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) \[-1350.50\,kJ\,mo{{l}^{-1}}\]
Correct Answer: A
Solution :
\[{{\Delta }_{r}}{{G}^{0}}={{\Delta }_{f}}{{G}^{0}}\](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 \[\Delta {{G}^{0}}=\Delta {{H}^{0}}-T\Delta {{G}^{0}}\Rightarrow \Delta {{H}^{0}}=\Delta {{G}^{0}}+T\Delta {{S}^{0}}\] \[=-257.2+300(-0.094)=-285.4kJ<0\]
You need to login to perform this action.
You will be redirected in
3 sec
