Checkpoint 1 — IQ1: Energy Changes in Chemical Reactions
Covering Lessons 01–05: energy profiles, endothermic & exothermic reactions, calorimetry, enthalpy of neutralisation & solution, and catalysts.
What’s Covered
- Exothermic vs endothermic
- Energy profile diagrams
- Activation energy Ea
- ΔH from profile diagrams
- Thermochemical equations
- State symbols & conditions
- ΔH sign conventions
- Reversing equations
- q = mcΔT
- Spirit burner experiment
- ΔHc = −q/n
- Sources of error
- ΔHn ≈ −57 kJ mol⁻¹
- Net ionic equation H⁺ + OH⁻ → H₂O
- Dissolution enthalpy
- Lattice vs hydration energy
- Homogeneous vs heterogeneous
- Catalyst lowers Ea only
- ΔH unchanged by catalyst
- Catalytic converter example
Multiple Choice — 10 marks
1. In an energy profile diagram for an exothermic reaction, the activation energy (Ea) is:
2. A reaction has Ea(forward) = 120 kJ mol⁻¹ and ΔH = −40 kJ mol⁻¹. What is Ea for the reverse reaction?
3. The thermochemical equation $\text{H}_2\text{O(l)} \to \text{H}_2\text{O(g)}$, $\Delta H = +44 \text{ kJ mol}^{-1}$. For the reverse reaction (condensation), ΔH is:
4. A student burns 0.500 g of ethanol (M = 46.07 g mol⁻¹, ΔHc = −1367 kJ mol⁻¹) in a spirit burner. Using q = mcΔT, which expression gives the theoretical temperature rise in 200.0 g of water?
5. In a spirit burner calorimetry experiment, the experimental ΔHc is always less negative than the literature value. The most significant reason is:
6. The molar enthalpy of neutralisation of any strong acid with any strong base is approximately −57 kJ mol⁻¹. This is because:
7. When potassium nitrate (KNO₃) dissolves in water, the solution temperature drops. The enthalpy of dissolution (ΔHsoln) is:
8. A catalyst is added to a reaction. Which statement correctly describes the effect on the energy profile?
9. In the industrial Haber process, an iron catalyst is used to synthesise ammonia from N₂(g) and H₂(g). This iron catalyst is:
10. Which combination is impossible for a chemical reaction at a given temperature?
Short Answer — 20 marks
Q11 (5 marks)
Draw and annotate an energy profile diagram for an exothermic reaction with a large activation energy. Label: (i) reactants and products; (ii) Ea(forward); (iii) Ea(reverse); (iv) ΔH; (v) transition state. Then explain what happens to Ea(forward) when a catalyst is added, and whether ΔH changes.
Energy profile description: x-axis = reaction progress; y-axis = energy (kJ mol⁻¹). Reactants begin at a higher energy than products (exothermic → products are lower). A peak (transition state) lies above the reactants. ΔH is the difference between reactant and product energy levels (negative for exothermic). (2 marks for correct diagram annotations)
Labels: (i) Reactants (higher) and Products (lower); (ii) Ea(forward) = energy from reactants to peak; (iii) Ea(reverse) = energy from products to peak (larger than Ea(forward) for exothermic); (iv) ΔH = difference between reactants and products (negative, downward arrow); (v) Transition state at the peak. (1 mark)
Catalyst effect: Adding a catalyst lowers the peak (transition state) → Ea(forward) decreases. The catalyst provides an alternative reaction pathway with lower energy requirements. (1 mark) ΔH does not change — the energies of the reactants and products are identical in the catalysed and uncatalysed pathways; only the height of the peak changes. (1 mark)
Q12 (5 marks)
A student burns 1.20 g of propan-1-ol (C₃H₇OH, M = 60.10 g mol⁻¹) in a spirit burner. The heat released raises the temperature of 300.0 g of water by 12.5°C. (a) Calculate q, the heat absorbed by the water. (1 mark) (b) Calculate the experimental molar enthalpy of combustion, ΔHc. (2 marks) (c) The literature value is −2020 kJ mol⁻¹. Calculate the percentage error and identify the most likely cause of the discrepancy. (2 marks)
(a) q = mcΔT = 300.0 × 4.18 × 12.5 = 15675 J = 15.68 kJ (1 mark)
(b) n = 1.20 ÷ 60.10 = 0.01996 mol; ΔHc = −q/n = −15.68 ÷ 0.01996 = −785 kJ mol⁻¹ (1 mark for n, 1 mark for ΔHc with sign and units)
(c) % error = |−785 − (−2020)| ÷ 2020 × 100 = 1235/2020 × 100 = 61.1% (1 mark). The large discrepancy is primarily due to heat loss to the surroundings (bench, air, copper beaker) — the spirit burner setup is thermally inefficient and most of the combustion energy does not reach the water. (1 mark)
Q13 (4 marks)
(a) Write the net ionic equation for the neutralisation of HCl(aq) with NaOH(aq) and state ΔHn. (2 marks)
(b) When LiOH is used instead of NaOH, ΔHn is slightly different (−56.2 kJ mol⁻¹ instead of −57.3 kJ mol⁻¹). Explain why, using your knowledge of hydration energies. (2 marks)
(a) H⁺(aq) + OH⁻(aq) → H₂O(l), ΔHn ≈ −57 kJ mol⁻¹. (1 mark for equation showing ionic form; 1 mark for ΔH with sign and units)
(b) All strong acid/strong base neutralisations involve the same reaction (H⁺ + OH⁻ → H₂O) and should give the same ΔHn. The slight difference arises because LiOH is a strong base but Li⁺ has a very high charge density (small ionic radius) — its hydration energy is particularly large. (1 mark) This means a small amount of energy is absorbed when the Li⁺(aq) ion slightly rearranges its hydration sphere during the mixing process, slightly offsetting the heat released. The Na⁺ and Li⁺ ions are spectator ions in the net ionic equation but their different hydration energies contribute a small amount to the overall energy balance of the mixing process. (1 mark)
Q14 (6 marks)
The decomposition of hydrogen peroxide: $2\text{H}_2\text{O}_2\text{(aq)} \to 2\text{H}_2\text{O(l)} + \text{O}_2\text{(g)}$ is catalysed by MnO₂(s).
(a) Identify MnO₂ as a homogeneous or heterogeneous catalyst. Justify. (2 marks)
(b) Draw two energy profile diagrams on the same axes — one for the uncatalysed reaction and one for the catalysed reaction. Annotate both with Ea values. (2 marks)
(c) The reaction is exothermic (ΔH = −196 kJ mol⁻¹). Explain why adding more MnO₂ catalyst increases the rate but does not change ΔH. (2 marks)
(a) MnO₂ is a heterogeneous catalyst. (1 mark) It is a solid, while H₂O₂ is in aqueous solution — the catalyst and reactants are in different physical phases (solid and aqueous/liquid). The reaction occurs at the surface of the solid MnO₂. (1 mark)
(b) Both profiles: same reactant energy level (2 H₂O₂(aq)) at the start; same product energy level (2 H₂O(l) + O₂(g)) at the end; products are lower than reactants (exothermic). Uncatalysed: high peak (large Ea). Catalysed: lower peak (smaller Ea). Both converge at the same product energy. (1 mark for correct diagram description; 1 mark for labelled Ea(uncatalysed) > Ea(catalysed) on same axes)
(c) Adding more MnO₂ provides more active surface area — more reaction sites are available simultaneously → more H₂O₂ molecules can react at once → rate increases. (1 mark) ΔH does not change because the reactants (2 H₂O₂(aq)) and products (2 H₂O(l) + O₂(g)) are identical regardless of whether MnO₂ is present or how much is used. The energy difference between these states is a fixed thermodynamic quantity — it depends only on the nature of the chemical bonds broken and formed, not on the pathway or the amount of catalyst. (1 mark)
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If you scored below 20/30, review Lessons 01–05 before moving on to IQ2.
Next: move on to Lesson 06: Bond Energy & Enthalpy Changes