A) \[\Delta S>0\]
B) \[\Delta S<0\]
C) \[\Delta G>0\]
D) \[\Delta E=0\]
Correct Answer: A
Solution :
Process is endothermic i.e., \[\Delta H=+ve\] and process is also spontaneous i.e., \[\Delta G=-ve\] Hence, from Gibbs-Helmholtz equation \[\Delta G=\Delta H-T.\Delta S\] \[\Delta S\] must be positive so that \[\Delta G\] may be negative.
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