Question

For the reaction, $a A + bB \rightarrow cC + dD ,$ the plot of log $k$ vs $\frac{1}{ T }$ is given below :

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

Answer 1

$\log K =\log A -\frac{ Ea }{2.303 RT }$

Slope $=\frac{ Ea }{2.303 R }=10,000$

$\log \left(\frac{ K _{2}}{ K _{1}}\right)=\frac{ Ea }{2.303 R }\left(\frac{1}{ T _{1}}-\frac{1}{ T _{2}}\right)$

$\log \left(\frac{10^{-4}}{10^{-5}}\right)=10,000\left[\frac{1}{500}-\frac{1}{ T _{2}}\right]$

$T _{2}=526.31 \simeq 526 \,K$