\( = \Delta {{\text{S}}^ \circ }_{X{Y_3}} - [\frac{1}{2}\Delta {S^ \circ }_{{X_2}} + \frac{3}{2}\Delta {S^ \circ }_{{Y_2}} = 50 - [\frac{1}{2} \times 60 + \frac{3}{2} \times 40]\, = 50 - [30 + 60]\, = - 40\,\,sq \div \,\,{{\text{K}}^{{\text{ - 1}}}}\,\)
\(\Delta {\text{G}} = \Delta H - T\Delta S\) સંતુલને \(\Delta {\text{G}} = {\text{0}}\,\,\,\,\,\therefore \,\,\Delta {\text{H}} = {\text{T}}\Delta {\text{S}}\,\,\,\,\,\,{\text{T}} = \frac{{\Delta {\text{H}}}}{{\Delta {\text{S}}}} = \frac{{ - 3000}}{{ - 40}} = 750\,K\)
$CaO_{(s)}\,\, + \,\,{H_2}O_{(l)}\,\, \to \,\,Ca{(OH)_2}_{(s)}\,;\,\,\,........(i)$ $\,\Delta {H_{1.8\,^oC}} = \,\, - \,\,15.26\,\,K\,cal$
$H_2O_{(l)}\,$ $ \to $ ${H_{2{(g)}}}$ $+$ $\frac{1}{2}O_{2(g)}$ $\,\Delta {H_{1.8\,^oC}} = \,\, - \,\,68.37\,\,K\,cal$
$Ca_{(s)} + \frac{1}{2}O_{2(g)} = CaO_{(s)}$ $\,\Delta {H_{1.8\,^oC}} = \,\, \,\,-151.80\,\,K\,cal$
કારણ : અચળ તાપમાન અને દબાણે રાસાયણિક પ્રક્રિયા ગીબ્સ ઉર્જાના ઘટાડાની દિશામાં સ્વયંભુ થાય છે.