Question

A first-order reaction has a half-life of 16.7 s . How long does it take for the concentration of the reactant in the reaction to fall to one-fourth of its initial value?

Answer 1

It takes 33.4 s for the concentration of A to fall to one-fourth of its original value.

A half-life is the time it takes for the concentration to fall to half its original value.

Assume the initial concentration is 1.00 mol/L. Then,

$\text{1.00 mol/L }\stackrel{\text{1st half-life }}{\longrightarrow}\text{ 0.50 mol/L } \stackrel{\text{ 2nd half-life }}{\longrightarrow}\text{0.25 mol/L}$

The concentration drops to one-fourth of its initial value in two half-lives.

∴ Time = 2 × 16.7 s = 33.4 s