Half-Life

Spec Mapping — OCR H432 Module 5.1.1 — Half-life, covering the definition of half-life ( $t_{1/2}$ ) as the time taken for a reactant concentration to fall to half its initial value, the diagnostic significance of constant half-life as the experimental signature of first-order kinetics, the derivation of $t_{1/2} = \ln 2 / k$ for first-order reactions, the contrasting half-life dependence on $[A]_0$ for zero ( $t_{1/2} \propto [A]_0$ ) and second ( $t_{1/2} \propto 1/[A]_0$ ) order reactions, and the application of half-life to compute $k$ or predict [A] at any later time (refer to the official OCR H432 specification document for exact wording).

Half-life is the most powerful single diagnostic of first-order kinetics in the OCR H432 corpus. The reasoning chain is short and elegant: for a first-order reaction $[A] = [A]_0 e^{-kt}$ , setting $[A] = [A]_0/2$ gives $t_{1/2} = \ln 2 / k$ — a result that depends only on $k$ , not on $[A]_0$ . So if you observe a reactant's concentration halving in equal time intervals throughout the reaction, you have proven first-order kinetics without needing any other data. Crucially, zero-order reactions show shortening half-lives (each successive halving takes less time, because the constant rate burns through smaller and smaller [A] proportions), and second-order reactions show lengthening half-lives (each successive halving takes twice as long, because rate $\propto [A]^2$ falls quadratically). This three-way diagnostic is the OCR examiner's favourite kinetics question style — a $[A]$ - $t$ table showing three or four successive half-lives is enough to identify order and compute $k$ in a single calculation. By the end of this lesson you will be able to: define half-life precisely, identify order from successive half-life behaviour, derive $k = \ln 2 / t_{1/2}$ , and compute [A] at any time using either successive halving or the exponential decay law.

Key Equation: for a first-order reaction, the half-life is constant and given by $t_{1/2} = \frac{\ln 2}{k} = \frac{0.693}{k}$ independent of $[A]_0$ . Equivalently, $k = \ln 2 / t_{1/2}$ . For zero order: $t_{1/2} = [A]_0 / 2k$ (shortens as reaction proceeds); for second order: $t_{1/2} = 1/(k[A]_0)$ (lengthens as reaction proceeds).

Definition of half-life

The half-life $t_{1/2}$ of a reactant is the time taken for its concentration to fall to half of its initial (or current) value. For example, if [A] starts at 0.80 mol dm $^{-3}$ and after 50 s has dropped to 0.40 mol dm $^{-3}$ , the half-life at that stage is 50 s. The crucial subtlety is that "initial" can refer to any starting point on the concentration-time curve — not just $t = 0$ . The half-life from $[A] = 0.80$ to $[A] = 0.40$ might differ from the half-life from $[A] = 0.40$ to $[A] = 0.20$ . The pattern of successive half-lives is the diagnostic of order.

How half-life depends on order

Order	Half-life formula	Direction of change as reaction proceeds
0	$t_{1/2} = [A]_0 / (2k)$	Shortens (halves each cycle)
1	$t_{1/2} = \ln 2 / k$	Constant
2	$t_{1/2} = 1/(k[A]_0)$	Lengthens (doubles each cycle)

The most important result for OCR exams: constant half-life proves first order. This is a specification bullet-point and appears on roughly half of every H432 kinetics paper.

Why first-order half-life is constant — derivation

For a first-order reaction, the integrated rate law (Lesson 3) is

$[A] = [A]_0\, e^{-kt}$

Setting $[A] = [A]_0/2$ at $t = t_{1/2}$ :

\begin{aligned} [A]_0/2 &= [A]_0\, e^{-k\, t_{1/2}} \\ \tfrac{1}{2} &= e^{-k\, t_{1/2}} \\ \ln\left(\tfrac{1}{2}\right) &= -k\, t_{1/2} \\ -\ln 2 &= -k\, t_{1/2} \\ t_{1/2} &= \frac{\ln 2}{k} = \frac{0.693}{k} \end{aligned}

The half-life depends only on $k$ , not on $[A]_0$ . So every successive half-life is the same. After one $t_{1/2}$ , [A] has fallen to 1/2 of its start; after two, to 1/4; after three, to 1/8. After $n$ half-lives, $[A] = [A]_0 \times (1/2)^n$ .

This is the same maths that governs radioactive decay (every radioactive nucleus is first order — rate of disintegration $= \lambda N$ , where $\lambda = k$ ). The radiochemical "half-life" of carbon-14 (5730 years) or uranium-238 ( $4.5 \times 10^9$ years) is mathematically identical to a chemical first-order half-life.

Why zero and second order have non-constant half-lives

Zero order ( $\text{rate} = k$ , constant): [A] falls linearly with $-d[A]/dt = k$ . Time to halve from $[A]_0$ to $[A]_0/2$ is $t_{1/2} = [A]_0/(2k)$ . Halving again ( $[A]_0/2 \to [A]_0/4$ ) takes only $t_{1/2}/2 = [A]_0/(4k)$ — shorter. Each successive half-life is half the previous.

Second order ( $\text{rate} = k[A]^2$ ): integrated form $1/[A] - 1/[A]_0 = kt$ . Setting $[A] = [A]_0/2$ gives $1/(t_{1/2}) - 1/[A]_0 = kt_{1/2} \Rightarrow 2/[A]_0 - 1/[A]_0 = kt_{1/2} \Rightarrow t_{1/2} = 1/(k[A]_0)$ . Each successive halving starts at half the previous [A], so the next $t_{1/2}$ is double the previous — half-life lengthens.

Visualisation — three orders, three half-life patterns

graph TD
  A["Read successive half-lives from [A]-t graph"] --> B{All the same?}
  B -->|Yes| C[First order; k = ln2/t1/2]
  B -->|No| D{Getting shorter?}
  D -->|Yes| E[Zero order; t1/2 propto [A]0]
  D -->|No| F{Getting longer?}
  F -->|Yes| G[Second order; t1/2 propto 1/[A]0]
  F -->|No| H[Re-check data accuracy]

Worked example 1 — finding $k$ from a half-life

Q: A first-order reaction has a half-life of 120 s. Calculate $k$ .

A: $k = \ln 2 / t_{1/2} = 0.693/120 = \mathbf{5.78 \times 10^{-3}\;\text{s}^{-1}}$ .

Worked example 2 — finding half-life from $k$

Q: A first-order reaction has $k = 2.0 \times 10^{-4}$ s $^{-1}$ . Calculate $t_{1/2}$ in seconds and minutes.

A: $t_{1/2} = \ln 2 / k = 0.693/(2.0 \times 10^{-4}) = \mathbf{3465\;\text{s} \approx 58\;\text{min}}$ .

Worked example 3 — using multiple half-lives

Q: A first-order reaction starts with $[A]_0 = 0.60$ mol dm $^{-3}$ . The half-life is 25 s. Find [A] after 100 s.

A: $100\,\text{s} = 4 \times t_{1/2}$ .

Cycle	$t$ / s	[A] / mol dm $^{-3}$
1	25	0.30
2	50	0.15
3	75	0.075
4	100	0.0375

In general, $[A] = [A]_0 \times (1/2)^n$ where $n = t/t_{1/2}$ . Verifying with the exponential: $k = 0.693/25 = 0.0277$ s $^{-1}$ ; $[A](100) = 0.60 \times e^{-0.0277 \times 100} = 0.60 \times e^{-2.77} = 0.60 \times 0.0625 = 0.0375$ — agreement.

Worked example 4 — reading half-life from a graph

Q: A concentration-time table:

$t$ / s	$[A]$ / mol dm $^{-3}$
0	1.00
40	0.50
80	0.25
120	0.125

Identify order and compute $k$ .

A: Successive half-lives: 40 s, 40 s, 40 s — constant. First order. $k = \ln 2/40 = \mathbf{1.73 \times 10^{-2}\;\text{s}^{-1}}$ .

Worked example 5 — second-order half-life lengthening

Q: A reaction $2NO_2 \rightarrow N_2O_4$ :

$t$ / s	$[NO_2]$ / mol dm $^{-3}$
0	0.100
10	0.050
30	0.025
70	0.0125

Identify order and find $k$ .

A: Successive half-lives: 10 s, 20 s, 40 s — each doubles. Second order. Using $t_{1/2} = 1/(k[A]_0)$ with $[A]_0 = 0.100$ , $t_{1/2} = 10$ s: $k = 1/(10 \times 0.100) = \mathbf{1.0\;\text{mol}^{-1}\,\text{dm}^{3}\,\text{s}^{-1}}$ .

Worked example 6 — zero-order half-life shortening

Q: A reaction shows:

$t$ / s	$[A]$ / mol dm $^{-3}$
0	1.00
25	0.50
37.5	0.25
43.75	0.125

Identify order and compute $k$ .

A: Half-lives: 25 s, 12.5 s, 6.25 s — each halves. Zero order. $t_{1/2} = [A]_0/(2k)$ , so $k = [A]_0/(2 t_{1/2}) = 1.00/(2 \times 25) = \mathbf{0.020\;\text{mol dm}^{-3}\,\text{s}^{-1}}$ .

Worked example 7 — counter-intuitive precision

Q: A first-order reaction has $t_{1/2} = 60.0$ s. After how many seconds will [A] have fallen to 1% of its initial value?

A: $[A]/[A]_0 = 0.01 = (1/2)^n$ , so $n = \log(0.01)/\log(0.5) = -2/-0.301 = 6.64$ . Time $= 6.64 \times 60.0 = \mathbf{398\;\text{s}}$ .

Alternative: $0.01 = e^{-kt}$ , $\ln(0.01) = -kt$ , $k = 0.693/60 = 0.01155$ , $t = 4.605/0.01155 = 399$ s — agreement.

The "five half-lives ≈ 3%" rule of thumb implies $\sim 5 \times 60 = 300$ s would already give 3.1%; another half-life and you are at 1.6%; a seventh gets you to 0.78%. So ~6.6 half-lives is correct for 1%.

Synoptic Links

Synoptic Links — Connects to:

ocr-alevel-chemistry-quantitative-rates-equilibrium / concentration-time-graphs (Lesson 3 — the $[A]$ - $t$ curve from which successive half-lives are read).

ocr-alevel-chemistry-quantitative-rates-equilibrium / orders-of-reaction (Lesson 2 — half-life behaviour is the cleanest order discriminator).

ocr-alevel-chemistry-quantitative-rates-equilibrium / rate-constant-k (Lesson 7 — $k = \ln 2 / t_{1/2}$ for first order; one of the most-used relations in 5.1.1).

ocr-alevel-chemistry-quantitative-rates-equilibrium / the-arrhenius-equation (Lesson 9 — Arrhenius gives the $T$ -dependence of $k$ , which through $t_{1/2} = \ln 2/k$ means $T$ -dependence of half-life too).

ocr-alevel-physics-nuclear / radioactive-decay (radioactive decay is mathematically identical first-order kinetics — half-life is the bridge concept).

Practical Activity Group anchor: PAG 9 (Continuous-monitoring rate) is the primary practical context. The colorimetry of $Br_2 + HCOOH$ , the gas-volume tracking of Mg + HCl, and the mass-loss of marble chips + HCl all produce $[A]$ vs $t$ curves with readable successive half-lives. Cohort-favourite OCR exam questions ask candidates to identify three successive half-lives from such a curve and use them to confirm first-order kinetics and compute $k$ .

Specimen question modelled on the OCR H432 paper format

Question (9 marks): The decomposition of $H_2O_2(aq)$ catalysed by potassium iodide was monitored continuously by colorimetry at 298 K. The following data were obtained:

$t$ / s	$[H_2O_2]$ / mol dm $^{-3}$
0	1.000
180	0.707
360	0.500
540	0.353
720	0.250
1080	0.125

(a) Determine three successive half-lives and use them to deduce the order of the reaction with respect to $H_2O_2$ . [3 marks]

(b) Calculate the rate constant $k$ with units. [2 marks]

(d) The student suggested that the catalyst (KI) increases the rate by increasing the half-life. Comment on this statement. [2 marks]

AO breakdown

Mark	AO	Awarded for
1	AO2	Half-life 1: 0 → 360 s (1.000 → 0.500) = 360 s
2	AO2	Half-life 2: 360 → 720 s (0.500 → 0.250) = 360 s; Half-life 3: 720 → 1080 s (0.250 → 0.125) = 360 s
3	AO3	Constant half-life → first order with respect to $H_2O_2$
4	AO2	$k = \ln 2 / 360 = 1.93 \times 10^{-3}$
5	AO3	Units: s $^{-1}$
6	AO2	1800 s = 5 × 360 s = 5 half-lives
7	AO2	$[H_2O_2] = 1.000 \times (1/2)^5 = 0.0313$ mol dm $^{-3}$
8	AO3	Statement is wrong: catalyst decreases half-life (increases $k$ )
9	AO3	Quantitative or mechanistic explanation: lowers $E_a$ , raises $k$ , shortens $t_{1/2}$

AO split: AO1 = 0, AO2 = 5, AO3 = 4.

Mid-band response (6/9):

Half-Life

Half-Life

Definition of half-life

How half-life depends on order

Why first-order half-life is constant — derivation

Why zero and second order have non-constant half-lives

Visualisation — three orders, three half-life patterns

Worked example 1 — finding kkk from a half-life

Worked example 2 — finding half-life from kkk

Worked example 3 — using multiple half-lives

Worked example 4 — reading half-life from a graph

Worked example 5 — second-order half-life lengthening

Worked example 6 — zero-order half-life shortening

Worked example 7 — counter-intuitive precision

Synoptic Links

Specimen question modelled on the OCR H432 paper format

AO breakdown

More in Chemistry

Worked example 1 — finding $k$ from a half-life

Worked example 2 — finding half-life from $k$