First-Order Integrated Rate Law
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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Grade 12Grade 12 Chemistry
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UniversityProcess & Water Chemistry
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First-Order Integrated Rate Law explained
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 × = 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
- = Concentration at time t (M)
- = Initial concentration (M)
- = First-order rate constant (Hz)
- = Elapsed time (s)
Missing one of these? Work it out first, then come back
- Concentration at time t — Zero-Order Integrated Rate Law, Second-Order Integrated Rate Law
- Initial concentration — Dilution Equation (C1V1 = C2V2), Zero-Order Integrated Rate Law
- First-order rate constant — CSTR Design Equation (First Order), PFR Design Equation (First Order)
- Elapsed time — Half-Life Decay, Zero-Order Integrated Rate Law