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. The reactions, rate constants and data are invented for practice. Temperature is constant in every question, and rates are in M s⁻¹.
Question 1 (multiple choice · foundation)
The equation 2 NO₂ → N₂O₄ is treated as an elementary step. Which is its rate law?
- (A) rate = k[NO₂]
- (B) rate = k[NO₂]²
- (C) rate = k[N₂O₄]
- (D) rate = k[N₂O₄] / [NO₂]²
Answer and explanation
Answer: (B). In an elementary step, each reactant’s coefficient is its order. Two NO₂ particles collide, so the step is bimolecular and rate = k[NO₂]².
- (A) ignores the coefficient 2. That would describe a unimolecular step with one NO₂.
- (C) puts the product in the rate law. The rate of a forward step depends only on the reactants that collide.
- (D) has the form of an equilibrium expression (products over reactants), not a rate law.
Question 2 (multiple choice · core)
Which statement about the elementary step A + 2B → AB₂ is correct?
- (A) It is bimolecular, because two different substances react.
- (B) Its rate law is rate = k[A][B]², and the step is third order overall.
- (C) Its rate law cannot be written until experiments have been done.
- (D) It cannot occur, because termolecular steps are impossible.
Answer and explanation
Answer: (B). Three particles (one A and two B) collide, so the step is termolecular. The coefficients give the orders: first order in A, second order in B, third order overall.
- (A) counts substances instead of particles. Molecularity is the number of particles: 3.
- (C) would be true for an overall reaction, but the question says this is an elementary step, so its rate law follows from the coefficients.
- (D) overstates the idea. Termolecular steps are rare, not impossible.
Question 3 (multiple choice · core)
For the elementary step A + B → products, [A] is doubled and [B] is tripled at the same time. By what factor does the rate change?
- (A) 5
- (B) 6
- (C) 12
- (D) 18
Answer and explanation
Answer: (B). The step is bimolecular, so rate = k[A][B]. The rate changes by 2 × 3 = 6.
- (A) adds the factors instead of multiplying them.
- (C) is 2² × 3, as if the rate law were k[A]²[B].
- (D) is 2 × 3², as if the rate law were k[A][B]².
Question 4 (calculation · core)
The elementary step 2C → D has k = 6.0 × 10³ M⁻¹ s⁻¹.
(a) Write the rate law and state the molecularity. (b) Calculate the rate when [C] = 2.0 × 10⁻³ M. (c) Calculate the rate when [C] has fallen to one third of this value.
Worked solution
(a) Two C particles collide: bimolecular, rate = k[C]². (The units M⁻¹ s⁻¹ also fit a second-order rate law.)
(b) rate = (6.0 × 10³ M⁻¹ s⁻¹)(2.0 × 10⁻³ M)² = (6.0 × 10³)(4.0 × 10⁻⁶) = 0.024 M s⁻¹.
(c) The rate is second order in C, so dividing [C] by 3 divides the rate by 3² = 9: 0.024 ÷ 9 = 2.7 × 10⁻³ M s⁻¹.
Suggested mark points (3): 1 for rate = k[C]² with “bimolecular”; 1 for 0.024 M s⁻¹; 1 for 2.7 × 10⁻³ M s⁻¹.
Common error: using rate = k[C] gives the number 12, but its units are M⁻¹ s⁻¹ × M = s⁻¹, which is not a rate. Checking units catches the mistake.
Question 5 (constructed response · core)
For the invented gas reaction 2G + H → G₂H, experiments give rate = k[G]²[H].
(a) Write the rate law that would apply if the reaction were a single elementary step, and compare it with the experimental one. (b) Student P says: “The rate laws match, so the reaction is definitely one elementary step.” Student Q says: “It cannot be one step, because three particles can never collide at once.” Evaluate both claims. (c) Give the units of k.
Worked solution and suggested Marlbridge rubric
(a) If elementary, rate = k[G]²[H] (coefficients as orders). This matches the experimental rate law.
(b) Student P is wrong. A match shows a single step is possible, but a sequence of steps could give the same rate law, so it does not prove it. Student Q is also wrong. Termolecular steps are rare because three particles seldom meet at the same instant, but they are not impossible. So a single step cannot be ruled out either. (A mechanism of two bimolecular steps is also possible; more evidence would be needed.)
(c) rate (M s⁻¹) = k × M² × M, so k is in M⁻² s⁻¹.
| Point | What earns it |
|---|---|
| 1 | Writes rate = k[G]²[H] for the single step and states it matches |
| 1 | Rejects P’s claim: agreement is consistent with one step but does not prove it |
| 1 | Rejects Q’s claim: termolecular steps are rare, not impossible |
| 1 | k in M⁻² s⁻¹ with reasoning or a units check |
Question 6 (constructed response · stretch)
A student proposes that the invented reaction between X and Z happens in a single elementary step, X + Z → products. Initial-rate experiments give:
| Experiment | [X] (M) | [Z] (M) | Initial rate (M s⁻¹) |
|---|---|---|---|
| 1 | 0.010 | 0.010 | 2.0 × 10⁻⁴ |
| 2 | 0.020 | 0.010 | 4.0 × 10⁻⁴ |
| 3 | 0.010 | 0.030 | 1.8 × 10⁻³ |
(a) Determine the order with respect to X and to Z. (b) Write the experimental rate law and calculate k with units. (c) Is the student’s proposal consistent with the data? Justify your answer. (d) Write an elementary step whose rate law would match the data, and comment on how likely it is.
Worked solution and suggested Marlbridge rubric
(a) Experiments 1 → 2: [X] doubles, [Z] constant, rate doubles (× 2.0). First order in X. Experiments 1 → 3: [Z] triples, [X] constant, rate × 9.0 = 3². Second order in Z.
(b) rate = k[X][Z]². From experiment 1: k = 2.0 × 10⁻⁴ ÷ (0.010 × 0.010²) = 200 M⁻² s⁻¹ (experiments 2 and 3 give the same value).
(c) Not consistent. The step X + Z → products would give rate = k[X][Z], first order in Z. Tripling [Z] would then triple the rate to 6.0 × 10⁻⁴ M s⁻¹, but the measured rate rose ninefold. So the reaction is not that single step.
(d) X + 2Z → products (termolecular) gives rate = k[X][Z]². It matches, but three-particle collisions are rare, so a series of steps (for example two bimolecular steps) is more likely. The data alone cannot decide.
| Point | What earns it |
|---|---|
| 1 | Both orders, each supported by a pair of experiments |
| 1 | rate = k[X][Z]², k = 200 M⁻² s⁻¹ |
| 1 | Rejects the proposal by comparing its predicted rate law (k[X][Z]) with the data |
| 1 | Writes X + 2Z → products and notes it would be termolecular, so rare |
Question 7 (explanation · stretch)
For the elementary step 2Q → Q₂, explain at the particle level why doubling [Q] multiplies the rate by four, not two. Then explain why a step needing four particles to collide at once is not expected.
Model answer and suggested Marlbridge rubric
Model answer. The step happens when two Q particles collide. If [Q] doubles, there are twice as many Q particles in the same volume. Each one also has twice as many other Q particles it could hit. The number of possible Q–Q collisions per second therefore increases by 2 × 2 = 4, so the rate, which is proportional to the collision frequency, increases fourfold: rate = k[Q]².
Even three particles arriving at the same point at the same moment is rare, because the time two particles spend in contact is extremely short. For four particles to meet at once, two more would have to arrive within that same instant, which is far less likely again. So steps with four or more particles are not expected; such reactions happen through several simpler steps.
| Point | What earns it |
|---|---|
| 1 | Links rate to the frequency of Q–Q collisions |
| 1 | Explains that doubling [Q] doubles both the number of particles and the partners for each, so × 4 |
| 1 | Explains that simultaneous collisions of more particles become much less probable, so four-particle steps are not expected |
How did you do?
- Q1 or Q2 wrong: re-read “Why the coefficients become the orders” and the molecularity table in the study guide.
- Q3 or Q4 wrong: review Worked example 2. Multiply the factor for each reactant raised to its order.
- Q5 or Q6 incomplete: compare the predicted and experimental rate laws explicitly, and remember that agreement is not proof.
- Q7 incomplete: your answer needs the particle-level reason (collision frequency), not only the rate law.
Then tick off the topic checklist.