Question

The equilibrium constant for the reaction
$H_2\left(g\right)+S\left(s\right)\rightleftharpoons H_{2}S\left(g\right)$
is $18.5$ at $925K$ and $9.25$ at $1000K$ respectively. Calculate the enthalpy of the reaction.

Answer 1

Using the relation,
$\log \frac{K_2}{K_1}=\frac{\Delta H}{2.303R}\left[\frac{T_2-T_1}{T_2{T}_1}\right]$
$\log \frac{9.25}{18.5}=\frac{\Delta H}{2.303\times 8.314J/mol/K}\times \frac{1000K-925K}{925K\times 1000K}$
$-0.301=\frac{\Delta H\times 75}{2.303\times 8.314\times 925\times 1000}$
$\Delta H=-71081J{mol}^{-1}$
Thus, the enthalpy change for the reaction is $-71081J{mol}^{-1}$.