AQA A-Level Chemistry Paper 1, 2018: Question 2

📐 Step-by-Step Calculation

1. Find partial pressure of NH₃:
   p(NH₃) = 0.80 × 150 kPa = 120 kPa
2. Find remaining pressure for N₂ and H₂:
   p(remaining) = 150 − 120 = 30 kPa
   (or remaining mole fraction = 1 − 0.80 = 0.20 )
3. Use reacting ratio logic:
   N₂ and H₂ were mixed in a 1:3 ratio and react in a 1:3 ratio. Therefore, the unreacted gases remain in a 1:3 ratio (1 part N₂ to 3 parts H₂, total 4 parts).
   p(N₂) = ¼ × 30 = 7.5 kPa
   p(H₂) = ¾ × 30 = 22.5 kPa (or 23 kPa )

✅ Final Answers

• Partial pressure of nitrogen: 7.5 kPa [1 mark]
• Partial pressure of hydrogen: 22.5 kPa (or 23 kPa) [1 mark]
• Partial pressure of ammonia: 120 kPa [1 mark]

🧠 Exam Technique: The Mole Ratio Shortcut

Alternatively, split the remaining mole fraction of 0.20 directly:

• Mole fraction N₂ = ¼ × 0.20 = 0.05 → 0.05 × 150 = 7.5 kPa
• Mole fraction H₂ = ¾ × 0.20 = 0.15 → 0.15 × 150 = 22.5 kPa

❌ Common Errors

• Dividing remaining 30 kPa equally (15 kPa each) by forgetting the 1:3 ratio.
• Using stoichiometric coefficients from the equation instead of the unreacted gas fractions.

✅ Correct Expression

K_p = ^(ppNH₃)² / _{(ppN₂) × (ppH₂)³}

Also accepted: K_p = ^p(NH₃)² / _{[p(N₂) × p(H₂)³]}

❌ The Square Bracket Penalty

Never use square brackets [ ] in a K_p expression!

Writing [NH₃]² / [N₂][H₂]³ represents K_c (concentration), not partial pressure. The mark scheme explicitly states: "Penalise [ ]". Always use p(NH₃) or pp(NH₃) .

📐 Step-by-Step Calculation

1. Substitute given values into expression:
   K_p = (1.10 × 10³)² / [ (1.20 × 10²) × (1.50 × 10²)³ ]
2. Evaluate numerator and denominator:
   Numerator: (1.10 × 10³)² = 1.21 × 10⁶
   Denominator: (1.20 × 10²) × (3.375 × 10⁶) = 4.05 × 10⁸
3. Divide to find numerical value:
   K_p = 1.21 × 10⁶ / 4.05 × 10⁸ = 2.98765... × 10⁻³
   = 2.99 × 10⁻³ (or 3.0 × 10⁻³ / 0.0030) [1 mark]
4. Determine units:
   Units = (kPa)² / [ (kPa) × (kPa)³ ] = kPa² / kPa⁴ = kPa⁻² [1 mark]

❌ Common Calculation Traps

• Calculator error: Forgetting to cube (1.50 × 10²) or failing to put brackets around the entire denominator in your calculator.
• Wrong unit signs: Writing kPa² instead of kPa⁻² .
• Units note: If partial pressures were converted to Pa, 2.99 × 10⁻⁹ Pa⁻² is allowed.

✅ 3-Step Model Answer

1. Effect on K_p:
   Decreases (or gets smaller / lower) [1 mark].
2. Equilibrium shift:
   The equilibrium shifts in the endothermic direction (which is to the left / reverse direction) [1 mark].
3. Le Chatelier justification:
   To oppose the increase in temperature (or to reduce the temperature / absorb heat) [1 mark].

💡 Key Knowledge

• ΔH = −92 kJ mol⁻¹: Negative ΔH means the forward reaction is exothermic, so the backward reaction is endothermic.
• Why K_p decreases: Shifting to the left decreases the partial pressure of products (numerator) and increases reactants (denominator), making the overall ratio smaller.
• Crucial Rule: Temperature is the only factor that changes the value of K_p!