For a reaction to be spontaneous, the required conditions are
For a reaction to be spontaneous, the required conditions are
$\Delta_r H^{\circ}=-$ ve, $\Delta_r S^{\circ}=-$ ve, at high $T$.
$\Delta_r H^{\circ}=+$ ve, $\Delta_r S^{\circ}=+$ ve, at high $T$.
$\Delta_r H^{\circ}=+$ ve, $\Delta_r S^{\circ}=+$ ve, at low $T$.
$\Delta_r H^{\circ}=+$ ve, $\Delta_r S^{\circ}=-$ ve, at all $T$.
Solution
Depending upon the sign and magnitude of $\Delta H$ and $-T \Delta S$, the sum of these terms determines the sign of $\Delta G$. This sign of $\Delta G$ will determine i.e. if $\Delta G$ is negative then the reaction is spontaneous and if $\Delta G$ is positive then the reaction is non-spontaneous.
\begin{array}{|c|c|c|c|c|}
\hline\Delta H & \Delta S & -T \Delta S & \Delta G & Spontaneous/ Non-spontaneous \\
\hline+ & - & + & + & Non-spontaneous \\
\hline- & + & - & - & Spontaneous \\
\hline - & - & + & & \begin{array}{l}
Spontaneous at low T \\
Non-spontaneous at high T
\end{array} \\
\hline+ & + & - & +o r & \begin{array}{l}
Non-spontaneous at low T \\
Spontaneous at high T
\end{array} \\
\hline
\end{array}