These are original Marlbridge practice questions, not past exam questions. The mark points are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. All reactions and data are hypothetical. Energies are in kJ mol⁻¹. For sketches, draw on paper and compare with the description in the solution.
Questions 1 to 3 refer to Figure 1.
Question 1 (multiple choice · foundation)
How many elementary steps and how many intermediates does the mechanism in Figure 1 have?
- (A) 3 steps and 3 intermediates
- (B) 3 steps and 2 intermediates
- (C) 2 steps and 3 intermediates
- (D) 6 steps and 2 intermediates
Answer and explanation
Answer: (B). There are three peaks (P, R, T), one transition state per elementary step, so three steps. The two valleys between peaks (Q and S) are intermediates.
- (A) counts the products (or reactants) as an intermediate. They are the start and end of the overall reaction.
- (C) swaps the counts: it treats valleys as steps.
- (D) counts every rise and every fall as a separate step. A rise and the fall after it are one step.
Question 2 (multiple choice · core)
Which step in Figure 1 is rate-limiting, and what is its activation energy?
- (A) Step 1, 35 kJ mol⁻¹
- (B) Step 2, 85 kJ mol⁻¹
- (C) Step 2, 70 kJ mol⁻¹
- (D) Step 3, 40 kJ mol⁻¹
Answer and explanation
Answer: (C). Measure each barrier from the valley where the step starts. Step 1: 75 − 40 = 35. Step 2: 125 − 55 = 70. Step 3: 70 − 30 = 40. Step 2 has the largest barrier.
- (A) picks step 1 because it comes first. The first step is not automatically the slowest.
- (B) measures step 2’s barrier from the reactants (125 − 40). Step 2 starts at intermediate Q, not at the reactants.
- (D) has the right value for step 3, but 40 is smaller than 70. The large drop after T is a large negative ΔH, not a large barrier.
Question 3 (multiple choice · core)
Which statement about the reaction in Figure 1 is correct?
- (A) ΔH overall = −40 kJ mol⁻¹, and step 1 is endothermic.
- (B) ΔH overall = +40 kJ mol⁻¹, and step 1 is endothermic.
- (C) ΔH overall = −40 kJ mol⁻¹, and every step is exothermic.
- (D) ΔH overall = −125 kJ mol⁻¹, the drop from the highest peak to the products.
Answer and explanation
Answer: (A). ΔH overall = products − reactants = 0 − 40 = −40 kJ mol⁻¹. Step 1 goes from 40 up to Q at 55, so its ΔH is +15 kJ mol⁻¹: endothermic. Check: +15 + (−25) + (−30) = −40.
- (B) subtracts the wrong way round (reactants − products).
- (C) has the right ΔH, but step 1 ends higher than it starts.
- (D) measures from a transition state. ΔH compares only the start and end of the reaction.
Question 4 (sketch · core)
A hypothetical two-step reaction has these step energetics (kJ mol⁻¹):
| Step | Ea (forward) | ΔH | Speed |
|---|---|---|---|
| 1 | 90 | +35 | slow |
| 2 | 20 | −80 | fast |
(a) Taking the reactants as 0 kJ mol⁻¹, calculate the energies of TS 1, the intermediate, TS 2 and the products. (b) Sketch and label the energy profile. (c) State the overall ΔH and whether the reaction is exothermic or endothermic.
Worked solution and suggested Marlbridge rubric
(a) TS 1 = 0 + 90 = 90. Intermediate = 0 + 35 = 35. TS 2 = 35 + 20 = 55. Products = 35 − 80 = −45 (kJ mol⁻¹).
(b) A good sketch shows: axes labelled energy and reaction coordinate; a flat start for the reactants; a tall first hump to TS 1; a valley for the intermediate above the reactants; a small second hump to TS 2 that stays well below TS 1; and products ending below the reactants. Label TS 1, TS 2, the intermediate, the reactants, the products, and Ea for step 1 drawn from the reactants up to TS 1.
(c) ΔH overall = −45 − 0 = −45 kJ mol⁻¹, exothermic. Check: +35 − 80 = −45.
| Point | What earns it |
|---|---|
| 1 | All four energies correct |
| 1 | Two humps, with the first clearly taller than the second (measured from their own starting levels) |
| 1 | Intermediate valley above the reactants; products below the reactants |
| 1 | Labels: transition states, intermediate, and Ea for step 1 from the reactants |
| 1 | ΔH = −45 kJ mol⁻¹, exothermic |
Accept sketches that are not to scale, as long as the relative heights are correct.
Question 5 (multiple choice · stretch)
A hypothetical reaction 2E + F → E₂F has the mechanism below. Its profile has energies (kJ mol⁻¹): reactants 0, TS 1 at 25, intermediate EF at 15, TS 2 at 95, products −30.
Step 1: E + F ⇌ EF Step 2: EF + E → E₂F
Which rate law is consistent with the profile and the mechanism?
- (A) rate = k[E][F]
- (B) rate = k[EF][E]
- (C) rate = k[E][F]²
- (D) rate = k[E]²[F]
Answer and explanation
Answer: (D). Step 1 has a barrier of 25; step 2 has a barrier of 95 − 15 = 80, so step 2 is slow. EF can go back over TS 1 with only 25 − 15 = 10 kJ mol⁻¹, so step 1 is a fast pre-equilibrium. Slow step: rate = k₂[EF][E]. Pre-equilibrium: [EF] = (k₁ / k₋₁)[E][F]. Rate = k[E]²[F].
- (A) is the rate law if step 1 were rate-limiting. The profile shows step 1 has the smaller barrier.
- (B) leaves the intermediate EF in the rate law.
- (C) puts the square on the wrong species. Two E particles (one in each step) and one F go into the slow step’s transition state.
Question 6 (constructed response · core)
A hypothetical two-step reaction has a profile with these energies (kJ mol⁻¹): reactants 0, TS 1 at 45, intermediate at 30, TS 2 at 110, products 10.
(a) Calculate the activation energy and ΔH for each step. (b) Calculate the overall ΔH and show that it equals the sum of the steps’ ΔH values. (c) Identify the rate-limiting step and explain, in terms of particle collisions, why it is slower. (d) Calculate the activation energy for the reverse of step 2.
Worked solution and suggested Marlbridge rubric
(a) Step 1: Ea = 45 − 0 = 45; ΔH = 30 − 0 = +30. Step 2: Ea = 110 − 30 = 80; ΔH = 10 − 30 = −20 (kJ mol⁻¹).
(b) ΔH overall = 10 − 0 = +10 kJ mol⁻¹ (endothermic). Sum: +30 + (−20) = +10. ✓
(c) Step 2 has the larger activation energy (80 > 45). At a given temperature, only a small fraction of collisions between particles have at least the activation energy. The larger the activation energy, the smaller that fraction, so fewer collisions per second succeed in step 2 and it is slower.
(d) The reverse of step 2 starts at the products (10) and climbs to TS 2 (110): 100 kJ mol⁻¹. (Or Ea − ΔH = 80 − (−20) = 100.)
| Point | What earns it |
|---|---|
| 1 | Both activation energies, step 2 measured from the intermediate |
| 1 | Both step ΔH values with correct signs |
| 1 | Overall ΔH = +10 kJ mol⁻¹ with the sum shown |
| 1 | Step 2 identified, with a reason based on its larger Ea |
| 1 | Collision explanation: smaller fraction of collisions with enough energy |
| 1 | Reverse Ea = 100 kJ mol⁻¹ |
Question 7 (explanation · stretch)
A hypothetical reaction A + B + C → P goes by two elementary steps: step 1, A + B → I; step 2, I + C → P. Experiments give rate = k[A][B].
(a) Identify the rate-limiting step and justify your answer using the rate law. (b) Describe, without numbers, the shape of an energy profile consistent with this, saying how each barrier is measured. (c) A student draws the intermediate’s valley below the products. Is that possible? Explain.
Model answer and suggested Marlbridge rubric
(a) Step 1. Its molecularity gives rate = k₁[A][B], which matches the experiment. If step 2 were slow, C would appear in the rate law (with I replaced by an expression in A and B), but the rate does not depend on [C].
(b) Two humps. The first climb, from the reactants up to TS 1, is the larger barrier. The second climb, measured from the intermediate’s valley up to TS 2, is smaller. There is one valley (intermediate I) between the humps.
(c) Yes. It just means step 2 is endothermic (the products are above I). The rate law only requires step 1 to have the larger barrier. Even with I in a deep valley, the climb from I to TS 2 can still be smaller than the climb in step 1. Overall, the products can be above or below the reactants: the rate law says nothing about ΔH.
| Point | What earns it |
|---|---|
| 1 | Step 1, with the rate law matched to its molecularity (or the absence of C explained) |
| 1 | Profile with two humps, first barrier larger, each barrier measured from its own starting level |
| 1 | Valley below the products is possible: step 2 is endothermic |
| 1 | Explains that only the relative sizes of the barriers decide the rate-limiting step |
How did you do?
- Q1 or Q3 wrong: re-read “Anatomy of a multistep profile” in the study guide.
- Q2 or Q6 wrong: practise measuring each barrier from its own starting valley (Worked example 1).
- Q4 wrong: work through Worked example 2 again, building the table of heights first.
- Q5 or Q7 wrong: revisit the pre-equilibrium approximation and the section “Which step is rate-limiting?”.
Then tick off the topic checklist.