Lessons · Kinetics

How fast does hydrogen peroxide decompose? A first-order clock

Species accounting machine-checked — the integrated rate law + half-lifek data-sourced (openstax-chemistry-2e)2 modeling assumptions (disclosed)

A bottle of hydrogen peroxide slowly falls apart on the shelf: 2 H₂O₂ → 2 H₂O + O₂. The reaction is first order in H₂O₂ — its rate is proportional to how much is left, $\text{rate} = k[\mathrm{H_2O_2}]$ , with $k = 3.21\times10^{-5}\ \mathrm{s^{-1}}$ . That one fact fixes the whole future: the amount left follows the integrated rate law $[\mathrm{H_2O_2}] = [\mathrm{H_2O_2}]_0\,e^{-kt}$ , a smooth exponential decay from 1.000 M. Its half-life — the time to lose half — is $t_{1/2} = \ln 2 / k$ , and for a first-order reaction it does not depend on how much you start with. This is the species ledger again, but with the extent marching in time.

2\,\mathrm{H_{2}O_{2}} \rightarrow 2\,\mathrm{H_{2}O} + \mathrm{O_{2}}

\text{rate} = k\,[\mathrm{H_{2}O_{2}}]

·k = 3.21×10⁻⁵ s⁻¹first order · dilute aqueous solution at 40 °C

Integrated rate lawOne constant fixes the whole future — $[\mathrm{H_{2}O_{2}}] = [\mathrm{H_{2}O_{2}}]_0\,e^{-kt}$ .

Starting at 1 M, the concentration decays smoothly. The rate = k[H₂O₂] falls as the reactant depletes, so the curve flattens — but it never quite reaches zero.

The decay curveEvery point is the integrated rate law — the ledger marched forward in time.

to halving #10.5 Mstep took 6 h · 50% gone

to halving #20.25 Mstep took 6 h · 75% gone

to halving #30.125 Mstep took 6 h · 87.5% gone

half-life t½6 h= 21600 s

t½ relation

t_{1/2} = \dfrac{\ln 2}{k}

k · t½ = 0.69314718 = ln 2 ✓

constant half-life6 → 6 → 6 heach step the same 6 h

The first-order tellBecause $t_{1/2} = \dfrac{\ln 2}{k}$ has no concentration in it, every half-life is the same 6 h: 1 M → 0.5 M → 0.25 M → 0.125 M. The reactant keeps halving on the same clock — the machine checks c(n·t½) = c₀/2ⁿ at each landmark. That constancy is the classic test for a first-order reaction.

VerificationProven at build time — not asserted.

✓ The reaction conserves every element [reaction balanced]
✓ Every curve point matches the order-1 integrated law — re-derived independently [integrated law]
✓ The first t½ = ln2/k (order 1) [half-life relation]
✓ Successive half-lives stay equal — the order-1 fingerprint [half-life progression]

✗ Common misconception: “As the peroxide runs low there is less of it, so the last half disappears faster than the first — the half-life shrinks as the reaction proceeds.”

That is true for some reactions — but not first order. For a first-order reaction the half-life $t_{1/2} = \dfrac{\ln 2}{k}$ depends only on k, not on how much is left. So the last half takes exactly as long as the first: 1 M → 0.5 M → 0.25 M → 0.125 M, each drop the same 6 h. The rate does slow as [H₂O₂] falls (rate = k[H₂O₂]), but the concentration and the time-to-halve fall together, so the clock never changes. A growing half-life would signal a higher-order reaction (second order), and a shrinking one an order below first (zero order) — the two cases this tier's other lessons show.

Modeling assumptions — author-asserted, disclosed not discharged

model The reaction is first order in H₂O₂ ( $\text{rate} = k[\mathrm{H_2O_2}]$ ) — an experimentally determined rate law, not read off the balanced equation. The order and the rate constant $k$ are the sourced data; the integrated law and half-life follow exactly.
model The rate constant is constant over the run — fixed temperature, dilute solution, unchanging conditions — so the integrated rate law and half-life hold throughout.

Concepts in this lesson

Linked into the Chemical Atlas where an entry exists; the rest fill in as the Atlas grows.

Practice this

The lesson goes deep on one scenario; the gym builds fluency by repetition. Drill these:

Kinetics: integrated rate laws, half-life, and reading the order →