First-Order Half-Life Calculator: Amount Left After a Time

This calculator applies first-order kinetics. Enter the starting amount (any unit: grams, moles or mol/L), the half-life and the time elapsed in the same time unit; you get the amount left, the percentage left and the rate constant k.

Formula

N = N₀ × 0.5^{(t ÷ t½)}, equivalent to N = N₀ × e^−kt with k = ln 2 ÷ t½.

• N₀: starting amount; N: amount left after time t, in the same unit
• t½: half-life, the time for half of the reactant to react; constant for a first-order reaction
• k: rate constant in reciprocal time units (per minute if the times are in minutes); ln 2 ≈ 0.693147

Worked Example

A reactant that decomposes by first-order kinetics with a half-life of 30 min; 80 g at the start, 75 min later.

1. Number of half-lives: 75 ÷ 30 = 2.5
2. Fraction left: 0.5^2.5 = 0.176777, that is 17.68 %
3. Amount left: 80 × 0.176777 = 14.142 g
4. Rate constant: 0.693147 ÷ 30 = 0.023105 per minute

Assumptions

• The reaction is first order in this reactant, so the half-life does not depend on how much is left; radioactive decay follows the same law.
• Temperature is constant; the half-life of a chemical reaction changes with temperature.
• Half-life and time are entered in the same unit.

How to read the result

After one half-life 50% is left, after two 25% and after three 12.5%; the example falls between two and three half-lives. The amount never reaches exactly zero in this model. The rate constant gives the initial rate: rate = k × N. For second-order or zero-order reactions the half-life changes as the reaction goes on, and this formula does not apply.

Limits and common mistakes

• Entering the half-life in hours and the time in minutes.
• Using the formula for a zero- or second-order reaction; check the order from data first.
• Confusing k with the half-life: they are inversely related, k × t½ = 0.693.

Frequently Asked Questions

How do you calculate the amount left from the half-life?

Multiply the starting amount by 0.5 raised to the number of half-lives elapsed, t ÷ t½.

How much of 80 g is left after 75 min if the half-life is 30 min?

About 14.142 g, or 17.68% of the start.

What is the rate constant for a half-life of 30 minutes?

k = ln 2 ÷ 30 ≈ 0.023105 per minute.

Is the half-life constant?

For a first-order reaction at constant temperature, yes; for other orders it depends on the concentration.

in n0 | Starting amount | any unit | 80 | 0
in th | Half-life | time | 30 | 0.000000001
in t | Time elapsed | time | 75 | 0
out n | Amount left | same unit as start | n0 * 0.5^(t / th) | 3
out pct | Fraction left | % | n / n0 * 100 | 2
out k | Rate constant | 1/time | ln(2) / th | 6