Question

For the reaction, $aA+bB \rightarrow cC+dD$ , the plot of $logk$ vs $\frac{1}{ T}$ is given below:

The temperature at which the rate constant of the reaction is $10^{- 4}s^{- 1}$ is [Rounded off to the nearest integer) [Given: The rate constant of the reaction is $10^{- 5}s^{- 1}$ at $500\,K$ ]

Answer 1

$log_{10}K=log_{10}A-\frac{E_{a}}{2 . 303 RT}$
Slope $=\frac{E_{a}}{2 . 303 R}=-10000$
$log_{10}\frac{K_{2}}{ K_{1}}=\frac{E_{a}}{2 . 303 R}\times \left[\frac{1}{ T_{1}} - \frac{1}{ T_{2}}\right]$
$log_{10}\frac{10^{- 4}}{10^{- 5}}=10000\times \left[\frac{1}{500} - \frac{1}{ T}\right]$
$1=10000\times \left[\frac{1}{500} - \frac{1}{ T}\right]$
$\frac{1}{10000}=\frac{1}{500}-\frac{1}{ T}$
$\frac{1}{T}=\frac{1}{500}-\frac{1}{10000}$
$\frac{1}{T}=\frac{20 - 1}{10000}=\frac{19}{10000}$
$T=\frac{10 , 000}{19}\Rightarrow 526\,K$