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The correct thermodynamic conditions for the spontaneous reaction at all temperatures is
A
$\Delta H > 0$ and $\Delta S < 0$
B
$\Delta H 0$
C
$\Delta H < 0$ and $\Delta S < 0$
D
$\Delta H < 0$ and $\Delta S = 0$
Explanation
ΔG is negative at all T when ΔH < 0 and ΔS ≥ 0.
Detailed Solution
According to the Gibbs–Helmholtz equation, $\Delta G = \Delta H - T\Delta S$.
When $\Delta H 0$: $\Delta G = (-ve) - T(+ve) = -ve$ at all temperatures, so the reaction is spontaneous.
When $\Delta H < 0$ and $\Delta S = 0$: $\Delta G = (-ve) - T(0) = -ve$ at all temperatures.
Note: the source key accepts both '$\Delta H 0$' and '$\Delta H < 0$ and $\Delta S = 0$'.
When $\Delta H 0$: $\Delta G = (-ve) - T(+ve) = -ve$ at all temperatures, so the reaction is spontaneous.
When $\Delta H < 0$ and $\Delta S = 0$: $\Delta G = (-ve) - T(0) = -ve$ at all temperatures.
Note: the source key accepts both '$\Delta H 0$' and '$\Delta H < 0$ and $\Delta S = 0$'.