• # question_answer For the complete combustion of ethanol amount of heat produced as measured in bomb calorimeter, is $1364.47\text{ kJ }mo{{l}^{-1}}$ at $25{}^\circ C$. Assuming ideality the enthalpy of combustion, $\Delta {{H}_{C}},$ for the reaction will be$(R=8.314J{{K}^{-1}}mo{{l}^{-1}})$ A) $=1366.95\,kJ\,mo{{l}^{-1}}$    B) $=1361.95\,kJ\,mo{{l}^{-1}}$ C) $-1460.50kJ\,mo{{l}^{-1}}$        D) $-1350.50\,kJ\,mo{{l}^{-1}}$

${{\Delta }_{r}}{{G}^{0}}={{\Delta }_{f}}{{G}^{0}}$(products)$-{{\Delta }_{f}}{{G}^{0}}$(reactants) $=-394.4-(-137.2)=-257.2kJ<0$ The above negative value of$\Delta G$indicates that the process is spontaneous. Also $\Delta {{G}^{0}}=\Delta {{H}^{0}}-T\Delta {{G}^{0}}\Rightarrow \Delta {{H}^{0}}=\Delta {{G}^{0}}+T\Delta {{S}^{0}}$ $=-257.2+300(-0.094)=-285.4kJ<0$