Chemistry • Year 11 • Module 3 • Lesson 11

Collision Theory & Reaction Rate

Build HSC Band 5–6 extended-response technique on collision theory, activation energy and the particle-level interpretation of experimental rate data.

Master · Extended Response

1. Extended response — collision theory and the Haber process data (Band 5–6)

8 marks Band 5–6

Stimulus. Incitec Pivot operates a Haber process plant at Gibson Island, Brisbane, producing ammonia for fertiliser production. The balanced equation is: N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = −92 kJ mol⁻¹. The table below shows experimental data from a research study measuring the rate of NH₃ formation under different conditions.

Trial	Temperature (°C)	[N₂] (mol L⁻¹)	[H₂] (mol L⁻¹)	Iron catalyst?	Initial rate (mol L⁻¹ s⁻¹)
1	25	0.50	1.50	No	~0 (negligible)
2	450	0.50	1.50	No	0.0003
3	450	0.50	1.50	Yes	0.014
4	450	1.00	1.50	Yes	0.028
5	450	1.00	3.00	Yes	0.056

Source: hypothetical research trial data. Rates illustrative of industrial Haber process kinetics literature.

Q1. Analyse and evaluate the experimental data above using collision theory to explain the rate differences observed across Trials 1–5. In your response you must:

Define effective collision and state both conditions required for one to occur.
Explain the rate difference between Trial 1 and Trial 2 using the Maxwell–Boltzmann distribution and E_a.
Explain the rate difference between Trial 2 and Trial 3 using E_a and the proportion of effective collisions.
Explain the rate differences across Trials 3, 4 and 5 using collision frequency.
Reach a justified conclusion about which factor — temperature, catalyst, or concentration — produces the greatest increase in rate in this dataset, and explain why that factor is so effective.

Stuck? Work through each pair of trials in order. Trials 1→2: same concentration, same catalyst (none), different T. Trials 2→3: same T, same concentration, different catalyst. Trials 3→4→5: same T, same catalyst, different concentrations.

2. Source critique — evaluate a student's claim (Band 5–6)

7 marks Band 5–6

"I understand now why a glow stick in hot water glows more brightly. The hot water lowers the activation energy, so more particles have energy above the threshold. The reaction speeds up. Also, adding a catalyst to a reaction must increase the enthalpy change, because the reaction releases more energy when it goes faster. And the rate of a reaction stays constant throughout as long as the temperature doesn't change."

— Year 11 student's class notes after the lesson on collision theory.

Q2. Evaluate this student's notes. Identify every scientific error, explain the correct chemistry for each, and propose how each error could be detected by analysing experimental data or an energy diagram. Then rewrite the student's notes as a scientifically accurate paragraph using correct collision theory language.

Identify errors first: (1) Does hot water lower E_a, or does it change the kinetic energy of particles? (2) Does a catalyst change ΔH? (3) Is rate constant when T is constant but reactants are being consumed? Then correct each one precisely.

Answers — Do not peek before attempting

Q1 — Sample Band 6 response (8 marks), annotated

An effective collision is one that results in the formation of products. For a collision to be effective, two conditions must be simultaneously met: (1) the colliding particles must have kinetic energy equal to or greater than the activation energy (E_a), and (2) the particles must collide in the correct orientation so that reactive sites face each other. [1 — defines effective collision with both conditions]

Trials 1 vs 2 (temperature effect): Both trials use identical concentrations and no catalyst, so the only variable is temperature (25 °C vs 450 °C). On the Maxwell–Boltzmann distribution for N₂ and H₂ at 25 °C, the area under the curve to the right of E_a (approximately 335 kJ mol⁻¹) is negligible — virtually no particles have sufficient kinetic energy for an effective collision, so the rate is approximately zero. At 450 °C the distribution shifts to higher kinetic energies and broadens; a measurable proportion of particles now exceed E_a, yielding a small but nonzero rate of 0.0003 mol L⁻¹ s⁻¹. [2 — Maxwell–Boltzmann correctly applied to both temperatures; rate linked to fraction exceeding E_a]

Trials 2 vs 3 (catalyst effect): Both trials run at 450 °C with identical concentrations; Trial 3 adds an iron catalyst. The catalyst provides an alternative reaction pathway with a lower E_a (approximately 163 kJ mol⁻¹). At 450 °C, the fraction of particles in the Maxwell–Boltzmann distribution with kinetic energy ≥ 163 kJ mol⁻¹ is far larger than the fraction exceeding 335 kJ mol⁻¹. Consequently, effective collisions per second increase dramatically — the rate jumps from 0.0003 to 0.014 mol L⁻¹ s⁻¹, an approximately 47-fold increase. The catalyst does not change the Maxwell–Boltzmann curve (i.e. does not change temperature or kinetic energy distribution); it only lowers the energy threshold. [2 — E_a lowering correctly explained; rate change linked to larger fraction exceeding lower E_a]

Trials 3, 4 and 5 (concentration effect): Temperature and catalyst are constant across these trials; only concentrations change. Trial 4 doubles [N₂] from 0.50 to 1.00 mol L⁻¹; rate doubles from 0.014 to 0.028 mol L⁻¹ s⁻¹. Trial 5 further doubles [H₂] from 1.50 to 3.00 mol L⁻¹; rate doubles again to 0.056 mol L⁻¹ s⁻¹. Higher concentration means more reactant particles per unit volume, which increases collision frequency between N₂ and H₂ molecules. Since temperature and E_a are unchanged, the proportion of those collisions that are effective stays the same, but more collisions per second means proportionally more effective collisions per second. [1 — collision frequency correctly linked to concentration; proportionality of rate to concentration noted]

Conclusion: Adding the iron catalyst (Trial 2 → 3) produced the greatest single rate increase (~47-fold) compared to doubling concentration (2-fold each time) or raising temperature to 450 °C from 25 °C (~negligible to 0.0003). The catalyst is so effective because it changes the proportion of collisions that are effective — the fundamental quality of each collision — by lowering E_a. Increasing concentration only increases the number of collisions without changing whether those collisions can succeed. The large E_a for the uncatalysed Haber reaction makes the catalyst essential: at realistic temperatures, the uncatalysed E_a is simply too high for a commercially viable rate. [1 — justified conclusion distinguishing catalyst from concentration effect; quality vs quantity of collisions reasoning]

Marking criteria:

1 mark — Defines effective collision and states both conditions (energy ≥ E_a AND correct orientation).
2 marks — Explains Trials 1 vs 2 correctly using the Maxwell–Boltzmann distribution and E_a: at 25 °C virtually no particles exceed E_a (rate ~0); at 450 °C a measurable fraction does (rate = 0.0003). Award 1 mark if only one of the two temperatures is fully explained.
2 marks — Explains Trials 2 vs 3 correctly: catalyst lowers E_a → larger fraction of particles in existing distribution exceeds the lower E_a → greater effective collision rate. Penalise if student states catalyst changes the distribution shape or ΔH.
1 mark — Explains Trials 3, 4, 5 correctly: concentration increases collision frequency → more effective collisions per second (proportion unchanged).
1 mark — Reaches a justified conclusion identifying the catalyst as the factor with the largest rate increase and correctly explaining why (changes quality of collisions by lowering E_a, not just quantity).
1 mark — Uses precise lesson terminology throughout: effective collision, activation energy, Maxwell–Boltzmann distribution, collision frequency, proportion of effective collisions, transition state.

Q2 — Sample Band 6 source-critique response (7 marks)

The student's notes contain three scientific errors.

Error 1: "hot water lowers the activation energy." This is incorrect. The activation energy (E_a) is a fixed property of a particular reaction that depends on the reaction mechanism — it is not changed by temperature. Hot water increases the average kinetic energy of the particles inside the glow stick. This means a larger proportion of those particles have kinetic energy ≥ E_a, so more collisions per second are effective. On an energy diagram, the transition state peak (which sets E_a) remains at exactly the same height whether the glow stick is placed in hot water or icy water — only the distribution of particle energies changes. [2 marks: 1 identifies error, 1 explains correct chemistry]

Error 2: "adding a catalyst increases the enthalpy change (ΔH)." This is incorrect. A catalyst provides an alternative reaction pathway with a lower activation energy (E_a). It does NOT change the enthalpy change (ΔH) of the reaction. ΔH depends on the energy levels of the reactants and products, which are unchanged by a catalyst. On an energy profile diagram, adding a catalyst lowers the transition state peak (lower E_a) but leaves the reactant and product energy levels — and therefore ΔH — exactly the same. The Common Error from the lesson applies here: faster rate does not mean more energy is released per mole of product; it means more product is formed per second. [2 marks: 1 identifies error, 1 explains correct chemistry]

Error 3: "the rate stays constant throughout as long as the temperature doesn't change." This is incorrect. Reaction rate decreases over time even at constant temperature, because the concentration of reactants falls as they are consumed. Lower reactant concentration means fewer collisions per second (lower collision frequency). Since temperature and E_a are constant, the proportion of effective collisions is unchanged, but fewer total collisions means fewer effective collisions per second — and rate falls. On a volume-of-gas-vs-time graph the curve would be steep early and flatten toward the end, confirming the rate is not constant. [2 marks: 1 identifies error, 1 explains correct chemistry]

Accurate rewrite: "A glow stick glows more brightly in hot water because the hot water increases the average kinetic energy of the reactant particles inside. A larger proportion of these particles now have kinetic energy ≥ E_a, so more effective collisions occur per second and the reaction rate increases — the glow is brighter. The activation energy (E_a) is not changed by temperature: it is a fixed property of the reaction. Adding a catalyst also speeds up the reaction, but by lowering E_a, not by changing ΔH; the enthalpy change of the reaction is unaffected by a catalyst. The rate of a reaction is not constant even at constant temperature — as reactants are consumed, their concentration falls, collision frequency decreases, and the rate slows progressively." [1 mark: accurate reformulation in correct collision theory language]

Marking criteria:

2 marks — Error 1: identifies that hot water does not lower E_a (1), and explains that hot water increases average kinetic energy of particles so a larger proportion exceed E_a — E_a itself is a fixed property of the reaction (1).
2 marks — Error 2: identifies that a catalyst does not change ΔH (1), and explains that a catalyst lowers E_a only; reactant and product energy levels (and therefore ΔH) are unchanged (1).
2 marks — Error 3: identifies that rate is not constant at constant temperature (1), and explains that falling reactant concentration reduces collision frequency, reducing rate over time (1).
1 mark — Accurately rewrites the student's notes using correct collision theory language throughout (E_a, kinetic energy, collision frequency, effective collision, ΔH, reaction rate).