Chemistry - Chemical Kinetics Question with Solution | TestHub

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

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

Given that the concentration decreases by $90\%$, the remaining concentration $[A{]}_t$ is $10\%$ of the initial concentration $[A{]}_0$.

$[A{]}_t=0.1[A{]}_0$

Substituting $t=30\text{ min}$ and the concentrations into the rate equation:

$k=\frac{2.303}{30}\log \left(\frac{[A{]}_0}{0.1[A{]}_0}\right)$

$k=\frac{2.303}{30}\log (10)$

Since $\log (10)=1$:

$k=\frac{2.303}{30}=0.07676{\text{ min}}^{-1}$

$k=7.67\times 10^{-2}{\text{ min}}^{-1}$

Answer: $7.67\times 10^{-2}\cdot {\min }^{-1}$