These are original Marlbridge practice questions, not past exam questions. All experimental data 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. Data for every question: C 12.01, O 16.00, Ca 40.08 g mol⁻¹; molar volume of a gas at the lab conditions used here = 24.5 L mol⁻¹.
Question 1 (multiple choice · foundation)
Ammonia burns in oxygen: 4 NH₃(g) + 5 O₂(g) → 4 NO(g) + 6 H₂O(g). At one moment, O₂ is being used up at 0.250 M s⁻¹. At what rate is H₂O being formed?
- (A) 0.208 M s⁻¹
- (B) 0.250 M s⁻¹
- (C) 0.300 M s⁻¹
- (D) 1.50 M s⁻¹
Answer and explanation
Answer: (C). For every 5 mol O₂ used, 6 mol H₂O form, so H₂O forms 6/5 as fast as O₂ is used: 0.250 × 6/5 = 0.300 M s⁻¹.
- (A) uses the ratio upside down (0.250 × 5/6).
- (B) assumes every species changes at the same rate, ignoring the coefficients.
- (D) multiplies by 6 without dividing by 5.
Question 2 (multiple choice · core)
A student reacts a 0.10 g strip of magnesium ribbon with 50 mL of 1.0 M hydrochloric acid at 20 °C. The acid is in excess. Which change would not increase the initial rate of reaction?
- (A) Using 0.10 g of magnesium powder instead of ribbon
- (B) Warming the acid to 40 °C before adding the magnesium
- (C) Using 100 mL of 1.0 M hydrochloric acid instead of 50 mL
- (D) Using 50 mL of 2.0 M hydrochloric acid instead of 1.0 M
Answer and explanation
Answer: (C). Doubling the volume of the same acid does not change its concentration. Each unit of magnesium surface is hit by H⁺ ions just as often as before, so the initial rate is the same.
- (A) increases the surface area, so more magnesium atoms are exposed to collisions.
- (B) raises the temperature: particles collide more often and more collisions have enough energy.
- (D) raises [H⁺], so collisions with the magnesium surface happen more often.
Question 3 (multiple choice · core)
For the reaction A → 2 B, a student records:
| Time (s) | 0 | 20 | 40 | 60 | 80 |
|---|---|---|---|---|---|
| [A] (M) | 0.600 | 0.480 | 0.390 | 0.320 | 0.265 |
What is the average rate of disappearance of A between 20 s and 60 s?
- (A) 0.0040 M s⁻¹
- (B) 0.0047 M s⁻¹
- (C) 0.0053 M s⁻¹
- (D) 0.0080 M s⁻¹
Answer and explanation
Answer: (A). −Δ[A]/Δt = (0.480 − 0.320) M ÷ (60 − 20) s = 0.160 M ÷ 40 s = 0.0040 M s⁻¹.
- (B) uses the interval from 0 s to 60 s instead (0.280 M ÷ 60 s).
- (C) divides a single concentration by a single time (0.320 M ÷ 60 s). A rate needs a change in concentration.
- (D) is the average rate of formation of B over the interval, which is twice the rate for A.
Notice also that the drop in [A] in each 20 s interval gets smaller (0.120, 0.090, 0.070, 0.055 M): the rate is falling.
Question 4 (calculation · core)
Bromide and bromate ions react in acid:
5 Br⁻(aq) + BrO₃⁻(aq) + 6 H⁺(aq) → 3 Br₂(aq) + 3 H₂O(l)
In the first 40.0 s, [Br₂] rises by 0.0180 M. Calculate (a) the average rate of formation of Br₂, (b) the average rates of disappearance of Br⁻, BrO₃⁻ and H⁺, and (c) the average rate of reaction.
Worked solution
- (a) Δ[Br₂]/Δt = 0.0180 M ÷ 40.0 s = 4.50 × 10⁻⁴ M s⁻¹.
- (b) Use the coefficients relative to Br₂ (3):
- Br⁻: 4.50 × 10⁻⁴ × 5/3 = 7.50 × 10⁻⁴ M s⁻¹
- BrO₃⁻: 4.50 × 10⁻⁴ × 1/3 = 1.50 × 10⁻⁴ M s⁻¹
- H⁺: 4.50 × 10⁻⁴ × 6/3 = 9.00 × 10⁻⁴ M s⁻¹
- (c) Rate of reaction = (1/3) Δ[Br₂]/Δt = 1.50 × 10⁻⁴ M s⁻¹. (The same as the rate for BrO₃⁻, whose coefficient is 1.)
Suggested mark points (3): 1 for (a); 1 for all three rates in (b) with correct ratios; 1 for (c) divided by the coefficient 3.
Common error: dividing by 5/3 instead of multiplying, which gives 2.70 × 10⁻⁴ M s⁻¹ for Br⁻, smaller than the Br₂ rate. Br⁻ has the larger coefficient, so its rate must be larger.
Question 5 (constructed response · core)
A student adds a little manganese(IV) oxide catalyst to 25.0 mL of hydrogen peroxide solution and collects the oxygen in a gas syringe. 2 H₂O₂(aq) → 2 H₂O(l) + O₂(g).
| Time (s) | 0 | 20 | 40 | 60 | 80 | 100 |
|---|---|---|---|---|---|---|
| Volume of O₂ (mL) | 0.0 | 18.0 | 31.0 | 40.0 | 46.0 | 49.0 |
(a) Calculate the average rate of O₂ production, in mL s⁻¹, from 0 to 40 s and from 40 to 80 s. (b) Convert the 0 to 40 s rate into the average rate of disappearance of H₂O₂, in M s⁻¹. (c) Explain, in terms of particles, why the rate falls during the experiment.
Worked solution and suggested Marlbridge rubric
(a) 0 to 40 s: 31.0 mL ÷ 40 s = 0.775 mL s⁻¹. 40 to 80 s: (46.0 − 31.0) mL ÷ 40 s = 0.375 mL s⁻¹.
(b) O₂: 0.775 mL s⁻¹ = 0.000775 L s⁻¹; ÷ 24.5 L mol⁻¹ = 3.16 × 10⁻⁵ mol s⁻¹. H₂O₂ is used twice as fast (2 : 1): 6.33 × 10⁻⁵ mol s⁻¹. In 0.0250 L: 6.33 × 10⁻⁵ ÷ 0.0250 = 2.53 × 10⁻³ M s⁻¹.
(c) As H₂O₂ decomposes, its concentration falls. Fewer H₂O₂ molecules per unit volume means fewer collisions with the catalyst surface each second, so fewer molecules react each second and O₂ is produced more slowly.
| Point | What earns it |
|---|---|
| 1 | Both average rates in (a) with units |
| 1 | Converts mL s⁻¹ to mol s⁻¹ using 24.5 L mol⁻¹ |
| 1 | Uses the 2 : 1 ratio of H₂O₂ to O₂ |
| 1 | Divides by 0.0250 L to give 2.53 × 10⁻³ M s⁻¹ |
| 1 | Links falling [H₂O₂] to less frequent collisions (with the catalyst) and a lower rate |
Accept carried-forward answers in (b). In (c), “the reactant is running out” alone is not enough; the point needs the link to collision frequency.
Question 6 (constructed response · core)
Two experiments each use 2.00 g of calcium carbonate and 50.0 mL of 1.0 M hydrochloric acid at the same temperature. Experiment 1 uses marble chips; experiment 2 uses powder. The time to collect 60.0 mL of CO₂ is 150 s with chips and 25 s with powder.
CaCO₃(s) + 2 HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g)
A student claims: “The powder reacts faster because powdered calcium carbonate has a higher concentration than the chips.”
(a) Calculate the average rate of CO₂ production in each experiment over the time measured. (b) Evaluate the student’s claim, using particle-level reasoning. (c) The chips experiment is repeated with 50.0 mL of 2.0 M acid. Predict, with a reason, how (i) the initial rate and (ii) the final volume of CO₂ compare with experiment 1.
Worked solution and suggested Marlbridge rubric
(a) Chips: 60.0 mL ÷ 150 s = 0.40 mL s⁻¹. Powder: 60.0 mL ÷ 25 s = 2.4 mL s⁻¹ (six times faster).
(b) The claim is incorrect. Calcium carbonate is a solid, so it does not have a concentration in the solution; both samples are the same pure substance with the same mass. The powder reacts faster because it has a much larger surface area. Only CaCO₃ at the surface can be hit by H⁺ ions, so more exposed surface means more collisions each second and a higher rate.
(c)(i) The initial rate is higher: doubling [HCl] means H⁺ ions hit the chip surface more often. (c)(ii) The final volume is the same. Moles of CaCO₃ = 2.00 g ÷ 100.09 g mol⁻¹ = 0.0200 mol. Even 1.0 M acid holds 0.0500 mol HCl, which needs only 0.0250 mol CaCO₃, so CaCO₃ is the limiting reactant in both cases. Both experiments give 0.0200 mol CO₂ (about 490 mL at the lab conditions).
| Point | What earns it |
|---|---|
| 1 | Both average rates in (a) |
| 1 | States the claim is wrong because a solid has no concentration in solution (or the samples are the same substance) |
| 1 | Explains the faster rate by larger surface area giving more frequent collisions |
| 1 | (c)(i) higher rate, linked to more frequent collisions at higher [H⁺] |
| 1 | (c)(ii) same final volume, supported by identifying CaCO₃ as limiting |
For the last point, a correct answer without the limiting-reactant reasoning earns no credit.
Question 7 (calculation and explanation · stretch)
For the reaction A → 3 B, [A] is 0.500 M at 0 s, 0.320 M at 100 s and 0.200 M at 200 s. A student draws the tangent to the [A] curve at 100 s. The tangent passes through the points (0 s, 0.460 M) and (250 s, 0.110 M).
(a) Calculate the instantaneous rate of disappearance of A at 100 s. (b) Calculate the instantaneous rate of formation of B at 100 s. (c) Calculate the average rate of disappearance of A from 0 to 200 s, and explain why the average rate from 0 to 100 s is larger than your answer to (a).
Worked solution and suggested Marlbridge rubric
(a) Slope of tangent = (0.110 − 0.460) M ÷ (250 − 0) s = −0.0014 M s⁻¹. Rate of disappearance of A = 1.4 × 10⁻³ M s⁻¹. (Check: the tangent passes through 0.460 − 0.0014 × 100 = 0.320 M at 100 s, which is on the curve.)
(b) B forms three times as fast: 3 × 1.4 × 10⁻³ = 4.2 × 10⁻³ M s⁻¹.
(c) Average 0 to 200 s = (0.500 − 0.200) M ÷ 200 s = 1.5 × 10⁻³ M s⁻¹. The average rate from 0 to 100 s is (0.500 − 0.320) ÷ 100 = 1.8 × 10⁻³ M s⁻¹. It is larger than the rate at exactly 100 s because it includes the early part of the reaction, when [A] was higher and the reaction was faster. The rate falls throughout, so the rate at the end of an interval is lower than the average over it.
| Point | What earns it |
|---|---|
| 1 | Tangent slope used, with the sign made positive: 1.4 × 10⁻³ M s⁻¹ |
| 1 | Multiplies by 3 for B |
| 1 | 1.5 × 10⁻³ M s⁻¹ for the average from 0 to 200 s |
| 1 | Explains that the earlier part of the interval had higher [A] and a higher rate |
How did you do?
- Q1 or Q4 wrong: re-read “Rates of different species” and Worked example 1 in the study guide.
- Q3, Q5 or Q7 wrong: revisit average and instantaneous rate (Figure 1) and Worked example 2.
- Q2 or Q6 incomplete: your explanation needs the particle-level link (collisions per second), not only “it is faster”.
Then tick off the topic checklist.