b
\(\Delta S\) = \(E\)\(\Delta S\) (નીપજ) - \(E\)\(\Delta S\) (પ્રક્રિયક)

\(\Delta S{\mkern 1mu} {\mkern 1mu}  = {\mkern 1mu} {\mkern 1mu} \Delta {S_{(xy_3)}} - {\mkern 1mu} {\mkern 1mu} \left[ {\frac{1}{2}\Delta {S_{(x_2)}}{\mkern 1mu}  + {\mkern 1mu} {\mkern 1mu} \frac{3}{2}\Delta {S_{({y_2})}}{\mkern 1mu} } \right]{\mkern 1mu} \)

\(\therefore {\mkern 1mu} {\mkern 1mu} \Delta S{\mkern 1mu} {\mkern 1mu}  = {\mkern 1mu} {\mkern 1mu} 50{\mkern 1mu} {\mkern 1mu}  - {\mkern 1mu} {\mkern 1mu} \left( {\frac{{60}}{2} + \frac{{3 \times 40}}{2}} \right){\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \)

\(\therefore {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \Delta S{\mkern 1mu} {\mkern 1mu}  = {\mkern 1mu} {\mkern 1mu} 50 - (30 + 60)\)

\(\Delta S = -40\) જૂલ / મોલ . કેલ્વિન સંતુલને \(\Delta G = 0\)

\(\therefore \Delta H = T \Delta S\)

\(\therefore T\,\, = \,\frac{{\Delta {\text{H}}}}{{\Delta {\text{S}}}}\,\)  પરથી \(T\,\, = \,\,750\) કેલ્વિન મળે