Question

The rate constant for a first–order reaction is $60\ \mathrm{s^{-1}$. How much time will it take to reduce the initial concentration of the reactant to $\dfrac{1}{16}$th?}

Hint:

For "reduce to a fraction" questions in first–order kinetics, use \(t=\dfrac{1}{k}\ln\!\left(\dfrac{1}{\text{fraction remaining}}\right)\); powers of 2 are quick via \(\ln 2\approx0.693\).

Answer 1

For a first–order reaction, \( \ln\!\left(\dfrac{[A]_0}{[A]_t}\right)=kt \).
Here \( \dfrac{[A]_0}{[A]_t}=16 $\Rightarrow$ \ln 16 = 4\ln 2 = 4\times 0.693 = 2.772 \).
Given \( k=60\ \mathrm{s^{-1}} \).
\[ t=\frac{\ln 16}{k}=\frac{2.772}{60}=4.62\times 10^{-2}\ \mathrm{s}\approx 0.046\ \mathrm{s}. \] \[ \boxed{t \approx 4.6\times 10^{-2}\ \text{s}} \]