AP Chemistry Rate Law

Reading orders from data, solving for k and its units, and using integrated rate laws, in four worked examples.

Kinetics questions on the exam are almost always about a table or a graph: you are given the data and asked to turn it into a rate law, a rate constant, or a half-life. The Unit 5 review shows one initial-rates example and one half-life example. This page adds the harder cases, including a reaction with a different order for each reactant, the units of k, and a data set that has to be identified by plotting.

The Rate Law and the Method of Initial Rates

A rate law has the form rate = k[A]^m[B]ⁿ, where m and n are the orders in A and B, their sum is the overall order, and k is the rate constant. The orders come from experiments, never from the coefficients of the overall balanced equation. The method of initial rates finds them from a table:

1. Choose two trials where only one concentration changes.
2. Compare the rates. If the concentration doubles and the rate is unchanged, the order is 0; if the rate doubles, the order is 1; if the rate quadruples, the order is 2; if it rises eightfold, the order is 3.
3. Repeat for each reactant with a different pair of trials.
4. Solve for k by substituting one trial into the finished rate law.
5. Write the units of k.

Worked Example 1: Finding Orders, k, and a Predicted Rate From Initial-Rate Data

For 2 NO + O₂ → 2 NO₂, these initial rates were measured:

Order in NO: trials 1 and 2 change only [NO], which doubles. The rate rises from 2.5 × 10⁻³ to 1.0 × 10⁻², a factor of 4 = 2². So the order in NO is 2.
Order in O₂: trials 1 and 3 change only [O₂], which doubles. The rate doubles (factor of 2 = 2¹), so the order in O₂ is 1.
Rate law: rate = k[NO]²[O₂], overall third order.
Solve for k with trial 1: 2.5 × 10⁻³ = k(0.010)²(0.010), so k = 2.5 × 10⁻³ ÷ 1.0 × 10⁻⁶ = 2.5 × 10³ M⁻²s⁻¹.
Predict a new rate: at [NO] = 0.030 M and [O₂] = 0.015 M, rate = (2.5 × 10³)(0.030)²(0.015) = 3.4 × 10⁻² M/s.

Here the orders happen to match the coefficients in the balanced equation, but that is not a rule: they had to be read from the table, and for many reactions they do not match.

Worked Example 2: Deriving the Units of the Rate Constant for Each Order

Find the units of k for zero-, first-, second-, and third-order reactions.

Rate has units of M/s, so k = rate ÷ (concentration terms). Divide M/s by the concentration units raised to the overall order:
Zero order: k = (M/s) ÷ M⁰ = M/s.
First order: k = (M/s) ÷ M = s⁻¹.
Second order: k = (M/s) ÷ M² = M⁻¹s⁻¹.
Third order: k = (M/s) ÷ M³ = M⁻²s⁻¹, matching Example 1.

In general the units are M¹⁻ⁿs⁻¹ for overall order n, so the units of k are also a quick check on the order you found.

Worked Example 3: Identifying a Reaction Order From Concentration-Time Data

The concentration of A is 0.80 M at 0 s, 0.40 M at 30 s, 0.20 M at 60 s, and 0.10 M at 90 s. Find the order and k.

Test each plot. [A] against time is curved, so not zero order. 1/[A] (1.25, 2.5, 5.0, 10 M⁻¹) against time is curved, so not second order. ln[A] (−0.223, −0.916, −1.609, −2.303) falls by exactly 0.693 every 30 s, a straight line, so the reaction is first order.
Rate constant: the slope of ln[A] against t is −0.693 ÷ 30 s = −0.0231 s⁻¹, so k = 0.0231 s⁻¹.

A quicker diagnostic: the concentration halves every 30 s, a constant half-life, which marks a first-order reaction.

Worked Example 4: Using Half-Life to Find How Long a First-Order Reaction Takes

For the reaction in Example 3, how long does it take for [A] to fall from 0.80 M to 0.10 M, and what fraction remains after two half-lives?

The half-life is t_1/2 = 0.693 ÷ k = 0.693 ÷ 0.0231 s⁻¹ = 30.0 s.
From 0.80 M to 0.10 M is a factor of 8 = 2³, so it takes 3 half-lives: 3 × 30.0 s = 90 s.
Check with the integrated law: t = ln(0.80/0.10) ÷ 0.0231 = 2.079 ÷ 0.0231 = 90 s. ✓
After two half-lives, (1/2)² = 25% remains.

The shortcut of counting half-lives works only for first-order reactions, whose half-life stays constant. The connection of rate laws to a proposed mechanism is covered in the Unit 5 review, and the temperature dependence of k is the subject of activation energy.

Five Rate Law Practice Questions With Answers

1. Doubling [A] doubles the rate, and doubling [B] has no effect. What is the rate law?
   Show answer

   Rate = k[A]. The order in A is 1 and the order in B is 0.
2. What are the units of k for a second-order reaction?
   Show answer

   M⁻¹s⁻¹, or L/(mol·s).
3. Doubling [X] makes the rate eight times larger. What is the order in X?
   Show answer

   Third order, since 2³ = 8.
4. A first-order reaction has k = 0.0462 s⁻¹. What is its half-life?
   Show answer

   t_1/2 = 0.693 ÷ 0.0462 = 15.0 s.
5. Which plot is a straight line for a second-order reaction?
   Show answer

   1/[A] against time.

Common Rate Law Mistakes

• Reading orders from the balanced equation. They come from experimental data.
• Comparing trials where two concentrations change. Only one reactant should differ in the pair you use for a given order.
• Leaving k without units. Units depend on the overall order and are often part of the answer.
• Using t_1/2 = 0.693/k for other orders. It applies only to first-order reactions.
• Plotting the wrong quantity. Zero order needs [A], first order ln[A], second order 1/[A] against time.

A set of kinetics problems is a good check on where Unit 5 stands. To see how that translates into an overall estimate, the AP Chem Score Calculator turns your practice scores into a 1–5 range.

Related Resources

• AP Chem Score Calculator
• Unit 5 Review: Kinetics
• AP Chemistry Activation Energy
• AP Chemistry Kinetic Molecular Theory
• AP Chemistry Molarity
• AP Chemistry Significant Figures Rules
• AP Chemistry Reference Sheet
• AP Chemistry Study Guide

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