In a first-order reaction the rate depends on the concentration of one reactant to the power 1. Its integrated rate law lets you find the rate constant k from two concentrations measured a known time apart.
This calculator uses k = (2.303/t) log([A]₀/[A]), as in Class 12 Chemical Kinetics, and gives the half-life and the time for 99% completion. You can use any quantity proportional to concentration, such as pressure or absorbance.
| Initial concentration [A]₀ | 0.8 mol/L |
| Concentration at time t, [A] | 0.2 mol/L |
| Time (t) | 30 min |
| Rate constant k (per time unit) | 0.0462098 |
| Half-life t½ (time unit) | 15 |
| Reacted so far | 75% |
| Time for 99% completion (time unit) | 99.6578 |
k = 0.0462098 per minute, and the half-life is 15 min (two half-lives in 30 min).
Try this example →| Initial concentration [A]₀ | 1.24 mol/L |
| Concentration at time t, [A] | 0.93 mol/L |
| Time (t) | 10 min |
| Rate constant k (per time unit) | 0.0287682 |
| Half-life t½ (time unit) | 24.0942 |
| Reacted so far | 25% |
| Time for 99% completion (time unit) | 160.078 |
k = 0.0287682 per minute.
Try this example →k = (2.303/t) log([A]₀/[A]), or equivalently k = (1/t) ln([A]₀/[A]).
t½ = 0.693/k. It does not depend on the starting concentration.
Time⁻¹, such as s⁻¹ or min⁻¹.
Yes. That's why it has a constant half-life.