Integrated First-Order Rate Law

For a first-order reaction, ln[A]t = −kt + ln[A]0 predicts the remaining concentration at any time t.

\ln[A]_t = -kt + \ln[A]_0

ln[A]t = −kt + ln[A]0

What it means

Where the rate law describes speed at one instant, the integrated form describes the whole history: how much reactant remains after a given time. Its linear form (y = mx + b with y = ln[A]t) is also the diagnostic test — plot ln[A] versus time and a straight line confirms first order, with slope −k. The equation's exponential heart means a fixed fraction of what's left decays per unit time, which is why half-life is constant for first-order reactions but not for others. When to use it: finding remaining concentration after time t, finding the time to reach a target concentration, or testing whether experimental data are first order via a ln[A]-versus-time plot.

Variables

Symbol	Name	Unit	Meaning
$[A]t$	Concentration at time t	M (mol/L)	Concentration of the reactant at time t.
$k$	Rate constant	s⁻¹ (time⁻¹)	The first-order rate constant at the reaction temperature.
$t$	Time	s (or min)	Elapsed time since the reaction started.
$[A]0$	Initial concentration	M (mol/L)	Concentration of the reactant at t = 0.

Assumptions

The reaction is truly first order in the reactant.
The rate constant k is fixed — temperature does not change during the reaction.
The reverse reaction is negligible over the time considered.

Worked example

Problem

A first-order reaction has k = 0.0500 s⁻¹ and [A]0 = 0.800 M. What is [A] after 20.0 s?

Given:

First-order reaction with k = 0.0500 s⁻¹
[A]0 = 0.800 M, t = 20.0 s

Write the integrated first-order law: ln[A]t = −kt + ln[A]0.
Compute −kt: −(0.0500 s⁻¹)(20.0 s) = −1.00.
Compute ln[A]0: ln(0.800) = −0.2231.
Add: ln[A]t = −1.00 + (−0.2231) = −1.2231.
Exponentiate: [A]t = e^(−1.2231) = 0.2943 M, which rounds to 0.294 M.

[A]t = 0.294 M

After 20 s the reactant has decayed from 0.800 M to 0.294 M — about 63% consumed, matching the exponential decay first-order reactions follow.