Question

For first order parallel reaction $k_1$ and $k_2$ are $4$ and $2mi{n}^{-1}$ respectively at $300K$. The activation energies for the formation of $B$ and $C$ are $30,000$ and $38,314J/mol$ respectively. The temperature at which $B$ and $C$ will be obtained in equimolar ratio is:

• A. $757.48K$
• B. $378.74K$
• C. $600K$
• D. None of these

Answer 1

The correct option is B $378.74K$

$\ln \frac{k_1^{{}^{\prime }}}{k_1}=\frac{E_1}{R}[\frac{1}{T_1}-\frac{1}{T_2}]...\left(i\right)$

$\ln \frac{k_2^{{}^{\prime }}}{k_2}=\frac{E_2}{R}[\frac{1}{T_1}-\frac{1}{T_2}]...\left(ii\right)$

Subtracting equation $\left(ii\right)$ from $\left(i\right)$,

$\ln \frac{k_2^{{}^{\prime }}}{k_2}-\ln \frac{k_1^{{}^{\prime }}}{k_1}=\left(\frac{E_2-E_1}{R}\right)(\frac{1}{T_1}-\frac{1}{T_2})$

For equimolar formation of B and C,
$k_2^{{}^{\prime }}=k_1^{{}^{\prime }}$

$\therefore \ln \left(\frac{k_1}{k_2}\right)=\left(\frac{8314}{8.314}\right)\left(\frac{T_2-300}{300\times T_2}\right)$

$\ln 2=\left(1000\right)\left(\frac{T_2-300}{300\times T_2}\right)$

$\therefore T_2=378.74K$