Question

The plot of ${\log }_{10}{K}_p$ against $1/T$ for the reaction:
$S{O}_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)\rightleftharpoons S{O}_3\left(g\right)$
is a straight line with slope $=4.95\times {10}^3$. If the $K_p$ at $25$ if standard entropies for $S{O}_2\left(g\right),O_2\left(g\right)$ and $S{O}_3\left(g\right)$ are $248.2,205.1$ and $256.8J{K}^{-1}{mol}^{-1}$ at $25$ respectively is $K_p=X\times {10}^{11}$ then $10X$ is___________.

Answer 1

$d\ln K_p=\frac{\Delta H}{R{T}^2}$
On integration $\ln K_p=-\frac{\Delta H}{RT}+C$
or ${\log }_{10}{K}_p=-\frac{\Delta H}{2.303RT}+C$
Thus, slope $=-\frac{\Delta H}{2.303R}=4.95\times {10}^3$
$\therefore \Delta H=-2.303\times 8.314\times 4.95\times {10}^3$
$=-9.48\times {10}^{4}J$
Also, $\Delta S=S_{S{O}_3}-S_{S{O}_2}-\frac{1}{2}\times S_{O_2}$
$=256.8-248.2-\frac{1}{2}\times 205.1$
$=-93.95J{K}^{-1}{mol}^{-1}$
$\therefore \Delta G=\Delta H-T\Delta S$
$=-9.48\times {10}^4-298\times (-93.95)$
$=-6.68\times {10}^4$
Now, $\Delta G=-2.303RT\log K_p$
$-6.68\times {10}^4=-2.303\times 8.314\times 298\times \log K_p$
$K_p=5.1\times {10}^{11}$