A) 285.7 K
B) 273 K
C) 450 K
D) 300 K
Correct Answer: A
Solution :
At equilibrium Gibbs free energy change \[(\Delta {{G}^{o}})\] is equal to zero. The following thermodynamic relation is used to show the relation of \[\Delta {{G}^{o}}\] with enthalpy change \[(\Delta {{H}^{o}})\]and entropy change \[(\Delta {{S}^{o}})\] \[\Delta {{G}^{o}}=\Delta {{H}^{o}}-T\Delta S\] \[=30\times {{10}^{3}}\,(J\,mo{{l}^{-1}})-T\times 105\,(J{{K}^{-1}}\,mo{{l}^{-1}})\] \[\therefore \] \[T=\frac{30\times {{10}^{3}}}{105}K=285.71\,K\]You need to login to perform this action.
You will be redirected in
3 sec