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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) $\Delta H < 0$ and $\Delta S < 0$

(c) $\Delta H < 0$ and $\Delta S = 0$

(d) $\Delta H > 0$ and $\Delta S < 0$
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Ans: (a, c)

Sol: $ΔG = ΔH - TΔS$

If $ΔH < 0$ and $ΔS > 0$

$ΔG = (-\text{ve}) - T(+\text{ve})$

then at all temperatures, $ΔG = -\text{ve}$, spontaneous reaction.

If $ΔH < 0$ and $ΔS = 0$

$ΔG = (-\text{ve}) - T(0) = -\text{ve}$ at all temperatures.
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As we know,

ΔG = ΔH - TΔS

For the reaction to be spontaneous:

ΔG = -ve

ΔG is negative at all temperature when:

ΔG(-ve) = ΔH(-ve)  - T ΔS(+ve)

i.e., ΔH < 0 and ΔS > 0

Also, when ΔS = 0, ΔG = -ve

ΔG(-ve) = ΔH(-ve) -T(0)

Ans: (A) , (C)

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