OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), June 2025: Question 17

✅ Mark Scheme Answer

Iron / Fe is an element (in its standard state).

💡 Key Knowledge

By definition, standard enthalpy change of formation (Δ_fH^⦵) is the enthalpy change when 1 mole of a compound is formed from its constituent elements under standard conditions in their standard states. Thus, an element in its standard state requires no transformation.

📐 Step-by-Step Calculation

1. Formula: Δ_rH = Σ Δ_fH(products) − Σ Δ_fH(reactants)
2. Sum of Products: [2 × 0] + [3 × (−394)] = −1182 kJ mol⁻¹
3. Sum of Reactants: [1 × (−824)] + [3 × (−111)] = −824 − 333 = −1157 kJ mol⁻¹
4. Combine: Δ_rH = −1182 − (−1157) = −25 kJ mol⁻¹

❌ Common Errors

• Missing stoichiometric multipliers: Forgetting to multiply CO₂ by 3 or CO by 3 (gives +763 or −247 kJ mol⁻¹).
• Sign inversions: Inverting products and reactants leads to +25 kJ mol⁻¹.
• Incorrect bracket handling: Adding −1157 rather than subtracting it gives −2339 kJ mol⁻¹.

✅ Mark Scheme Answer

• Mark 1: Fe₂O₃ is a solid and CO₂ is a gas.
• Mark 2: CO₂ / gas is more disordered OR has a greater number of ways energy/particles can be arranged / dispersed.

🧠 Exam Technique

Always state the states of matter explicitly first! Do not just say "CO₂ is more disordered" without identifying that CO₂ is a gas while Fe₂O₃ is an ionic solid lattice. A complete answer connects the physical state directly to entropy.

📐 Step-by-Step Calculation

1. Formula: ΔS^⦵ = ΣS^⦵(products) − ΣS^⦵(reactants)
2. Insert known data:
   +15 = [2 × S^⦵(Fe) + 3(214)] − [87 + 3(198)]
3. Simplify known terms:
   +15 = 2S^⦵(Fe) + 642 − (87 + 594)
   +15 = 2S^⦵(Fe) + 642 − 681
   +15 = 2S^⦵(Fe) − 39
4. Solve for 2S^⦵(Fe):
   2S^⦵(Fe) = 15 + 39 = 54
5. Divide by 2: S^⦵(Fe) = 54 / 2 = 27 J K⁻¹ mol⁻¹

❌ Common Errors

• Forgetting dividing by 2: Quoting 54 J K⁻¹ mol⁻¹ instead of dividing by the 2 moles of Fe.
• Reversing reactants and products: Calculating reactants minus products produces 12 J K⁻¹ mol⁻¹.
• Omitting stoichiometry: Neglecting 3 × 214 or 3 × 198 gives answers like 241 or −171 J K⁻¹ mol⁻¹.

📐 Step-by-Step Calculation

Step 1: Convert Units to match

• T = 100 °C = 100 + 273 = 373 K
• ΔS = +543 J K⁻¹ mol⁻¹ = +0.543 kJ K⁻¹ mol⁻¹ (or ΔH = 491 000 J mol⁻¹)

Step 2: Show reaction is not feasible at 100 °C

Since ΔG > 0, the forward reaction is not feasible.

Step 3: Minimum Temperature for Feasibility

At the point of feasibility, ΔG = 0, so T = ΔH / ΔS:

To the nearest whole number: 904 K (or 905 K if using rounded values)

❌ Common Calculation Traps

• Temperature unit slip: Using 100 instead of converting to Kelvin (373 K).
• Entropy unit mismatch: Forgetting to divide ΔS by 1000 or multiply ΔH by 1000, resulting in answers off by three orders of magnitude.
• Missing feasibility statement: Stating the value of ΔG (+288) without explicitly concluding that it is not feasible because ΔG > 0.
• Rounding error: Question asks specifically for the nearest whole number; failing to round (e.g. leaving 904.2) loses the final mark.

✅ Mark Scheme Answer

There are an equal number of (gaseous) reactant and product molecules / moles (2 on each side)
OR pressure units cancel out: (kPa × kPa) / (kPa × kPa) .

🧠 Exam Technique

Always refer to the number of moles / molecules or write out the units cancelling algebraically. Do NOT say "equal concentrations" (since this is K_p, not K_c) and do not simply say "same number of reactants and products" without specifying moles/molecules.

📐 Step-by-Step Calculation

1. Write K_p expression:
   K_p = [p(H₂) × p(CO₂)] / [p(H₂O) × p(CO)] = 8.13
2. Identify stoichiometric equality:

   Since equimolar amounts of H₂O and CO react in a 1:1 ratio, at equilibrium:
   p(H₂O) = p(CO) = x

3. Substitute known values into expression:
   8.13 = (211 × 211) / x²
   x² = (211 × 211) / 8.13 = 44521 / 8.13 = 5476.138...
4. Solve for x:
   x = √(5476.138) = 74.0009... kPa
5. Final Answer: Partial pressure of H₂O = 74 kPa (or 74.0 kPa)
   Partial pressure of CO = 74 kPa (or 74.0 kPa)

🧠 Examiner Insight

Many students struggle to see why p(H₂O) = p(CO). The question states: "An equal number of moles of steam and carbon monoxide are left... until equilibrium is reached." Because they react in a 1:1 ratio, their remaining moles—and therefore their mole fractions and partial pressures—must be identical.

✅ Mark Scheme Answer

• The second temperature is lower.
• The forward reaction is exothermic (ΔH is negative) / reverse reaction is endothermic.
• Since K_p increased (from 8.13 to 129), the equilibrium position shifted to the right / products to oppose the decrease in temperature.

💡 Key Principle: Temperature and K

• Temperature is the only factor that alters the value of an equilibrium constant (K_c or K_p).
• When K_p increases, the partial pressure of products increases relative to reactants (equilibrium shifts right).
• By Le Chatelier's principle, lowering temperature favours the exothermic direction (+heat produced). Since ΔH < 0, a temperature lower than 700 K shifts equilibrium to the right, increasing K_p.