What this diagnostic is for
Use this page to find which Unit 5 topics to revisit. Answer every question before opening any answer, then use the table at the end.
These are original Marlbridge practice questions, not past exam questions. Species written with letters are fictional and all data are invented. They are not calibrated and give no predicted score: treat each wrong answer as a pointer to a topic.
Calculator and data. A scientific calculator with ln is allowed. [A]ₜ − [A]₀ = −kt; ln[A]ₜ − ln[A]₀ = −kt; 1/[A]ₜ − 1/[A]₀ = kt; t½ = 0.693/k. Energies are in kJ mol⁻¹.
Question 1 (multiple choice · 5.1)
Hydrogen sulfide burns: 2 H₂S(g) + 3 O₂(g) → 2 SO₂(g) + 2 H₂O(g). At one moment, SO₂ is forming at 0.18 M s⁻¹. At what rate is O₂ being used up?
- (A) 0.09 M s⁻¹
- (B) 0.12 M s⁻¹
- (C) 0.18 M s⁻¹
- (D) 0.27 M s⁻¹
Answer and explanation
Answer: (D). 3 mol O₂ are used per 2 mol SO₂ formed: 0.18 × 3/2 = 0.27 M s⁻¹.
- (A) is the rate of reaction (0.18 ÷ 2).
- (B) uses the ratio upside down (× 2/3).
- (C) ignores the coefficients.
If you missed this: Topic 5.1 study guide: “Rates of different species: use the coefficients”.
Question 2 (multiple choice · 5.2)
Initial rates for P + Q → R at one temperature:
| Experiment | [P] (M) | [Q] (M) | Initial rate (M s⁻¹) |
|---|---|---|---|
| 1 | 0.010 | 0.020 | 1.5 × 10⁻⁵ |
| 2 | 0.030 | 0.020 | 1.35 × 10⁻⁴ |
| 3 | 0.030 | 0.040 | 1.35 × 10⁻⁴ |
Which gives the rate law and the rate constant?
- (A) rate = k[P]², k = 0.15 M⁻¹ s⁻¹
- (B) rate = k[P]²[Q], k = 7.5 M⁻² s⁻¹
- (C) rate = k[P], k = 1.5 × 10⁻³ s⁻¹
- (D) rate = k[P]², k = 0.15 s⁻¹
Answer and explanation
Answer: (A). Experiments 1 → 2: [P] × 3, rate × 9, so second order in P. Experiments 2 → 3: [Q] × 2, rate unchanged, so zero order in Q. k = 1.5 × 10⁻⁵ ÷ (0.010)² = 0.15 M⁻¹ s⁻¹.
- (B) includes Q, but doubling [Q] had no effect.
- (C) treats a ninefold rise as first order.
- (D) has the wrong units for a second-order k.
If you missed this: Topic 5.2 study guide: “The method of initial rates”.
Question 3 (multiple choice · 5.3)
A reaction is first order in A. [A] falls from 0.600 M to 0.150 M in 25.0 min. What is the rate constant?
- (A) 0.0180 M min⁻¹
- (B) 0.0277 min⁻¹
- (C) 0.0555 min⁻¹
- (D) 0.200 M⁻¹ min⁻¹
Answer and explanation
Answer: (C). k = ln(0.600 ÷ 0.150) ÷ 25.0 = 0.0555 min⁻¹. Check: 0.600 → 0.150 M is two half-lives, so t½ = 12.5 min.
- (A) is the zero-order slope, (0.600 − 0.150) ÷ 25.0.
- (B) treats 25.0 min as one half-life.
- (D) uses the second-order law.
If you missed this: Topic 5.3 study guide: “Half-life”.
Question 4 (short answer · 5.3)
Compound A decomposes at constant temperature. The times are not equally spaced.
| t (min) | 0 | 10 | 30 | 60 |
|---|---|---|---|---|
| [A] (M) | 0.500 | 0.4094 | 0.2744 | 0.1506 |
(a) Show, with calculated values, which plot is linear, and state the order in A. (b) Find k and the half-life. (c) How long after the start does [A] reach 0.0500 M?
Worked answer
(a) The time steps are unequal, so compare slopes.
| Interval (min) | Slope of [A] (M min⁻¹) | Slope of ln[A] (min⁻¹) | Slope of 1/[A] (M⁻¹ min⁻¹) |
|---|---|---|---|
| 0–10 | −0.00906 | −0.0200 | 0.0443 |
| 10–30 | −0.00675 | −0.0200 | 0.0601 |
| 30–60 | −0.00413 | −0.0200 | 0.0999 |
Only ln[A] has a constant slope: first order in A.
(b) Slope = −k, so k = 0.0200 min⁻¹. t½ = 0.693 ÷ 0.0200 = 34.7 min.
(c) t = ln(0.500 ÷ 0.0500) ÷ 0.0200 = 2.303 ÷ 0.0200 = 115 min.
If you missed this: Topic 5.3 study guide: “Testing the data: which plot is straight?”
Question 5 (multiple choice · 5.4)
An elementary step has a rate constant with units M⁻² s⁻¹ (rate in M s⁻¹). Which could the step be?
- (A) X → products
- (B) X + Y → products
- (C) 2X → products
- (D) 2X + Y → products
Answer and explanation
Answer: (D). M⁻² s⁻¹ means third order overall (k × M³ = M s⁻¹). For an elementary step the orders are the coefficients: 2X + Y gives rate = k[X]²[Y].
- (A) is unimolecular: k in s⁻¹.
- (B) and (C) are bimolecular (M⁻¹ s⁻¹).
If you missed this: Topic 5.4 study guide: “Where the units of k come from”.
Question 6 (multiple choice · 5.5)
A gas that reacts in one elementary step is compressed to half its volume at constant temperature. Which is correct?
- (A) Collisions are more frequent, and more of them have at least Eₐ because particles hit harder.
- (B) Collisions are more frequent; the fraction with at least Eₐ is unchanged.
- (C) Collision frequency is unchanged; more collisions have at least Eₐ.
- (D) Collisions are more frequent, and Eₐ falls.
Answer and explanation
Answer: (B). Halving the volume doubles every concentration, so particles meet more often. The temperature, and so the fraction beyond Eₐ, is unchanged.
- (A) links concentration to collision energy.
- (C) reverses the two effects.
- (D) Compression does not change Eₐ.
If you missed this: Topic 5.5 study guide: “Collision frequency and concentration”.
Question 7 (multiple choice · 5.6)
An elementary step has Eₐ(forward) = 48 kJ mol⁻¹ and ΔH = −62 kJ mol⁻¹. Both directions have similar Arrhenius factors A. At one temperature, which is correct?
- (A) Eₐ(reverse) = 14 kJ mol⁻¹; the reverse step is faster.
- (B) Eₐ(reverse) = 48 kJ mol⁻¹; both directions share one transition state.
- (C) Eₐ(reverse) = 110 kJ mol⁻¹; the reverse step is slower.
- (D) Eₐ(reverse) = 62 kJ mol⁻¹, the size of ΔH.
Answer and explanation
Answer: (C). The transition state is 48 kJ mol⁻¹ above the reactants and the products 62 below them, so the climb back is 48 + 62 = 110 kJ mol⁻¹ (Eₐ(forward) − ΔH). With similar A, a higher barrier means fewer collisions reach the transition state: the reverse step is slower.
- (A) subtracts the wrong way (62 − 48).
- (B) The two directions climb to the shared transition state from different levels.
- (D) uses ΔH alone and forgets the forward climb.
If you missed this: Topic 5.6 study guide: “Temperature and the Arrhenius equation”.
Question 8 (multiple choice · 5.7)
A₂ + 2B → 2AB is proposed to go by four steps:
Step 1: A₂ + Q → A₂Q Step 2: A₂Q → A + AQ Step 3: AQ + B → AB + Q Step 4: A + B → AB
Which list gives all the intermediates?
- (A) A₂Q, AQ and A
- (B) A₂Q, AQ, A and Q
- (C) A₂Q and AQ only
- (D) Q only
Answer and explanation
Answer: (A). Each is made, then used: A₂Q (made in 1, used in 2), AQ (2, then 3), A (2, then 4).
- (B) includes Q, which is used first (step 1) and given back (step 3): a catalyst.
- (C) misses A, made in step 2 and used in step 4.
- (D) names the catalyst, not an intermediate.
If you missed this: Topic 5.7 study guide: “The parts of a mechanism”.
Question 9 (multiple choice · 5.8)
For X₂ + 2Y → 2XY, a proposed mechanism is Step 1 (slow): X₂ + Y → X₂Y; Step 2 (fast): X₂Y + Y → 2XY. Experiments give rate = k[X₂][Y]. A student says: “This proves the mechanism.” Which is the best evaluation?
- (A) Proven, because the predicted and experimental rate laws match.
- (B) Consistent but not proven: another mechanism could predict the same rate law.
- (C) Rejected: the overall equation has 2Y, so the reaction must be second order in Y.
- (D) Rejected: the intermediate X₂Y is not in the rate law.
Answer and explanation
Answer: (B). The steps add up and the slow first step predicts rate = k[X₂][Y]. A match shows the mechanism is possible, not unique.
- (A) overstates the evidence.
- (C) uses overall coefficients as orders; the second Y reacts after the slow step.
- (D) An intermediate should not appear in a testable rate law.
If you missed this: Topic 5.8 study guide: “Testing a mechanism against experiment”.
Question 10 (short answer · 5.9)
2G + J → P is proposed to go by:
Step 1 (fast, reversible): G + J ⇌ GJ Step 2 (fast, reversible): GJ + G ⇌ G₂J Step 3 (slow): G₂J → P
(a) Identify the intermediates. (b) Derive the rate law in terms of [G] and [J]. (c) Predict the effect on the rate of tripling [G].
Worked answer
(a) GJ (made in step 1, used in step 2) and G₂J (made in step 2, used in step 3).
(b)
- Slow step: rate = k₃[G₂J], but G₂J is an intermediate.
- Step 2 at equilibrium: [G₂J] = (k₂/k₋₂)[GJ][G]. Step 1 at equilibrium: [GJ] = (k₁/k₋₁)[G][J].
- Substitute: rate = k₃(k₂/k₋₂)(k₁/k₋₁)[G]²[J], so rate = k[G]²[J].
(c) Second order in G, so the rate rises by 3² = 9 times.
If you missed this: Topic 5.9 study guide: “The pre-equilibrium recipe”.
Question 11 (multiple choice · 5.10)
A two-step reaction has step 1: Eₐ = 40, ΔH = +15; step 2: Eₐ = 60, ΔH = −70. With the reactants at 0, which gives the energy of TS 2, the energy of the products and the rate-limiting step?
- (A) +60; −55; step 2
- (B) +75; −55; step 1, because it is uphill
- (C) +75; −70; step 2
- (D) +75; −55; step 2
Answer and explanation
Answer: (D). Intermediate = 0 + 15 = 15; TS 2 = 15 + 60 = 75; products = 15 − 70 = −55. Step 2 has the larger barrier (60 > 40).
- (A) measures step 2’s barrier from the reactants instead of from the intermediate.
- (B) An uphill step is not necessarily slow; barriers decide.
- (C) uses step 2’s ΔH as the overall ΔH.
If you missed this: Topic 5.10 study guide: “Reading the numbers”.
Question 12 (multiple choice · 5.11)
Heating and adding a catalyst both increase the fraction of collisions that can react. How?
- (A) Heating lowers Eₐ; the catalyst changes the energy distribution.
- (B) Both lower the Eₐ of the original path.
- (C) Heating changes the energy distribution; the catalyst gives a new path with a lower Eₐ.
- (D) Both shift the energy distribution to higher energies.
Answer and explanation
Answer: (C). Heating flattens and right-shifts the Maxwell–Boltzmann curve, so more area lies beyond the same Eₐ. A catalyst leaves the curve alone and lowers the threshold by changing the mechanism.
- (A) swaps the two effects.
- (B) Neither changes the original path’s barrier.
- (D) A catalyst gives the particles no extra energy.
If you missed this: Topic 5.11 study guide: “Why a lower barrier means a faster reaction”.
Your next step
| Topic | Question(s) | If you missed it, read |
|---|---|---|
| 5.1 Reaction rates | 1 | Topic 5.1 study guide |
| 5.2 Rate law | 2 | Topic 5.2 study guide |
| 5.3 Concentration over time | 3, 4 | Topic 5.3 study guide |
| 5.4 Elementary reactions | 5 | Topic 5.4 study guide |
| 5.5 Collision model | 6 | Topic 5.5 study guide |
| 5.6 Energy profile | 7 | Topic 5.6 study guide |
| 5.7 Reaction mechanisms | 8 | Topic 5.7 study guide |
| 5.8 Mechanism and rate law | 9 | Topic 5.8 study guide |
| 5.9 Pre-equilibrium | 10 | Topic 5.9 study guide |
| 5.10 Multistep energy profile | 11 | Topic 5.10 study guide |
| 5.11 Catalysis | 12 | Topic 5.11 study guide |
How to use your result
- Look at topics, not the total. Count a guess, or any wrong part of a short answer, as a miss.
- Fix gaps in order. Topics 5.1 to 5.3 measure rates; 5.4 to 5.6 explain them; 5.7 to 5.11 build mechanisms on both. Start with the earliest topic you missed.
- Then practise with each topic’s practice questions and the Unit 5 mixed review.