Unit 5 practice test: Kinetics
12 exam-style questions from the unit, free, with an explanation for every option. Four answer choices each, and data sets that share one table, graph or particle diagram, as on the real exam.
Covers: Reaction Rates; Introduction to Rate Law; Concentration Changes Over Time; Elementary Reactions; Collision Model; Reaction Energy Profile; Introduction to Reaction Mechanisms; Reaction Mechanism and Rate Law; Pre-Equilibrium Approximation; Multistep Reaction Energy Profile; Catalysis.
Data table
Testing mechanisms for hydrogen iodide
For H₂(g) + I₂(g) → 2 HI(g) at a fixed high temperature, initial rates were measured:
Mechanism P: H₂ + I₂ → 2 HI (one step).
Mechanism Q: (1) I₂ ⇌ 2 I (fast, reversible); (2) H₂ + 2 I → 2 HI (slow).
| Trial | [H₂] (M) | [I₂] (M) | Initial rate (M/s) |
|---|---|---|---|
| 1 | 0.050 | 0.050 | 1.2 × 10⁻⁵ |
| 2 | 0.100 | 0.050 | 2.4 × 10⁻⁵ |
| 3 | 0.050 | 0.100 | 2.4 × 10⁻⁵ |
1. What is the experimental rate law?
- rate = k[H₂][I₂]
- rate = k[H₂]²[I₂]
- rate = k[H₂][I₂]²
- rate = k[I₂]
Show the answer
Doubling [H₂] doubles the rate (trials 1, 2); doubling [I₂] doubles the rate (trials 1, 3). First order in each.
- Correct: rate = k[H₂][I₂]: Right: first order in both.
- rate = k[H₂]²[I₂]: Second order in H₂ would quadruple the rate in trial 2.
- rate = k[H₂][I₂]²: Second order in I₂ would quadruple the rate in trial 3.
- rate = k[I₂]: The rate also depends on [H₂].
2. What rate law does mechanism Q predict?
- rate = k[H₂][I₂]
- rate = k[H₂][I]²
- rate = k[H₂][I₂]²
- rate = k[I₂]
Show the answer
Slow step: rate = k₂[H₂][I]². Fast reversible step: k₁[I₂] = k₋₁[I]², so [I]² = (k₁/k₋₁)[I₂]. Substituting: rate = (k₂k₁/k₋₁)[H₂][I₂].
- Correct: rate = k[H₂][I₂]: Right: [I]² is replaced by a term in [I₂].
- rate = k[H₂][I]²: This still contains the intermediate I.
- rate = k[H₂][I₂]²: Only one I₂ is broken in step 1, so [I₂] is first order.
- rate = k[I₂]: H₂ is a reactant in the slow step, so it must appear.
3. Which conclusion is justified by the data?
- Both fit; rate data do not separate them
- Q fits and P does not, because Q has a reversible step
- P fits and Q does not, because P is a single step
- Neither fits, because I atoms are missing from the rate law
Show the answer
P predicts k[H₂][I₂] directly; Q predicts k[H₂][I₂] after substitution. Both match the measured rate law, so the rate data alone cannot decide between them.
- Correct: Both fit; rate data do not separate them: Right: two mechanisms, one rate law.
- Q fits and P does not, because Q has a reversible step: Having a reversible step is not evidence for a mechanism.
- P fits and Q does not, because P is a single step: Being a single step is not evidence either.
- Neither fits, because I atoms are missing from the rate law: An intermediate should not appear in the final rate law; its absence is expected.
4. In mechanism Q, why can the first step be treated as having equal forward and back rates?
- It runs fast both ways compared with step 2
- I atoms do not react with H₂ here
- The slow step uses up the I₂ as soon as it forms
- The first step has no activation energy
Show the answer
Because I₂ ⇌ 2 I runs quickly both ways while step 2 removes I only slowly, the first step stays balanced: as many I₂ break apart each second as I pairs rejoin.
- Correct: It runs fast both ways compared with step 2: Right: the fast step stays balanced while the slow step trickles.
- I atoms do not react with H₂ here: I atoms do react with H₂; that is step 2.
- The slow step uses up the I₂ as soon as it forms: The slow step uses I atoms, not I₂ directly, and slowly.
- The first step has no activation energy: Every step has some activation energy; being fast does not mean zero.
Graph
Two reactions at the same temperature
Energy profiles for two different one-step reactions, P and Q, run at the same temperature with the same starting concentrations. Assume their orientation needs are similar.
Reaction PReaction Q
Data table
| Reaction progress | Reaction P | Reaction Q |
|---|---|---|
| 0 | 50 | 50 |
| 1 | 50 | 50 |
| 2 | 50 | 50 |
| 3 | 62.5 | 77.5 |
| 4 | 87.5 | 132.5 |
| 5 | 100 | 160 |
| 6 | 93.8 | 125 |
| 7 | 81.3 | 55 |
| 8 | 75 | 20 |
| 9 | 75 | 20 |
| 10 | 75 | 20 |
5. Which reaction is faster, and why?
- P, because its activation energy is smaller
- Q, because it releases more energy
- Q, because its products are lower in energy
- They are equally fast, because they start at the same energy
Show the answer
P has Ea = 100 − 50 = 50 kJ/mol; Q has Ea = 160 − 50 = 110 kJ/mol. At the same temperature, a larger fraction of collisions clears the smaller barrier, so P is faster. How much energy a reaction releases does not set its rate.
- Correct: P, because its activation energy is smaller: Right: the barrier height controls the rate.
- Q, because it releases more energy: Releasing more energy says nothing about how fast the reaction starts; Q has the much higher barrier.
- Q, because its products are lower in energy: The product level sets the overall energy change, not the rate.
- They are equally fast, because they start at the same energy: Equal starting energies do not mean equal barriers.
6. Which statement describes the overall energy change of each reaction?
- P absorbs energy (+25 kJ/mol); Q releases energy (−30 kJ/mol)
- P releases energy (−25 kJ/mol); Q absorbs energy (+30 kJ/mol)
- P absorbs 50 kJ/mol; Q absorbs 110 kJ/mol
- Both release energy, since both start at 50 kJ/mol
Show the answer
P: 75 − 50 = +25 kJ/mol, products higher, energy absorbed. Q: 20 − 50 = −30 kJ/mol, products lower, energy released.
- Correct: P absorbs energy (+25 kJ/mol); Q releases energy (−30 kJ/mol): Right: products minus reactants for each.
- P releases energy (−25 kJ/mol); Q absorbs energy (+30 kJ/mol): The signs are reversed: P's products are higher than its reactants.
- P absorbs 50 kJ/mol; Q absorbs 110 kJ/mol: Those are the activation energies, not the overall changes.
- Both release energy, since both start at 50 kJ/mol: Where a reaction starts does not tell you whether it releases energy; compare the product level.
7. A student claims, "Q releases energy, so it must be the faster reaction." Which evidence from the graph best refutes this?
- Q's barrier is 110 kJ/mol, more than double P's 50 kJ/mol
- Q's products are at 20 kJ/mol
- P's products are at 75 kJ/mol
- Both reactions start at 50 kJ/mol
Show the answer
Rate depends on the activation energy, not on the overall energy change. Q has the larger barrier, so fewer collisions succeed and it is slower, even though it releases energy.
- Correct: Q's barrier is 110 kJ/mol, more than double P's 50 kJ/mol: Right: compares the barriers, which set the rates.
- Q's products are at 20 kJ/mol: This is the product level, which sets the energy change, not the rate.
- P's products are at 75 kJ/mol: P's product level is also about the energy change, not the rate.
- Both reactions start at 50 kJ/mol: Equal starting points do not compare the barriers.
8. On a graph of [reactant] against time, how do you find the instantaneous rate at t = 50 s?
- Take the negative of the slope of the tangent at 50 s
- Divide the concentration at 50 s by 50 s
- Join the first and last points and take the slope of that line
- Read the concentration at 50 s; that value is the rate
Show the answer
The instantaneous rate is the slope of the curve at one moment, which is the slope of the tangent line there. The reactant falls, so the slope is negative and the rate is its negative.
- Correct: Take the negative of the slope of the tangent at 50 s: Right: the slope of the tangent at that time, made positive for a reactant.
- Divide the concentration at 50 s by 50 s: A rate is a change divided by a time, not a concentration divided by a time.
- Join the first and last points and take the slope of that line: That gives the average rate over the whole run, not the rate at 50 s.
- Read the concentration at 50 s; that value is the rate: A concentration is not a rate; the rate is how fast the concentration changes.
9. For A + B → C, the rate is first order in B. When [A] and [B] are both doubled at the same time, the initial rate becomes 8 times larger. What is the order in A?
- Zero order
- First order
- Second order
- Third order
Show the answer
Doubling [B] alone doubles the rate (first order). The rest of the factor of 8 comes from A: 8 ÷ 2 = 4 = 2², so the order in A is 2.
- Zero order: Zero order in A would make the total change only 2.
- First order: First order in both would give 2 × 2 = 4, not 8.
- Correct: Second order: Right: 2ᵐ × 2¹ = 8 gives m = 2.
- Third order: This puts the whole factor of 8 on A and forgets the doubling from B.
10. Iodine-131 decays by a first-order process with a half-life of 8.02 days. How many days pass until 12.5% of a sample is left?
Type a number in days.
Show the answer
100% → 50% → 25% → 12.5% is three half-lives: 3 × 8.02 days = 24.1 days. Radioactive decay is first order, so every half-life is the same length.
- Answer: 24.1 days
11. The elementary step NO + O₃ → NO₂ + O₂ has k = 1.8 × 10⁷ M⁻¹ s⁻¹ at a certain temperature. What is the rate, in M/s, when [NO] = 2.0 × 10⁻⁸ M and [O₃] = 5.0 × 10⁻⁷ M?
Type a number in M/s.
Show the answer
rate = k[NO][O₃] = (1.8 × 10⁷ M⁻¹ s⁻¹)(2.0 × 10⁻⁸ M)(5.0 × 10⁻⁷ M) = 1.8 × 10⁻⁷ M/s.
- Answer: 1.8 × 10-7 M/s
12. Food spoils more slowly in a refrigerator. Which explanation uses the collision model correctly?
- At lower temperature fewer collisions have enough energy to react
- Cold temperatures raise the activation energy of the spoiling reactions
- Cold molecules stop moving, so they no longer collide
- Cold air contains fewer molecules of oxygen per liter
Show the answer
Lower temperature shifts the energy distribution to lower energies, so a much smaller fraction of collisions reaches Ea, and the spoiling reactions slow down.
- Correct: At lower temperature fewer collisions have enough energy to react: Right: fewer energetic collisions.
- Cold temperatures raise the activation energy of the spoiling reactions: The activation energy is set by the reaction pathway; temperature does not change it.
- Cold molecules stop moving, so they no longer collide: Molecules still move at refrigerator temperatures; they just move more slowly.
- Cold air contains fewer molecules of oxygen per liter: Cold air is slightly denser, which would speed reactions with oxygen, not slow them.
Keep going
Practice has every question in the unit, with feedback after each one; the notes for every topic are free. Other units: Unit 1 · Unit 2 · Unit 3 · Unit 4 · Unit 5 · Unit 6 · Unit 7 · Unit 8 · Unit 9.