AP Chemistry Kinetics FRQ — Practice Questions & Scoring Guide 2026
Kinetics is one of the highest-frequency FRQ topics on AP Chemistry. It appears in at least one free response question on almost every exam. This guide covers the exact concepts you need, how AP readers score kinetics FRQs, and practice questions with solutions.
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What AP Chem Kinetics FRQs Test
AP Chemistry kinetics FRQs draw from Unit 5 (Kinetics). The most frequently tested concepts:
| Concept | Appears |
|---|---|
| Determining rate law from experimental data | Every year |
| Units of rate constant $k$ | Nearly every year |
| Reaction mechanism consistency with rate law | Every 1–2 years |
| Effect of temperature on rate (Arrhenius) | Every 1–2 years |
| Half-life calculations (first-order) | Every 1–2 years |
| Integrated rate laws (zero, first, second order) | Every 2–3 years |
| Catalysts and activation energy | Every 2–3 years |
Practice FRQ 1: Rate Law from Experimental Data
This is the most tested kinetics FRQ type.
Problem: A student conducts experiments to determine the rate law for:
$$\text{A} + \text{B} \rightarrow \text{products}$$
| Experiment | [A] (M) | [B] (M) | Initial Rate (M/s) |
|---|---|---|---|
| 1 | 0.10 | 0.10 | 2.0 × 10⁻³ |
| 2 | 0.20 | 0.10 | 4.0 × 10⁻³ |
| 3 | 0.10 | 0.30 | 1.8 × 10⁻² |
(a) Determine the order with respect to A and with respect to B. (b) Write the rate law. (c) Calculate the rate constant $k$ with appropriate units.
Solution:
(a) Order with respect to A:
Compare experiments 1 and 2 ([B] constant, [A] doubles):
$$\frac{\text{Rate}_2}{\text{Rate}_1} = \frac{k[0.20]^m[0.10]^n}{k[0.10]^m[0.10]^n} = \left(\frac{0.20}{0.10}\right)^m = 2^m$$
$$\frac{4.0 \times 10^{-3}}{2.0 \times 10^{-3}} = 2 = 2^m \implies m = 1$$
First order in A.
Order with respect to B:
Compare experiments 1 and 3 ([A] constant, [B] triples):
$$\frac{\text{Rate}_3}{\text{Rate}_1} = \left(\frac{0.30}{0.10}\right)^n = 3^n$$
$$\frac{1.8 \times 10^{-2}}{2.0 \times 10^{-3}} = 9 = 3^n \implies n = 2$$
Second order in B.
(b) Rate law:
$$\text{rate} = k[\text{A}][\text{B}]^2$$
(c) Rate constant:
Using experiment 1: $2.0 \times 10^{-3} = k(0.10)(0.10)^2 = k(1.0 \times 10^{-3})$
$$k = \frac{2.0 \times 10^{-3}}{1.0 \times 10^{-3}} = 2.0 \text{ M}^{-2}\text{s}^{-1}$$
Units check: Overall order = 1 + 2 = 3. Units of $k$ = M^(1−3) s^(−1) = M^(−2) s^(−1). ✓
AP Reader note: You must show your ratio calculation explicitly. Stating "the order is 1" without showing the ratio earns 0 points for that part.
Practice FRQ 2: Reaction Mechanism
Problem: The proposed mechanism for a reaction is:
Step 1 (slow): $\text{NO}_2 + \text{NO}_2 \rightarrow \text{NO}_3 + \text{NO}$
Step 2 (fast): $\text{NO}_3 + \text{CO} \rightarrow \text{NO}_2 + \text{CO}_2$
(a) Write the overall reaction. (b) Identify any intermediates. (c) Write the rate law predicted by this mechanism.
Solution:
(a) Add the two steps and cancel species that appear on both sides:
$$\text{NO}_2 + \cancel{\text{NO}_2} + \cancel{\text{NO}_3} + \text{CO} \rightarrow \cancel{\text{NO}_3} + \text{NO} + \cancel{\text{NO}_2} + \text{CO}_2$$
Overall: $\text{NO}_2 + \text{CO} \rightarrow \text{NO} + \text{CO}_2$
(b) Intermediates are species produced in one step and consumed in another: NO₃ is an intermediate.
(c) The rate law is determined by the slow (rate-determining) step:
From Step 1: $\text{rate} = k[\text{NO}_2]^2$
Intermediates cannot appear in the final rate law. Since NO₃ is an intermediate, not a reactant, Step 2 does not change the rate law expression.
Practice FRQ 3: Integrated Rate Laws
Problem: The decomposition of N₂O₅ is first order with $k = 4.8 \times 10^{-4}$ s⁻¹ at 45°C.
(a) What is the half-life of the reaction? (b) If the initial concentration is 0.200 M, what is the concentration after 2000 s?
Solution:
(a) Half-life for first-order reaction:
$$t_{1/2} = \frac{0.693}{k} = \frac{0.693}{4.8 \times 10^{-4}} = 1440 \text{ s} \approx 1400 \text{ s}$$
(b) First-order integrated rate law: $\ln[\text{A}]_t = \ln[\text{A}]_0 - kt$
$$\ln[\text{A}]_{2000} = \ln(0.200) - (4.8 \times 10^{-4})(2000)$$
$$= -1.609 - 0.960 = -2.569$$
$$[\text{A}]_{2000} = e^{-2.569} = 0.0768 \text{ M}$$
Practice FRQ 4: Arrhenius Equation and Temperature
Problem: A reaction has an activation energy of 85 kJ/mol. At 25°C ($T_1 = 298$ K) the rate constant is $k_1 = 1.5 \times 10^{-3}$ M⁻¹s⁻¹. What is $k_2$ at 50°C ($T_2 = 323$ K)?
Solution:
Use the two-temperature Arrhenius equation:
$$\ln\frac{k_2}{k_1} = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$$
$$= \frac{85000}{8.314}\left(\frac{1}{298} - \frac{1}{323}\right)$$
$$= 10221 \times (3.356 \times 10^{-3} - 3.096 \times 10^{-3})$$
$$= 10221 \times 2.60 \times 10^{-4} = 2.657$$
$$\frac{k_2}{k_1} = e^{2.657} = 14.3$$
$$k_2 = 14.3 \times 1.5 \times 10^{-3} = 2.14 \times 10^{-2} \text{ M}^{-1}\text{s}^{-1}$$
A 25°C temperature increase more than 14× the rate constant — this illustrates why temperature control matters in chemical reactions.
AP Chem Kinetics FRQ Scoring Tips
Show every ratio. For rate law determination, always write out the ratio of two experiments. Never just state the order — show the math.
Units of k matter. AP readers award a separate point for correct units. Derive them from the rate law: if rate = k[A][B]², then units of k = (M/s) / (M · M²) = M⁻²s⁻¹.
Distinguish rate from rate constant. Rate changes with concentration and temperature. $k$ only changes with temperature.
Mechanism questions: The rate-determining step determines the rate law. If an intermediate appears in the slow step, you must eliminate it by expressing it in terms of reactants.
Catalyst questions: A catalyst lowers activation energy, increases $k$, but does not change $\Delta H$ or the equilibrium constant.