Chemistry - Chemical Kinetics Question with Solution | TestHub

Determine 'a': Since $T_{3/4}=2\times T_{1/2}$ in both data sets, the reaction is first-order with respect to the complex, so $a=1$.

Formulate $T_{1/2}$: For a first-order process, the half-life is $T_{1/2}=\frac{\ln 2}{k^{\prime }}=\frac{\ln 2}{k[H^{+}{]}^b}$, making it independent of the complex concentration.

Determine 'b': When $[H^{+}]$ doubles (from $0.01$ to $0.02M$), $T_{1/2}$ is halved (from $1$ to $0.5$), meaning $T_{1/2}\propto \frac{1}{[H^{+}{]}^1}$, so $b=1$.

Final Answer: Combining $a=1$ and $b=1$ gives the final answer as 11.