Edexcel A-Level Chemistry AS Paper 2, November 2021: Question 9

📐 Step-by-Step Calculation

1. Reactants Bonds Broken: 1 mole of N≡N and 3 moles of H–H.
   Energy input = 945 + (3 × 436) = +2253 kJ mol⁻¹
2. Products Bonds Formed: 6 moles of N–H bonds (since stoichiometry forms 2NH₃, and each NH₃ has 3 N–H bonds).
   Energy release = 6 × 391 = -2346 kJ mol⁻¹
3. Overall Enthalpy Change (ΔH):
   ΔH = Bonds broken – Bonds formed (or sum of bonds)
   -2346 + 2253 = -93 kJ mol⁻¹

❌ Common Calculation Traps

• Stoichiometry errors: Forgetting that 3 moles of H₂ means multiplying the H–H bond energy by 3.
• Product bonds: Forgetting that 2 moles of NH₃ yields 6 N–H bonds in total.
• Sign omission: A positive or missing sign will cost you the final mark. Always include the negative sign for exothermic reactions.

✅ Correct Answers (Any Two)

• The equation calculated in (a)(i) is for the formation of two moles of ammonia (whereas standard enthalpy of formation is defined for one mole).
• Mean bond enthalpies are average values derived from a range of different compounds, not specific to the actual bonding environment within ammonia.

🧠 Exam Technique & Examiner Insight

Examiners note that candidates frequently waste time discussing differing industrial conditions (like temperature and pressure in the Haber process) or heat losses. Stick strictly to definitions and data limitations: stoichiometry differences and average vs. specific bond energies.

✅ Correct Answers

(iii) Atom Economy: D (100%)

Reasoning: The equation N₂(g) + 3H₂(g) ⇌ 2NH₃(g) features only a single product, meaning all reactant atoms are converted into the desired product.

(iv) Equilibrium Expression (Kc): C

Kc = [NH₃]² / ([N₂][H₂]³)

❌ Distractor Analysis

• For (iii): A is hydrogen percentage, B is half atom economy, C is nitrogen percentage.
• For (iv): Option A uses molar quantities instead of concentrations and is inverted; Option B uses square brackets with no powers; Option D is written for the reverse reaction.

💡 Key Knowledge: Why remove product early?

• Removing product shifts the equilibrium position to the right (Le Chatelier's principle), favouring the forward reaction to replace what was lost.
• Removing product decreases the rate of the back reaction, maintaining a high net forward rate.
• Waiting to reach equilibrium can take an impractically long time as the reaction slows down near equilibrium.
• Unreacted reactants can be continuously separated and recycled back into the reactor.

🧠 Exam Technique

This is a standard 3-mark industrial chemistry application question. Ensure you write distinct, scientifically rigorous statements covering equilibrium position, reaction rates, and recycling logistics.

✅ (i) Catalyst Warming & Melting

• Warming: To provide/overcome the activation energy (or to speed up the slow room-temperature reaction). Note: Do not write "to lower activation energy" as catalysts provide an alternative pathway with lower Ea, they don't alter the activation energy of the original pathway.
• Melting: The reaction is sufficiently/very exothermic ( ΔH = -905.2 kJ mol⁻¹ ) that the heat released raises the temperature enough to melt the copper.

✅ (ii) Reaction Profile Requirements

• Must show an intermediate peak (transition state/intermediate energy level) higher than the reactants.
• The final products line must be drawn strictly below the reactant line to visually represent the exothermic nature ( ΔH negative ).
• Clearly label ΔH or show a downward vertical arrow from reactants to products.

💡 Step-by-Step Heterogeneous Catalysis

1. Adsorption: Reactant molecules ( NH₃ and O₂ ) adsorb onto the active sites on the surface of the catalyst. (Examiner tip: Use "adsorb", not "absorb"!)
2. Reaction: Bonds in the reactant molecules are weakened and broken, and new bonds form as the reaction takes place on the surface.
3. Desorption: Product molecules ( H₂O and NO ) desorb from the active sites and leave the catalyst surface.

❌ Common Errors

• Writing "absorb" instead of "adsorb" will lose you the first marking point. Adsorption involves surface attachment; absorption involves entering the bulk material.
• Missing the desorption stage at the end of the sequence.