A) 55.0
B) 60.0
C) 68.5
D) 120.0
Correct Answer: B
Solution :
Gibbs free energy. \[\Delta G=\Delta H-T\Delta S\] But, at equilibrium, \[\Delta G=0\] \[\therefore \] \[\Delta H=T\Delta S\] or \[\Delta S=\frac{\Delta H}{T}=\frac{21.3\times {{10}^{3}}}{355}=60\,J{{K}^{-1}}mo{{l}^{-1}}\]You need to login to perform this action.
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