$\frac{{\log \,\,{K_2}}}{{\log \,\,{K_1}}}\, = \,\frac{{ - \,{E_a}}}{{2.303\,R}}\left( {\frac{1}{{{T_2}}}\, - \,\frac{1}{{{T_1}}}} \right)$
Given $\frac{{\log \,\,{K_2}}}{{\log \,\,{K_1}}}\, = 2$ $T_2\,=\,308$: $T_1\,=\,298$
$\therefore \,\,\log \,\,2\,\, = \frac{{ - \,{E_a}}}{{2.303\, \times 8.314}}\left( {\frac{1}{{308}}\, - \,\frac{1}{{298}}} \right)$
${E_a}\, = \,52.9\,\,kJ\,\,mo{l^{ - 1}}$
[લો; $R =8.314 \,J\, mol ^{-1}\, K ^{-1}$ In $3.555=1.268$]
$2X \rightleftharpoons {X_2}$
${X_2} + Y \to {X_2}Y\,\left( {slow} \right)$
તો પ્રક્રિયાકમ જણાવો.