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