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A reaction occurs spontaneously if -
A
$T\Delta S \gt \Delta H$ and $\Delta H$ is +ve and $\Delta S$ is $-$ve
B
$T\Delta S = \Delta H$ and both $\Delta H$ and $\Delta S$ are +ve
C
$T\Delta S \lt \Delta H$ and both $\Delta H$ and $\Delta S$ are +ve
D
$T\Delta S \gt \Delta H$ and both $\Delta H$ and $\Delta S$ are +ve
Detailed Solution
A reaction is spontaneous when $\Delta G = \Delta H - T\Delta S \lt 0$.
If both $\Delta H$ and $\Delta S$ are positive, $\Delta G$ is negative only when $T\Delta S \gt \Delta H$, i.e. at a sufficiently high temperature.
So '$T\Delta S \gt \Delta H$ and both $\Delta H$ and $\Delta S$ are +ve' gives a spontaneous reaction.
If $T\Delta S = \Delta H$, then $\Delta G = 0$ and the system is at equilibrium.
If $T\Delta S \lt \Delta H$ with both positive, $\Delta G \gt 0$ and the reaction is non-spontaneous.
If $\Delta H$ is positive and $\Delta S$ is negative, $\Delta G$ is always positive, and $T\Delta S \gt \Delta H$ is impossible.
So the reaction occurs spontaneously if $T\Delta S \gt \Delta H$ and both $\Delta H$ and $\Delta S$ are positive.
If both $\Delta H$ and $\Delta S$ are positive, $\Delta G$ is negative only when $T\Delta S \gt \Delta H$, i.e. at a sufficiently high temperature.
So '$T\Delta S \gt \Delta H$ and both $\Delta H$ and $\Delta S$ are +ve' gives a spontaneous reaction.
If $T\Delta S = \Delta H$, then $\Delta G = 0$ and the system is at equilibrium.
If $T\Delta S \lt \Delta H$ with both positive, $\Delta G \gt 0$ and the reaction is non-spontaneous.
If $\Delta H$ is positive and $\Delta S$ is negative, $\Delta G$ is always positive, and $T\Delta S \gt \Delta H$ is impossible.
So the reaction occurs spontaneously if $T\Delta S \gt \Delta H$ and both $\Delta H$ and $\Delta S$ are positive.