Question

Prove that `[A] = [A]_0 (1/2)^n`, where [A]₀ = initial concentration, [A] = concentration after time t, and n = number of half-lives.

Answer 1

Given:

[A]₀ = initial concentration,

[A] = concentration after time t, and

n = number of half-lives

For a first‑order reaction,

[A] = [A]₀​e^−kt

Half-life is `t_(1//2) = (ln 2)/k`

If n half‑lives have passed, t = n t_1/2​. 

Substitute

`[A] = [A]_0​e^(−k(nt_(1//2)​))`

= `[A]_0 e^(-kn(ln 2)//k)`

= `[A]_0e^(-n ln 2)`

= `[A]_0 (e^(ln 2))^-( n)`

= [A]₀2⁻ⁿ

= `[A]_0 (1/2)^n`

Hence proved.