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. Reactions and data are invented unless a real substance is named. Energies are in kJ mol⁻¹ and are measured from the reactants unless stated.
Question 1 (multiple choice · foundation)
A small amount of a catalyst is added to a reaction mixture kept at constant temperature. Which statement describes how the catalyst increases the rate?
- (A) It lowers the enthalpy change, ΔH, so less energy is needed for the reaction.
- (B) It provides an alternative reaction path with a lower activation energy, so a larger fraction of collisions are effective.
- (C) It increases the average kinetic energy of the reactant particles, so they collide harder.
- (D) It reacts completely with one reactant, so it must be added in a stoichiometric amount.
Answer and explanation
Answer: (B). A catalyst changes the mechanism. The new path has a lower highest barrier, so at the same temperature more collisions have enough energy to react.
- (A) ΔH depends only on the energies of reactants and products, which a catalyst does not change.
- (C) At constant temperature the average kinetic energy is unchanged. A catalyst moves the energy threshold, not the energy distribution.
- (D) A catalyst is regenerated in a later step, so a small amount can convert a large amount of reactant.
Question 2 (multiple choice · core)
The reaction X₂ + 2 Y → 2 XY is thought to follow this mechanism:
- Step 1: X₂ + M → X₂M
- Step 2: X₂M + Y → XY + XM
- Step 3: XM + Y → XY + M
Which statement is correct?
- (A) M is a catalyst; X₂M and XM are intermediates.
- (B) M is an intermediate; X₂M is a catalyst.
- (C) XM is a catalyst because it appears in two steps.
- (D) M is a reactant because it is on the left side of step 1.
Answer and explanation
Answer: (A). M is a reactant in step 1 and a product in step 3, so it is used up and regenerated: a catalyst. X₂M is made in step 1 and used in step 2; XM is made in step 2 and used in step 3. Both are intermediates. Adding the steps and cancelling X₂M, XM and M gives X₂ + 2 Y → 2 XY.
- (B) reverses the definitions: an intermediate is produced first and consumed later, which describes X₂M, not M.
- (C) Appearing in two steps is not the test. XM is produced before it is consumed, so it is an intermediate.
- (D) M is on the left in step 1 but is given back in step 3, so it cancels and is not a reactant overall.
Question 3 (multiple choice · core)
An uncatalysed reaction has a forward activation energy of 120 kJ mol⁻¹ and ΔH = +30 kJ mol⁻¹. With a catalyst, the highest transition state on the path is 70 kJ mol⁻¹ above the reactants. What is the activation energy for the reverse reaction on the catalysed path?
- (A) 40 kJ mol⁻¹
- (B) 70 kJ mol⁻¹
- (C) 90 kJ mol⁻¹
- (D) 100 kJ mol⁻¹
Answer and explanation
Answer: (A). The products are 30 kJ mol⁻¹ above the reactants. The reverse reaction starts at the products and must reach the highest point, 70 kJ mol⁻¹ above the reactants: 70 − 30 = 40 kJ mol⁻¹.
- (B) is the forward barrier on the catalysed path.
- (C) is the reverse barrier on the uncatalysed path (120 − 30). The catalyst lowers this too.
- (D) adds ΔH instead of subtracting it. That would be right only if the products were 30 kJ mol⁻¹ below the reactants.
Question 4 (calculation · core)
The reaction A + B → P is catalysed by a dissolved metal ion, Cat. Initial rates were measured at constant temperature.
| Experiment | [A] (M) | [B] (M) | [Cat] (M) | Initial rate (M s⁻¹) |
|---|---|---|---|---|
| 1 | 0.10 | 0.10 | 0.0020 | 3.0 × 10⁻⁵ |
| 2 | 0.20 | 0.10 | 0.0020 | 6.0 × 10⁻⁵ |
| 3 | 0.10 | 0.30 | 0.0020 | 3.0 × 10⁻⁵ |
| 4 | 0.10 | 0.10 | 0.0050 | 7.5 × 10⁻⁵ |
(a) Find the order with respect to A, B and Cat. (b) Write the rate law and calculate k, with units. (c) Propose a two-step mechanism consistent with the rate law, with Cat consumed in the slow step. (d) Calculate the initial rate when [A] = 0.25 M, [B] = 0.20 M and [Cat] = 0.0040 M.
Worked solution and suggested Marlbridge rubric
(a) Experiments 1 → 2: [A] doubles, rate doubles: first order in A. Experiments 1 → 3: [B] triples, rate unchanged: zero order in B. Experiments 1 → 4: [Cat] × 2.5, rate × 2.5 (7.5 ÷ 3.0): first order in Cat.
(b) rate = k[A][Cat]. From experiment 1: k = 3.0 × 10⁻⁵ ÷ (0.10 × 0.0020) = 0.15 M⁻¹ s⁻¹. (Every experiment gives the same value.)
(c) Step 1 (slow): A + Cat → ACat. Step 2 (fast): ACat + B → P + Cat. The steps add to A + B → P; the slow first step gives rate = k[A][Cat]; B enters only after the slow step, so it is not in the rate law.
(d) rate = 0.15 × 0.25 × 0.0040 = 1.5 × 10⁻⁴ M s⁻¹.
| Point | What earns it |
|---|---|
| 1 | All three orders, each supported by a pair of experiments |
| 1 | Rate law rate = k[A][Cat] |
| 1 | k = 0.15 with units M⁻¹ s⁻¹ |
| 1 | Any valid mechanism: one A and one Cat in the slow first step, Cat regenerated later, B after the slow step, steps add up |
| 1 | 1.5 × 10⁻⁴ M s⁻¹ (carry forward an incorrect k from (b)) |
Question 5 (constructed response · core)
A compound S reacts with water: S + H₂O → P + Q. The reaction is very slow in pure water but fast when acid is added. A proposed mechanism is:
- Step 1 (fast, reversible): S + H₃O⁺ ⇌ SH⁺ + H₂O
- Step 2 (slow): SH⁺ + H₂O → T⁺
- Step 3 (fast): T⁺ + H₂O → P + Q + H₃O⁺
(a) Identify the catalyst and every intermediate. Justify your choices. (b) Explain, in terms of the particles, why adding acid increases the rate. (c) The experimental rate law is rate = k[S][H₃O⁺], with k = 2.4 M⁻¹ s⁻¹. Calculate the rate when [S] = 0.050 M and [H₃O⁺] = 0.010 M, and again when [H₃O⁺] = 0.0010 M. (d) A student says: “[H₃O⁺] should fall during the reaction, because H₃O⁺ is a reactant in step 1.” Evaluate this claim.
Worked solution and suggested Marlbridge rubric
(a) Catalyst: H₃O⁺. It is a reactant in step 1 and is regenerated in step 3. Intermediates: SH⁺ and T⁺. Each is produced in one step and consumed in a later step. (Adding the steps gives S + H₂O → P + Q.)
(b) In acid, S gains a proton to form SH⁺. This is a new intermediate that reacts with water by a new elementary step (step 2) with a lower activation energy than any step available to neutral S. More collisions with water are therefore effective, so the rate increases. More H₃O⁺ means more S is converted to SH⁺ at any moment.
(c) rate = 2.4 × 0.050 × 0.010 = 1.2 × 10⁻³ M s⁻¹. With [H₃O⁺] = 0.0010 M: rate = 2.4 × 0.050 × 0.0010 = 1.2 × 10⁻⁴ M s⁻¹, ten times slower.
(d) The claim is incorrect. H₃O⁺ is used in step 1 but made again in step 3, so it cancels from the overall equation. Its total concentration stays constant during the reaction (apart from the small amount tied up as SH⁺ and T⁺ at any moment).
| Point | What earns it |
|---|---|
| 1 | H₃O⁺ as catalyst, with “used in step 1, regenerated in step 3” |
| 1 | SH⁺ and T⁺ as intermediates, with “made, then used” |
| 1 | Protonation gives a new intermediate and a new path with a lower activation energy, so more effective collisions |
| 1 | Both rates correct (1.2 × 10⁻³ and 1.2 × 10⁻⁴ M s⁻¹) |
| 1 | Rejects the claim and links it to regeneration in step 3 |
Question 6 (constructed response · core)
Ethene reacts with hydrogen to form ethane, C₂H₄ + H₂ → C₂H₆. The reaction is extremely slow as a gas mixture but fast over finely divided nickel metal.
(a) Describe, in terms of bonds and surface-bound species, how the nickel surface provides a new reaction path. (b) Name one new intermediate that is present on the catalysed path but not on the uncatalysed path. (c) Two reaction vessels each hold 1.0 g of nickel: one as a fine powder, one as a single small cube. Predict which gives the faster reaction and justify your prediction at the particle level. (d) A student says: “Nickel is a reactant, because nickel atoms form bonds to hydrogen.” Explain why nickel is classed as a catalyst instead.
Worked solution and suggested Marlbridge rubric
(a) H₂ molecules bind to nickel atoms on the surface, and the H–H bond breaks, leaving H atoms bonded to the surface. Ethene also binds to the surface. Surface H atoms then add to the ethene carbons in separate, low-barrier elementary steps, and ethane leaves. No high-barrier H₂–C₂H₄ collision is needed.
(b) Any one: H atoms bonded to the nickel surface; ethene bound to the surface; a surface-bound C₂H₅ group (ethene after one H has been added).
(c) The powder reacts faster. Only nickel atoms on the surface can bind reactants. The powder exposes far more surface atoms for the same mass, so more molecules bind and react at once.
(d) Nickel bonds to H and ethene, but these bonds break when ethane leaves. The nickel is regenerated unchanged and binds more reactant, so it is not in the overall equation: it is a catalyst.
| Point | What earns it |
|---|---|
| 1 | H₂ (or ethene) bonds to the surface, weakening or breaking the H–H bond |
| 1 | H atoms added in new elementary steps with lower barriers |
| 1 | A correct surface-bound intermediate |
| 1 | Powder faster, because more surface atoms (sites) per gram |
| 1 | Nickel regenerated / unchanged at the end, so not consumed overall |
Question 7 (explanation · stretch)
An enzyme, E, catalyses the conversion of a substrate into a product in a cell. Without the enzyme the activation energy is 84 kJ mol⁻¹. With the enzyme, the substrate first binds to the active site to form an enzyme–substrate complex, and the highest barrier on the path is 50 kJ mol⁻¹. The temperature is the same in both cases.
(a) A student claims: “The enzyme speeds up the reaction by giving the substrate molecules more kinetic energy.” Evaluate this claim. (b) Give two separate particle-level reasons why binding to the enzyme increases the rate. (c) Describe how the energy profile of the enzyme-catalysed path differs from the uncatalysed profile. Include the barrier values and what stays the same.
Model answer and suggested Marlbridge rubric
(a) The claim is incorrect. At the same temperature the substrate molecules have the same spread of kinetic energies. The enzyme lowers the barrier from 84 to 50 kJ mol⁻¹ (a drop of 34 kJ mol⁻¹), so a much larger fraction of collisions has enough energy.
(b)
- Orientation: the active site holds the substrate in place, so reacting groups meet in the right orientation and more encounters are effective.
- Lower activation energy: binding forms a new intermediate (the enzyme–substrate complex) and a new path whose elementary steps have lower barriers, for example because bonds in the bound substrate are strained or weakened.
(c) The catalysed profile has at least two humps with a dip between them for the enzyme–substrate complex (a new intermediate). Its highest point is 50 kJ mol⁻¹ above the reactants, compared with a single hump at 84 kJ mol⁻¹. Reactants and products are at the same energies on both, so ΔH is unchanged; the reverse barrier also drops by 34 kJ mol⁻¹.
| Point | What earns it |
|---|---|
| 1 | Rejects the claim: same temperature, same energy distribution |
| 1 | Lower barrier means a larger fraction of collisions have enough energy |
| 1 | Orientation reason linked to binding in the active site |
| 1 | New intermediate (enzyme–substrate complex) and new elementary steps |
| 1 | Profile: extra hump and dip, highest point 50 vs 84, reactant and product energies (ΔH) unchanged |
How did you do?
- Q1 or Q7(a) wrong: re-read “Why a lower barrier means a faster reaction” and Figure 2 in the study guide.
- Q2 or Q5(a) wrong: catalyst in first; intermediate out first.
- Q3 wrong: redo Worked example 2; measure every barrier from where that direction starts.
- Q4 wrong: redo Worked example 1.
- Q5(b), Q6 or Q7 incomplete: say how: new intermediate, new step, lower barrier or better orientation.
Then tick off the topic checklist.