Half-Life

Spec Mapping — OCR H432 Module 5.1.1 — Half-life, covering the definition of half-life ($t_{1/2}$t1/2​) 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 $t_{1/2} = \ln 2 / k$t1/2​=ln2/k for first-order reactions, the contrasting half-life dependence on $[A]_0$[A]0​ for zero ($t_{1/2} \propto [A]_0$t1/2​∝[A]0​) and second ($t_{1/2} \propto 1/[A]_0$t1/2​∝1/[A]0​) order reactions, and the application of half-life to compute $k$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 $[A] = [A]_0 e^{-kt}$[A]=[A]0​e−kt, setting $[A] = [A]_0/2$[A]=[A]0​/2 gives $t_{1/2} = \ln 2 / k$t1/2​=ln2/k — a result that depends only on $k$k, not on $[A]_0$[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 $\propto [A]^2$∝[A]2 falls quadratically). This three-way diagnostic is the OCR examiner's favourite kinetics question style — a $[A]$[A]-$t$t table showing three or four successive half-lives is enough to identify order and compute $k$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 $k = \ln 2 / t_{1/2}$k=ln2/t1/2​, 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 $t_{1/2} = \frac{\ln 2}{k} = \frac{0.693}{k}$t1/2​=kln2​=k0.693​ independent of $[A]_0$[A]0​. Equivalently, $k = \ln 2 / t_{1/2}$k=ln2/t1/2​. For zero order: $t_{1/2} = [A]_0 / 2k$t1/2​=[A]0​/2k (shortens as reaction proceeds); for second order: $t_{1/2} = 1/(k[A]_0)$t1/2​=1/(k[A]0​) (lengthens as reaction proceeds).

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Definition of half-life

The half-life $t_{1/2}$t1/2​ 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$^{-3}$−3 and after 50 s has dropped to 0.40 mol dm$^{-3}$−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 $t = 0$t=0. The half-life from $[A] = 0.80$[A]=0.80 to $[A] = 0.40$[A]=0.40 might differ from the half-life from $[A] = 0.40$[A]=0.40 to $[A] = 0.20$[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	$t_{1/2} = [A]_0 / (2k)$t1/2​=[A]0​/(2k)	Shortens (halves each cycle)
1	$t_{1/2} = \ln 2 / k$t1/2​=ln2/k	Constant
2	$t_{1/2} = 1/(k[A]_0)$t1/2​=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

$[A] = [A]_0\, e^{-kt}$[A]=[A]0​e−kt

Setting $[A] = [A]_0/2$[A]=[A]0​/2 at $t = t_{1/2}$t=t1/2​:

[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 <svg viewBox="0 0 720 380" xmlns="http://www.w3.org/2000/svg" role="img" aria-label="Half-life patterns: zero, first, second order"> <rect x="0" y="0" width="720" height="380" fill="#fafafa" stroke="#bbb" stroke-width="1"/> <text x="360" y="22" font-family="Helvetica, Arial, sans-serif" font-size="14" font-weight="bold" fill="#222" text-anchor="middle">Successive half-lives — three orders, three patterns</text> <line x1="80" y1="340" x2="680" y2="340" stroke="#333" stroke-width="1.4"/> <line x1="80" y1="50" x2="80" y2="340" stroke="#333" stroke-width="1.4"/> <text x="380" y="365" font-family="Helvetica, Arial, sans-serif" font-size="12" fill="#333" text-anchor="middle">time t / s</text> <text x="40" y="200" font-family="Helvetica, Arial, sans-serif" font-size="12" fill="#333" transform="rotate(-90 40 200)">[A] / mol dm^-3</text> <!-- zero order: linear --> <line x1="80" y1="60" x2="560" y2="340" stroke="#0b3d91" stroke-width="2.4"/> <text x="450" y="180" font-family="Helvetica, Arial, sans-serif" font-size="12" font-weight="bold" fill="#0b3d91">order 0 (t1/2 shortens)</text> <!-- first order: exponential --> <path d="M 80 60 Q 140 130 200 180 Q 280 240 380 290 Q 480 320 680 335" fill="none" stroke="#28793a" stroke-width="2.4"/> <text x="500" y="280" font-family="Helvetica, Arial, sans-serif" font-size="12" font-weight="bold" fill="#28793a">order 1 (t1/2 constant)</text> <!-- second order: steeper then long tail --> <path d="M 80 60 Q 110 210 200 270 Q 350 320 680 335" fill="none" stroke="#a8323a" stroke-width="2.4"/> <text x="540" y="328" font-family="Helvetica, Arial, sans-serif" font-size="12" font-weight="bold" fill="#a8323a">order 2 (t1/2 lengthens)</text> <!-- first-order constant t1/2 dashed marker lines --> <line x1="200" y1="340" x2="200" y2="180" stroke="#28793a" stroke-width="0.8" stroke-dasharray="3 3"/> <line x1="320" y1="340" x2="320" y2="260" stroke="#28793a" stroke-width="0.8" stroke-dasharray="3 3"/> <line x1="440" y1="340" x2="440" y2="305" stroke="#28793a" stroke-width="0.8" stroke-dasharray="3 3"/> <text x="170" y="355" font-family="Helvetica, Arial, sans-serif" font-size="9" fill="#28793a">t1/2</text> <text x="290" y="355" font-family="Helvetica, Arial, sans-serif" font-size="9" fill="#28793a">2 t1/2</text> <text x="410" y="355" font-family="Helvetica, Arial, sans-serif" font-size="9" fill="#28793a">3 t1/2</text> </svg> ```mermaid 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]** (c) Predict $[H_2O_2]$ after 1800 s, showing your working. **[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):**