The Rate Constant and Half-Life

The rate constant k is arguably the single most important number in a rate equation. It encapsulates everything about the reaction's speed that is not captured by the concentrations -- including the effect of temperature, the nature of the reactants, and the activation energy. Understanding what k tells you, what its units are, and how to calculate it is essential.

The Rate Constant k

In the rate equation Rate = k[A]^m[B]^n, the rate constant k is specific to a particular reaction at a particular temperature. If you change the temperature, k changes. If you change the concentration, k does not change -- that is the whole point of having concentrations explicitly in the equation.

Key points:

k increases with temperature (reactions get faster at higher temperatures).
k is larger for faster reactions at the same temperature.
A catalyst increases k by providing a pathway with lower Ea.
The value of k is independent of concentration.

Units of the Rate Constant

The units of k depend on the overall order of the reaction. Since Rate = k × [concentration terms], the units of k must make the equation dimensionally consistent.

Rate always has units of mol dm⁻³ s⁻¹.

Deriving Units for Each Order

Overall order	Rate equation form	Units of k	Derivation
0	Rate = k	mol dm⁻³ s⁻¹	Same as rate
1	Rate = k[A]	s⁻¹	(mol dm⁻³ s⁻¹) / (mol dm⁻³)
2	Rate = k[A]²	mol⁻¹ dm³ s⁻¹	(mol dm⁻³ s⁻¹) / (mol dm⁻³)²
3	Rate = k[A]²[B]	mol⁻² dm⁶ s⁻¹	(mol dm⁻³ s⁻¹) / (mol dm⁻³)³

General Formula

For a reaction of overall order n:

Units of k = mol^(1−n) dm^(3(n−1)) s⁻¹

This is a very common exam question -- you must be able to derive the units of k from the rate equation.

Calculating k from Experimental Data

Worked Example 1

Given: Rate = k[A]²[B], and from Experiment 1:

[A] = 0.10 mol dm⁻³
[B] = 0.10 mol dm⁻³
Rate = 2.0 × 10⁻⁴ mol dm⁻³ s⁻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

You should always check the units: for a third-order reaction, k should have units of mol⁻² dm⁶ s⁻¹. ✓

Worked Example 2: Verifying Consistency

To confirm your orders are correct, calculate k from every experiment:

Experiment	[A]	[B]	Rate	Calculated k
1	0.10	0.10	2.0 × 10⁻⁴	0.20
2	0.20	0.10	8.0 × 10⁻⁴	8.0×10⁻⁴ / (0.04 × 0.10) = 0.20
3	0.10	0.20	4.0 × 10⁻⁴	4.0×10⁻⁴ / (0.01 × 0.20) = 0.20

All three give k = 0.20 mol⁻² dm⁶ s⁻¹ -- this confirms the rate equation is correct.

Half-Life

The half-life (t½) of a reaction is the time taken for the concentration of a reactant to fall to half its initial value.

Half-Life and Zero-Order Reactions

For a zero-order reaction, [A] = [A]₀ − kt. Setting [A] = [A]₀/2:

[A]₀/2 = [A]₀ − kt½

t½ = [A]₀ / (2k)

The half-life depends on the initial concentration. As the reaction proceeds and [A] decreases, successive half-lives get shorter.

Half-Life and First-Order Reactions

For a first-order reaction, the integrated rate equation is:

ln[A] = −kt + ln[A]₀

Setting [A] = [A]₀/2:

ln([A]₀/2) = −kt½ + ln[A]₀

ln[A]₀ − ln 2 = −kt½ + ln[A]₀

t½ = ln 2 / k = 0.693 / k

This is a critically important result: the half-life of a first-order reaction is constant -- it does not depend on the initial concentration. This means every successive half-life is the same duration, regardless of what concentration you start with.

Half-Life and Second-Order Reactions

For a second-order reaction, t½ = 1 / (k[A]₀). The half-life depends on the initial concentration and increases as [A]₀ decreases (successive half-lives get longer).

Using Half-Life to Confirm First Order

If you measure the concentration of a reactant over time and find that the half-life is constant, this confirms that the reaction is first order. This is a quick and powerful diagnostic:

Pick any starting concentration on your graph and note the time.
Find the time at which the concentration has halved.
Repeat from the halved concentration.
If the time intervals are equal, the reaction is first order.

Worked Example 1: Confirming First Order

A student monitors the decomposition of N₂O₅ and records:

Time / s	[N₂O₅] / mol dm⁻³
0	0.400
50	0.200
100	0.100
150	0.050
200	0.025

t½ from 0.400 to 0.200: 50 s
t½ from 0.200 to 0.100: 50 s
t½ from 0.100 to 0.050: 50 s
t½ from 0.050 to 0.025: 50 s

The half-life is constant at 50 s, confirming first-order kinetics.

k = ln 2 / t½ = 0.693 / 50 = 0.0139 s⁻¹

Worked Example 2: Identifying Second Order from Half-Lives

A different experiment gives:

Time / s	[Y] / mol dm⁻³
0	0.400
25	0.200
75	0.100
175	0.050

t½ from 0.400 to 0.200: 25 s
t½ from 0.200 to 0.100: 50 s (75 − 25)
t½ from 0.100 to 0.050: 100 s (175 − 75)

The half-lives are doubling: 25, 50, 100 s. This is characteristic of second-order kinetics (when [A] halves, t½ doubles because t½ = 1/(k[A]₀)).

From the first half-life: k = 1/(t½ × [A]₀) = 1/(25 × 0.400) = 0.10 mol⁻¹ dm³ s⁻¹

Worked Example 3: Remaining Concentration After Multiple Half-Lives

A first-order reaction has t½ = 20 s. If [A]₀ = 1.60 mol dm⁻³, what is [A] after 80 s?

Number of half-lives = 80 / 20 = 4

[A] = [A]₀ × (½)^4 = 1.60 × 1/16 = 0.10 mol dm⁻³

Comparing Half-Lives Across Orders

Order	Half-life formula	Depends on [A]₀?	Successive half-lives	How to recognise
Zero	[A]₀ / 2k	Yes	Get shorter	Each t½ is shorter than the previous
First	ln 2 / k	No	Constant	All t½ values are equal
Second	1 / (k[A]₀)	Yes	Get longer (double)	Each t½ is double the previous

This pattern -- shorter, constant, longer -- is a useful way to distinguish orders from experimental data.

Common Mistakes to Avoid

Confusing k with rate. The rate changes as concentrations change; k is constant at a given temperature. A larger k means a faster reaction, but the rate also depends on concentration.
Wrong units for k. Always derive units from the rate equation. A common error is giving k in mol dm⁻³ s⁻¹ for a first-order reaction (should be s⁻¹).
Forgetting that k changes with temperature. Students sometimes calculate k at one temperature and use it at another.
Using t½ = 0.693/k for non-first-order reactions. This formula ONLY applies to first-order kinetics.
Miscounting half-lives. After 3 half-lives, the fraction remaining is (½)³ = 1/8, not 1/6 or 1/3.

Summary

The rate constant k captures the intrinsic speed of a reaction at a given temperature. Its units depend on the overall order and must be derived from the rate equation. For first-order reactions, the half-life is constant (t½ = ln 2/k), which provides both a diagnostic tool for confirming first-order behaviour and a straightforward way to calculate k. Zero-order half-lives decrease with time; second-order half-lives increase.

A-Level Deep Dive: The Rate Constant and Half-Life

Spec mapping

Edexcel 9CH0 specification, Topic 16 — Kinetics II requires candidates to determine the rate constant $k$ from initial-rate data, to recognise constant successive half-lives as the diagnostic of first-order behaviour, and to use $t_{1/2} = \ln 2 / k$ for first-order reactions (refer to the official specification document for exact wording). Topic 16 also requires candidates to derive units of $k$ from the rate equation by dimensional analysis and to use these units when comparing rate constants across reactions or temperatures. The half-life concept is examined principally on Paper 2 (Topics 11–19) and Paper 3 (synoptic with Core Practicals), and also surfaces in Topic 5 (mass spectrometry / radioactive decay) because nuclear decay is also a first-order process — the same $t_{1/2}$ formula applies. The data booklet does not list $t_{1/2} = \ln 2 / k$ .

Worked example with full mark scheme

Question (8 marks):

A first-order reaction $A \rightarrow B$ has the following concentration–time data at 25 °C.

t / s	0	100	200	300	400	500
[A] / mol dm⁻³	1.000	0.707	0.500	0.354	0.250	0.177

(a) Use the data to verify the reaction is first order with respect to A. (2)

(b) Determine the half-life $t_{1/2}$ and hence the rate constant $k$ . (3)

(d) State, with reasoning, the units of $k$ . (1)

Solution with mark scheme:

(a) M1 — read off successive half-lives: [A] falls from 1.000 to 0.500 in 200 s; from 0.500 to 0.250 in (400 − 200) = 200 s; from 0.250 to 0.125 (extrapolated) in another ~200 s. A1 — successive half-lives are constant (≈ 200 s), confirming the reaction is first order with respect to A.

(b) M1 — $t_{1/2} = 200$ s. M1 — $k = \ln 2 / t_{1/2} = 0.693 / 200$ . A1 — $k = 3.47 × 10⁻³$ s⁻¹ (3 sig fig).

(c) M1 — 25% of initial = two half-lives (100% → 50% → 25%). A1 — time = $2 \times 200 = 400$ s. (Verified directly from the table at t = 400 s.)

(d) M1 — for first order, rate = $k[A]$ , so $k = \text{rate} / [A]$ has units (mol dm⁻³ s⁻¹) / (mol dm⁻³) = s⁻¹.

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

Specimen question modelled on the Edexcel 9CH0 Paper 2 format

Question (6 marks): The decomposition of dinitrogen pentoxide $2\text{N}_2\text{O}_5(g) \rightarrow 4\text{NO}_2(g) + \text{O}_2(g)$ is first order in N₂O₅. At 65 °C, $k = 5.20 × 10^{-3}$ s⁻¹.

(a) Calculate the half-life. (2)

(b) Starting with [N₂O₅] = 0.200 mol dm⁻³, calculate [N₂O₅] after 5.00 minutes. (3)

Mark scheme decomposition by AO:

Part	Mark	AO	Earned by
(a) M1	1	AO1	$t_{1/2} = \ln 2 / k$ .
(a) A1	1	AO2	$t_{1/2} = 0.693 / 5.20 × 10^{-3} = 133$ s.
(b) M1	1	AO2	5.00 min = 300 s = 300/133 ≈ 2.25 half-lives.
(b) M1	1	AO2	Use $[A] = [A]_0 \cdot (1/2)^{t/t_{1/2}}$ or $\ln([A]/[A]_0) = -kt$ .
(b) A1	1	AO2	$[N_2O_5] = 0.200 \times \exp(-5.20 × 10^{-3} \times 300) = 0.200 × 0.211 = 0.0421$ mol dm⁻³.
(c) A1	1	AO3	87.5% decomposed = 12.5% remaining = three half-lives = $3 × 133 = 399$ s.

Total: 6 marks split AO1 = 1, AO2 = 4, AO3 = 1.

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