For the reaction, $\mathrm{X}\to \mathrm{Y}$,   $\mathrm{\Delta H}=-36\mathrm{kJ}/\mathrm{mole}\text{\hspace{0.17em}; \hspace{0.17em}}\mathrm{\Delta S}=-90{\mathrm{Jk}}^{-1}{\mathrm{mole}}^{-1}$ the reaction is spontaneous ( $\mathrm{\Delta H}\&\mathrm{\Delta S}$ do not vary with temperature) at

$\mathrm{\Delta G}=\mathrm{\Delta H}-\mathrm{T\Delta S}$
At equilibrium, $∆\mathrm{G}=0$

$\begin{array}{l}\therefore {\mathrm{T}}_{\mathrm{eq}}=\frac{\mathrm{\Delta H}}{\mathrm{\Delta S}}\\ {\mathrm{T}}_{\mathrm{eq}}=\frac{-36\times {10}^3{\mathrm{Jmole}}^{-1}}{-90{\mathrm{JK}}^{-1}{\mathrm{mole}}^{-1}}\\ {\mathrm{T}}_{\mathrm{eq}}=400\mathrm{K}\\ \mathrm{\Delta G}=\mathrm{\Delta H}-\mathrm{T\Delta S}\\ \mathrm{\Delta G}=\left(-\mathrm{ve}\right)-\mathrm{T}\left(-\mathrm{ve}\right)\end{array}$
$∆\mathrm{G}$  will be negative only when T is less than ${\mathrm{T}}_{\mathrm{eq}}$