College Board AP 2022 (style)-style practice: Section II (long FRQ, part). For A + B → products, the following initial-rate data are collected. Trial 1: [A] = 0.10, [B] = 0.10, rate = 2.0 × 10^-3. Trial 2: [A] = 0.20, [B] = 0.10, rate = 8.0 × 10^-3. Trial 3: [A] = 0.10, [B] = 0.20, rate = 2.0 × 10^-3 (all in M and M s^-1). (a) Determine the order with respect to A and to B. (b) Write the rate law. (c) Calculate the rate constant with units. (d) Justify the order in B.

A 4-point quantitative FRQ on the method of initial rates. (a) Orders (1 point): comparing trials 1 and 2, [A] doubles and the rate goes up by a factor of 4 (2^2), so the order in A is 2; comparing trials 1 and 3, [B] doubles but the rate is unchanged (2^0), so the order in B is 0. (b) Rate law (1 point): rate = k[A]^2. (c) Rate constant (1 point): k = rate[A]^2 = 2.0 × 10^-3(0.10)^2 = 0.20\ M^-1s^-1. (d) Justify (1 point): doubling [B] left the rate unchanged, which means the rate does not depend on [B]; the exponent is 0. Markers reward both orders from the data, the rate law, the rate constant with correct units, and the reasoning for zero order in B.

College Board AP 2021 (style)-style practice: Section I (multiple choice). A reaction has the rate law rate = k[X][Y]^2. The units of k are (A) s^-1 (B) M s^-1 (C) M^-1s^-1 (D) M^-2s^-1. Justify your choice.

A 1-point conceptual MCQ. The answer is (D). The overall order is 1 + 2 = 3. For an order-n rate law, k has units M^1-ns^-1, so for n = 3 that is M^-2s^-1. Check: M^-2s^-1 × M × M^2 = M s^-1, which is a rate. The trap is using the coefficients instead of the experimentally found orders.

United StatesChemistrySyllabus dot point

How does the rate of a reaction depend on the concentrations of the reactants, and how do we find that dependence?

Topic 5.2 Introduction to Rate Law: write the rate law of a reaction, determine the reaction orders and the rate constant from initial-rate data, and interpret the meaning of order and the units of the rate constant.

A focused answer to AP Chemistry Topic 5.2, covering the rate law, reaction order, the rate constant and its units, and how to find orders and k from initial-rate (method of initial rates) data, with full worked examples.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-04

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

A rate law expresses the reaction rate as the rate constant times reactant concentrations raised to powers called orders: $\text{rate} = k[\text{A}]^m[\text{B}]^n$ . The orders $m$ and $n$ are found by experiment, not from the coefficients; they say how the rate responds when a concentration changes (doubling a first-order reactant doubles the rate, doubling a second-order reactant quadruples it, and a zero-order reactant has no effect). The overall order is the sum of the individual orders. To find an order, change one concentration at a time and see how the rate responds (the method of initial rates). The rate constant $k$ depends on temperature and has units that make the whole expression come out in $\text{M s}^{-1}$ : for overall order $n$ , the units are $\text{M}^{1-n}\text{s}^{-1}$ .

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What this topic is asking
The rate law
Finding orders by the method of initial rates
The rate constant and its units
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What this topic is asking

The College Board (Topic 5.2) wants you to write a rate law, find the reaction orders and the rate constant from experimental initial-rate data, and understand what the order tells you and how the units of the rate constant depend on the overall order. The central idea is that orders are determined by experiment, not read off the balanced equation.

The rate law

The order tells you how the rate scales. A reaction first order in A doubles its rate when $[\text{A}]$ doubles; second order in A quadruples it ( $2^2$ ); zero order in A is unaffected. Orders are usually small integers (0, 1, 2) on the AP exam.

Finding orders by the method of initial rates

Concretely, pick two trials in which $[\text{B}]$ is held constant and $[\text{A}]$ changes. The ratio of rates equals the ratio of $[\text{A}]$ raised to the order $m$ : $\dfrac{\text{rate}_2}{\text{rate}_1} = \left(\dfrac{[\text{A}]_2}{[\text{A}]_1}\right)^m$ . Solve for $m$ . Repeat with a pair where $[\text{A}]$ is fixed to find $n$ .

The rate constant and its units

Once the orders are known, substitute any one trial into the rate law and solve for $k$ . The rate constant is independent of concentration but depends strongly on temperature (Topic 5.5). Its units follow from the overall order $n$ :

\text{units of } k = \text{M}^{1-n}\,\text{s}^{-1}

So a first-order $k$ has units $\text{s}^{-1}$ , a second-order $k$ has $\text{M}^{-1}\text{s}^{-1}$ , and a zero-order $k$ has $\text{M s}^{-1}$ . Quoting the right units is itself a marking point on the FRQ.

Finding orders, the rate law and k

For $2\text{NO}(g) + \text{O}_2(g) \rightarrow 2\text{NO}_2(g)$ , the data are: Trial 1 $[\text{NO}] = 0.010$ , $[\text{O}_2] = 0.010$ , rate $= 2.5 \times 10^{-5}$ ; Trial 2 $[\text{NO}] = 0.020$ , $[\text{O}_2] = 0.010$ , rate $= 1.0 \times 10^{-4}$ ; Trial 3 $[\text{NO}] = 0.010$ , $[\text{O}_2] = 0.020$ , rate $= 5.0 \times 10^{-5}$ (M, $\text{M s}^{-1}$ ).

step 1 Order in NO (trials 1 and 2)

$[\text{O}_2]$ is fixed; $[\text{NO}]$ doubles and the rate goes up $\times 4$ . Since $2^m = 4$ , $m = 2$ .

step 2 Order in $\text{O}_2$ (trials 1 and 3)

$[\text{NO}]$ is fixed; $[\text{O}_2]$ doubles and the rate doubles. Since $2^n = 2$ , $n = 1$ .

step 3 Write the rate law

$\text{rate} = k[\text{NO}]^2[\text{O}_2]$ ; overall order $= 3$ .

step 4 Solve for k with units

$k = \dfrac{2.5 \times 10^{-5}}{(0.010)^2(0.010)} = \dfrac{2.5 \times 10^{-5}}{1.0 \times 10^{-6}} = 25\ \text{M}^{-2}\text{s}^{-1}$ .

Try this

Q1. A reaction is first order in A and second order in B. By what factor does the rate change if $[\text{A}]$ is tripled and $[\text{B}]$ is doubled? [2 points]

Cue. $3^1 \times 2^2 = 3 \times 4 = 12$ times faster.

Q2. State the units of $k$ for a reaction that is overall first order. [1 point]

Cue. $\text{M}^{1-1}\text{s}^{-1} = \text{s}^{-1}$ .

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2022 (style)4 marksSection II (long FRQ, part). For

\text{A} + \text{B} \rightarrow \text{products}

, the following initial-rate data are collected. Trial 1:

[\text{A}] = 0.10

[\text{B}] = 0.10

, rate

= 2.0 \times 10^{-3}

. Trial 2:

[\text{A}] = 0.20

[\text{B}] = 0.10

, rate

= 8.0 \times 10^{-3}

. Trial 3:

[\text{A}] = 0.10

[\text{B}] = 0.20

, rate

= 2.0 \times 10^{-3}

(all in M and

\text{M s}^{-1}

). (a) Determine the order with respect to A and to B. (b) Write the rate law. (c) Calculate the rate constant with units. (d) Justify the order in B.

Show worked answer →

A 4-point quantitative FRQ on the method of initial rates.

(a) Orders (1 point): comparing trials 1 and 2, $[\text{A}]$ doubles and the rate goes up by a factor of 4 ( $2^2$ ), so the order in A is 2; comparing trials 1 and 3, $[\text{B}]$ doubles but the rate is unchanged ( $2^0$ ), so the order in B is 0.
(b) Rate law (1 point): $\text{rate} = k[\text{A}]^2$ .
(c) Rate constant (1 point): $k = \dfrac{\text{rate}}{[\text{A}]^2} = \dfrac{2.0 \times 10^{-3}}{(0.10)^2} = 0.20\ \text{M}^{-1}\text{s}^{-1}$ .
(d) Justify (1 point): doubling $[\text{B}]$ left the rate unchanged, which means the rate does not depend on $[\text{B}]$ ; the exponent is 0.

Markers reward both orders from the data, the rate law, the rate constant with correct units, and the reasoning for zero order in B.

AP 2021 (style)1 marksSection I (multiple choice). A reaction has the rate law

\text{rate} = k[\text{X}][\text{Y}]^2

. The units of

k

are (A)

\text{s}^{-1}

(B)

\text{M s}^{-1}

(C)

\text{M}^{-1}\text{s}^{-1}

(D)

\text{M}^{-2}\text{s}^{-1}

. Justify your choice.

Show worked answer →

A 1-point conceptual MCQ. The answer is (D).

The overall order is $1 + 2 = 3$ . For an order- $n$ rate law, $k$ has units $\text{M}^{1-n}\text{s}^{-1}$ , so for $n = 3$ that is $\text{M}^{-2}\text{s}^{-1}$ . Check: $\text{M}^{-2}\text{s}^{-1} \times \text{M} \times \text{M}^2 = \text{M s}^{-1}$ , which is a rate. The trap is using the coefficients instead of the experimentally found orders.

Related dot points

Sources & how we know this

AP Chemistry Course and Exam Description — College Board (2020)