Chemistry • Year 12 • Module 5 • Lesson 3
Collision Theory Applied to Equilibrium
Apply collision theory to real rate-vs-time graphs, experimental data, and industrial scenarios — the reasoning skills that separate Band 3 from Band 5 answers.
1. Interpret a rate-vs-time graph for an approaching equilibrium
The graph below shows the forward rate (red) and reverse rate (blue) for the reaction H2(g) + I2(g) ⇌ 2HI(g) as it approaches dynamic equilibrium from pure reactants at constant temperature. 9 marks
1.1 Describe the trend in the forward rate from the start of the reaction to equilibrium. Use collision theory language in your answer. 2 marks
1.2 Explain why the reverse rate starts at zero and increases over time. 2 marks
1.3 Estimate the equilibrium rate from the graph. What does this value represent physically? 2 marks
1.4 A student claims that at equilibrium “the reaction has stopped.” Use the graph to explain why this claim is incorrect. 3 marks
2. Interpret experimental data — catalyst and equilibrium rate
A chemist investigated the effect of a vanadium(V) oxide (V2O5) catalyst on the reaction 2SO2(g) + O2(g) ⇌ 2SO3(g), ΔH = −196 kJ mol−1. Three sealed vessels at the same initial conditions were run: without catalyst, with 1 g catalyst, and with 5 g catalyst. Data after reaching equilibrium are shown below. 7 marks
| Condition | Time to reach equilibrium (min) | [SO3] at equilibrium (mol L−1) | Keq at 450°C |
|---|---|---|---|
| No catalyst | 85 | 0.62 | 2.8 × 102 |
| 1 g V2O5 | 18 | 0.62 | 2.8 × 102 |
| 5 g V2O5 | 7 | 0.62 | 2.8 × 102 |
2.1 Identify the trend in time to reach equilibrium as more catalyst is added. 1 mark
2.2 Using the data, what conclusion can be drawn about the effect of catalyst on equilibrium position? Justify using two pieces of data. 2 marks
2.3 Using collision theory, explain why the catalyst decreases the time to reach equilibrium without changing [SO3] at equilibrium. 3 marks
2.4 What variable would need to be changed to increase [SO3] at equilibrium? Explain. 1 mark
3. Case study — Incitec Pivot and the Haber process catalyst
Incitec Pivot Limited operates one of Australia’s largest ammonia plants at Gibson Island, Queensland, producing ammonia via the Haber process: N2(g) + 3H2(g) ⇌ 2NH3(g), ΔH = −92 kJ mol−1. The process uses an iron (Fe) catalyst promoted with potassium oxide (K2O) and alumina (Al2O3). Operating conditions are approximately 400–500°C and 15–25 MPa. 6 marks
3.1 Explain, using collision theory and activation energy, why the Fe catalyst is essential at Incitec Pivot even though it does not improve the equilibrium yield of NH3. 3 marks
3.2 Predict and justify: if Incitec Pivot increased its operating temperature from 450°C to 550°C while keeping pressure constant, would the equilibrium yield of NH3 increase, decrease, or stay the same? Use Ea reasoning. 3 marks
4. Sequence the steps — approaching dynamic equilibrium
The events below describe how a reversible reaction approaches dynamic equilibrium from pure reactants, but they are shuffled. Write the correct order (1–6) in the “Order” column. 5 marks
| Order | Event |
|---|---|
| The frequency of effective reverse collisions increases as product concentration grows. | |
| Only reactant particles are present; the reverse rate is zero. | |
| The forward rate and reverse rate are equal and non-zero; concentrations stop changing macroscopically. | |
| Reactants begin colliding effectively; products start to form; the forward rate is at its maximum. | |
| Reactant concentration decreases; the forward rate begins to fall below its initial maximum. | |
| The forward rate is still greater than the reverse rate; net conversion of reactants to products continues. |
Q1.1 — Forward rate trend
The forward rate starts at its maximum value (approximately 5 on the graph) because the concentration of reactants (H2 and I2) is highest at the start — the frequency of effective forward collisions is highest. As the reaction proceeds, H2 and I2 are consumed, their concentrations decrease, the frequency of effective forward collisions decreases, and the forward rate decreases until it reaches the equilibrium plateau (approximately 2). [1 mark for describing the decreasing trend; 1 mark for explaining using collision theory language: decreasing reactant concentration → fewer effective collisions → lower forward rate.]
Q1.2 — Reverse rate from zero
The reverse rate starts at zero because at t = 0, no products (HI) are present — there are no product particles to collide effectively in the reverse direction. As the reaction proceeds forward, HI accumulates; the frequency of effective reverse collisions increases; so the reverse rate increases from zero. [1 mark for no products at start; 1 mark for HI accumulation → increasing reverse rate.]
Q1.3 — Equilibrium rate estimate
From the graph, the equilibrium rate plateau is approximately 2 mol L−1 s−1. This value represents the rate at which both the forward reaction (H2 + I2 → 2HI) and the reverse reaction (2HI → H2 + I2) are proceeding simultaneously. Reactions are occurring continuously in both directions at equal speed — macroscopic concentrations do not change, but molecular activity continues. [1 mark for reading ~2 from graph; 1 mark for explaining that both forward and reverse rates equal this value simultaneously.]
Q1.4 — Refuting “reaction has stopped”
The student’s claim is incorrect. The graph shows that at equilibrium, both the forward rate and reverse rate plateau at a non-zero value (approximately 2 mol L−1 s−1). If the reaction had stopped, both curves would reach zero. Instead, they reach a common positive value, indicating that molecular collisions and bond rearrangements are still occurring in both directions simultaneously at equal rates. This is the defining feature of dynamic equilibrium — it is dynamic at the molecular level even though concentrations appear static at the macroscopic level. [1 mark for citing the non-zero plateau; 1 mark for identifying both curves as equal and non-zero; 1 mark for explaining ‘dynamic’ meaning molecular activity continues.]
Q2 — Catalyst and equilibrium data
2.1 As the amount of V2O5 catalyst increases, the time to reach equilibrium decreases (from 85 min to 18 min to 7 min).
2.2 The catalyst has no effect on equilibrium position. [SO3] at equilibrium is 0.62 mol L−1 in all three conditions, and Keq is 2.8 × 102 in all three conditions — both are unchanged despite the catalyst quantity increasing. [1 mark for correct conclusion; 1 mark for citing two identical values across rows.]
2.3 The V2O5 catalyst provides an alternative lower-energy reaction pathway, reducing the activation energy for both the forward (SO2 + O2 → SO3) and reverse (SO3 → SO2 + O2) reactions by the same amount. More particles now have sufficient energy to undergo effective collisions in both directions — both forward and reverse rates increase by the same factor. Equilibrium is therefore reached more quickly. Since both rates increase by the same factor, their ratio (which determines Keq and equilibrium position) is unchanged, so [SO3] at equilibrium remains 0.62 mol L−1. [1 mark for lowering Ea for both directions; 1 mark for both rates increasing by same factor; 1 mark for ratio of rates unchanged → equilibrium position unchanged.]
2.4 Temperature. Temperature is the only factor that changes Keq. Decreasing temperature (for this exothermic forward reaction) would shift equilibrium right, increasing [SO3].
Q3 — Incitec Pivot case study
3.1 Without a catalyst, the Ea for N2 + 3H2 → 2NH3 is very high (the N≡N triple bond is extremely strong — ~945 kJ mol−1). At economically practical temperatures, the fraction of particles with sufficient energy to cross this barrier is too small to give a commercially useful rate. The Fe catalyst provides an alternative reaction pathway with a lower Ea for both the forward and reverse reactions. This increases the frequency of effective forward collisions at the same temperature, allowing equilibrium to be reached faster. Because Ea is lowered equally for both directions, the equilibrium position (yield) is unchanged. The catalyst makes the process economically viable by increasing rate without sacrificing yield. [3 marks: 1 for citing high Ea without catalyst; 1 for catalyst lowering Ea both directions; 1 for unchanged equilibrium position.]
3.2 The equilibrium yield of NH3 would decrease. N2 + 3H2 ⇌ 2NH3 is exothermic (ΔH = −92 kJ mol−1), so Ea(forward) < Ea(reverse). Increasing temperature from 450°C to 550°C increases the average kinetic energy of all particles, raising both forward and reverse rates. However, the reverse (endothermic) rate increases proportionally more because it has the higher Ea — a greater proportion of particles can cross the higher reverse barrier at the new temperature. Reverse rate > forward rate — equilibrium shifts left — less NH3 is produced — Keq decreases. [1 mark for correct prediction (decrease); 1 mark for exothermic → Ea(rev) > Ea(fwd); 1 mark for reverse rate increasing more → shift left.]
Q4 — Sequence
Correct order: 2 → 4 → 5 → 1 → 6 → 3 (Only reactants present / reverse rate = 0; reactants collide effectively, products begin forming, forward rate at maximum; reactant concentration decreases, forward rate falls; product concentration grows, reverse rate increases; forward still > reverse, net forward conversion continues; forward = reverse, equilibrium established). Award 1 mark per correctly placed event (5 transitions = 5 marks; accept adjacency scoring).