Question

The time required for 10% completion of a first order reaction at 298 K is equal to that required for its 25% completion at 308 K. If the pre-exponential factor for the reaction is $3.56\times {10}^9{s}^{-1}$, the value of rate constant at 318 K will be:

• A. $k=9.3\times {10}^{-4}{s}^{-1}$
• B. $k=8.1\times {10}^{-4}{s}^{-1}$
• C. $k=6.9\times {10}^{-4}{s}^{-1}$
• D. None of these

Answer 1

The correct option is A $k=9.3\times {10}^{-4}{s}^{-1}$

$t_1=\frac{2.3}{k_1}log\left(\frac{100}{100-10}\right)$

$t_2=\frac{2.3}{k_2}log\left(\frac{100}{100-25}\right)$

As $t_1=t_2$

$\frac{2.3}{k_1}log\left(\frac{10}{9}\right)=\frac{2.3}{k_2}log\left(\frac{4}{3}\right)$

$\therefore \frac{k_2}{k_1}=\left(\frac{log\frac{4}{3}}{log\frac{10}{9}}\right)=2.73$

Also, $log\frac{k_2}{k_1}=\frac{E_a}{2.3R}\left(\frac{T_2-T_1}{T_1{T}_2}\right)$

$log\left(2.73\right)=$ $\frac{E_a}{2.3\times 8.314J{K}^{-1}mo{l}^{-1}}\times \left(\frac{10}{298\times 308}\right)$

$\therefore E_a=76.6\times {10}^{3}Jmo{l}^{-1}=76.6kJmo{l}^{-1}$

Further, $k=A{e}^{-E_a/RT}$

or In k = In A - $\frac{E_a}{RT}$

In k = In $(3.56\times {10}^9{s}^{-1})-\frac{76.6KJmo{l}^{-1}}{8.314\times {10}^{-3}KJmo{l}^{-1}\times 318}$

$\therefore k=9.3\times {10}^{-4}{s}^{-1}$