Kinetics
Unit 5 of AP Chemistry, worth 7–9% of the exam. 14 questions below, each with the working. Every answer was checked by a second pass before it was published.
Reaction rates, rate laws, concentration changes over time, elementary reactions, collision model, reaction energy profiles, mechanisms, catalysis.
How this unit is tested
Start by identifying what type of kinetics question you're facing: (1) a data table of initial rates, (2) a concentration-vs-time relationship, (3) a mechanism, or (4) an energy-diagram/catalysis question. Each type has its own toolkit, and mixing them up is the most common source of errors on this unit.
For rate-law questions, never assume the exponents match the balanced overall equation — determine each order experimentally by comparing two trials where only one concentration changes, then find the change in rate. Once you have the rate law, solve for k using any single trial and always report units, since they depend on the overall reaction order.
For concentration-over-time questions, recognize the three integrated rate laws by which plot is linear: [A] vs t (zero order), ln[A] vs t (first order), 1/[A] vs t (second order). A constant half-life across trials is the signature of a first-order reaction — use that as a quick diagnostic before doing algebra.
For mechanisms, remember that only elementary steps have rate laws you can write directly from stoichiometry. The overall rate law is set by the slowest step (rate-determining step). If that step contains an intermediate, you must eliminate it — either using the previous fast step directly (if it's the source of the intermediate) or by applying the fast pre-equilibrium approximation before you can compare your predicted rate law to the experimental one. Finally, for energy diagrams and catalysis, keep straight what a catalyst changes (Ea, mechanism, rate) versus what it never changes (ΔH, K_eq).
What you have to know
Differential rate law
For reaction aA + bB → products, rate = $k[A]^m[B]^n$, where m and n (the orders) must be determined experimentally, not read from the balanced equation, unless the step is elementary.
Integrated rate laws
Zero order: $[A]_t=-kt+[A]_0$. First order: $\ln[A]_t=-kt+\ln[A]_0$. Second order: $\frac{1}{[A]_t}=kt+\frac{1}{[A]_0}$.
Half-life formulas
Zero order: $t_{1/2}=\frac{[A]_0}{2k}$. First order: $t_{1/2}=\frac{0.693}{k}$ (independent of concentration). Second order: $t_{1/2}=\frac{1}{k[A]_0}$.
Arrhenius equation
$k=Ae^{-E_a/RT}$, or in linear form $\ln k = -\frac{E_a}{R}\left(\frac{1}{T}\right)+\ln A$; relates the rate constant to temperature and activation energy.
Elementary steps and molecularity
For a single elementary step, the rate law exponents equal the stoichiometric coefficients of that step; molecularity is the number of species colliding (uni-, bi-, or termolecular).
Rate-determining step principle
The overall rate law equals the rate law of the slowest elementary step; if that step contains a reaction intermediate, substitute using the preceding fast equilibrium step so the final rate law contains only species from the overall equation.
14 practice questions
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For the reaction 2A + B → 3C, the rate of disappearance of A is measured as 0.040 M/s. What is the rate of appearance of C?
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Answer. 0.060 M/s
Using rate=-1/2·Δ[A]/Δt=1/3·Δ[C]/Δt, the reaction rate itself is 0.040/2=0.020 M/s, so Δ[C]/Δt=3×0.020=0.060 M/s. -
A reaction follows rate=k[A]^2[B]. If [A] is tripled and [B] is halved, by what factor does the rate change?
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Answer. 4.5
The factor is (3)^2×(0.5)^1=9×0.5=4.5, since rate depends on the square of [A] and the first power of [B]. -
Given: Exp1 [X]=0.050,[Y]=0.050,rate=2.5×10⁻4; Exp2 [X]=0.100,[Y]=0.050,rate=2.5×10⁻4; Exp3 [X]=0.100,[Y]=0.100,rate=1.0×10⁻3. What is the rate law?
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Answer. rate = k[Y]^2 (zero order in X)
Doubling [X] (Exp1→2) with [Y] fixed causes no change in rate, so order in X is 0. Doubling [Y] (Exp2→3) with [X] fixed quadruples the rate, so order in Y is 2. -
A first-order reaction has k=0.045 s⁻1. What is its half-life?
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Answer. ≈15.4 s
For first order, $t_{1/2}=0.693/k=0.693/0.045\approx15.4$ s, independent of the starting concentration. -
A second-order reaction has k=0.50 M⁻1s⁻1 and starts at [A]0=0.20 M. What is the half-life?
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Answer. 10 s
For second order, $t_{1/2}=1/(k[A]_0)=1/(0.50\times0.20)=10$ s. -
Which plot is linear for a zero-order reaction?
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Answer. [A] vs t
Zero-order integrated rate law is $[A]_t=-kt+[A]_0$, which is linear when [A] is plotted against t, with slope -k. -
The step 2NO(g) + O2(g) → 2NO2(g) is proposed as a single elementary reaction. Write its rate law.
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Answer. rate = k[NO]^2[O2]
For an elementary step, the rate law exponents equal the stoichiometric coefficients directly, giving a termolecular reaction with orders 2 and 1. -
Mechanism: Step 1 (slow) NO2+NO2→NO3+NO; Step 2 (fast) NO3+CO→NO2+CO2; overall NO2+CO→NO+CO2. Predict the rate law and identify the intermediate.
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Answer. rate = k[NO2]^2; intermediate is NO3
The overall rate law matches the slow (rate-determining) step, giving rate=k[NO2]^2 from its stoichiometry. NO3 is produced in step 1 and consumed in step 2, so it never appears in the overall equation, making it an intermediate. -
Mechanism: Step 1 (fast equilibrium) 2NO ⇌ N2O2, with forward rate constant k1 and reverse k-1; Step 2 (slow) N2O2+O2→2NO2, rate constant k2. Derive the overall rate law in terms of [NO] and [O2] only.
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Answer. rate = k[NO]^2[O2], where k = k2k1/k-1
The rate depends on the slow step: rate=k2[N2O2]. Since step 1 is a fast equilibrium, forward and reverse rates are equal: k1[NO]^2=k-1[N2O2], so [N2O2]=(k1/k-1)[NO]^2. Substituting gives rate=k2(k1/k-1)[NO]^2[O2]. -
On a reaction energy profile for an exothermic reaction, what do the reactant-to-peak gap and the reactant-to-product gap represent?
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Answer. Reactant-to-peak = activation energy (Ea); reactant-to-product = ΔH (negative for exothermic)
The peak represents the transition state, the highest-energy, unstable arrangement during the reaction. The vertical distance from reactants up to that peak is Ea, while the net vertical drop from reactants to products is the enthalpy change, ΔH. -
What effect does adding a catalyst have on a reaction?
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Answer. It decreases the activation energy for both the forward and reverse reactions
A catalyst provides an alternative mechanism with a lower-energy transition state, speeding up both directions equally. It does not change ΔH or the equilibrium constant. -
According to collision theory, why does raising the temperature typically increase reaction rate more dramatically than raising concentration does?
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Answer. Because higher temperature exponentially increases the fraction of molecules with kinetic energy at or above Ea
Per the Boltzmann distribution embedded in the Arrhenius equation, a modest temperature increase sharply raises the number of sufficiently energetic, correctly oriented collisions, while increasing concentration only linearly raises total collision frequency without changing the energy distribution. -
A reaction has k1=1.0×10⁻3 s⁻1 at 300 K and k2=1.0×10⁻2 s⁻1 at 350 K. Estimate the activation energy Ea.
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Answer. ≈40.2 kJ/mol
Using $\ln(k_2/k_1)=-\frac{E_a}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$: $\ln(10)=2.303=-\frac{E_a}{8.314}(0.002857-0.003333)$, giving $E_a\approx2.303/0.00005729\approx40190$ J/mol, or about 40.2 kJ/mol. -
Which statement best describes a catalyst's role within a reaction mechanism?
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Answer. It appears as a reactant in an early step and is regenerated in a later step
A catalyst is consumed in one elementary step but reproduced in a subsequent step, so it speeds up the mechanism without being used up overall or appearing in the net balanced equation.
What people get wrong
- Assuming reaction orders match the coefficients of the overall balanced equation — this is only true for elementary steps. Always determine orders from experimental data for the overall reaction.
- Leaving a reaction intermediate in the final predicted rate law after using the rate-determining step. Instead, use the fast pre-equilibrium (or the step that produced the intermediate) to rewrite it in terms of original reactants.
- Confusing which linear plot corresponds to which order (mixing up ln[A] vs t with 1/[A] vs t). Instead, memorize: zero order is linear as [A] vs t, first order as ln[A] vs t, second order as 1/[A] vs t.
- Believing a catalyst changes ΔH or shifts the equilibrium position. Instead, remember a catalyst only lowers Ea for both directions and speeds up reaching the same equilibrium, without changing K or ΔH.
- Forgetting to divide or multiply by stoichiometric coefficients when relating the rate of disappearance of one species to the rate of appearance of another. Instead, always write $\text{rate}=-\frac{1}{a}\frac{\Delta[A]}{\Delta t}=\frac{1}{c}\frac{\Delta[C]}{\Delta t}$ before solving.
Drill this unit until it sticks
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