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Lessons · Kinetics

When the half-life keeps growing: a second-order dimerization

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

Two butadiene molecules snap together into a ring: 2 C₄H₆ → C₈H₁₂. The reaction is second order in butadiene — its rate depends on two C₄H₆ molecules colliding, rate=k[C4H6]2\text{rate} = k[\mathrm{C_4H_6}]^2, with k=5.76×10−2 M−1 min−1k = 5.76\times10^{-2}\ \mathrm{M^{-1}\,min^{-1}}. The amount left follows the second-order integrated law 1[C4H6]=1[C4H6]0+kt\frac{1}{[\mathrm{C_4H_6}]} = \frac{1}{[\mathrm{C_4H_6}]_0} + kt (it is 1/[C4H6]1/[\mathrm{C_4H_6}], not [C4H6][\mathrm{C_4H_6}], that climbs in a straight line). Its half-life is t1/2=1/(k[C4H6]0)t_{1/2} = 1/(k[\mathrm{C_4H_6}]_0) — and because [C4H6]0[\mathrm{C_4H_6}]_0 sits in the denominator, the half-life grows as the reactant thins out: the second half takes longer than the first. Same species ledger, extent marching in time — but a very different clock from first order.

2 C4H6→C8H122\,\mathrm{C_{4}H_{6}} \rightarrow \mathrm{C_{8}H_{12}}
rate=k [C4H6]2\text{rate} = k\,[\mathrm{C_{4}H_{6}}]^2·k = 5.76×10⁻² M⁻¹ min⁻¹second order · gas phase (source states no temperature)
Integrated rate lawOne constant fixes the whole future — 1[C4H6]=1[C4H6]0+kt\dfrac{1}{[\mathrm{C_{4}H_{6}}]} = \dfrac{1}{[\mathrm{C_{4}H_{6}}]_0} + kt.

Starting at 0.2 M, the concentration drops fast, then crawls. The rate = k[C₄H₆]² depends on two molecules meeting, so it collapses as the reactant thins out — the tail drags on and never quite reaches zero.

The decay curveEvery point is the integrated rate law — the ledger marched forward in time.
0.000.05000.1000.1500.2000.01.42.94.35.87.38.71012131516171920hours[C₄H₆] (M)½ left¼ left⅛ left
to halving #10.1 Mstep took 1.45 h · 50% gone
to halving #20.05 Mstep took 2.89 h · 75% gone
to halving #30.025 Mstep took 5.79 h · 87.5% gone
first half-life t½1.45 h= 5210 s
t½ relationt1/2=1k [C4H6]0t_{1/2} = \dfrac{1}{k\,[\mathrm{C_{4}H_{6}}]_0}t½ ∝ 1/[A]₀ — grows as [A] falls ✓
the half-life doubles1.45 → 2.89 → 5.79 heach half takes ~2× as long

The second-order tellBecause t1/2=1k [C4H6]0t_{1/2} = \dfrac{1}{k\,[\mathrm{C_{4}H_{6}}]_0} has [C₄H₆]₀ in the denominator, each half-life is longer than the last (1.45 → 2.89 → 5.79 h — roughly doubling): as the reactant thins out, collisions between two C₄H₆ molecules get rarer, so the tail drags on. The machine checks each successive halving takes about twice as long. A constant half-life would be the sign of a first-order reaction, not this.

VerificationProven at build time — not asserted.
  • ✓ The reaction conserves every element [reaction balanced]
  • ✓ Every curve point matches the order-2 integrated law — re-derived independently [integrated law]
  • ✓ The first t½ = 1/k[A]₀ (order 2) [half-life relation]
  • ✓ Successive half-lives double — the order-2 fingerprint [half-life progression]
✗ Common misconception: “The half-life is a fixed property of the reaction, so using up the second half takes the same time as the first — a half-life is a half-life.”

That would be true for a first-order reaction — but this one is second order. Here t1/2=1k [C4H6]0t_{1/2} = \dfrac{1}{k\,[\mathrm{C_{4}H_{6}}]_0} grows as [C₄H₆]₀ falls, so each successive half takes longer, not the same: 1.45 → 2.89 → 5.79 h. A second-order rate ∝ [C₄H₆]² needs two molecules to meet; as they thin out the reaction crawls, and the machine checks each halving takes about twice the last.

Modeling assumptions — author-asserted, disclosed not discharged
  • model The reaction is second order in C₄H₆ (rate=k[C4H6]2\text{rate} = k[\mathrm{C_4H_6}]^2) — an experimentally determined rate law, not read off the balanced equation. The order and rate constant kk are the sourced data; the integrated law and half-life follow exactly.
  • model The rate constant is constant over the run — fixed temperature, gas phase, unchanging conditions — so the integrated rate law and half-life hold throughout, and the reverse reaction is negligible.

Concepts in this lesson

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

Practice this

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