Chemistry - Chemical Kinetics Question with Solution | TestHub

For a first order reaction, the rate constant $k$ is given by

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

Given $t=2700\text{ s}$ and $99\%$ of the reactant decomposes.

Let $[A{]}_0=100$, then $[A{]}_t=100-99=1$.

Substituting the values into the equation:

$k=\frac{2.303}{2700}\log \left(\frac{100}{1}\right)$

$k=\frac{2.303}{2700}\times 2$

$k=\frac{4.606}{2700}$

$k\approx 1.7\times 10^{-3}{\text{ s}}^{-1}$

Answer: $1.7\times 10^{-3}{\text{ s}}^{-1}$.