Question

If $75\%$ of a first order reaction was completed in $90\mathrm{minutes},60\%$ of the same reaction would be completed in approximately $(\mathrm{in}\mathrm{minutes})$_______. (take:$\log 2=0.30;\log 2.5=0.40)$.

Answer 1

The same reaction would be completed in approximately $\mathbf{60}\mathbf{}\mathbf{min}$.

Explanation:

$$\begin{gathered} \mathbf{Step}\mathbf{}\mathbf{1}\mathbf{:}\mathbf{}\mathbf{Calculation}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{t}}_{\mathbf{1}\mathbf{/}\mathbf{2}} \\ {\mathrm{t}}_{75\%}=90\min =2\times {\mathrm{t}}_{1/2} \\ {\mathrm{t}}_{1/2}=45\min \\ \mathbf{Step}\mathbf{}\mathbf{2}\mathbf{:}\mathbf{}\mathbf{Calculation}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{t}}_{\mathbf{60}\mathbf{\%}\mathbf{}\mathbf{}} \\ \left[\frac{\ln \left(2\right)}{45}\right]\times 60\%=\ln \left[\frac{100}{40}\right] \\ {\mathrm{t}}_{60\%}=45\times \left[\frac{0.4}{0.3}\right] \\ {\mathbf{t}}_{\mathbf{60}\mathbf{\%}\mathbf{}\mathbf{}}\mathbf{=}\mathbf{}\mathbf{60}\mathbf{}\mathbf{min} \\ \mathrm{Hence},\mathrm{the}\mathrm{time}\mathrm{required}\mathrm{is}\mathbf{60}\mathbf{}\mathbf{min}\mathbf{.} \end{gathered}$$