First-Order Integrated Rate Law

[A]=[A]0 e−kt[\mathrm{A}] = [\mathrm{A}]_0\,e^{-kt}

Worked example: 0.100 M first-order, k = 0.0231 /s, 30.0 s → 0.0500074 M — press Try an example to run it live, then adjust anything.

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First-Order Integrated Rate Law explained

[A]0t[A]k

In a first-order reaction each molecule decomposes independently with a fixed probability per second, so the rate is proportional to how much is left and the concentration falls exponentially. Taking logarithms gives ln[A] = ln[A]₀ − kt, a straight line — the diagnostic test that distinguishes first order from every other order. The defining property is a constant half-life, t½ = ln2/k = 0.693/k, independent of where you start.

The maths is identical to radioactive decay, which is why the same equation covers the elimination of most drugs from the bloodstream, the fading of a chemiluminescent glow stick, and the isomerisation of cyclopropane to propene. Worked case: with k = 0.0231 s⁻¹, a 0.100 M solution falls to 0.100 × e−0.0231×30e^{-0.0231 \times 30} = 0.0500 M in 30 s — the half-life, since 0.693/0.0231 = 30.0 s. The usual arithmetic slip is mixing k in s⁻¹ with a time in minutes; the calculator converts both to SI first, so enter each with its real unit rather than pre-converting.

First-Order Integrated Rate Law formula

[A]=[A]0 e−kt[\mathrm{A}] = [\mathrm{A}]_0\,e^{-kt}
Where
  • [A][\mathrm{A}]= Concentration at time t (M)
  • [A]0[\mathrm{A}]_0= Initial concentration (M)
  • kk= First-order rate constant (Hz)
  • tt= Elapsed time (s)

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