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Which of the following condition depicts a favourable condition for a spontaneous reaction?


(a) $T\Delta S > \Delta H, \Delta H = +ve, \Delta S = +ve$
(b) $T\Delta S > \Delta H, \Delta H = +ve, \Delta S = -ve$
(c) $T\Delta S = \Delta H, \Delta H = -ve, \Delta S = -ve$
(d) $T\Delta S = \Delta H, \Delta H = +ve, \Delta S = +ve$

1 Answer

Toolbox:
  • $\Delta G = \Delta H - T\Delta S$
  • For a process to be spontaneous, $\Delta G$ should be negative.
Answer: $T\Delta S > \Delta H, \Delta H = +ve, \Delta S = +ve$
 
We know, $\Delta G = \Delta H - T\Delta S$
For $\Delta G$ to be negative, $0>\Delta G$ i.e. or $0>\Delta H-T\Delta S$
for $0>\Delta H-T\Delta S$ , the following conditions must exist:
(i) $\Delta H = +ve$
(ii) $\Delta S = +ve$
(iii) $T\Delta S > \Delta H$
answered Feb 25, 2014 by mosymeow_1
edited Feb 25, 2014 by mosymeow_1
 

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