A) \[-25.1\,\,kJ\,\,mo{{l}^{-1}}\]
B) \[+25.1\,\,kJ\,\,mo{{l}^{-1}}\]
C) \[18.7\,\,kJ\,\,mo{{l}^{-1}}\]
D) \[-69.4\,\,kJ\,\,mo{{l}^{-1}}\]
Correct Answer: D
Solution :
\[\because \,\,\Delta {{H}^{o}}\] and \[\Delta {{S}^{o}}\] remain constant in the given temperature range. \[\therefore \] \[\Delta {{S}^{o}}=-\frac{\Delta {{G}^{o}}-\Delta {{H}^{-}}}{T}\] \[=-\left( \frac{-66.9+4.18}{300} \right)\] \[=0.08367\,kJ\,{{K}^{-1}}mo{{l}^{-1}}\] \[\therefore \] \[\Delta G_{330}^{o}=\Delta {{H}^{o}}-T\Delta {{S}^{o}}\] \[=-41.8-330\times 0.08367\] \[=-69.4\,kJ\,mo{{l}^{-1}}\]
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