These are original Marlbridge practice questions, not past exam questions. All rate data and rate constants are invented for practice. The mark points are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. In every question, all the experiments in one table are at the same temperature.
Question 1 (multiple choice · foundation)
A reaction has the rate law rate = k[A]²[B]. By what factor does the rate change if [A] is tripled and [B] is halved at the same time?
- (A) 1.5
- (B) 3.0
- (C) 4.5
- (D) 9.0
Answer and explanation
Answer: (C). Tripling [A] multiplies the rate by 3² = 9. Halving [B] multiplies it by ½. Together: 9 × ½ = 4.5.
- (A) treats A as first order (3 × ½).
- (B) ignores [B] and treats A as first order.
- (D) includes the effect of A but forgets that [B] was halved.
Question 2 (multiple choice · foundation)
The rate law for a reaction is rate = k[X][Y]². What are the units of k when concentrations are in M and time is in seconds?
- (A) s⁻¹
- (B) M⁻¹ s⁻¹
- (C) M⁻² s⁻¹
- (D) M² s⁻¹
Answer and explanation
Answer: (C). The overall order is 1 + 2 = 3. k must turn M³ into M s⁻¹, so k has units M s⁻¹ ÷ M³ = M⁻² s⁻¹.
- (A) is for an overall first-order reaction.
- (B) is for an overall second-order reaction (it ignores the order in X).
- (D) gets the sign of the power wrong: (M² s⁻¹)(M³) = M⁵ s⁻¹, not M s⁻¹.
Question 3 (multiple choice · core)
Which statement about the rate constant k for a reaction is correct?
- (A) k gets larger when the concentration of a reactant is increased.
- (B) k has the same value at every temperature.
- (C) At a fixed temperature, k has the same value whatever the starting concentrations.
- (D) k always has units of M s⁻¹, the same as the rate.
Answer and explanation
Answer: (C). k is a constant for a given reaction at a given temperature. Changing concentrations changes the rate, not k.
- (A) confuses k with the rate. The rate rises with concentration (for order above zero); k does not.
- (B) is false: k depends on temperature, and for almost all reactions it increases as temperature rises.
- (D) is true only for a zero-order reaction. The units of k depend on the overall order.
Question 4 (calculation · core)
Hydrogen peroxide oxidises iodide ions in acid: H₂O₂(aq) + 2 I⁻(aq) + 2 H⁺(aq) → I₂(aq) + 2 H₂O(l). [H⁺] is the same in every experiment.
| Experiment | [H₂O₂] (M) | [I⁻] (M) | Initial rate (M s⁻¹) |
|---|---|---|---|
| 1 | 0.010 | 0.020 | 3.4 × 10⁻⁶ |
| 2 | 0.030 | 0.020 | 1.02 × 10⁻⁵ |
| 3 | 0.030 | 0.050 | 2.55 × 10⁻⁵ |
(a) Determine the order with respect to H₂O₂ and to I⁻. (b) Write the rate law and calculate k with units. (c) Calculate the initial rate when [H₂O₂] = 0.020 M and [I⁻] = 0.040 M.
Worked solution
(a) Experiments 1 → 2: [I⁻] fixed, [H₂O₂] × 3, rate × (1.02 × 10⁻⁵ ÷ 3.4 × 10⁻⁶) = × 3. So first order in H₂O₂. Experiments 2 → 3: [H₂O₂] fixed, [I⁻] × 2.5, rate × (2.55 × 10⁻⁵ ÷ 1.02 × 10⁻⁵) = × 2.5. So first order in I⁻.
(b) rate = k[H₂O₂][I⁻]. From experiment 1: k = 3.4 × 10⁻⁶ ÷ (0.010 × 0.020) = 0.017 M⁻¹ s⁻¹.
(c) rate = 0.017 × 0.020 × 0.040 = 1.4 × 10⁻⁵ M s⁻¹ (1.36 × 10⁻⁵ before rounding).
Suggested mark points (4): 1 for each order with the experiments compared; 1 for k with correct units; 1 for the rate in (c).
Common error: taking the coefficient 2 as the order in I⁻. That would predict a rate × 2.5² = 6.25 between experiments 2 and 3, but the data show × 2.5.
Question 5 (calculation · core)
Nitrogen dioxide decomposes: 2 NO₂(g) → 2 NO(g) + O₂(g). At a certain temperature the rate law is rate = k[NO₂]² with k = 0.75 M⁻¹ s⁻¹, where “rate” is the rate of reaction, −½ Δ[NO₂]/Δt.
(a) Calculate the rate of reaction when [NO₂] = 0.0200 M. (b) At that moment, how fast is NO₂ being used up? (c) What [NO₂] gives a rate of reaction of 1.2 × 10⁻³ M s⁻¹? (d) By what factor does the rate change if [NO₂] is halved?
Worked solution
(a) rate = 0.75 M⁻¹ s⁻¹ × (0.0200 M)² = 3.0 × 10⁻⁴ M s⁻¹.
(b) −Δ[NO₂]/Δt = 2 × rate of reaction = 6.0 × 10⁻⁴ M s⁻¹. (O₂ forms at 3.0 × 10⁻⁴ M s⁻¹, the same as the rate of reaction, because its coefficient is 1.)
(c) [NO₂]² = 1.2 × 10⁻³ ÷ 0.75 = 1.6 × 10⁻³ M², so [NO₂] = √(1.6 × 10⁻³) = 0.040 M.
(d) (½)² = ¼. The rate falls to a quarter.
Suggested mark points (4): one for each part. In (c), award the point only if the square root is taken.
Check for (c): the target rate is 4 times the rate in (a), so [NO₂] must be 2 times 0.0200 M. It is.
Question 6 (constructed response · core)
A student studies the fading of a blue food dye by hypochlorite ions, OCl⁻, using a spectrophotometer with a 1.00 cm cell. The dye is the only coloured species. The initial rate is recorded as the rate at which absorbance falls.
| Experiment | [dye] (M) | [OCl⁻] (M) | Starting absorbance | Initial rate of absorbance decrease (s⁻¹) |
|---|---|---|---|---|
| 1 | 2.0 × 10⁻⁵ | 0.010 | 1.00 | 0.0030 |
| 2 | 3.0 × 10⁻⁵ | 0.010 | 1.50 | 0.0045 |
| 3 | 2.0 × 10⁻⁵ | 0.020 | 1.00 | 0.0060 |
(a) Explain why the rate of absorbance decrease can be used in place of the rate of decrease of [dye] when finding the orders. (b) Determine the order with respect to the dye and to OCl⁻, and write the rate law. (c) The product εb for the dye is 5.0 × 10⁴ M⁻¹. Convert the rate in experiment 1 to M s⁻¹ and calculate k with units. (d) A student says: “In experiment 3 the rate doubled, so k doubled.” Evaluate this claim.
Worked solution and suggested Marlbridge rubric
(a) By the Beer–Lambert law, A = εbc. ε and b are the same in every experiment, so absorbance is directly proportional to [dye]. Any ratio of absorbance rates equals the same ratio of concentration rates.
(b) Experiments 1 → 2: [OCl⁻] fixed, [dye] × 1.5, rate × 1.5 (0.0045 ÷ 0.0030). First order in dye. Experiments 1 → 3: [dye] fixed, [OCl⁻] × 2, rate × 2. First order in OCl⁻. Rate law: rate = k[dye][OCl⁻].
(c) Δ[dye]/Δt = (ΔA/Δt) ÷ εb = 0.0030 s⁻¹ ÷ 5.0 × 10⁴ M⁻¹ = 6.0 × 10⁻⁸ M s⁻¹. k = 6.0 × 10⁻⁸ ÷ (2.0 × 10⁻⁵ × 0.010) = 0.30 M⁻¹ s⁻¹.
(d) The claim is incorrect. The rate doubled because [OCl⁻] doubled and the reaction is first order in OCl⁻. k is the same at a fixed temperature: experiment 3 gives (0.0060 ÷ 5.0 × 10⁴) ÷ (2.0 × 10⁻⁵ × 0.020) = 0.30 M⁻¹ s⁻¹, the same as experiment 1. Only a temperature change (or a catalyst) would change k.
| Point | What earns it |
|---|---|
| 1 | A ∝ [dye] because ε and b are constant (Beer–Lambert law) |
| 1 | Both orders with the experiments compared, and the rate law |
| 1 | Converts to 6.0 × 10⁻⁸ M s⁻¹ by dividing by εb |
| 1 | k = 0.30 M⁻¹ s⁻¹ with units |
| 1 | Rejects the claim: the rate change comes from [OCl⁻]; k is unchanged at constant temperature (supported by the same k from experiment 3) |
Question 7 (constructed response · stretch)
For a reaction 2 A + B → C, a student claims the rate law must be rate = k[A]²[B] “because of the balanced equation”. Invented initial-rate data:
| Experiment | [A] (M) | [B] (M) | Initial rate (M s⁻¹) |
|---|---|---|---|
| 1 | 0.10 | 0.10 | 4.0 × 10⁻⁴ |
| 2 | 0.20 | 0.10 | 8.0 × 10⁻⁴ |
| 3 | 0.30 | 0.20 | 4.8 × 10⁻³ |
(a) Determine the order with respect to A and with respect to B. (b) Use the data to evaluate the student’s claim. (c) Write the correct rate law and calculate k with units. (d) Which would raise the rate more: doubling [A] or doubling [B]? Justify.
Worked solution and suggested Marlbridge rubric
(a) Experiments 1 → 2: [B] fixed, [A] × 2, rate × 2. First order in A. Experiments 1 → 3: [A] × 3 and [B] × 2; rate × (4.8 × 10⁻³ ÷ 4.0 × 10⁻⁴) = × 12. The A change accounts for × 3 (first order), so B accounts for 12 ÷ 3 = × 4. [B] doubled and the rate rose 4 times: second order in B.
(b) The claim is incorrect. If the rate law were k[A]²[B], doubling [A] in experiments 1 → 2 would multiply the rate by 4, but the data show × 2. Orders must come from experiment; they do not have to match the coefficients.
(c) rate = k[A][B]². k = 4.0 × 10⁻⁴ ÷ (0.10 × 0.10²) = 0.40 M⁻² s⁻¹.
(d) Doubling [B] raises the rate more: × 2² = × 4, compared with × 2 for doubling [A].
| Point | What earns it |
|---|---|
| 1 | First order in A from experiments 1 and 2 |
| 1 | Second order in B, after removing the effect of A in experiment 3 |
| 1 | Rejects the claim using a specific data comparison |
| 1 | Correct rate law and k = 0.40 M⁻² s⁻¹ with units |
| 1 | Doubling [B] (× 4) beats doubling [A] (× 2), with the reason |
How did you do?
- Q1, Q5 or Q7(d) wrong: revisit “What each order means” and Figure 1 in the study guide.
- Q2, Q3 or Q6(d) wrong: re-read “The rate constant, k”.
- Q4, Q6(b) or Q7(a) wrong: work through the method of initial rates and Worked example 2 again.
Then tick off the topic checklist.