Kinetics — AP Chemistry Chem Study Guide

For: AP Chemistry candidates sitting AP Chemistry.

Covers: Reaction rate measurement methods, rate law and reaction order calculations, integrated rate equations and half-life formulas, activation energy calculations with the Arrhenius equation, and reaction mechanism analysis including intermediate identification.

You should already know: High-school chemistry, Algebra 2.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Chemistry style for educational use. They are not reproductions of past College Board papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official College Board mark schemes for grading conventions.

1. What Is Kinetics?

Kinetics is the study of the speed of chemical reactions, the factors that alter reaction rate, and the step-by-step pathways reactants follow as they convert to products. It appears in Unit 5 of the AP Chemistry CED, accounts for 7-9% of your total exam score, and is tested in both multiple-choice and free-response sections, often paired with thermodynamics questions to contrast reaction favorability (ΔG) and observable reaction speed. The term is sometimes referred to as "chemical kinetics" in textbooks, but the AP exam consistently uses the shorthand "kinetics" for this topic.

2. Rate of reaction — measurement

Reaction rate is defined as the change in concentration of a reactant or product per unit time, with standard units of $M/s$ ($mol\cdot L^{-1}\cdot s^{-1}$). Because rate is always reported as a positive value, you add a negative sign when calculating rate from decreasing reactant concentrations: $$rate=-\frac{\Delta [\text{reactant}]}{\Delta t}=\frac{\Delta [\text{product}]}{\Delta t}$$ For reactions with non-1 stoichiometric coefficients, normalize rate by dividing by the coefficient of the species you are measuring. For the general reaction $aA+bB\to cC+dD$: $$rate=-\frac{1}{a}\frac{\Delta [A]}{\Delta t}=-\frac{1}{b}\frac{\Delta [B]}{\Delta t}=\frac{1}{c}\frac{\Delta [C]}{\Delta t}=\frac{1}{d}\frac{\Delta [D]}{\Delta t}$$ AP exam questions frequently test 4 common rate measurement methods: pressure tracking for gas-phase reactions, colorimetry for reactions with colored reactants/products, titration of reaction aliquots for acid-base/redox reactions, and mass change for reactions producing solid or gaseous products.

Worked Example: For the reaction $2N_2{O}_5(g)\to 4N{O}_2(g)+O_2(g)$, $[N_2{O}_5]$ decreases from 0.400 M to 0.320 M over 100 seconds. Calculate the overall reaction rate and the rate of formation of $N{O}_2$.

1. Calculate $\Delta [N_2{O}_5]=0.320M-0.400M=-0.080M$
2. Overall rate: $rate=-\frac{1}{2}\frac{\Delta [N_2{O}_5]}{\Delta t}=-\frac{1}{2}\frac{-0.080M}{100s}=4.0\times 10^{-4}M/s$
3. Rate of $N{O}_2$ formation: $4\times rate=4\times 4.0\times 10^{-4}M/s=1.6\times 10^{-3}M/s$

Exam tip: Examiners regularly test stoichiometric rate conversions, so never forget to divide by the species coefficient when calculating overall reaction rate.

3. Rate laws and orders

A rate law is an experimentally determined equation that relates reaction rate to reactant concentrations, with the general form: $$rate=k[A{]}^m[B{]}^n$$ Where $k$ = temperature-dependent rate constant, $m$ = reaction order with respect to $A$, $n$ = reaction order with respect to $B$, and overall reaction order = $m+n$. Critical note: reaction orders are not equal to the stoichiometric coefficients of the overall reaction, and can only be determined from experimental data or the rate law of the slow step of a reaction mechanism. Orders can be 0, 1, 2, or fractional.

To determine orders from initial rate data, compare two trials where only one reactant concentration changes, and calculate how the rate scales with that concentration change.

Worked Example: Use the data below for the reaction $A+B\to C$ to find the rate law, overall order, and rate constant $k$.

1. Compare trials 1 and 2: $[A]$ doubles, $[B]$ constant, rate doubles → $m=1$ (first order in A)
2. Compare trials 1 and 3: $[B]$ doubles, $[A]$ constant, rate quadruples → $n=2$ (second order in B)
3. Rate law: $rate=k[A][B{]}^2$
4. Overall order: $1+2=3$
5. Solve for $k$ using trial 1 data: $k=\frac{rate}{[A][B{]}^2}=\frac{2.0\times 10^{-3}}{0.1\times (0.1{)}^2}=2.0M^{-2}{s}^{-1}$

Exam tip: Units of $k$ depend on overall order: 0 order = $M\cdot s^{-1}$, 1st order = $s^{-1}$, 2nd order = $M^{-1}\cdot s^{-1}$, 3rd order = $M^{-2}\cdot s^{-1}$. You can use this to quickly verify your order calculations on the exam.

4. Integrated rate equations and half-life

Integrated rate laws relate reactant concentration to time, so you can calculate how much reactant remains after a given time, or how long it takes for a reactant concentration to drop to a target value. The three key integrated rate laws tested on the AP exam are:

1. Zero order: Integrated form: $[A{]}_t=-kt+[A{]}_0$; linear plot of $[A]$ vs $t$, slope = $-k$; half-life $t_{1/2}=\frac{[A{]}_0}{2k}$ (increases as initial concentration increases)
2. First order: Integrated form: $\ln [A{]}_t=-kt+\ln [A{]}_0$; linear plot of $\ln [A]$ vs $t$, slope = $-k$; half-life $t_{1/2}=\frac{\ln 2}{k}\approx \frac{0.693}{k}$ (independent of initial concentration, a frequently tested property)
3. Second order: Integrated form: $\frac{1}{[A{]}_t}=kt+\frac{1}{[A{]}_0}$; linear plot of $\frac{1}{[A]}$ vs $t$, slope = $k$; half-life $t_{1/2}=\frac{1}{k[A{]}_0}$ (decreases as initial concentration increases)

Worked Example: The first-order decomposition of $H_2{O}_2$ has a rate constant $k=7.30\times 10^{-4}{s}^{-1}$ at 25°C. If initial $[H_2{O}_2]=0.150M$, calculate $[H_2{O}_2]$ after 1000 s and the reaction half-life.

1. Substitute into first-order integrated law: $\ln [A{]}_t=-(7.30\times 10^{-4}{s}^{-1})(1000s)+\ln (0.150)=-0.730-1.897=-2.627$
2. Solve for $[A{]}_t$: $[A{]}_t=e^{-2.627}=0.072M$
3. Calculate half-life: $t_{1/2}=\frac{0.693}{7.30\times 10^{-4}{s}^{-1}}=949s$

Exam tip: Examiners frequently ask you to identify reaction order from linear plots, so memorize the plot for each order: zero = $[A]$ vs $t$, first = $\ln [A]$ vs $t$, second = $\frac{1}{[A]}$ vs $t$.

5. Activation energy and Arrhenius

Activation energy ($E_a$) is the minimum energy reactant molecules must have to undergo a successful collision and form products, measured in $kJ/mol$. The Arrhenius equation relates the rate constant $k$ to absolute temperature $T$ (Kelvin) and $E_a$: $$k=A{e}^{-\frac{E_a}{RT}}$$ Where $A$ = frequency factor (accounts for collision frequency and correct orientation), $R=8.314J\cdot mo{l}^{-1}\cdot K^{-1}$ (use this value for kinetics calculations, not the $0.0821L\cdot atm\cdot mo{l}^{-1}\cdot K^{-1}$ value for gas laws).

For calculations using two $k$ values at different temperatures, use the two-point rearranged form of the Arrhenius equation: $$\ln \left(\frac{k_2}{k_1}\right)=\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)$$ The linear form of the equation $\ln k=-\frac{E_a}{R}\left(\frac{1}{T}\right)+\ln A$ means a plot of $\ln k$ vs $\frac{1}{T}$ is linear, with slope = $-\frac{E_a}{R}$.

Worked Example: The rate constant for a reaction is $1.3\times 10^{-3}{M}^{-1}{s}^{-1}$ at 200 K, and $7.0\times 10^{-3}{M}^{-1}{s}^{-1}$ at 300 K. Calculate $E_a$ in $kJ/mol$.

1. Substitute into the two-point equation: $\ln \left(\frac{7.0\times 10^{-3}}{1.3\times 10^{-3}}\right)=\frac{E_a}{8.314J\cdot mo{l}^{-1}\cdot K^{-1}}\left(\frac{1}{200K}-\frac{1}{300K}\right)$
2. Simplify: $1.684=\frac{E_a}{8.314}\times 0.001667$
3. Solve for $E_a$: $E_a=\frac{1.684\times 8.314}{0.001667}=8390J/mol=8.4kJ/mol$

Exam tip: Always convert $E_a$ to $J/mol$ when using $R=8.314J\cdot mo{l}^{-1}\cdot K^{-1}$ to avoid unit mismatch errors.

6. Reaction mechanisms and intermediates

A reaction mechanism is the sequence of elementary (single-collision) steps that make up an overall reaction. Unlike the overall reaction, the rate law of an elementary step is directly determined by its stoichiometry: unimolecular steps (1 reactant) are first order, bimolecular steps (2 reactants) are second order, and rare termolecular steps (3 reactants) are third order.

An intermediate is a species produced in one elementary step and consumed in a later step, so it does not appear in the overall reaction equation. The rate-determining step (RDS) is the slowest elementary step, so its rate law matches the experimentally determined overall rate law. For a mechanism to be valid: 1) the sum of elementary steps equals the overall reaction, and 2) the rate law of the RDS matches the experimental rate law.

Worked Example: The overall reaction $2NO(g)+O_2(g)\to 2N{O}_2(g)$ has an experimental rate law $rate=k[NO{]}^2[O_2]$. Prove the proposed mechanism below is valid, and identify the intermediate. Step 1 (fast equilibrium): $NO+O_2\rightleftharpoons N{O}_3$ Step 2 (slow, RDS): $N{O}_3+NO\to 2N{O}_2$

1. Sum the elementary steps: $NO+O_2+N{O}_3+NO\to N{O}_3+2N{O}_2$. Cancel the intermediate $N{O}_3$ to get $2NO+O_2\to 2N{O}_2$, matching the overall reaction.
2. Write the RDS rate law: $rate=k_2[N{O}_3][NO]$. Substitute the intermediate $N{O}_3$ using the fast equilibrium: forward rate = reverse rate → $k_1[NO][O_2]=k_{-1}[N{O}_3]$ → $[N{O}_3]=\frac{k_1}{k_{-1}}[NO][O_2]$.
3. Substitute into the RDS rate law: $rate=k_2\times \frac{k_1}{k_{-1}}[NO][O_2][NO]=k[NO{]}^2[O_2]$, matching the experimental rate law. The intermediate is $N{O}_3$.

Exam tip: Intermediates never appear in the overall rate law, so always substitute intermediates using the fast equilibrium step before comparing to the experimental rate law.

7. Common Pitfalls (and how to avoid them)

• Wrong move: Using overall reaction stoichiometric coefficients to determine rate law orders. Why students do it: Confuse overall reactions with elementary steps. Correct move: Only use experimental initial rate data or the RDS rate law to determine overall reaction orders.
• Wrong move: Using the $0.0821L\cdot atm\cdot mo{l}^{-1}\cdot K^{-1}$ gas constant for Arrhenius calculations. Why students do it: Overgeneralize the R value used for gas law problems. Correct move: Always use $R=8.314J\cdot mo{l}^{-1}\cdot K^{-1}$ for Arrhenius calculations, and convert $E_a$ to joules to match units.
• Wrong move: Assuming half-life is always independent of initial concentration. Why students do it: Most AP kinetics problems test first-order reactions, leading to overgeneralization. Correct move: Only first-order reactions have concentration-independent half-lives; zero-order half-lives increase with initial $[A]$, second-order half-lives decrease with initial $[A]$.
• Wrong move: Including intermediates in the overall rate law for a reaction mechanism. Why students do it: Forget intermediates are consumed before the reaction completes. Correct move: Substitute intermediates using the fast equilibrium step that produces them, so the final rate law only includes species in the overall reaction.
• Wrong move: Using Celsius temperature values directly in Arrhenius calculations. Why students do it: Rush through problems and skip unit conversion. Correct move: Always add 273 to Celsius temperatures to convert to Kelvin before plugging into the Arrhenius equation.

8. Practice Questions (AP Chemistry Style)

Question 1

For the reaction $3I^{-}(aq)+S_2{O}_8^{2-}(aq)\to I_3^{-}(aq)+2S{O}_4^{2-}(aq)$, the following initial rate data was collected:

Solution

a) Compare trials 1 and 2: $[I^{-}]$ doubles, rate doubles → first order in $I^{-}$. Compare trials 1 and 3: $[S_2{O}_8^{2-}]$ doubles, rate doubles → first order in $S_2{O}_8^{2-}$. Rate law: $rate=k[I^{-}][S_2{O}_8^{2-}]$. b) $k=\frac{rate}{[I^{-}][S_2{O}_8^{2-}]}=\frac{1.5\times 10^{-5}}{0.10\times 0.050}=3.0\times 10^{-3}{M}^{-1}{s}^{-1}$. c) Overall order = $1+1=2$.

Question 2

Radioactive decay of C-14 is a first-order process with a half-life of 5730 years. A fossil sample has 12.5% of the C-14 concentration found in living organisms. How old is the fossil?

Solution

First calculate $k$: $k=\frac{0.693}{5730yr}=1.21\times 10^{-4}y{r}^{-1}$. 12.5% remaining means $\frac{[A{]}_t}{[A{]}_0}=0.125$. Use the first-order integrated law: $\ln (0.125)=-(1.21\times 10^{-4}y{r}^{-1})t$ → $-2.079=-1.21\times 10^{-4}t$ → $t=17,200$ years. (Shortcut: 12.5% = 1/8 = 3 half-lives, so $3\times 5730=17,190$ years, consistent with the calculation.)

Question 3

The rate constant for a reaction is $2.5\times 10^{-2}{s}^{-1}$ at 15°C, and $7.8\times 10^{-2}{s}^{-1}$ at 35°C. Calculate the activation energy $E_a$ in $kJ/mol$.

Solution

Convert temperatures to Kelvin: $T_1=15+273=288K$, $T_2=35+273=308K$. Substitute into the two-point Arrhenius equation: $\ln \left(\frac{7.8\times 10^{-2}}{2.5\times 10^{-2}}\right)=\frac{E_a}{8.314}\left(\frac{1}{288}-\frac{1}{308}\right)$ $1.138=\frac{E_a}{8.314}\times 2.25\times 10^{-4}$ $E_a=\frac{1.138\times 8.314}{2.25\times 10^{-4}}=42,000J/mol=42kJ/mol$.

9. Quick Reference Cheatsheet

10. What's Next

Kinetics connects directly to multiple later topics in the AP Chemistry syllabus. You will apply your understanding of rate laws to analyze acid-base reaction kinetics, enzyme catalysis in biological systems, and nuclear decay (which follows first-order rate laws exclusively). Kinetics is also frequently paired with thermodynamics questions on the AP exam, where you will be asked to distinguish between thermodynamically favorable reactions (negative $\Delta G$) and kinetically slow reactions (high $E_a$), explaining why some favorable reactions do not proceed at observable rates at room temperature.

To reinforce your understanding of kinetics, practice working through more initial rate problems, integrated rate law calculations, and reaction mechanism analysis questions. If you get stuck on any problem, or have questions about specific kinetics concepts tested on the AP Chemistry exam, you can ask Ollie, our AI tutor, for personalized explanations and additional practice problems tailored to your weak points. You can also find more AP Chemistry study resources on the homepage to prepare for other units in the syllabus.