Question

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

Answer 1

Given that the rate constant for a first order reaction is $60s^{-1}$.

Let $a$ M be the initial concentration.

Final concentration will be $\frac{a}{16}$ M.

Rate constant, $k=60/s$

$t=\frac{2.303}{k}log\left(\frac{[A{]}_0}{\left[A\right]}\right)$

$t=\frac{2.303}{60}log\left(\frac{a}{\frac{a}{16}}\right)$

$t=4.6\times {10}^{-2}$ seconds