Rate Equations and Orders

The rate equation is the mathematical relationship between the rate of a reaction and the concentrations of the reactants. It is one of the most important tools in chemical kinetics, and unlike a balanced equation, it can only be determined experimentally -- it cannot be deduced from the stoichiometry of the reaction.

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The General Rate Equation

For a reaction involving reactants A and B, the rate equation takes the form:

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

Where:

• Rate is the rate of reaction (mol dm⁻³ s⁻¹)
• k is the rate constant -- a proportionality constant that depends on temperature
• [A] and [B] are the concentrations of reactants (mol dm⁻³)
• m is the order with respect to A
• n is the order with respect to B
• (m + n) is the overall order of the reaction

The orders m and n are usually 0, 1, or 2 (though fractional and negative orders exist, they are beyond A-level).

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Understanding Orders of Reaction

Zero Order (m = 0)

If the reaction is zero order with respect to A, then changing [A] has no effect on the rate.

Rate = k[A]⁰ = k

This means the rate is constant regardless of the concentration of A. On a concentration-time graph, [A] decreases linearly with time (a straight line). On a rate-concentration graph, the line is horizontal.

First Order (m = 1)

If the reaction is first order with respect to A, then the rate is directly proportional to [A].

Rate = k[A]

Doubling [A] doubles the rate. Tripling [A] triples the rate. On a concentration-time graph, the curve shows exponential decay. On a rate-concentration graph, the plot is a straight line through the origin.

Second Order (m = 2)

If the reaction is second order with respect to A, then the rate is proportional to [A]².

Rate = k[A]²

Doubling [A] quadruples the rate (2² = 4). Tripling [A] increases the rate by a factor of 9 (3² = 9). On a rate-concentration graph, the plot is a curve (parabola) passing through the origin.

Quick Reference: How Rate Changes with Concentration

Concentration change	Zero order effect	First order effect	Second order effect
×2	×1 (no change)	×2	×4
×3	×1 (no change)	×3	×9
×4	×1 (no change)	×4	×16
×½	×1 (no change)	×½	×¼

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Determining Order from Initial Rate Data (Table Method)

The most common exam method. You are given a table of experiments with different initial concentrations and the corresponding initial rates. You compare pairs of experiments where only one concentration changes.

flowchart TD
    A["Initial rate data table"] --> B["Find two experiments where<br/>only ONE concentration changes"]
    B --> C["Calculate the factor by which<br/>the concentration changed"]
    C --> D["Calculate the factor by which<br/>the rate changed"]
    D --> E{"rate factor =<br/>(conc factor)^n"}
    E -->|"rate unchanged"| F["n = 0 (zero order)"]
    E -->|"rate factor = conc factor"| G["n = 1 (first order)"]
    E -->|"rate factor = (conc factor)²"| H["n = 2 (second order)"]
    F --> I["Repeat for each reactant"]
    G --> I
    H --> I
    I --> J["Write rate equation:<br/>Rate = k[A]^m[B]^n"]
    J --> K["Calculate k using<br/>any experiment's data"]

Worked Example 1: Two Reactants

Experiment	[A] / mol dm⁻³	[B] / mol dm⁻³	Initial rate / mol dm⁻³ s⁻¹
1	0.10	0.10	2.0 × 10⁻⁴
2	0.20	0.10	8.0 × 10⁻⁴
3	0.10	0.20	4.0 × 10⁻⁴

Finding the order with respect to A: Compare Experiments 1 and 2 (where [B] is constant):

• [A] doubles (0.10 → 0.20): factor = 2
• Rate increases by factor: 8.0 × 10⁻⁴ / 2.0 × 10⁻⁴ = 4
• Since 2^m = 4, m = 2. A is second order.

Finding the order with respect to B: Compare Experiments 1 and 3 (where [A] is constant):

• [B] doubles (0.10 → 0.20): factor = 2
• Rate increases by factor: 4.0 × 10⁻⁴ / 2.0 × 10⁻⁴ = 2
• Since 2^n = 2, n = 1. B is first order.

Rate equation: Rate = k[A]²[B]

Overall order: 2 + 1 = 3 (third order)

Calculating k: Using Experiment 1:

k = Rate / ([A]²[B]) = 2.0 × 10⁻⁴ / (0.10² × 0.10) = 2.0 × 10⁻⁴ / 1.0 × 10⁻³ = 0.20 mol⁻² dm⁶ s⁻¹

Worked Example 2: Non-Integer Concentration Ratios

Sometimes concentration changes are not neat doublings. For example:

Experiment	[P] / mol dm⁻³	[Q] / mol dm⁻³	Rate / mol dm⁻³ s⁻¹
1	0.10	0.20	1.5 × 10⁻³
2	0.30	0.20	1.35 × 10⁻²
3	0.10	0.60	4.5 × 10⁻³

Order in P: Expts 1 & 2: [P] triples (0.10 → 0.30), rate increases by 1.35 × 10⁻² / 1.5 × 10⁻³ = 9 = 3². Second order in P.

Order in Q: Expts 1 & 3: [Q] triples (0.20 → 0.60), rate increases by 4.5 × 10⁻³ / 1.5 × 10⁻³ = 3 = 3¹. First order in Q.

Rate = k[P]²[Q]

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Determining Order from Concentration-Time Graphs

You can also determine orders by examining the shape of a concentration-time graph for a single experiment:

• Zero order: straight line (concentration decreases linearly). A plot of [A] vs t is linear.
• First order: exponential decay curve. A plot of ln[A] vs t is linear (with gradient = −k).
• Second order: a steeper curve than first order. A plot of 1/[A] vs t is linear (with gradient = k).

These "straight-line tests" are used when continuous monitoring data is available. If you plot [A] vs t and get a straight line, the reaction is zero order. If that plot is curved, try ln[A] vs t -- if that is linear, it is first order. If not, try 1/[A] vs t for second order.

Worked Example: Straight-Line Test

A student collects the following data for the decomposition of X:

t / s	[X] / mol dm⁻³	ln[X]	1/[X] / mol⁻¹ dm³
0	0.400	−0.916	2.50
20	0.286	−1.251	3.50
40	0.222	−1.505	4.50
60	0.182	−1.705	5.50
80	0.154	−1.871	6.50

Checking [X] vs t: not a straight line (values don't decrease by equal amounts). Not zero order.

Checking ln[X] vs t: the intervals are −0.335, −0.254, −0.200, −0.166 -- not constant. Not first order.

Checking 1/[X] vs t: the intervals are 1.00, 1.00, 1.00, 1.00 -- constant! Second order confirmed.

The gradient = (6.50 − 2.50) / (80 − 0) = 4.00 / 80 = k = 0.050 mol⁻¹ dm³ s⁻¹

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Worked Example: Three Reactants

Experiment	[X] / mol dm⁻³	[Y] / mol dm⁻³	[Z] / mol dm⁻³	Rate / mol dm⁻³ s⁻¹
1	0.10	0.10	0.10	3.0 × 10⁻³
2	0.20	0.10	0.10	6.0 × 10⁻³
3	0.10	0.30	0.10	3.0 × 10⁻³
4	0.10	0.10	0.20	1.2 × 10⁻²

• Experiments 1 & 2: [X] doubles, rate doubles → first order in X.
• Experiments 1 & 3: [Y] triples, rate unchanged → zero order in Y.
• Experiments 1 & 4: [Z] doubles, rate quadruples → second order in Z.

Rate = k[X][Z]² (overall order = 3)

Notice that Y appears in the balanced equation but not in the rate equation -- Y is not involved in the rate-determining step.

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Important Points

1. The rate equation is determined experimentally, not from the balanced equation. A reaction might have the equation 2A + B → C but be first order with respect to A. The stoichiometric coefficient (2) does not determine the order.

2. Species not in the rate equation do not affect the rate. If a species appears in the balanced equation but not in the rate equation (order = 0), it is not involved in the rate-determining step.

3. The rate-determining step is the slowest step in a multi-step mechanism. Only species involved in or before the rate-determining step appear in the rate equation. This links kinetics to mechanism.

4. Catalysts can appear in the rate equation if they are involved in or before the rate-determining step, even though they do not appear in the overall balanced equation.

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Common Mistakes to Avoid

1. Assuming the order equals the stoichiometric coefficient. For 2NO + O₂ → 2NO₂, the rate equation happens to be Rate = k[NO]²[O₂], but this is coincidence. Always determine orders experimentally.

2. Comparing experiments where two concentrations change. You must isolate one variable at a time. If both [A] and [B] change between two experiments, you cannot determine either order from that pair.

3. Forgetting to verify k is constant. After finding orders, calculate k from each experiment. If the values agree (within rounding), your orders are correct. If they don't match, recheck your analysis.

4. Confusing the rate equation with the equilibrium expression. The rate equation uses experimentally determined orders; the equilibrium expression always uses stoichiometric coefficients.

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Summary

The rate equation Rate = k[A]^m[B]^n must be found experimentally. Orders are determined by comparing how rates change when concentrations change (table method) or by plotting concentration-time data and testing for straight lines. The overall order is the sum of individual orders. The connection between the rate equation and the reaction mechanism (via the rate-determining step) is a fundamental concept in kinetics.

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A-Level Deep Dive: Rate Equations and Orders

Spec mapping

Edexcel 9CH0 specification, Topic 16 — Kinetics II sets out rate equations of the form $\text{rate} = k[A]^m[B]^n$rate=k[A]m[B]n, the meaning of "order with respect to a given reactant" and "overall order", and methods (initial-rate, half-life, linearised plot) for determining orders from experimental data (refer to the official specification document for exact wording). Topic 16 is built on the qualitative collision-theory foundation of Topic 9 (Lesson 1) and the rate-measurement techniques of Lesson 2. Examined principally on Paper 2 (Topics 11–19) and Paper 3 (synoptic with Core Practicals). The rate equation is the gateway to the rate-determining step and reaction mechanisms — orders are not predictable from stoichiometry; they reveal mechanism. The data booklet does not list the rate equation form.

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Worked example with full mark scheme

Question (8 marks):

The reaction $2\text{NO}(g) + \text{O}_2(g) \rightarrow 2\text{NO}_2(g)$2NO(g)+O2​(g)→2NO2​(g) was studied at 25 °C. The following initial-rate data were obtained.

Experiment	[NO]₀ / mol dm⁻³	[O₂]₀ / mol dm⁻³	Initial rate / mol dm⁻³ s⁻¹
1	0.020	0.010	2.40 × 10⁻⁴
2	0.040	0.010	9.60 × 10⁻⁴
3	0.020	0.020	4.80 × 10⁻⁴
4	0.040	0.020	1.92 × 10⁻³

(a) Determine the order of reaction with respect to NO and the order with respect to O₂. (4)

(b) State the overall order and write the rate equation. (2)

(c) Calculate the rate constant $k$k, giving its value with appropriate units. (2)

Solution with mark scheme:

(a) M1 — comparing experiments 1 and 2: [NO] doubles, [O₂] is constant; rate increases from 2.40 × 10⁻⁴ to 9.60 × 10⁻⁴, a factor of 4. A1 — $\text{rate} \propto [\text{NO}]^m$rate∝[NO]m with $2^m = 4$2m=4, so $m = 2$m=2; second order in NO.

M1 — comparing experiments 1 and 3: [O₂] doubles, [NO] is constant; rate increases from 2.40 × 10⁻⁴ to 4.80 × 10⁻⁴, a factor of 2. A1 — $2^n = 2$2n=2, so $n = 1$n=1; first order in O₂.

(Cross-check with experiment 4: doubling both [NO] and [O₂] should multiply the rate by $2^2 \cdot 2^1 = 8$22⋅21=8. From experiment 1 → 4: $1.92 × 10^{-3} / 2.40 × 10^{-4} = 8.0$1.92×10−3/2.40×10−4=8.0. Consistent.)

(b) M1 — overall order = $m + n = 2 + 1 = 3$m+n=2+1=3. A1 — rate equation: $\text{rate} = k[\text{NO}]^2[\text{O}_2]$rate=k[NO]2[O2​].

(c) M1 — substitute experiment 1: $2.40 × 10^{-4} = k(0.020)^2(0.010) = k \times 4.0 × 10^{-6}$2.40×10−4=k(0.020)2(0.010)=k×4.0×10−6. A1 — $k = 60$k=60 mol⁻² dm⁶ s⁻¹ (units derived from rearranging: rate / (mol dm⁻³)³ = (mol dm⁻³ s⁻¹) / (mol dm⁻³)³ = mol⁻² dm⁶ s⁻¹).

Total: 8 marks (M5 A3, split as shown).

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Specimen question modelled on the Edexcel 9CH0 Paper 2 format

Question (6 marks): A student claims that for the reaction $\text{H}_2\text{O}_2 + 2\text{I}^- + 2\text{H}^+ \rightarrow 2\text{H}_2\text{O} + \text{I}_2$H2​O2​+2I−+2H+→2H2​O+I2​, the rate equation must be $\text{rate} = k[\text{H}_2\text{O}_2][\text{I}^-]^2[\text{H}^+]^2$rate=k[H2​O2​][I−]2[H+]2 because the equation has 1 mole of H₂O₂, 2 moles of I⁻ and 2 moles of H⁺.

(a) State why the student's reasoning is incorrect. (2)

(b) Initial-rate experiments give: rate is first order in H₂O₂, first order in I⁻, and zero order in H⁺. Write the rate equation and explain what zero order in H⁺ implies about the mechanism. (4)

Mark scheme decomposition by AO: