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, $T \Delta S=\Delta H$

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

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

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

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