The half-life of a first-order reaction is 14 s. What time is required for the concentration of the reactant to fall to 1/8 of its initial value?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 42 s
The half-life of a first-order reaction is the time in which the concentration of the reactant falls to half its value. It does not depend on the starting concentration, so every half-life here is 14 s.
After 1 half-life: [R] = [R]0/2
After 2 half-lives: [R] = [R]0/4
After 3 half-lives: [R] = [R]0/8
Falling to 1/8 takes three half-lives:
\( \displaystyle t=3\times t_{1/2} \)
\( \displaystyle t=3\times14\,\mathrm{s}=42\,\mathrm{s} \)