Free MCAT-style practice, 5E
MCAT Chemical Kinetics Practice Questions
Ten free MCAT-style kinetics practice questions on rate laws from data, rate constants and units, mechanisms, the Arrhenius equation, half-life, catalysts and reaction coordinate diagrams, with full explanations.
Quick concept primer
A rate law, rate = k[A]m[B]n, must be found from experiment, not read from the balanced equation. Compare two trials in which only one concentration changes: if doubling [A] doubles the rate, m = 1; if it quadruples the rate, m = 2; if nothing changes, m = 0. The units of k follow from the overall order, and for a first-order reaction the half-life, 0.693/k, does not depend on concentration.
For a multistep mechanism, the slowest step limits the overall rate, and the rate law matches that step's molecularity once intermediates are accounted for. On a reaction coordinate diagram, transition states are energy maxima and intermediates are local minima; the step with the highest transition state relative to the starting point is rate-determining. The Arrhenius equation, k = Ae-Ea/RT, shows that raising the temperature or lowering the activation energy increases k exponentially. A catalyst lowers Ea for the forward and reverse reactions equally, so it speeds the approach to equilibrium without changing ΔG, ΔH or K.
Traps: stoichiometric coefficients equal reaction orders only for elementary steps, and the product that forms fastest (kinetic control) is not always the most stable one (thermodynamic control).
10 practice questions
Question 1
A student studies the reaction 2A + B → C in three experiments at the same temperature and records the starting rate of each:
| Experiment | Starting [A], mol/L | Starting [B], mol/L | Starting rate, mol/(L s) |
|---|---|---|---|
| Run 1 | 0.050 | 0.020 | 1.5 × 10-5 |
| Run 2 | 0.100 | 0.020 | 6.0 × 10-5 |
| Run 3 | 0.050 | 0.060 | 4.5 × 10-5 |
Which rate law fits these data?
- rate = k[A][B]
- rate = k[A]2
- rate = k[A]2[B]
- rate = k[A][B]2
Show answer and explanation
Answer: C. rate = k[A]2[B]
From run 1 to run 2, [A] doubles with [B] fixed and the rate rises fourfold, so the order in A is 2. From run 1 to run 3, [B] triples with [A] fixed and the rate triples, so the order in B is 1. The rate law is k[A]2[B]. Choice B omits B even though changing [B] changed the rate.
Question 2
A reaction obeys rate = k[X]2[Y]. When [X] = 0.10 M and [Y] = 0.10 M, the initial rate is 2.0 × 10-4 M/s. What is the rate constant?
- 0.20 M-2 s-1
- 0.020 M-1 s-1
- 2.0 × 10-3 M-2 s-1
- 20 M-2 s-1
Show answer and explanation
Answer: A. 0.20 M-2 s-1
Rearranging, k = rate / ([X]2[Y]) = 2.0 × 10-4 / (0.010 × 0.10) = 0.20. For a third-order rate law, k has units of M-2 s-1 so that k times M3 gives M/s. Choice B forgets to square [X] and gives second-order units, and choices C and D are off by a factor of 100.
Question 3
A proposed mechanism for NO2 + CO → NO + CO2 is:
Step 1 (slow): NO2 + NO2 → NO3 + NO
Step 2 (fast): NO3 + CO → NO2 + CO2
Which statement is consistent with this mechanism?
- rate = k[NO2][CO], and NO3 is a catalyst
- rate = k[NO2]2, and NO3 is an intermediate
- rate = k[NO2][CO], and NO3 is an intermediate
- rate = k[NO2]2[CO], and NO is a catalyst
Show answer and explanation
Answer: B. rate = k[NO2]2, and NO3 is an intermediate
The slow first step is elementary and bimolecular in NO2, so it sets rate = k[NO2]2, and CO does not appear because it reacts only after the bottleneck. NO3 is made in step 1 and used in step 2, so it is an intermediate. A catalyst would be consumed first and regenerated later, and NO is simply a product.
Question 4
A reaction has an activation energy of 50 kJ/mol. Assuming the pre-exponential factor stays constant, about how many times faster is it at 310 K than at 300 K? (R = 8.314 J/(mol K))
- 1.03 times
- 1.2 times
- 10 times
- 1.9 times
Show answer and explanation
Answer: D. 1.9 times
From the Arrhenius equation, k2/k1 = exp[(Ea/R)(1/T1 - 1/T2)] = exp[(50,000/8.314)(1/300 - 1/310)] = exp(0.647), which is about 1.9. Choice A assumes the rate scales linearly with absolute temperature, which ignores the exponential dependence. A tenfold increase would require a much larger activation energy or temperature change.
Question 5
All of the following are true of a catalyst added to a reversible reaction EXCEPT:
- It increases the equilibrium constant, shifting the equilibrium toward products
- It lowers the activation energy of the forward and reverse reactions by the same amount
- It does not change the overall ΔG or ΔH of the reaction
- It is regenerated at the end of the reaction cycle
Show answer and explanation
Answer: A. It increases the equilibrium constant, shifting the equilibrium toward products
A catalyst provides a lower-energy pathway that speeds the forward and reverse reactions equally, so equilibrium is reached sooner but at the same position. K depends on ΔG°, which is a difference between reactant and product energies that the catalyst does not alter. Being regenerated is what distinguishes a catalyst from a reactant.
Question 6
A two-step reaction has the energies below, relative to the reactants.
| Point on the pathway | Energy (kJ/mol) |
|---|---|
| Reactants | 0 |
| Transition state 1 | +80 |
| Intermediate | +30 |
| Transition state 2 | +60 |
| Products | -20 |
Which statement is correct?
- Step 2 is rate-determining, and the reaction is endothermic
- Step 1 is rate-determining, and the reaction is endothermic
- Step 1 is rate-determining, and the reaction is exothermic by 20 kJ/mol
- Step 2 is rate-determining, and the reaction is exothermic by 20 kJ/mol
Show answer and explanation
Answer: C. Step 1 is rate-determining, and the reaction is exothermic by 20 kJ/mol
The highest point on the pathway is transition state 1 at +80 kJ/mol, and step 1 also has the larger individual barrier (80 versus 60 - 30 = 30 kJ/mol), so step 1 is rate-determining. Products lie 20 kJ/mol below reactants, so the reaction is exothermic. Treating the intermediate energy as the products' energy would wrongly suggest an endothermic reaction.
Question 7
A drug is eliminated by first-order kinetics with a rate constant of 0.0231 min-1. About what percentage of the initial amount remains after 90 minutes?
- 25 percent
- 12.5 percent
- 6.25 percent
- 33 percent
Show answer and explanation
Answer: B. 12.5 percent
The half-life is 0.693/k = 0.693/0.0231 = 30 minutes. Ninety minutes is three half-lives, so the fraction remaining is (1/2)3 = 1/8, or 12.5 percent. Choice A stops after two half-lives, choice C counts four, and choice D treats the decay as linear.
Question 8
A starting material R can form either of two products. The pathways are reversible when heated, but at low temperature the reverse reactions are negligibly slow.
| Product | Activation energy from R (kJ/mol) | Energy relative to R (kJ/mol) |
|---|---|---|
| X | 55 | -15 |
| Y | 70 | -45 |
Which outcome is expected?
- Y dominates under both conditions, because it is more stable
- X dominates under both conditions, because it forms faster
- Y dominates at low temperature and short times, and X dominates at high temperature and long times
- X dominates at low temperature and short times, and Y dominates at high temperature and long times
Show answer and explanation
Answer: D. X dominates at low temperature and short times, and Y dominates at high temperature and long times
When the reactions cannot reverse, the product ratio reflects relative rates, and X has the lower barrier, so X accumulates (kinetic control). When heating makes both pathways reversible, the system equilibrates and the lower-energy product, Y, dominates (thermodynamic control). Choices A and B each ignore one of the two regimes, and choice C reverses them.
Question 9
The rate constant for a reaction with a single reactant has units of M-1 s-1. If the reactant's concentration is doubled, what happens to the initial rate?
- It quadruples
- It doubles
- It does not change
- It is halved
Show answer and explanation
Answer: A. It quadruples
For rate in M/s, k must have units of M1-n s-1, where n is the overall order. M-1 s-1 means n = 2, so the rate law is rate = k[A]2 and doubling [A] multiplies the rate by 22 = 4. Doubling would match first order (k in s-1), and no change would match zero order (k in M/s).
Question 10
Which statement correctly distinguishes a reaction intermediate from a transition state?
- Both are energy maxima, but an intermediate lasts longer
- An intermediate sits in a local energy minimum and can sometimes be detected or trapped, while a transition state is an energy maximum with partially formed bonds that cannot be isolated
- A transition state appears in the rate law, while an intermediate never affects the rate
- An intermediate appears in the overall balanced equation, while a transition state does not
Show answer and explanation
Answer: B. An intermediate sits in a local energy minimum and can sometimes be detected or trapped, while a transition state is an energy maximum with partially formed bonds that cannot be isolated
On a reaction coordinate diagram an intermediate occupies a valley between two transition states, so it has a finite lifetime and is sometimes observable. A transition state is the peak of an energy barrier, a fleeting arrangement with partial bonds. Intermediates are formed and consumed, so they cancel out of the overall equation, and their formation often controls the rate.