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 reactions and data are invented unless a real isotope is named. Data for every question: [A]ₜ − [A]₀ = −kt; ln[A]ₜ − ln[A]₀ = −kt; 1/[A]ₜ − 1/[A]₀ = kt; t½ = 0.693/k. Temperature is constant in every question.
Question 1 (multiple choice · foundation)
For the decomposition of an invented compound A, a plot of 1/[A] against time is a straight line with a slope of 0.15 M⁻¹ min⁻¹. Which of the following is correct?
- (A) rate = k[A], with k = 0.15 min⁻¹
- (B) rate = k[A]², with k = 0.15 M⁻¹ min⁻¹
- (C) rate = k[A]², with k = −0.15 M⁻¹ min⁻¹
- (D) rate = k, with k = 0.15 M min⁻¹
Answer and explanation
Answer: (B). A straight 1/[A] against t plot means the reaction is second order in A. For second order the slope is +k, so k = 0.15 M⁻¹ min⁻¹, and these units match a second-order rate constant.
- (A) is the conclusion for a straight ln[A] plot (first order). The units min⁻¹ also do not match a 1/[A] slope.
- (C) gets the order right but the sign wrong. 1/[A] rises as [A] falls, so the slope is positive, and k is never negative.
- (D) is the conclusion for a straight [A] plot (zero order).
Question 2 (multiple choice · core)
An invented reaction is first order in reactant B, with k = 0.0231 min⁻¹. What percentage of the starting B is left after 90.0 min?
- (A) 0%
- (B) 6.25%
- (C) 12.5%
- (D) 25.0%
Answer and explanation
Answer: (C). t½ = 0.693 ÷ 0.0231 min⁻¹ = 30.0 min. 90.0 min is 3 half-lives, so the fraction left is (½)³ = 1/8 = 12.5%.
- (A) assumes each half-life removes half of the starting amount, so the reaction would be finished after 2 half-lives. Each half-life removes half of what is left.
- (B) is (½)⁴, four half-lives: a counting error.
- (D) is (½)², two half-lives: a counting error.
Question 3 (multiple choice · core)
The concentration of an invented reactant C is measured during one experiment.
| t (s) | 0 | 20 | 40 | 60 |
|---|---|---|---|---|
| [C] (M) | 0.240 | 0.200 | 0.160 | 0.120 |
Which statement is correct?
- (A) The reaction is first order in C, with k = 2.0 × 10⁻³ s⁻¹.
- (B) The reaction is zero order in C, with k = 2.0 × 10⁻³ M s⁻¹.
- (C) The reaction is zero order in C, with k = −2.0 × 10⁻³ M s⁻¹.
- (D) The reaction is second order in C, because the half-life gets shorter.
Answer and explanation
Answer: (B). [C] falls by exactly 0.040 M every 20 s, so the [C] against t plot is straight: zero order. slope = (0.120 − 0.240) M ÷ 60 s = −2.0 × 10⁻³ M s⁻¹, and k = −slope = 2.0 × 10⁻³ M s⁻¹.
- (A) uses the right number but the wrong order and units. The ln[C] values (−1.427, −1.609, −1.833, −2.120) do not change by a constant amount, so the ln plot is not straight.
- (C) gives a negative rate constant. The slope is negative; k is positive.
- (D) is backwards. Second-order half-lives get longer as the reactant is used up; shorter half-lives point to zero order.
Question 4 (calculation · core)
An invented reactant D decomposes in a first-order reaction with k = 4.5 × 10⁻³ s⁻¹. The starting concentration is 0.0800 M.
(a) Calculate [D] after 3.00 minutes. (b) Calculate the time, in seconds, for [D] to fall to 0.0100 M.
Worked solution
(a) Convert time to match k: 3.00 min = 180 s. ln[D]ₜ = ln(0.0800) − (4.5 × 10⁻³ s⁻¹)(180 s) = −2.526 − 0.810 = −3.336 [D]ₜ = e^−3.336 = 0.0356 M
(b) ln(0.0800 / 0.0100) = kt, so t = ln 8 ÷ 4.5 × 10⁻³ s⁻¹ = 2.079 ÷ 0.0045 s⁻¹ = 462 s.
Check: t½ = 0.693 ÷ 0.0045 = 154 s. 0.0800 → 0.0100 M is three half-lives, and 3 × 154 = 462 s.
Suggested mark points (3): 1 for converting 3.00 min to 180 s; 1 for 0.0356 M; 1 for 462 s (by the integrated law or by three half-lives).
Common error: substituting t = 3.00 with k in s⁻¹ gives 0.0789 M, which is barely below the start. That should look wrong: 3 minutes is more than one half-life.
Question 5 (constructed response · core)
A student follows the fading of an invented dye, D, by measuring its concentration with a spectrophotometer (absorbance is proportional to [D]).
| t (min) | 0 | 20 | 40 | 60 | 80 |
|---|---|---|---|---|---|
| [D] (× 10⁻⁵ M) | 4.00 | 3.12 | 2.43 | 1.89 | 1.47 |
(a) Determine the order of the reaction in D. Justify your answer with values calculated from the data. (b) Determine the rate constant, with units. (c) Calculate the half-life. (d) Explain why the student could have plotted absorbance instead of concentration and still found the same order and k.
Worked solution and suggested Marlbridge rubric
(a) Calculate ln[D] (with [D] in M): −10.127, −10.375, −10.625, −10.876, −11.128. The differences are −0.248, −0.250, −0.251, −0.252, which are constant within the rounding of the data. ln[D] against t is straight, so the reaction is first order in D. For comparison, [D] falls by 0.88, 0.69, 0.54, 0.42 (× 10⁻⁵ M) and 1/[D] rises by 0.705, 0.910, 1.18, 1.51 (× 10⁴ M⁻¹). Neither is constant.
(b) slope = (−11.128 − (−10.127)) ÷ (80 − 0) min = −0.0125 min⁻¹, so k = 0.0125 min⁻¹ (1.25 × 10⁻² min⁻¹).
(c) t½ = 0.693 ÷ 0.0125 min⁻¹ = 55.4 min.
(d) If absorbance A = (constant) × [D], then ln A = ln(constant) + ln[D]. The constant only shifts the line up or down; it does not change the slope. So ln A against t is straight whenever ln[D] against t is, with the same slope −k.
| Point | What earns it |
|---|---|
| 1 | Calculates ln[D] values (or plots ln[D] against t) and shows they change by a constant amount in equal times |
| 1 | Concludes first order and shows at least one other plot is not straight |
| 1 | k = 0.0125 min⁻¹ with correct units, from the slope |
| 1 | t½ = 55.4 min using k from (b) |
| 1 | Explains that a proportionality constant changes only the intercept of the ln plot, not its slope |
Accept k from a best-fit line between 0.0124 and 0.0126 min⁻¹. Accept “the time to halve is constant” as evidence in (a) if supported by values.
Question 6 (constructed response · core)
An invented compound Y decomposes in a reaction that is second order in Y, with k = 0.540 M⁻¹ s⁻¹. The starting concentration is 0.0250 M.
(a) Calculate the time for [Y] to fall from 0.0250 M to 0.0125 M. (b) Calculate the time for [Y] to fall from 0.0125 M to 0.00625 M. (c) A student says: “The second half-life is longer because k gets smaller as the reaction goes on.” Evaluate this claim.
Worked solution and suggested Marlbridge rubric
(a) 1/[Y]ₜ − 1/[Y]₀ = kt, so t = (1/0.0125 − 1/0.0250) M⁻¹ ÷ 0.540 M⁻¹ s⁻¹ = (80.0 − 40.0) ÷ 0.540 = 74.1 s.
(b) t = (1/0.00625 − 1/0.0125) ÷ 0.540 = (160.0 − 80.0) ÷ 0.540 = 148 s.
(c) The claim is incorrect. k depends on temperature, not on concentration, so it stays at 0.540 M⁻¹ s⁻¹ throughout. The second half-life is longer because the rate law is rate = k[Y]². When [Y] halves, the rate falls to one quarter (from 3.38 × 10⁻⁴ to 8.44 × 10⁻⁵ M s⁻¹), so it takes longer to remove the next half.
| Point | What earns it |
|---|---|
| 1 | 74.1 s using the second-order integrated law |
| 1 | 148 s (about double the first half-life) |
| 1 | States k is constant at constant temperature |
| 1 | Explains the longer half-life by the rate depending on [Y]², so the rate falls as [Y] falls |
Question 7 (explanation · stretch)
Carbon-14 is radioactive with a half-life of about 5700 years. A student says: “A 2.0 g sample of carbon-14 takes twice as long to lose half its carbon-14 as a 1.0 g sample, because there is twice as much to decay.”
(a) Explain why radioactive decay follows a first-order rate law. (b) Evaluate the student’s claim. (c) What mass of carbon-14 remains from the 2.0 g sample after 17 100 years?
Model answer and suggested Marlbridge rubric
(a) Each nucleus decays on its own, without colliding with anything, and has the same chance of decaying in a given time. So the number of decays per unit time is proportional to the number of nuclei present: rate = kN, which is first order.
(b) The claim is incorrect. The larger sample has twice as many nuclei, but it also decays twice as fast, because the rate is proportional to the amount present. For a first-order process t½ = 0.693/k, which does not depend on the starting amount. Both samples lose half their carbon-14 in about 5700 years.
(c) 17 100 ÷ 5700 = 3 half-lives, so (½)³ = 1/8 remains: 2.0 g × 1/8 = 0.25 g.
| Point | What earns it |
|---|---|
| 1 | Links rate proportional to the number of nuclei to a first-order rate law |
| 1 | States the half-life is independent of the starting amount (t½ = 0.693/k has no amount in it) |
| 1 | Explains why: double the nuclei gives double the rate |
| 1 | 0.25 g, using three half-lives |
How did you do?
- Q1 or Q3 wrong: re-read the table “The three integrated rate laws” in the study guide, especially the slope signs.
- Q2, Q4 or Q7 wrong: review “Half-life” and Worked example 2. Count half-lives on paper.
- Q5 or Q6 incomplete: your answer needs evidence (calculated values) and a reason, not only the conclusion.
Then tick off the topic checklist.