Rate equations and order of reaction

Chemistry· Unit 20: Further reaction kinetics· 20 min read

1. 1. Core Definitions and the General Rate Equation★★☆☆☆⏱ 5 min

📘 Definition

Rate Equation

For the general reaction $a A + b B \to c C + d D$

An expression that describes how reaction rate depends on reactant concentrations, determined experimentally (not from reaction stoichiometry)

rate = k[A]^m[B]^n

Where $k$ = the rate constant, $[A], [B]$ = concentrations of A and B, $m$ = order with respect to A, $n$ = order with respect to B.

📘 Definition

Order of Reaction

The power to which a reactant's concentration term is raised in the rate equation. Orders can be 0, 1, 2, or rarely fractions/ higher values

Example:

If $m = 1$ , the reaction is first order with respect to A

📐 Worked Example

For the reaction $2 H I (g) \to H_{2} (g) + I_{2} (g)$ , the rate equation is $r a t e = k [H I]^{2}$ . State the order with respect to HI and the overall order of reaction.

1
1. The order with respect to a reactant equals the power of its concentration term:
2
$[HI]^2$
3
The power is 2, so the order of reaction with respect to HI is 2.
4
1. Overall order is the sum of all individual orders. There is only one reactant term, so overall order = $2$ .

2. 2. Deducing Order from Initial Rate Data★★★☆☆⏱ 7 min

The most common exam question asks you to deduce the rate equation from a table of experimental initial rate data. The method compares two experiments where only one concentration is changed, to find its effect on rate:

Zero order: If [X] doubles and rate stays the same, order = 0 ( $2^{0} = 1$ , rate × 1)
First order: If [X] doubles and rate doubles, order = 1 ( $2^{1} = 2$ , rate × 2)
Second order: If [X] doubles and rate quadruples, order = 2 ( $2^{2} = 4$ , rate × 4)

📐 Worked Example

Deduce the rate equation for the reaction $A + 2 B \to C + D$ using the data below:
1: [A] = 0.1, [B] = 0.1, rate = 0.001
2: [A] = 0.2, [B] = 0.1, rate = 0.002
3: [A] = 0.1, [B] = 0.2, rate = 0.004 (all units mol dm⁻³/s)

1
1. Find order with respect to A: compare Experiments 1 and 2 (constant [B])
2
[A] doubles, rate doubles, so order with respect to A = 1.
3
1. Find order with respect to B: compare Experiments 1 and 3 (constant [A])
4
[B] doubles, rate quadruples. $2^{n} = 4 \to n = 2$ , so order with respect to B = 2.
5
1. Substitute orders into the general rate equation:
6
$rate = k[A]^1[B]^2 = k[A][B]^2$

Exam tip:

Always confirm only one reactant concentration changes between the two experiments you compare. If two change, you cannot directly calculate order.

3. 3. Deducing Order from Rate-Concentration Graphs★★★☆☆⏱ 5 min

Order can also be found from continuous rate data plotted as rate against concentration. Each order gives a characteristic graph shape:

Order	Shape of rate vs [X] graph	Key property
0	Horizontal line	Gradient = 0, rate independent of [X]
1	Straight line through origin	Gradient equals the rate constant $k$
2	Upwards curving parabola	Gradient increases as [X] increases

📐 Worked Example

A rate-concentration graph is a straight line through the origin with gradient 0.032 s⁻¹. State the order of reaction and the value of the rate constant.

1
1. A straight line through the origin for a rate-concentration graph is the characteristic shape of a first order reaction.
2
1. For first order reactions, the gradient of the rate-concentration graph equals the rate constant $k$ .
3
Therefore: order = 1, $k = 0.032$ s⁻¹

4. 4. Rate Equations and Reaction Mechanisms★★★★☆⏱ 3 min

The rate equation gives information about the rate-determining step (RDS), the slowest step in a reaction mechanism that controls the overall reaction rate. The order with respect to a reactant equals the number of molecules of that reactant that participate in the rate-determining step.

📐 Worked Example

The reaction $N O_{2} (g) + C O (g) \to N O (g) + C O_{2} (g)$ has the rate equation $r a t e = k [N O_{2}]^{2}$ . What does this tell us about the rate-determining step?

1
1. The exponent of each reactant equals the number of its molecules in the RDS.
2
$[N O_{2}]$ is raised to the power 2, so 2 molecules of $N O_{2}$ are present in the rate-determining step.
3
CO does not appear in the rate equation, so its order is 0. This means CO is not involved in the rate-determining step, and reacts in a faster, later step after the RDS.

5. Common Pitfalls

Wrong move:

Assuming order equals the stoichiometric coefficient in the balanced equation

Why:

Rate equations are determined experimentally, not from reaction stoichiometry. Only elementary single-step reactions have orders matching coefficients.

Correct move:

Always deduce order from experimental data, never assume it equals the balancing number.

Wrong move:

Treating the rate constant $k$ as constant regardless of temperature

Why:

The rate constant always increases with increasing temperature, even for the same reaction.

Correct move:

Remember that $k$ is temperature dependent, and always quote the temperature when giving a value of $k$ .

Wrong move:

Comparing two experiments where two reactant concentrations change to find order

Why:

If two concentrations change, you cannot isolate the effect of each reactant on the rate.

Correct move:

Always select two experiments where all concentrations except one are constant to calculate each order individually.

Wrong move:

Miscalculating order when concentration changes by a factor other than 2, e.g. calling order 3 when concentration triples and rate increases 9x

Why:

Order is the exponent, not the factor of increase in concentration.

Correct move:

Calculate order as $n = \frac{l n ( r a t e _{2} / r a t e _{1} )}{l n ([ X ] _{2} / [ X ] _{1} )}$ . For 3x concentration / 9x rate, $n = \frac{l n 9}{l n 3} = 2$ .

6. Quick Reference Cheatsheet

Change in [X]	Change in rate	Order of X
[X] doubles	No change	0
[X] doubles	Rate doubles	1
[X] doubles	Rate × 4	2
[X] triples	Rate × 3	1
[X] triples	Rate × 9	2
X not in rate equation	No change	0
Overall order		Sum of all individual orders

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.

2022 · 2
Deduce rate equation from data
2023 · 4
Calculate order and rate constant
2021 · 1
Interpret rate equation meaning

Going deeper

syllabusCIE 9701 2025-2026 SyllabusReview kinetics in section 11

What's Next

Rate equations are the foundation of all further work in reaction kinetics for CIE A-Level Chemistry. Mastering how to find order and construct rate equations is critical for scoring high marks in kinetics questions, which appear regularly across all papers (1, 2 and 4). The relationship between order and the rate-determining step that you learned here also forms the basis for understanding how reaction mechanisms are determined from experimental data. Next, you will build on this core knowledge to explore half-life for first order reactions, how to identify the rate-determining step, and how to calculate activation energy using the Arrhenius equation.