Question

For a reaction, consider the plot of $logK$ versus $\frac{1}{T}$ given in the figure.
If rate constant of this reaction at ${\text{400 K is10}}^{-5}se{c}^{-1}$ then the rate constant at $500K$ is :-

• A. $2\times 10^{-4}se{c}^{-1}$
• B. $10^{-6}se{c}^{-1}$
• C. $10^{-4}se{c}^{-1}$
• D. $4\times 10^{-4}se{c}^{-1}$

Answer 1

Answer: (C)

$E_a=2000\times 2.303R$
$log\frac{K_2}{K_1}=\frac{E_a}{2.303R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)$
$log\frac{K_2}{{\left(10\right)}^{-5}}=\frac{2000\times 2.303R}{2.303R}\left(\frac{1}{400}-\frac{1}{500}\right)$
$log\frac{K_2}{10^{-5}}=2000\times \frac{1}{2000}$
$log\frac{K_2}{10^{-5}}=1$
$\frac{K_2}{10^{-5}}=10\Rightarrow K_2=10\times 10^{-5}$
$=10^{-4}se{c}^{-1}$