The plot of logk _ f versus (1)/(T) for a reversible reaction A ( g...

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

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

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

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

$log \dfrac{{k}_{2}}{{k}_{1}} = \dfrac{- {E}_{a}}{R}\left(\dfrac{1}{{T}_{2}} - \dfrac{1}{{T}_{1}}\right)$

${k}_{b} = {A}_{b}e\dfrac{- {E}_{ab}}{RT}$

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

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

$2 . 303{logk}_{b} = 2 . 303{logA}_{b} - \dfrac{{E}_{ab}}{500R}$

$\dfrac{{E}_{a}}{500R} = 2 . 303\left({logA}_{b} - {logk}_{b}\right)$

$\dfrac{{E}_{a}}{500R} = 2 . 303\left(log{10}^{11} - 3\right)$

$\dfrac{{E}_{a}}{500R} = 2 . 303 ( 11 - 3 ) = 2 . 303 \times 8$

${E}_{a} = 2 . 303 \times 8 \times 500R$

$\ln \left(\dfrac{{k}_{2}}{{k}_{1}}\right) = \dfrac{- {E}_{a}}{R}\left(\dfrac{1}{{T}_{2}} - \dfrac{1}{{T}_{1}}\right)$

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

$\ln \left(\dfrac{{k}_{250K}}{{k}_{500K}}\right) = \dfrac{- 2 . 303 \times 8 \times 500R}{R}\left(\dfrac{1}{500}\right)$

$2 . 303\left({logk}_{250K} - {logk}_{500K}\right) = - 2 . 303 \times 8$

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

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

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