How to use this review
Use this page after the Unit 5 diagnostic. Answer all seven questions on paper before opening any answer.
These are original Marlbridge practice questions, not past exam questions. Species written with letters are fictional and all data are invented. Each rubric is a suggested Marlbridge rubric, not official scoring, and your total gives no predicted score.
Big ideas of the unit
- Rates of different species are linked by coefficients and are always positive (Topic 5.1).
- An overall rate law is measured, not read from the equation. Initial rates give the orders and k (Topic 5.2); so does whichever of [A], ln[A] or 1/[A] against t is straight (Topic 5.3).
- For one elementary step, the coefficients are the orders. Three-particle collisions are rare (Topic 5.4).
- Rate depends on collision frequency, energy and orientation. Concentration changes the first; temperature mainly the second (Topic 5.5).
- Eₐ is the climb to the transition state. A higher barrier gives a slower, more temperature-sensitive step (Topic 5.6).
- A mechanism must add up, use reasonable steps and match the rate law. Intermediates are made then used; catalysts used then given back (Topics 5.7 and 5.8).
- If a fast reversible step comes before the slow step, substitute out the intermediate using forward rate = reverse rate (Topic 5.9).
- A multistep profile has one hump per step, each barrier measured from where its step starts (Topic 5.10). A catalyst gives a new path with a lower highest point, leaves ΔH unchanged and can appear in the rate law (Topic 5.11).
Key relationships and methods
| Idea | Relationship or method | Topic |
|---|---|---|
| Rates of species | rate = −(1/a) Δ[A]/Δt = (1/c) Δ[C]/Δt | 5.1 |
| Rate law | rate = k[A]ᵐ[B]ⁿ; rate ratio = (concentration ratio)^order; k in M^(1 − overall order) s⁻¹ | 5.2 |
| Integrated laws | zero: [A] vs t slope −k; first: ln[A] vs t slope −k, constant t½ = 0.693/k; second: 1/[A] vs t slope +k | 5.3 |
| Elementary step | orders = coefficients; molecularity = particles colliding | 5.4 |
| Temperature | k = A e^(−Eₐ/RT): used to explain, never to calculate | 5.5, 5.6 |
| Profile | Eₐ(reverse) = Eₐ(forward) − ΔH; overall ΔH = sum of step ΔH | 5.6, 5.10 |
| Mechanism rate law | slow first step: its own rate law; fast equilibrium first: forward rate = reverse rate, then substitute | 5.8, 5.9 |
| Catalyst | used then regenerated; in the rate law if it reacts at or before the slow step | 5.11 |
Question 1 (multiple choice · mixed)
The reaction 2A → B is first order in A, with −Δ[A]/Δt = k[A] and a half-life of 150 s. At what rate is B forming when [A] = 0.40 M?
- (A) 3.7 × 10⁻³ M s⁻¹
- (B) 2.7 × 10⁻³ M s⁻¹
- (C) 1.8 × 10⁻³ M s⁻¹
- (D) 9.2 × 10⁻⁴ M s⁻¹
Answer and explanation
Answer: (D). k = 0.693 ÷ 150 s = 4.62 × 10⁻³ s⁻¹ (Topic 5.3). −Δ[A]/Δt = 4.62 × 10⁻³ × 0.40 = 1.85 × 10⁻³ M s⁻¹ (Topic 5.2). B forms half as fast: 9.2 × 10⁻⁴ M s⁻¹ (Topic 5.1).
- (A) doubles the rate for A instead of halving it.
- (B) is 0.40 ÷ 150.
- (C) is the rate for A.
Question 2 (multiple choice · mixed)
A reaction goes by two steps: step 1 (slow, Eₐ = 90 kJ mol⁻¹) then step 2 (fast, Eₐ = 25 kJ mol⁻¹). It is warmed by 10 K at the same concentrations. Why does the rate rise?
- (A) Step 2’s rate constant increases by the larger factor.
- (B) A larger fraction of step 1 collisions can reach TS 1; step 1 stays rate-limiting.
- (C) Heating lowers step 1’s Eₐ towards step 2’s.
- (D) Collisions are more frequent; the fraction with enough energy hardly changes.
Answer and explanation
Answer: (B). The slow step sets the rate (Topic 5.8). Its barrier lies far out in the Maxwell–Boltzmann tail, so warming multiplies the fraction of step 1 collisions with enough energy (Topics 5.5 and 5.6). k₁ grows by a larger factor than k₂, so step 1 stays the bottleneck (Topic 5.10).
- (A) Step 2 is not the bottleneck, and its lower barrier makes it less temperature-sensitive.
- (C) Temperature does not change Eₐ.
- (D) Collision frequency rises only slightly.
Question 3 (multiple choice · mixed)
For 2A + B → A₂B, experiments give rate = k[A]²[B]. Which mechanism adds up to the overall equation, predicts this rate law and uses only unimolecular or bimolecular steps?
- (A) One step: 2A + B → A₂B
- (B) A + B → AB (slow); AB + A → A₂B (fast)
- (C) A + A ⇌ A₂ (fast, reversible); A₂ + B → A₂B (slow)
- (D) A + A ⇌ A₂ (fast, reversible); A₂ + B → AB + A (slow)
Answer and explanation
Answer: (C). The steps add to 2A + B → A₂B (Topic 5.7). The slow step gives rate = k₂[A₂][B]; the pre-equilibrium gives [A₂] = (k₁/k₋₁)[A]², so rate = k[A]²[B] (Topic 5.9). Both steps are bimolecular.
- (A) fits, but it is termolecular, which is rare (Topic 5.4).
- (B) adds up but predicts rate = k[A][B].
- (D) has the right rate law, but its steps add to A + B → AB.
Question 4 (constructed response · mixed)
Initial rates for M₂ + 2N → 2MN at one temperature:
| Experiment | [M₂] (M) | [N] (M) | Initial rate (M s⁻¹) |
|---|---|---|---|
| 1 | 0.040 | 0.050 | 3.0 × 10⁻⁵ |
| 2 | 0.160 | 0.050 | 6.0 × 10⁻⁵ |
| 3 | 0.160 | 0.150 | 1.8 × 10⁻⁴ |
(a) Determine the order with respect to M₂ and to N. (b) Write the rate law and calculate k with units. (c) Explain why this reaction cannot be a single elementary step. (d) Mechanism I: M₂ + N → MN + M (slow); M + N → MN (fast). Mechanism II: M₂ ⇌ 2M (fast, reversible); M + N → MN (slow, happens twice). Which is consistent with the data? (e) Predict the initial rate when [M₂] = 0.090 M and [N] = 0.20 M.
Worked solution and suggested Marlbridge rubric
(a) Experiments 1 → 2: [M₂] × 4, rate × 2. 4ᵐ = 2, so m = ½. Experiments 2 → 3: [N] × 3, rate × 3, so first order in N.
(b) rate = k[M₂]^½[N]. k = 3.0 × 10⁻⁵ ÷ (0.20 × 0.050) = 3.0 × 10⁻³ M^(−½) s⁻¹ (overall order 1½). All three experiments agree.
(c) An elementary step’s orders are its whole-number coefficients, so an order of ½ cannot come from one collision.
(d) Both mechanisms add up to M₂ + 2N → 2MN.
- Mechanism I: slow first step, so rate = k[M₂][N]. First order in M₂: rejected.
- Mechanism II: slow step rate = k₂[M][N], with M an intermediate. k₁[M₂] = k₋₁[M]², so [M] = (K₁[M₂])^½ and rate = k₂K₁^½[M₂]^½[N]. Consistent (not proven).
(e) rate = 3.0 × 10⁻³ × (0.090)^½ × 0.20 = 3.0 × 10⁻³ × 0.30 × 0.20 = 1.8 × 10⁻⁴ M s⁻¹.
| Point | What earns it |
|---|---|
| 1 | (a) Half order in M₂ from experiments 1 and 2 |
| 1 | (a) First order in N from experiments 2 and 3 |
| 1 | (b) Rate law and k = 3.0 × 10⁻³ |
| 1 | (b) Units M^(−½) s⁻¹ |
| 1 | (c) Elementary orders are whole-number coefficients |
| 1 | (d) Mechanism I rejected because it predicts first order in M₂ |
| 1 | (d) Mechanism II: [M] substituted to give [M₂]^½[N] |
| 1 | (e) 1.8 × 10⁻⁴ M s⁻¹ (carry forward from (b)) |
Total: 8 points.
Question 5 (constructed response · mixed)
Compound A decomposes: A → 2B. Data at 300 K:
| t (s) | 0 | 50 | 100 | 150 |
|---|---|---|---|---|
| [A] (M) | 0.0800 | 0.0566 | 0.0400 | 0.0283 |
At 320 K, with the same starting concentration, the half-life is 25 s.
(a) Show with calculated values that the reaction is first order in A. (b) Find k at 300 K, with units. (c) Calculate the rate of formation of B at t = 100 s at 300 K. (d) Find k at 320 K, and the time at 320 K for [A] to fall from 0.0800 M to 0.00500 M. (e) Use a sketch of Maxwell–Boltzmann curves to explain why k is four times larger at 320 K. Evaluate a student’s claim that Eₐ is lower at 320 K.
Worked solution and suggested Marlbridge rubric
(a) ln[A]: −2.526, −2.872, −3.219, −3.565, falling by a constant 0.346–0.347 per 50 s, so ln[A] against t is straight. 1/[A] rises by 5.2, 7.3, 10.3 M⁻¹: not constant. First order. (Also: [A] halves every 100 s.)
(b) Slope = −0.00693 s⁻¹, so k = 6.93 × 10⁻³ s⁻¹ (0.693 ÷ 100 s).
(c) −Δ[A]/Δt = k[A] = 6.93 × 10⁻³ × 0.0400 = 2.77 × 10⁻⁴ M s⁻¹. B forms twice as fast: 5.54 × 10⁻⁴ M s⁻¹.
(d) k = 0.693 ÷ 25 s = 0.0277 s⁻¹, four times the 300 K value. 0.0800 → 0.00500 M is four half-lives: 100 s.
(e) Sketch number of particles against kinetic energy. The 320 K curve has a lower peak, shifted to higher energy, with a thicker tail and the same area. A larger area lies beyond Eₐ at 320 K, so a larger fraction of collisions reach the transition state; collisions are only slightly more frequent. The student is wrong: Eₐ does not change with temperature.
| Point | What earns it |
|---|---|
| 1 | (a) ln[A] values with constant differences (or constant half-life) |
| 1 | (a) 1/[A] (or [A]) shown not to be linear |
| 1 | (b) k = 6.93 × 10⁻³ s⁻¹ with units |
| 1 | (c) 5.54 × 10⁻⁴ M s⁻¹, doubling the rate for A |
| 1 | (d) 0.0277 s⁻¹ and 100 s |
| 1 | (e) Sketch: lower, right-shifted peak at 320 K |
| 1 | (e) Larger fraction beyond Eₐ named as the main cause |
| 1 | (e) Rejects the claim: Eₐ does not change with temperature |
Total: 8 points. Do not award the sketch point if the 320 K peak is drawn taller.
Question 6 (constructed response · mixed)
Uncatalysed, R + T → RT is one step with Eₐ = 120 kJ mol⁻¹ and ΔH = −30 kJ mol⁻¹. A trace of Z makes it much faster, by this proposed mechanism:
Step 1: T + Z ⇌ TZ Step 2: TZ + R → RTZ Step 3: RTZ → RT + Z
On the catalysed profile, relative to the reactants at 0 kJ mol⁻¹: TS 1 is at 34, TZ at 20, TS 2 at 72, RTZ at −10, TS 3 at 15 and the products at −30.
(a) Show that the steps add up, and identify the catalyst and the intermediates with reasons. (b) Calculate Eₐ for each catalysed step and identify the rate-limiting step. (c) Use the profile to explain why step 1 can be treated as a fast equilibrium, then derive the rate law. (d) Predict the effect on the rate of doubling [T] while halving [Z]. (e) A student says: “Step 1 is uphill, so it must be the slow step.” Evaluate this, and show that ΔH is the same with and without Z.
Worked solution and suggested Marlbridge rubric
(a) Cancel TZ, RTZ and Z from both sides: R + T → RT. Z is the catalyst: used in step 1, given back in step 3. TZ and RTZ are intermediates: each made in one step, used in a later one.
(b) Step 1: 34 − 0 = 34; step 2: 72 − 20 = 52; step 3: 15 − (−10) = 25 kJ mol⁻¹. Step 2 has the largest barrier, and TS 2 is the highest point: step 2 is rate-limiting.
(c) From TZ, the way back over TS 1 is 34 − 20 = 14 kJ mol⁻¹, far less than the 52 kJ mol⁻¹ climb over TS 2. So TZ usually falls apart before it reacts with R: step 1 stays near equilibrium. Slow step: rate = k₂[TZ][R]. k₁[T][Z] = k₋₁[TZ], so [TZ] = (k₁/k₋₁)[T][Z] and rate = k[R][T][Z].
(d) rate ∝ [T][Z]: 2 × ½ = 1, no change.
(e) Wrong. Barriers decide speed, not a step’s ΔH: step 1’s barrier (34) is smaller than step 2’s (52). Step ΔH values +20, −30 and −20 kJ mol⁻¹ sum to −30 kJ mol⁻¹, the same as without Z.
| Point | What earns it |
|---|---|
| 1 | (a) Correct sum with cancellations |
| 1 | (a) Z catalyst; TZ and RTZ intermediates, with reasons |
| 1 | (b) 34, 52 and 25 kJ mol⁻¹; step 2 rate-limiting |
| 1 | (c) Small return barrier from TZ (14 against 52) justifies the equilibrium |
| 1 | (c) rate = k[R][T][Z] with TZ substituted |
| 1 | (d) No change (2 × ½) |
| 1 | (e) Rejects the claim (barriers decide) and shows ΔH = −30 kJ mol⁻¹ on both paths |
Total: 7 points.
Question 7 (constructed response · mixed)
Gas A dimerises: 2A → A₂. The rate law is written −Δ[A]/Δt = k[A]². The times are not equally spaced.
| t (s) | 0 | 40 | 160 | 240 |
|---|---|---|---|---|
| [A] (M) | 0.0250 | 0.0200 | 0.0125 | 0.0100 |
(a) Show that the data fit a second-order rate law, and find k. (b) Predict the time at which [A] reaches 0.00500 M. (c) A student says: “[A] halved in the first 160 s, so it will be 0.00625 M at 320 s.” Evaluate this. (d) If the reaction is one elementary step, explain at the particle level why rate ∝ [A]². What can the data show about this proposal?
Worked solution and suggested Marlbridge rubric
(a) 1/[A]: 40.0, 50.0, 80.0, 100.0 M⁻¹. Slopes: 10 ÷ 40, 30 ÷ 120, 20 ÷ 80 = 0.250 M⁻¹ s⁻¹ each, so 1/[A] against t is straight. ln[A] slopes are −0.00558, −0.00392, −0.00279 s⁻¹, not constant. Second order; slope = +k, so k = 0.250 M⁻¹ s⁻¹.
(b) t = (1/0.00500 − 1/0.0250) ÷ 0.250 = (200 − 40) ÷ 0.250 = 640 s.
(c) Wrong. Only a first-order half-life is constant. The next half-life (0.0125 → 0.00625 M) is 1 ÷ (0.250 × 0.0125) = 320 s, so [A] reaches 0.00625 M at 480 s. At 320 s, 1/[A] = 40 + 0.250 × 320 = 120 M⁻¹, so [A] = 0.00833 M. Half-lives lengthen as [A] falls.
(d) Two A particles must collide. Doubling [A] doubles the A particles and the partners each can hit, so A–A collisions per second rise fourfold; at constant temperature the successful fraction is unchanged, so rate ∝ [A]². The data are consistent with a bimolecular step but do not prove it: a multistep mechanism could give the same rate law.
| Point | What earns it |
|---|---|
| 1 | (a) Constant 1/[A] slope shown, and ln[A] slope not constant |
| 1 | (a) k = 0.250 M⁻¹ s⁻¹ |
| 1 | (b) 640 s |
| 1 | (c) Rejects the claim: second-order half-lives get longer |
| 1 | (d) Collision-counting reason for [A]² |
| 1 | (d) Consistent with one step, but not proof |
Total: 6 points.
How did you do?
Total: 32 points (3 multiple choice, 29 constructed response), but where you lost them matters more.
- Rates, rate laws or integrated laws (Questions 1, 4(a)–(b), 5(a)–(d), 7(a)–(c)): revisit Topics 5.1 to 5.3.
- Particle-level explanations (Questions 2, 5(e), 7(d)): revisit Topics 5.4 to 5.6.
- Mechanisms and rate laws (Questions 3, 4(c)–(d), 6(a), 6(c)–(d)): revisit Topics 5.7 to 5.9.
- Profiles and catalysts (Questions 6(b), 6(e)): revisit Topics 5.10 and 5.11.
If several topics were weak, return to the Unit 5 diagnostic. Then tick off each topic checklist: 5.1, 5.2, 5.3, 5.4, 5.5, 5.6, 5.7, 5.8, 5.9, 5.10, 5.11.