c
For a reaction to be at equilibrium \(\Delta G=0\)

since \(\Delta G=\Delta H-T \Delta S\)

so at equilibrium \(\Delta H-T \Delta S=0\)

or \(\quad \Delta H=T \Delta S\)

For the reaction

\(\frac{1}{2} X_{2}+\frac{3}{2} Y_{2} \rightarrow X Y_{3}\)

\(\Delta H=-30 \,k J\) (given)

Calculating \(\Delta S\) for the above reaction, we get

\(\Delta S=50-\left[\frac{1}{2} \times 60+\frac{3}{2} \times 40\right] \,J K^{-1}\)

\(=50-(30+60) \,J \,K^{-1}\)

\(=-40 \,J \,K^{-1}\)

At equilibrium, \(T \Delta S=\Delta H\)

\([\text { as } \quad \Delta G=0]\)

\(\therefore \quad T \times(-40)=-30 \times 1000\)

\([\text { as } 1\, k J=1000 \,J]\)

\(\quad T=\frac{-30 \times 1000}{-40}=\quad 750 \,K\)