Thank you for reporting, we will resolve it shortly
Q.
For a reaction consider the plot of $ln \, k$ versus $1/T$ given in the figure. If the rate constant of this reaction at $400\, K$ is $10^{-5} \,s^{-1}$, then the rate constant at $500\, K$ is
We know,
$
\ln \frac{ K _{2}}{ K _{1}}=\frac{ E _{ a }}{ R }\left[\frac{1}{ T _{1}}-\frac{1}{ T _{2}}\right]
$
Given that,
$
K _{1}=10^{-5} ; T _{1}=400 K ; T _{2}=500 K ; -\frac{ E _{ a }}{ R }=-4606 K
$
Now,
$
2.303 \log \frac{ K _{2}}{10^{-5}}=4606\left[\frac{1}{400}-\frac{1}{500}\right]
$
$
\Rightarrow K _{2}=10^{-4} s ^{-1}
$
