[A]ₜ = [A]₀ × e^(−kt) — fill in every field except the one you want; the empty field is solved for.
s⁻¹
Concentration at time t ([A]ₜ) = 0.12579 M
- Convert t: 2 min =
120 s - Rearrange:
[A]ₜ = [A]₀ × e^(−k × t) - Substitute:
(0.5 M) × e^(−(0.0115 s⁻¹) × (120 s))=0.12579 M
Half-life (t½ = ln 2 ÷ k)
t½ = 60.27 s
A first-order half-life is constant: it does not depend on how much is left.
How it works
The integrated rate laws link concentration and time: zero order [A]ₜ = [A]₀ − kt; first order [A]ₜ = [A]₀e^(−kt); second order 1/[A]ₜ = 1/[A]₀ + kt. Times are converted to seconds, so k must be per second (or M s⁻¹, M⁻¹ s⁻¹).
To find the order from data, plot [A], ln[A] and 1/[A] against time: whichever is a straight line gives the order, and its slope gives k. Radioactive decay and many decompositions are first order.
Frequently asked questions
- How do I know which order to use?
- From experiment: if ln[A] against time is a straight line, the reaction is first order; if 1/[A] is straight, second order; if [A] itself is straight, zero order.
- Why is a first-order half-life constant?
- Because t½ = ln 2 ÷ k contains no concentration term — the same fraction reacts in each equal time interval.
- What are the units of k?
- Zero order: M s⁻¹; first order: s⁻¹; second order: M⁻¹ s⁻¹. The units always make the rate come out in M s⁻¹.
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