Introduction to Rate Law

AP Chemistry· AP Chemistry CED — Kinetics· 14 min read

1. Core Definition of a Rate Law★★☆☆☆⏱ 2 min

A rate law (or rate equation) is a mathematical relationship that connects the instantaneous rate of a chemical reaction to the concentration of reactants (and rarely catalysts, which are not commonly tested in introductory problems).

A key AP-exam tested distinction: rate laws are exclusively experimentally derived. You cannot determine a rate law from the balanced reaction stoichiometry alone. This topic is the foundation of all kinetics, which makes up 7-9% of total AP Chemistry exam score, with rate law questions appearing regularly in both MCQ and FRQ sections.

📘 Definition

Rate Law

$rate = k [A]^{m} [B]^{n} ...$

An experimentally derived equation relating instantaneous reaction rate to reactant concentrations at a given temperature.

Example:

For $2 N O (g) + O_{2} (g) \to 2 N O_{2} (g)$ , the rate law is $rate = k [N O]^{2} [O_{2}]$

2. Rate Law Structure and Reaction Order★★☆☆☆⏱ 4 min

For a general overall reaction $a A + b B \to c C$ , the general form of the rate law is:

\text{rate} = k [A]^m [B]^n

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

Reaction order describes how changing a reactant's concentration impacts overall rate: zero order means concentration change has no effect, first order means rate is directly proportional to concentration, second order means rate is proportional to the square of concentration. Critically, individual orders do not need to match the stoichiometric coefficients of the balanced overall reaction.

📐 Worked Example

For the reaction $2 N O (g) + O_{2} (g) \to 2 N O_{2} (g)$ , the experimentally determined rate law is $rate = k [N O]^{2} [O_{2}]$ . 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 $[N O]$ is tripled and $[O_{2}]$ is held constant? (c) What are the units of $k$ for this reaction, if time is measured in seconds?

1
The exponent of each reactant gives its order. The exponent of $[N O]$ is 2, and the implicit exponent of $[O_{2}]$ is 1. Calculate overall order:
$2 + 1 = 3$
2
Compare original and new rate, with new $[N O] = 3 [N O]_{1}$ :
$\text{rate}_2 = k (3[NO]_1)^2 [O_2] = 9 k [NO]_1^2 [O_2] = 9 \text{rate}_1$
3
Rearrange to isolate $k$ and substitute units to find units of $k$ :
$k = \frac{\text{rate}}{[NO]^2 [O_2]} = \frac{\text{M} \cdot \text{s}^{-1}}{\text{M}^2 \cdot \text{M}} = \text{M}^{-2} \cdot \text{s}^{-1}$

Exam tip:

When asked for units of $k$ , always match the time units given in the problem. The general formula for units of $k$ is $M^{- (overall order - 1)} time^{- 1}$ .

3. Method of Initial Rates★★★☆☆⏱ 5 min

The method of initial rates is the primary experimental technique tested on the AP exam to determine a rate law from raw experimental data. Initial rate is the instantaneous rate measured immediately after the reaction starts, before any significant change in reactant concentration occurs. Data is provided as a table of multiple reaction runs, each with different starting concentrations and a measured initial rate.

To find the order for a reactant, compare two runs where only that reactant's concentration changes, and all other concentrations are held constant, using the relationship:

\frac{\text{rate}_2}{\text{rate}_1} = \left( \frac{[A]_2}{[A]_1} \right)^m

On the AP exam, ratios almost always result in whole-number orders that can be solved by inspection, no logarithms required. Once all orders are found, substitute data from any run to calculate $k$ , then write the final rate law.

Run	$[A]$ (M)	$[B]$ (M)	Initial Rate (M s⁻¹)
1	0.10	0.10	$2.0 \times 1 0^{- 4}$
2	0.20	0.10	$4.0 \times 1 0^{- 4}$
3	0.10	0.20	$8.0 \times 1 0^{- 4}$

📐 Worked Example

Determine the rate law and calculate the rate constant $k$ for the reaction $A + 2 B \to C$ using the experimental data above.

1
Find order with respect to A by comparing Run 1 and Run 2, where $[B]$ is constant:
$\frac{\text{rate}_2}{\text{rate}_1} = 2 = \left( \frac{0.20}{0.10} \right)^m = 2^m \rightarrow m=1$
2
Find order with respect to B by comparing Run 1 and Run 3, where $[A]$ is constant:
$\frac{\text{rate}_3}{\text{rate}_1} = 4 = \left( \frac{0.20}{0.10} \right)^n = 2^n \rightarrow n=2$
3
Write the rate law with confirmed orders:
$\text{rate} = k [A][B]^2$
4
Substitute data from Run 1 to calculate $k$ :
$k = \frac{2.0 \times 10^{-4}}{(0.10)(0.10)^2} = 0.20 \ M^{-2} s^{-1}$
5
Final rate law:
$\text{rate} = (0.20 \ M^{-2} s^{-1}) [A][B]^2$

Exam tip:

Always verify your $k$ value by checking with data from another run. A consistent $k$ (within rounding) confirms your reaction orders are correct.

4. Zero-Order Rate Laws★★★☆☆⏱ 3 min

A zero-order reaction with respect to a reactant means changing that reactant's concentration has no effect on the overall reaction rate. This is most common for saturated enzyme-catalyzed reactions and heterogeneous reactions catalyzed on solid surfaces.

Since any value raised to the power of zero equals 1, a zero-order reactant is omitted from the final rate law expression. If all reactants are zero order, the rate is simply equal to $k$ , so rate is constant regardless of concentration.

Run	$[N_{2} O]$ (M)	Initial Rate (M s⁻¹)
1	0.25	$4.1 \times 1 0^{- 3}$
2	0.50	$4.1 \times 1 0^{- 3}$
3	1.00	$4.1 \times 1 0^{- 3}$

📐 Worked Example

The decomposition of $N_{2} O (g)$ on a solid platinum catalyst follows $2 N_{2} O (g) \to 2 N_{2} (g) + O_{2} (g)$ . Use the data above to determine the rate law.

1
Compare Run 1 and Run 2: $[N_{2} O]$ doubles from 0.25 M to 0.50 M, but initial rate remains unchanged at $4.1 \times 1 0^{- 3} M s^{- 1}$ .
2
Apply the initial rate relationship to find order $m$ :
$\frac{\text{rate}_2}{\text{rate}_1} = 1 = \left( \frac{0.50}{0.25} \right)^m = 2^m \rightarrow m=0$
3
Confirm with Run 2 and Run 3: $[N_{2} O]$ doubles again, rate is still unchanged, so $m = 0$ is confirmed.
4
Simplify the rate law, since $[N_{2} O]^{0} = 1$ :
$\text{rate} = k [N_2O]^0 = k = 4.1 \times 10^{-3} \ M s^{-1}$

✓ Quick check

Test your understanding with this AP-style multiple choice question:

The reaction $X + Y \to Z$ was studied with the method of initial rates, giving the data below. What is the rate law for this reaction?
Experiment $[X]$ (M) $[Y]$ (M) Initial Rate (M min⁻¹)
1 0.10 0.10 0.0030
2 0.20 0.10 0.0120
3 0.20 0.20 0.0120
- A) $rate = k [X] [Y]$
- B) $rate = k [X]^{2} [Y]$
- C) $rate = k [X]^{2}$
- D) $rate = k [Y]^{2}$
Reveal answer
C —
Correct! Order with respect to X is 2, order with respect to Y is 0, so Y is omitted from the final rate law.

Experiment	$[X]$ (M)	$[Y]$ (M)	Initial Rate (M min⁻¹)
1	0.10	0.10	0.0030
2	0.20	0.10	0.0120
3	0.20	0.20	0.0120

Exam tip:

If changing a reactant's concentration leaves rate unchanged, immediately assign order 0 and omit it from the final rate law, unless explicitly asked to write the full general form.

5. Common Pitfalls

Wrong move:

Equating reaction order to stoichiometric coefficients from the balanced overall reaction.

Why:

Students assume coefficients translate directly to exponents, as they do for equilibrium constants and elementary reactions, so they skip the experimental ratio step.

Correct move:

Always determine reaction order from experimental data for an overall reaction; only use stoichiometry if explicitly told the reaction is elementary.

Wrong move:

Reversing the ratio of rates or concentrations when solving for order (e.g. $\frac{[ A ] _{1}}{[ A ] _{2}}$ instead of matching $\frac{r a t e _{2}}{r a t e _{1}}$ ).

Why:

Mixing up which run is which when comparing leads to incorrect order calculations.

Correct move:

Label numerator and denominator consistently: always put the second run's values in the numerator and first run's in the denominator for both rate and concentration.

Wrong move:

Memorizing units of $k$ for one order and applying it to all reactions, or using the wrong time unit.

Why:

Units of $k$ change with overall reaction order and depend on the time unit given in the problem.

Correct move:

Always derive units by rearranging the rate law: $k = \frac{r a t e}{\prod [ X ] ^{n}}$ , substitute units, and simplify.

Wrong move:

Including pure solids or pure liquid solvents in the rate law expression.

Why:

Students remember to omit these from equilibrium expressions but forget the same rule applies to rate laws, as 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 incorrectly extend this to all conditions.

Correct move:

Remember $k$ is only constant at a given temperature; $k$ always increases when temperature increases, or when a catalyst is added.

6. Quick Reference Cheatsheet

Category	Formula/Relationship	Key Notes
General rate law form	$rate = k [A]^{m} [B]^{n} ...$	For overall reactions; $m, n$ are experimentally determined
Method of initial rates (find order $m$ )	$\frac{rate _{2}}{rate _{1}} = (\frac{[ A ] _{2}}{[ A ] _{1}})^{m}$	Use when only $[A]$ changes between runs
Overall reaction order	$Overall order = \sum individual orders$	Determines units of the rate constant $k$
Units of $k$ (overall order $n$ )	$M^{- (n - 1)} time^{- 1}$	Match time unit to the problem (seconds/minutes)
Zero order rate law	$rate = k$	Reactant concentration does not affect rate; common for saturated catalysis
First order (one reactant)	$rate = k [A]$	Doubling $[A]$ doubles the reaction rate
Second order (one reactant)	$rate = k [A]^{2}$	Doubling $[A]$ quadruples the reaction rate
Rate constant dependency	$k = constant at fixed T$	$k$ increases with increasing $T$ and catalyst addition

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Determine rate law from initial rate data
2022 · FRQ
Calculate $k$ and reaction order

What's Next

This topic is the foundational prerequisite for all further kinetics topics on the AP Chemistry syllabus. The skills you learn here to determine reaction order, calculate the rate constant, and write a rate law from experimental data are required for all subsequent kinetics topics, which make up the majority of kinetics points on the AP exam. Next, you will apply these concepts to derive integrated rate laws, which describe how reactant concentration changes over time for different reaction orders. You will also use the rate constant $k$ you calculate here to find reaction half-lives, then connect rate laws to reaction mechanisms to derive rate laws from elementary reaction steps. Mastery of this introductory content is critical for success on all kinetics exam questions.