s2
(b) The second step involves passing carbon monoxide over impure nickel at 323K.
The impurities do not react. The nickel reaction is
Ni(s) + 4CO(g) → Ni(CO) (g) Δ H d = –191 kJ mol–1
4 r
(i) Calculate the total entropy change, ΔS d , for this reaction.
total
Include a sign and units in your answer.
Substance S d / J mol–1 K–1
Ni(s) +29.9
CO(g) +197.6
Ni(CO)4(g) +313.4
(5)
(ii) Predict the sign of the Gibbs Free Energy change, ΔG, for this reaction and
justify your choice. No calculation is required.
(1)
(iii) 50.0molof carbon monoxide is mixed with excess impure solid nickel at 323K
in an industrial reactor.
At equilibrium, 0.750molof carbon monoxide remains. The pressure is
maintained at 1.5atm throughout.
Calculate the value of Kp at 323K. Include units with your answer.
(6)
(c) The final stage of the Mond Process is the thermal decomposition of the
nickel carbonyl gas, Ni(CO)*P71914A02932*4, to give pure nickel and carbon monoxide.
The reaction mixture is heated to 523K.
Explain, in qualitative terms, why the entropy change of the system, ΔS d , for
system
this decomposition reaction is positive.
(2)
(Total for Question 10 = 15 marks)
Mark scheme
Show the mark scheme
Question
Answer Additional Guidance Mark
Number
(1)
10(a) • 1s22s22p63s23p63d8 Allow omission of superscripts
Allow [Ar] for 1s22s22p63s23p6
Allow 2p and 3p split into x, y and z
Ignore 4s0
Question
Answer Additional Guidance Mark
Number
Example of calculation (5)
10(b)(i)
• expression for entropies of reactants and products (1) ∆Ssystem = (313.4) – ((4 × 197.6) + 29.9)
• calculation of ∆S (1) ∆S = − 506.9 (J mol−1 K−1)
system system
• expression of ∆Ssurroundings (1) ∆Ssurroundings = − (∆H ÷ T) = −(−191 000 ÷ 323)
∆S = (+) 591.3 (J mol−1 K−1)
• calculation of ∆Ssurroundings (1) surroundings
∆S = (591.3 − 506.9) = +84.4 J mol−1 K−1
• calculation of ∆Stotal with sign and units (1) total
Ignore SF except 1SF
Allow +0.0844 kJ mol−1 K−1
Accept units in any order
Allow mol− for mol−1
TE throughout
Correct answer with no working scores (5)
Question
Answer Additional Guidance Mark
Number
An answer that makes reference to the following points: (1)
10(b)(ii)
• negative (sign)
and
the reaction is feasible (since it is an industrial process)
or
negative (sign)
and
∆Stotal is positive
(so reaction is feasible because the enthalpy change is negative)
Ignore just ∆G<0 and ∆Stotal >0
Question
Answer Additional Guidance Mark
Number
An example of calculation (6)
10(b)(iii)
• (M1) calculation of equilibrium moles of Ni(CO)4 (1) n(Ni(CO)4) = ((50 – 0.75) ÷ 4 =) 12.3125 (mol)
• (M2) calculation of CO and Ni(CO)4 mole fractions (1) Total number of moles = 0.750 + 12.3125 = 13.0625
ꭓCO = (0.75 ÷ 13.0625 =) 0.057416
ꭓNi(CO)4 = (12.3125 ÷ 13.0625 =) 0.942584
• (M3) calculation of CO and Ni(CO)4 partial pressures (1) p(CO) = 0.057416 × 1.5 = 0.086124 (atm)
p(Ni(CO)4) = 0.942584 × 1.5 = 1.413876 (atm)
• (M4) expression of Kp (1) K = (p(Ni(CO) ) ÷ (p(CO)4))
p 4
Do not award square brackets
• (M5) calculation of Kp (1) K = (1.413876 ÷ 0.0861244)
p
= 25698.9/25699/25700
TE throughout
Ignore SF except 1SF
Correct answer with or without working scores (5)
• (M6) units (1) atm−3
Question
Answer Additional Guidance Mark
Number
An explanation that makes reference to the following points: (2)
10(c)
• number of moles of gases increases (which have greater entropy) (1) Allow number of gaseous moles goes from 1 to 4
Allow particles for moles
Allow more gaseous molecules
Ignore just more molecules
Ignore just equation
Do not allow reference to nickel as molecule(s)
• increase from forming 4CO(g) is larger in magnitude than the Allow comparison such as
decrease from forming solid Ni (1) ‘entropy change is positive even though a solid made’
(Total Question 10 = 15 marks)
How to answer it
The Mond Process & Thermodynamics Study Guide
What this question tests
This synoptic question evaluates transition metal electronic configurations, entropy calculations (system, surroundings, and total), thermodynamic feasibility via Gibbs Free Energy, equilibrium constant expressions and calculations ($K_p$) involving mole fractions and partial pressures, and qualitative reasoning about gas-phase entropy changes.
Part (a) — Electronic Configuration of Ions
Complete the electronic configuration of the Ni²⁺ ion.
✅ Correct Answer
1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁸
1 Mark
💡 Key Knowledge
- Neutral nickel has atomic number 28: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁸ 4s²
- When transition metals form ions, electrons are always removed from the highest energy principal quantum shell first—meaning the 4s electrons leave before the 3d electrons.
❌ Common Errors
Students frequently forget to empty the 4s subshell and instead incorrectly remove electrons from the 3d subshell, writing 3d⁶ 4s² .
Part (b)(i) — Total Entropy Change Calculation
Calculate the total entropy change, ΔS_total, for the reaction. Include a sign and units.
📐 Step-by-Step Calculation
- Calculate system entropy (ΔS_system):
ΔS_system = ΣS(products) - ΣS(reactants)
ΔS_system = 313.4 - (4 × 197.6 + 29.9) = -506.9 J mol⁻¹ K⁻¹ - Calculate surroundings entropy (ΔS_surroundings):
ΔS_surroundings = -(ΔH / T)
Note that ΔH is given in kJ mol⁻¹, so convert to J mol⁻¹ by multiplying by 1000 (-191000 J mol⁻¹).
ΔS_surroundings = -(-191000 / 323) = +591.33 J mol⁻¹ K⁻¹ - Calculate total entropy (ΔS_total):
ΔS_total = ΔS_system + ΔS_surroundings
ΔS_total = -506.9 + 591.33 = +84.4 J mol⁻¹ K⁻¹
5 Marks total (1 for system expression, 1 for system value, 1 for surroundings expression, 1 for surroundings value, 1 for total with sign and units).
🧠 Exam Technique & Traps
- Unit mismatch trap: ΔH is typically given in kJ mol⁻¹ while entropy values are in J mol⁻¹ K⁻¹. You must convert ΔH to joules by multiplying by 1000.
- Stoichiometry: Remember to multiply the entropy of CO by its stoichiometric coefficient (4) when finding ΣS(reactants).
- Units: Ensure your final unit is stated clearly as J mol⁻¹ K⁻¹ .
Part (b)(ii) — Gibbs Free Energy Feasibility
Predict the sign of the Gibbs Free Energy change, ΔG, for this reaction and justify your choice.
✅ Correct Answer
Sign: Negative (ΔG < 0)
Justification: The reaction is feasible (as it occurs industrially), which corresponds to a positive total entropy change (ΔS_total > 0) and therefore a negative ΔG.
1 Mark
❌ Common Errors
Many students simply state "ΔG is negative because ΔS_total is positive" without linking it back to thermodynamic feasibility or the context of the industrial process provided in the stem.
Part (b)(iii) — Equilibrium Constant Calculation (K_p)
Calculate the value of K_p at 323 K given 50.0 mol CO reacting with excess Ni, leaving 0.750 mol CO at equilibrium under 1.5 atm total pressure.
📐 Step-by-Step Calculation
- Equilibrium moles of Ni(CO)₄:
Moles of CO reacted = 50.0 - 0.750 = 49.25 mol.
From equation, 4 moles of CO produce 1 mole of Ni(CO)₄.
Equilibrium moles of Ni(CO)₄ = 49.25 / 4 = 12.3125 mol (M1) - Total moles at equilibrium:
Total moles = moles(CO) + moles(Ni(CO)₄) = 0.750 + 12.3125 = 13.0625 mol (Solid nickel is ignored as it is a pure solid phase). - Mole fractions (χ):
χ(CO) = 0.750 / 13.0625 = 0.057416
χ(Ni(CO)₄) = 12.3125 / 13.0625 = 0.942584 (M2) - Partial pressures (p):
p = mole fraction × total pressure (1.5 atm)
p(CO) = 0.057416 × 1.5 = 0.086124 atm
p(Ni(CO)₄) = 0.942584 × 1.5 = 1.413876 atm (M3) - K_p expression & calculation:
K_p = p(Ni(CO)₄) / (p(CO))⁴ (M4)
K_p = 1.413876 / (0.086124)⁴ = 25698.9 ( rounds to 25700 ) (M5) - Units:
Units = atm / atm⁴ = atm⁻³ (M6)
6 Marks total
🧠 Exam Technique & Guidance
- Solid omission: Pure solids like Ni(s) have a constant activity of 1 and are strictly omitted from the K_p expression and total mole calculations.
- Significant figures: Examiner accepts standard rounding as long as intermediate values aren't prematurely rounded too heavily.
- Error carried forward (TE): If you miscalculate the mole of CO, subsequent mole fraction and partial pressure marks can still be accessed using your figures.
Part (c) — Qualitative Entropy of Decomposition
Explain, in qualitative terms, why the entropy change of the system, ΔS_system, for the thermal decomposition of Ni(CO)₄ is positive.
✅ Correct Answer
- The number of moles of gas increases during the reaction (from 1 mole of gas on the left to 4 moles of gas on the right).
- Gases have significantly higher entropy than solids or liquids, and the creation of more gaseous moles outweighs the disappearance of solid nickel.
2 Marks
💡 Key Knowledge
Disorder increases whenever a reaction produces a greater number of gaseous molecules because gas particles are widely spaced and possess high translational freedom compared to condensed states.
Topics
Physical Chemistry · Inorganic Chemistry · Topic 1: Atomic Structure and the Periodic Table · Topic 11: Equilibrium II · Topic 13: Energetics II · Topic 15: Transition Metals
Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 3, June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.