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First Order Rate Law

Find the first-order rate constant k = (2.303/t)log([A]₀/[A]), with symbol legend and real-world examples.

k = \frac{2.303}{t}\,\log\frac{[A]₀}{[A]}

Frequently asked questions

What is a first-order reaction?

One whose speed depends on the amount of a single reactant: double that reactant and the rate doubles. It decays exponentially - losing the same fraction in each equal time slice. Radioactivity is the classic case.

What is special about its half-life?

It is constant - the time to lose half is the same whether you start with a lot or a little. That is why the example uses [A]₀ = 1, [A] = 0.5 (exactly half left) with t = 693 s, giving k ≈ 0.001 s⁻¹.

Why the 2.303 and the log?

Exponential decay leads to a natural-log relationship. The 2.303 converts a base-10 log (easy on a calculator) into the natural log the maths actually needs, since ln(x) = 2.303 × log₁₀(x).

How is it used in medicine?

Drugs usually clear first-order, giving a half-life that sets how often you must take them. This rate-constant calculation is how dosing schedules are designed.

How does carbon dating use it?

Carbon-14 decays first-order with a 5,730-year half-life. Measuring how much is left and applying this maths gives the age of a once-living sample.