Chemistry - Chemical Equilibrium Question with Solution | TestHub

$\mathrm{\Delta }\mathrm{G}°=-2.303\mathrm{R}\mathrm{T}\log \mathrm{K}$
$\log \mathrm{K}=-\frac{\mathrm{\Delta }\mathrm{G}°}{2.303\mathrm{R}\mathrm{T}}=-\frac{\left(-1.7\times {10}^3\right)}{2.303\times 8.314\times 298}=\log \mathrm{K}=+0.298$
${\mathrm{K}}_1=\sqrt{\mathrm{K}}=0.298$
${\mathrm{H}}_2+{\mathrm{I}}_2\rightleftharpoons 2\mathrm{H}\mathrm{I} {\mathrm{K}}_1=(0.504{)}^2$
$2\mathrm{H}\mathrm{I}\rightleftharpoons {\mathrm{H}}_2+{\mathrm{I}}_2 {\text{K}}_2=?$
${\mathrm{K}}_2=\frac{1}{{\mathrm{K}}_1}=\frac{1}{(0.504{)}^2}=3.6$