For a chemical process $\mathrm{A}\to \mathrm{B}$; $∆\mathrm{H}=45 \mathrm{KJ}/\mathrm{mole} \mathrm{and} ∆\mathrm{S}=15 {\mathrm{JK}}^{-1}{\mathrm{mole}}^{-1}$. The process is spontaneous at

At equilibrium, $∆\mathrm{G}=0$

$$\begin{gathered} ∆\mathrm{H}-\mathrm{T}∆\mathrm{S}=∆\mathrm{G}; \\ ∆\mathrm{H}=\mathrm{T}∆\mathrm{S} \\ {\mathrm{T}}_{\mathrm{eq}}=\frac{∆\mathrm{H}}{∆\mathrm{S}}=\frac{45\times {10}^3 \mathrm{J} {\mathrm{mol}}^{-1}}{15 \mathrm{J} {\mathrm{K}}^{-1}{\mathrm{mol}}^{-1}}=3000 \mathrm{K} \\ ∆\mathrm{G}=∆\mathrm{H}-\mathrm{T}∆\mathrm{S} \\ ∆\mathrm{G}=+\mathrm{ve} - \mathrm{T}\left(+\mathrm{ve}\right) \end{gathered}$$

$∆\mathrm{G}$ will be negative only at a  temperature higher than equilibrium temperature.