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 the reactions use fictional species written with letters, so you cannot rely on memory: reason from the steps.
Question 1 (multiple choice · foundation)
A fictional reaction, 2J + K₂ → 2JK, is proposed to occur by this mechanism:
Step 1: J + K₂ → JK + K Step 2: K + J → JK
Which species is an intermediate?
- (A) J
- (B) K₂
- (C) K
- (D) JK
Answer and explanation
Answer: (C). K is made in step 1 and used in step 2, and it cancels when you add the steps. It is not in the overall equation.
- (A) J is a reactant: it is used in both steps and never made.
- (B) K₂ is a reactant: it is used in step 1 and never made back.
- (D) JK is a product: it is made in both steps and never used. It appears in the overall equation, so it cannot be an intermediate.
Question 2 (multiple choice · core)
Consider this two-step mechanism for a fictional reaction:
Step 1: M + Z → MZ Step 2: MZ + W → MW + Z
Which statement is correct?
- (A) Z is a catalyst and MZ is an intermediate.
- (B) MZ is a catalyst and Z is an intermediate.
- (C) Z and MZ are both intermediates.
- (D) Z is a reactant, because it appears on the left of step 1.
Answer and explanation
Answer: (A). Adding the steps gives M + Z + MZ + W → MZ + MW + Z. Cancel Z and MZ: the overall equation is M + W → MW.
- Z is used first (step 1) and given back later (step 2): a catalyst.
- MZ is made first (step 1) and used later (step 2): an intermediate.
(B) swaps the two roles. (C) ignores the order of appearance; both species cancel, but only MZ is made first. (D) forgets that Z is regenerated in step 2, so it is not used up overall and does not appear in the overall equation.
Question 3 (multiple choice · core)
The overall equation for a fictional reaction is 2T + U → T₂U. Which proposed mechanism is consistent with this equation?
- (A) Step 1: T + U → TU. Step 2: TU + T → T₂U.
- (B) Step 1: T + U → TU. Step 2: TU + U → TU₂.
- (C) Step 1: T + T → T₂. Step 2: T₂ + U₂ → T₂U + U.
- (D) Step 1: T + U → TU. Step 2: TU + T₂ → T₂U + T.
Answer and explanation
Answer: (A). Adding: T + U + TU + T → TU + T₂U. Cancel TU: 2T + U → T₂U. This matches.
- (B) adds up to T + 2U → TU₂, a different reaction.
- (C) adds up to 2T + U₂ → T₂U + U. It uses U₂, which is not a reactant in the overall equation.
- (D) adds up to T₂ + U → T₂U (TU and one T cancel). The reactant is T₂, not two separate T particles. Every step in (D) is balanced, which is the trap: balanced steps are not enough.
Question 4 (constructed response · core)
A fictional compound AG₂ reacts with V. The proposed mechanism is:
Step 1: AG₂ → AG + G Step 2: G + V → GV Step 3: AG + V → A + GV
(a) Write the overall balanced equation. (b) Identify every intermediate. Explain how you identified them. (c) State the molecularity of each step. (d) Describe how the concentration of G changes from the start of the reaction to the end.
Worked solution and suggested Marlbridge rubric
(a) Left sides: AG₂ + G + V + AG + V. Right sides: AG + G + GV + A + GV. Cancel AG and G: AG₂ + 2V → A + 2GV.
(b) AG (made in step 1, used in step 3) and G (made in step 1, used in step 2). Each is made before it is used and cancels from the overall equation. There is no catalyst: nothing is used first and given back.
(c) Step 1: unimolecular (one AG₂ particle falls apart). Steps 2 and 3: bimolecular.
(d) G starts at zero, because it is not in the starting mixture. It builds up while the reaction runs (made by step 1, used by step 2) and falls back to zero when the reaction is over.
| Point | What earns it |
|---|---|
| 1 | Correct overall equation, with 2V and 2GV |
| 1 | Both AG and G identified as intermediates |
| 1 | Reason: each is made in an earlier step and used in a later one (so it cancels) |
| 1 | Correct molecularity for all three steps |
| 1 | G starts at zero, rises, then returns to zero (present only while the reaction runs) |
Do not award the point in (b) for “they cancel” alone if a catalyst is also named; the order of appearance must be used.
Question 5 (constructed response · stretch)
A fictional reaction 2RS → R₂ + S₂ is proposed to occur by this mechanism, in which Z is added to the reaction mixture:
Step 1: RS + Z → R + ZS Step 2: ZS + ZS → 2Z + S₂ Step 3: R + R → R₂
(a) A student adds the three steps exactly as written and finds that they do not give the overall equation. Explain why, and show how the steps do combine to give 2RS → R₂ + S₂. (b) The student also claims: “Z is an intermediate because it appears in more than one step.” Evaluate this claim. (c) Name the intermediates in this mechanism.
Worked solution and suggested Marlbridge rubric
(a) Steps 2 and 3 each use two of a species (two ZS, two R) that step 1 makes only one of. So step 1 must happen twice for every one time steps 2 and 3 happen.
2 × Step 1: 2RS + 2Z → 2R + 2ZS Step 2: 2ZS → 2Z + S₂ Step 3: 2R → R₂
Add: 2RS + 2Z + 2ZS + 2R → 2R + 2ZS + 2Z + S₂ + R₂. Cancel 2Z, 2ZS and 2R: 2RS → R₂ + S₂.
(b) The claim is incorrect. Appearing in more than one step is true of intermediates and catalysts. Z is used in step 1 and given back in step 2, and it was added to the mixture at the start, so Z is a catalyst, not an intermediate.
(c) R and ZS: each is made in step 1 and used in a later step (step 3 and step 2 respectively).
| Point | What earns it |
|---|---|
| 1 | Recognises that step 1 must occur twice (doubling the step) |
| 1 | Shows the cancellation leading to 2RS → R₂ + S₂ |
| 1 | States the claim is wrong and uses the order: Z used first, regenerated later, so catalyst |
| 1 | Names R and ZS as the intermediates |
An answer that halves steps 2 and 3 instead (giving half-particles) is not a valid elementary step; do not award the first point for it.
Question 6 (constructed response · core)
A fictional reaction has the overall equation AX + B → A + BX. Two mechanisms are proposed.
Mechanism 1: Step 1: AX → A + X. Step 2: X + B → BX. Mechanism 2: Step 1: AX + B → AXB. Step 2: AXB → A + BX.
(a) Show that both mechanisms are consistent with the overall equation. (b) Identify the intermediate in each mechanism. (c) A spectroscopic study detects the species AXB in the reacting mixture, but no free X atoms. Explain which mechanism this evidence supports, and why it does not prove that mechanism.
Worked solution and suggested Marlbridge rubric
(a) Mechanism 1: AX + X + B → A + X + BX; cancel X: AX + B → A + BX. Mechanism 2: AX + B + AXB → AXB + A + BX; cancel AXB: AX + B → A + BX. Both match.
(b) Mechanism 1: X. Mechanism 2: AXB.
(c) AXB is predicted only by mechanism 2, so detecting it supports mechanism 2. It does not prove it: another mechanism that has not been proposed might also form AXB, and the fact that X was not seen is weaker evidence, because X might be too short-lived or too dilute to detect.
| Point | What earns it |
|---|---|
| 1 | Both sums shown, with the cancelled species identified |
| 1 | Correct intermediate for each mechanism |
| 1 | Detected AXB is predicted only by mechanism 2, so it supports mechanism 2 |
| 1 | A valid limitation: support is not proof, or non-detection of X is weak evidence |
Question 7 (explanation · core)
The overall equation for a fictional reaction is 2A + 2B + C → products.
(a) Explain why this reaction is very unlikely to happen in a single elementary step. (b) State two conditions that any proposed multistep mechanism must meet before it is worth testing further.
Model answer and suggested Marlbridge rubric
(a) As a single step, five particles (2A, 2B and C) would have to collide at exactly the same moment, with the right orientation and enough energy. Even three-particle collisions are rare, so a five-particle collision is extremely unlikely. The reaction almost certainly happens as a series of steps, each involving one or two particles.
(b) Any two of: the steps, when added (with any multipliers) and cancelled, give the overall balanced equation; every step is a reasonable elementary step (normally uni- or bimolecular); every intermediate is made before it is used and cancels out; each step is balanced.
| Point | What earns it |
|---|---|
| 1 | Five particles would need to collide at once, and collisions of three or more particles are rare |
| 1 | First valid condition |
| 1 | Second valid condition (different from the first) |
How did you do?
- Q1, Q2 or Q5(b) wrong: re-read “The parts of a mechanism” and Figure 1 in the study guide. Check the order of appearance.
- Q3, Q4(a) or Q5(a) wrong: practise “Adding the steps” and Worked example 2, especially multipliers.
- Q6 or Q7 incomplete: your answer needs the reasoning about evidence and collision likelihood, not just the equations.
Then tick off the topic checklist.