Introduction to rate law — AP Chemistry Study Guide

For: AP Chemistry candidates sitting AP Chemistry.

Covers: The general form of rate laws, reaction order, rate constant properties, and the method of initial rates to determine rate law expressions from experimental data. Includes practice identifying components and calculating rate constants.

You should already know: Definitions of average and instantaneous reaction rate, stoichiometry of balanced chemical reactions, concentration units for solutions and gases.

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 / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Introduction to rate law?

A rate law (also called a rate equation) is a mathematical relationship that connects the instantaneous rate of a chemical reaction to the concentration of reactants (and occasionally catalysts, which are rarely tested in introductory problems). It is an exclusively experimentally derived expression—you cannot determine a rate law from the balanced reaction stoichiometry alone, a key distinction tested repeatedly on the AP exam. This topic is the foundation of all kinetics in AP Chemistry, and Unit 5 Kinetics accounts for 7-9% of total AP exam weight. Rate law questions appear in both MCQ (typically 1-2 standalone questions) and FRQ (almost always as the opening component of a multi-part kinetics FRQ, worth 2-4 points). Standard notation for rate laws uses $k$ for the rate constant, $[A]$ for the molar concentration of reactant $A$, and $n$ for the reaction order with respect to $A$. Even if a coefficient is 2 for A in the balanced equation, n may not equal 2—that must be found from experimental data.

2. Rate Law Structure and Reaction Order

The general form of a rate law for the overall reaction $aA+bB\to cC$ is: $$\text{rate}=k[A{]}^m[B{]}^n$$ Where each term is defined as:

• $\text{rate}$ = instantaneous reaction rate at the given starting concentrations, with units of ${\text{M time}}^{-1}$ (most commonly ${\text{M s}}^{-1}$ or ${\text{M min}}^{-1}$)
• $k$ = the rate constant, a proportionality constant unique to a specific reaction at a specific temperature. $k$ does not change with reactant concentration, only with temperature or addition of a catalyst.
• $[A],[B]$ = molar concentrations of reactants A and B; only gaseous and aqueous reactants are included (pure solids and liquids are omitted because their concentration is constant)
• $m,n$ = individual reaction orders with respect to A and B, almost always small positive integers (0, 1, 2) on the AP exam. The overall reaction order is the sum of all individual orders: $\text{overall order}=m+n$.

Reaction order tells you how changing a reactant’s concentration impacts the overall reaction rate. Zero order means changing the concentration has no effect on rate, first order means rate is directly proportional to concentration, and second order means rate is proportional to the square of concentration. Critically, individual orders do not need to match the stoichiometric coefficients in the balanced overall reaction—this is only true for elementary reactions, which are covered later in the course.

Worked Example

Problem: For the reaction $2NO(g)+O_2(g)\to 2N{O}_2(g)$, the experimentally determined rate law is $\text{rate}=k[NO{]}^2[O_2]$. Answer the following: (a) What is the order with respect to each reactant, and the overall reaction order? (b) How will the rate change if $[NO]$ is tripled and $[O_2]$ is held constant? (c) What are the units of $k$ for this reaction, if time is measured in seconds?

Solution steps:

1. The exponent of $[NO]$ is 2, so order with respect to NO is 2. The implicit exponent of $[O_2]$ is 1, so order with respect to O₂ is 1. Overall order = $2+1=3$.
2. Original rate: ${\text{rate}}_1=k[NO{]}^2[O_2]$. New $[NO]=3[NO{]}_1$, so ${\text{rate}}_2=k(3[NO{]}_1{)}^2[O_2]=9k[NO{]}_1^2[O_2]=9{\text{rate}}_1$. The rate increases by a factor of 9.
3. Units of rate are ${\text{M s}}^{-1}$. Rearrange to solve for units of $k$: $k=\frac{\text{rate}}{[NO{]}^2[O_2]}$, so units are $\frac{\text{M}\cdot {\text{s}}^{-1}}{{\text{M}}^2\cdot \text{M}}={\text{M}}^{-2}{\text{s}}^{-1}$.

Exam tip: When asked for units of $k$, always match the time units given in the problem. If concentrations are in M and time in minutes, your units must be $M^{-(overallorder-1)}mi{n}^{-1}$, not seconds.

3. Method of Initial Rates

The method of initial rates is the core experimental technique you will be tested on to determine a rate law from raw experimental data. Initial rate is the instantaneous reaction rate measured immediately after the reaction starts, before any significant change in reactant concentration occurs. Experimental data for this method is presented as a table of multiple runs of the same reaction, each with different starting reactant concentrations and a measured initial rate.

To find the order with respect to a given reactant, you compare two runs where only the concentration of that reactant changes, and all other reactant concentrations are held constant. The relationship used to solve for order is: $$\frac{{\text{rate}}_2}{{\text{rate}}_1}={\left(\frac{[A{]}_2}{[A{]}_1}\right)}^m$$ Where $m$ is the order with respect to A. On the AP exam, the ratio will almost always give a whole number order that can be solved by inspection, no logarithms required. Once you have all orders, you plug data from any run into the rate law to calculate $k$, then write the final rate law.

Worked Example

Problem: Use the experimental data below to determine the rate law for the reaction $A+2B\to C$:

Solution steps:

1. Find order with respect to A by comparing Run 1 and Run 2 (where $[B]$ is constant): $\frac{{\text{rate}}_2}{{\text{rate}}_1}=2$, $\frac{[A{]}_2}{[A{]}_1}=2$, so $2=2^m\to m=1$. Order with respect to A is 1.
2. Find order with respect to B by comparing Run 1 and Run 3 (where $[A]$ is constant): $\frac{{\text{rate}}_3}{{\text{rate}}_1}=4$, $\frac{[B{]}_3}{[B{]}_1}=2$, so $4=2^n\to n=2$. Order with respect to B is 2.
3. Write the rate law with confirmed orders: $\text{rate}=k[A][B{]}^2$.
4. Calculate $k$ from Run 1 data: $k=\frac{2.0\times 10^{-4}}{(0.10)(0.10{)}^2}=0.20M^{-2}{s}^{-1}$. Final rate law: $\text{rate}=(0.20M^{-2}{s}^{-1})[A][B{]}^2$.

Exam tip: Always check that you get the same $k$ value (within rounding) if you use data from any other run. If not, you made a mistake in calculating order. A quick check confirms the orders above are correct: Run 3 gives $k=8.0\times 10^{-4}/(0.10\times 0.20^2)=0.20M^{-2}{s}^{-1}$, matching the result from Run 1.

4. Zero Order Rate Laws

A zero order reaction with respect to a reactant means that changing the concentration of that reactant has no effect on the overall reaction rate. This most commonly occurs for enzyme-catalyzed reactions where the enzyme is saturated (all active sites are occupied at all times), or for heterogeneous reactions catalyzed on a solid surface. Since any number raised to the power of zero is 1, a zero order reactant does not appear in the final rate law expression. If all reactants are zero order, the rate is simply equal to $k$, so the rate is constant regardless of concentration.

Worked Example

Problem: The decomposition of $N_{2}O(g)$ on a solid platinum catalyst has the reaction $2N_{2}O(g)\to 2N_2(g)+O_2(g)$. Experimental data is below. Determine the rate law.

Solution steps:

1. Compare Run 1 and Run 2: $[N_{2}O]$ doubles from 0.25 M to 0.50 M, but the initial rate remains identical at $4.1\times 10^{-3}M{s}^{-1}$.
2. Apply the initial rate relationship: $\frac{{\text{rate}}_2}{{\text{rate}}_1}=1={\left(\frac{0.50}{0.25}\right)}^m=2^m$. The only value of $m$ that satisfies this is $m=0$.
3. Confirm with Run 2 and Run 3: $[N_{2}O]$ doubles again, rate is still unchanged, so $m=0$ is confirmed.
4. The final rate law simplifies to $\text{rate}=k[N_{2}O{]}^0=k$, so $k=4.1\times 10^{-3}M{s}^{-1}$, and the rate is constant at all $N_{2}O$ concentrations.

Exam tip: If you see that changing a reactant’s concentration leaves the rate unchanged, immediately assign an order of 0 and remove that reactant from the final rate law. Do not leave it in with an exponent of 0 unless explicitly asked to write the full general form.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Equating reaction order to stoichiometric coefficients from the balanced overall reaction. Why: Students assume coefficients translate directly to exponents because this works for equilibrium constants and elementary reactions, so they skip the experimental ratio step. Correct move: Always assume reaction order must be found from experimental data for an overall reaction; only use stoichiometry if explicitly told the reaction is elementary.
• Wrong move: Getting the ratio backwards when solving for order, e.g. writing $\frac{[A{]}_1}{[A{]}_2}$ instead of $\frac{[A{]}_2}{[A{]}_1}$ to match $\frac{rat{e}_2}{rat{e}_1}$. Why: Students mix up which run is which when comparing, leading to the wrong order. Correct move: Label numerator and denominator consistently: rate of the run with higher concentration divided by rate of the run with lower concentration, same order for concentration.
• Wrong move: Writing wrong units for the rate constant $k$. Why: Students memorize units for one order and apply it to all, or forget to match the time unit given in the problem. Correct move: Always derive units by rearranging the rate law: $k=rate/(\prod [X{]}^n)$, substitute units for each term, and simplify.
• Wrong move: Including pure solids or pure liquid solvents in the rate law expression. Why: Students remember to omit them from equilibrium expressions but forget the same rule applies to rate laws, because their concentration is constant. Correct move: Only include gaseous and aqueous reactants in the rate law; omit pure solids and liquids.
• Wrong move: Assuming the rate constant $k$ is constant at all temperatures. Why: Students learn $k$ is constant for a set of initial rate experiments, so they incorrectly assume it is constant for all conditions. Correct move: Always note that $k$ is only constant at a given temperature; if temperature increases, $k$ increases for all reactions.

6. Practice Questions (AP Chemistry Style)

Question 1 (Multiple Choice)

The reaction $X+Y\to Z$ was studied with the method of initial rates. The data table is below:

What is the rate law for this reaction? A) $\text{rate}=k[X][Y]$ B) $\text{rate}=k[X{]}^2[Y]$ C) $\text{rate}=k[X{]}^2$ D) $\text{rate}=k[Y{]}^2$

Worked Solution: First find the order with respect to X by comparing Experiments 1 and 2, where $[Y]$ is constant. The ratio of rates is $\frac{0.0120}{0.0030}=4$, and the ratio of $[X]$ is $\frac{0.20}{0.10}=2$, so $4=2^m\to m=2$. Next find the order with respect to Y by comparing Experiments 2 and 3, where $[X]$ is constant. The ratio of rates is $\frac{0.0120}{0.0120}=1$, and the ratio of $[Y]$ is $\frac{0.20}{0.10}=2$, so $1=2^n\to n=0$. Since Y is zero order, it does not appear in the rate law. Correct answer is C.

Question 2 (Free Response)

The decomposition of hydrogen peroxide in the presence of iodide ion is given by the reaction: $2H_2{O}_2(aq)\to 2H_{2}O(l)+O_2(g)$. Experimental initial rate data is below:

(a) Determine the rate law for this reaction. Show your working. (b) Calculate the value of the rate constant $k$, including correct units. (c) Predict the initial rate for a reaction where $[H_2{O}_2]=0.300M$ and $[I^{-}]=0.100M$.

Worked Solution: (a) Compare Run 1 and Run 2, where $[I^{-}]$ is constant: $\frac{rat{e}_2}{rat{e}_1}=2=(\frac{0.200}{0.100}{)}^m=2^m$, so $m=1$ (order with respect to $H_2{O}_2$ is 1). Compare Run 1 and Run 3, where $[H_2{O}_2]$ is constant: $\frac{rat{e}_3}{rat{e}_1}=1.5=(\frac{0.150}{0.100}{)}^n=1.5^n$, so $n=1$ (order with respect to $I^{-}$ is 1). Final rate law: $\boxed{\text{rate}=k[H_2{O}_2][I^{-}]}$ (b) Substitute Run 1 data: $k=\frac{rate}{[H_2{O}_2][I^{-}]}=\frac{1.2\times 10^{-4}M{s}^{-1}}{(0.100M)(0.100M)}=\boxed{0.012M^{-1}{s}^{-1}}$ (c) Substitute new concentrations: $\text{rate}=(0.012M^{-1}{s}^{-1})(0.300M)(0.100M)=\boxed{3.6\times 10^{-4}M{s}^{-1}}$

Question 3 (Application / Real-World Style)

The breakdown of glucose in blood by the saturated enzyme hexokinase follows zero-order kinetics with respect to glucose. At 37 °C, the rate constant for the reaction is $0.025mMmi{n}^{-1}$. If the initial concentration of glucose in a blood plasma sample is 5.0 mM, how long will it take for the glucose concentration to drop to 3.0 mM, and what is the reaction rate when glucose concentration is 4.0 mM? Explain your result in context.

Worked Solution: For a zero-order reaction, rate is constant and equal to $k$, regardless of glucose concentration. So the rate at 4.0 mM glucose is $0.025mMmi{n}^{-1}$. The change in glucose concentration is $\Delta [glucose]=5.0mM-3.0mM=2.0mM$. Rearrange the rate definition: $\Delta t=\frac{-\Delta [glucose]}{rate}=\frac{2.0mM}{0.025mMmi{n}^{-1}}=80min$. In context, because the enzyme is fully saturated, it breaks down glucose at a constant rate of 0.025 mM per minute, even as glucose concentration drops, as long as glucose concentration remains above the saturation threshold.

7. Quick Reference Cheatsheet

8. What's Next

This chapter is the foundational prerequisite for all further kinetics topics on the AP Chemistry syllabus. Next you will apply the reaction orders and rate constants you learn here to derive integrated rate laws, which describe how reactant concentration changes over time for different reaction orders. You will also use the rate constant $k$ calculated here to find half-lives of reactions, and then connect rate laws to reaction mechanisms, where you learn how to derive a rate law from the elementary steps of a reaction. Without mastering the skills of finding reaction order and rate law from initial rate data, you will not be able to correctly interpret integrated rate law data or connect rate laws to mechanisms, which make up the majority of kinetics points on the AP exam.

• Integrated rate laws
• Reaction mechanisms
• Reaction half-life
• Arrhenius equation and activation energy