Chemistry • Year 11 • Module 3 • Lesson 11
Collision Theory & Reaction Rate
Apply collision theory and rate calculations to real experimental data, Maxwell–Boltzmann graphs, cause-and-effect chains, and an Australian industrial chemistry context.
1. Interpret rate data — H2O2 decomposition experiment
A student monitored the decomposition of hydrogen peroxide (H2O2 → H2O + ½O2) at 25 °C by collecting oxygen gas in a gas syringe. The table below shows results from two trial runs at different hydrogen peroxide concentrations. 8 marks
| Time (s) | Volume O2 — Trial A, 0.5 mol L−1 H2O2 (mL) | Volume O2 — Trial B, 1.0 mol L−1 H2O2 (mL) |
|---|---|---|
| 0 | 0 | 0 |
| 20 | 7 | 14 |
| 40 | 12 | 24 |
| 60 | 15 | 30 |
| 80 | 17 | 33 |
| 100 | 17 | 34 |
1.1 Calculate the average rate of O2 production for Trial A between 0 and 40 s. Express your answer in mL s−1. 1 mark
1.2 Calculate the average rate of O2 production for Trial B between 0 and 40 s. 1 mark
1.3 Describe the trend in rate over time for Trial B (from 0–20 s to 80–100 s). 2 marks
1.4 Using collision theory, explain why Trial B has a higher average rate (0–40 s) than Trial A. Your answer must use the terms collision frequency and effective collision. 3 marks
1.5 Both trials produce approximately the same total volume of O2 (about 17 mL and 34 mL, consistent with their different concentrations). Explain why Trial B produces roughly twice as much total O2 as Trial A. 1 mark
2. Interpret a Maxwell–Boltzmann distribution graph
The figure below shows the Maxwell–Boltzmann distribution of particle kinetic energies for the same sample of gas at two different temperatures, T1 (lower) and T2 (higher). The dashed vertical line marks the activation energy (Ea) for a particular reaction. 8 marks
Stylised Maxwell–Boltzmann distribution based on standard physical chemistry principles. Adapted from standard HSC Chemistry representations.
2.1 What does the shaded area to the right of the Ea line represent on this graph? Using the graph, explain what the difference in shaded area between T1 and T2 tells you about the proportion of particles capable of an effective collision at each temperature. 3 marks
2.2 Using the graph, explain why a higher temperature leads to a faster reaction rate. Your answer must reference the shaded region to the right of Ea. 3 marks
2.3 Sketch on the diagram (or describe in words) what the effect of a catalyst would be on this graph. Does the catalyst move the Ea line or change the curve shape? Explain. 2 marks
3. Cause-and-effect chain — temperature and glow stick brightness
Complete the cause-and-effect chain below to explain why placing a cracked glow stick into a hot water bath at 45 °C makes it glow more brightly than leaving it at room temperature (20 °C). Fill in each effect box. 5 marks
| Cause (given) | → | Effect (you fill in) |
|---|---|---|
| Temperature rises from 20 °C to 45 °C | → | Average kinetic energy of reactant particles ________________________ |
| Average kinetic energy of reactant particles increases | → | Proportion of particles with KE ≥ Ea ________________________ |
| More particles have KE ≥ Ea | → | Number of effective collisions per second ________________________ |
| Number of effective collisions per second increases | → | Rate of the chemiluminescent reaction ________________________ |
| Rate of chemiluminescent reaction increases | → | Brightness of glow ________________________ (glow stick glows more brightly) |
Overall outcome (so…): A glow stick in hot water at 45 °C glows more brightly because ________________________
4. Australian context — the Haber process at Incitec Pivot, Brisbane
The Haber process (N2(g) + 3H2(g) ⇌ 2NH3(g), ΔH = −92 kJ mol−1) is used by Incitec Pivot at their Gibson Island facility in Brisbane to manufacture ammonia for fertilisers. The process uses an iron catalyst and is typically run at around 450 °C and 200 atm. The activation energy for the uncatalysed reaction is approximately 335 kJ mol−1; the iron catalyst reduces this to around 163 kJ mol−1. 6 marks
4.1 The Haber process is thermodynamically favourable (ΔH = −92 kJ mol−1), yet the reaction does not proceed rapidly at room temperature without a catalyst. Using collision theory and Ea, explain why. 2 marks
4.2 The iron catalyst lowers Ea from 335 kJ mol−1 to 163 kJ mol−1. Using the Maxwell–Boltzmann distribution, explain how this change increases the reaction rate at 450 °C. 2 marks
4.3 A process engineer proposes increasing the temperature to 600 °C to further speed up the Haber process. Using collision theory, predict how this would affect the rate. Would this also affect ΔH or Ea (uncatalysed)? Justify. 2 marks
5. Predict and justify — food spoilage in an Australian summer
In Australian summer heat (40 °C), leftover cooked chicken spoils noticeably faster than when refrigerated (4 °C). The biochemical reactions responsible for spoilage (bacterial metabolism and enzymatic decomposition) all require particles to exceed an activation energy before they can proceed. 4 marks
5.1 Using collision theory, predict and justify why food spoils faster at 40 °C than at 4 °C. Your response must reference average kinetic energy, Ea, and the proportion of effective collisions. 3 marks
5.2 Predict what would happen to the rate of food spoilage if the cooked chicken were frozen at −18 °C. Does the food spoilage reaction stop or just slow dramatically? Explain using Ea. 1 mark
Q1.1 — Trial A rate, 0–40 s
Rate = (12 − 0) mL ÷ (40 − 0) s = 12 ÷ 40 = 0.30 mL s−1.
Q1.2 — Trial B rate, 0–40 s
Rate = (24 − 0) mL ÷ (40 − 0) s = 24 ÷ 40 = 0.60 mL s−1.
Q1.3 — Trend for Trial B over time (2 marks)
The rate is highest in the 0–20 s interval (0.70 mL s−1) and decreases progressively at each subsequent time interval. By 80–100 s the rate has fallen to approximately 0.05 mL s−1 (only 1 mL collected) [1 — identifies decreasing trend]. The rate falls because the concentration of H2O2 decreases over time as it is consumed, reducing collision frequency [1 — links to reason].
Q1.4 — Why Trial B is faster (3 marks)
Trial B uses twice the concentration of H2O2 (1.0 vs 0.5 mol L−1). A higher concentration means more H2O2 molecules per unit volume [1 — links concentration to particle density]. This increases the collision frequency between H2O2 molecules (and between H2O2 and the catalyst surface if a catalyst is used) [1 — collision frequency]. Since temperature and Ea are the same, the proportion of collisions that are effective is unchanged, but the greater number of collisions per second means more effective collisions per second [1 — effective collision rate increases], giving a faster reaction rate.
Q1.5 — Why Trial B produces ~twice the total O2 (1 mark)
Trial B starts with twice the concentration of H2O2, so there are twice as many moles of reactant to decompose. The total volume of O2 produced depends on the total amount of reactant, not on the rate. Rate determines how fast the O2 is produced; the total amount of H2O2 determines how much O2 is ultimately produced [1].
Q2.1 — Shaded area and proportion of effective collisions (3 marks)
The shaded area to the right of the Ea line represents the fraction (proportion) of particles that have kinetic energy equal to or greater than Ea — these are the particles capable of undergoing an effective collision [1]. At T1, the shaded area is small, meaning only a small proportion of particles have KE ≥ Ea and very few collisions are effective [1]. At T2, the shaded area is larger, meaning a greater proportion of particles have KE ≥ Ea; more collisions per second are effective, so the reaction rate is higher at the higher temperature [1].
Q2.2 — Why higher T leads to faster rate (3 marks)
At T2 the distribution shifts to higher kinetic energies, so the area under the curve to the right of the Ea line (the shaded region) is larger than at T1 [1]. This larger shaded area represents a greater proportion of particles that have kinetic energy ≥ Ea [1]. More particles exceeding Ea means more effective collisions per second, so the reaction rate increases [1].
Q2.3 — Effect of catalyst on the graph (2 marks)
A catalyst does NOT change the curve shape (which depends only on temperature) [1]. Instead, a catalyst lowers Ea, so the dashed vertical Ea line would shift to the left on the graph. With a lower Ea, a larger proportion of particles in the existing distribution exceed the (now lower) energy barrier — the shaded area increases — and the rate increases, without any change in the distribution of kinetic energies [1].
Q3 — Cause-and-effect chain answers
Row 1 effect: increases. Row 2 effect: increases (a larger proportion exceeds Ea). Row 3 effect: increases. Row 4 effect: increases. Row 5 effect: increases / the glow stick glows more brightly. Overall: the hot water bath increases the average kinetic energy of the reactant particles, raising the proportion that exceed Ea, increasing effective collisions per second, which increases the chemiluminescent reaction rate, producing more light per second.
Q4.1 — Why Haber process slow at room temperature (2 marks)
Although ΔH is negative (thermodynamically favourable), the Ea for the uncatalysed reaction is approximately 335 kJ mol−1 [1]. At room temperature, the average kinetic energy of N2 and H2 molecules is far below 335 kJ mol−1, so only a tiny fraction of collisions have sufficient energy to reach the transition state and result in an effective collision. Reaction rate is negligible even though the reaction is thermodynamically favourable — Ea controls rate, not ΔH [1].
Q4.2 — Catalyst lowers Ea, increases rate (2 marks)
At 450 °C, the Maxwell–Boltzmann distribution shows a spread of kinetic energies among the reactant particles. Reducing Ea from 335 to 163 kJ mol−1 moves the Ea threshold to the left on the distribution [1]. A much larger fraction of the particles in the existing distribution now have kinetic energy ≥ 163 kJ mol−1 (compared to the fraction that exceeded 335 kJ mol−1), so many more collisions per second are effective and the reaction rate increases substantially [1].
Q4.3 — Effect of raising temperature to 600 °C (2 marks)
Raising the temperature to 600 °C would increase the average kinetic energy of all particles, shifting the Maxwell–Boltzmann distribution to higher energies and increasing the proportion of particles exceeding Ea. This would increase the rate of the Haber process [1]. However, ΔH and Ea (uncatalysed) are fixed properties of the reaction and would NOT change with temperature. Temperature changes how many particles exceed Ea — it does not lower Ea itself (that requires a catalyst) and does not alter the energy difference between reactants and products (ΔH) [1].
Q5.1 — Food spoilage at 40 °C vs 4 °C (3 marks)
At 40 °C, the average kinetic energy of the bacterial and enzymatic reactant particles is substantially higher than at 4 °C [1]. A larger proportion of these particles have kinetic energy ≥ Ea for the spoilage reactions [1]. This means more effective collisions per second, so the biochemical spoilage reactions proceed at a much higher rate — food deteriorates visibly faster in the heat [1].
Q5.2 — Freezing at −18 °C (1 mark)
At −18 °C, particle kinetic energies are so low that virtually no particles exceed Ea for the spoilage reactions. The rate slows to an almost undetectable level but does not stop completely — slow reactions still occur. Food can still spoil eventually in a freezer, but it takes far longer than at refrigerator or room temperature [1].