The plot of logk f versus 1 T for a reversible reaction A ( g ) ⇌ P ( g ) is shown. Pre-exponential…

From the question
$\begin{array}{ll}{\mathrm{A}}_{\mathrm{f}}={10}^{15},{\mathrm{A}}_{\mathrm{b}}={10}^{11},& \mathrm{f}=\text{ Forward reaction }\\ \mathrm{logK}\text{ at }500\mathrm{K}=6& \mathrm{b}=\text{ Backward reaction }\\ {\mathrm{logk}}_{\mathrm{f}}\text{ at }500\mathrm{K}=9\text{ (from graph) }& \\ {\mathrm{logk}}_{\mathrm{b}}\text{ at }500\mathrm{K}:& \\ \mathrm{logK}=\log \left(\frac{{\mathrm{k}}_{\mathrm{f}}}{{\mathrm{k}}_{\mathrm{b}}}\right) \text{ since }\Rightarrow \mathrm{K}=\frac{{\mathrm{k}}_{\mathrm{f}}}{{\mathrm{k}}_{\mathrm{b}}}& \end{array}$

$6={\mathrm{logk}}_{\mathrm{f}}-{\mathrm{logk}}_{\mathrm{b}}$

$6=9-{\mathrm{logk}}_{\mathrm{b}}$

$\log {\mathrm{k}}_{\mathrm{b}}=3\text{ at }500\mathrm{K}$

$\log \frac{{\mathrm{k}}_2}{{\mathrm{k}}_1}=\frac{-{\mathrm{E}}_{\mathrm{a}}}{\mathrm{R}}\left(\frac{1}{{\mathrm{T}}_2}-\frac{1}{{\mathrm{T}}_1}\right)$

${\mathrm{k}}_{\mathrm{b}}={\mathrm{A}}_{\mathrm{b}}{\mathrm{e}}^{\frac{-{\mathrm{E}}_{\mathrm{ab}}}{\mathrm{RT}}}$

$\ln {\mathrm{k}}_{\mathrm{b}}=\ln \left({\mathrm{A}}_{\mathrm{b}}{\mathrm{e}}^{\frac{-{\mathrm{E}}_{\mathrm{ab}}}{\mathrm{RT}}}\right)$

${\mathrm{lnk}}_{\mathrm{b}}={\mathrm{lnA}}_{\mathrm{b}}-\frac{{\mathrm{E}}_{\mathrm{ab}}}{\mathrm{RT}}$

$2.303{\mathrm{logk}}_{\mathrm{b}}=2.303{\mathrm{logA}}_{\mathrm{b}}-\frac{{\mathrm{E}}_{\mathrm{ab}}}{500\mathrm{R}}$

$\frac{{\mathrm{E}}_{\mathrm{a}}}{500\mathrm{R}}=2.303\left({\mathrm{logA}}_{\mathrm{b}}-{\mathrm{logk}}_{\mathrm{b}}\right)$

$\frac{{\mathrm{E}}_{\mathrm{a}}}{500\mathrm{R}}=2.303\left(\log {10}^{11}-3\right)$

$\frac{{\mathrm{E}}_{\mathrm{a}}}{500\mathrm{R}}=2.303(11-3)=2.303\times 8$

${\mathrm{E}}_{\mathrm{a}}=2.303\times 8\times 500\mathrm{R}$

$\ln \left(\frac{{\mathrm{k}}_2}{{\mathrm{k}}_1}\right)=\frac{-{\mathrm{E}}_{\mathrm{a}}}{\mathrm{R}}\left(\frac{1}{ {\mathrm{T}}_2}-\frac{1}{ {\mathrm{T}}_1}\right)$

$\ln \left(\frac{{\mathrm{k}}_{250 \mathrm{K}}}{{\mathrm{k}}_{500 \mathrm{K}}}\right)=\frac{-{\mathrm{E}}_{\mathrm{a}}}{\mathrm{R}}\left(\frac{1}{250}-\frac{1}{500}\right)$

$\ln \left(\frac{{\mathrm{k}}_{250\mathrm{K}}}{{\mathrm{k}}_{500\mathrm{K}}}\right)=\frac{-2.303\times 8\times 500\mathrm{R}}{\mathrm{R}}\left(\frac{1}{500}\right)$

$2.303\left({\mathrm{logk}}_{250\mathrm{K}}-{\mathrm{logk}}_{500 \mathrm{K}}\right)=-2.303\times 8$

${\mathrm{logk}}_{250\mathrm{K}}-3=-8$

${\mathrm{logk}}_{250\mathrm{K}}=-5$

$\left|{\mathrm{logk}}_{250\mathrm{K}}\right|=5$