For complete combustion of ethanol, `C_(2)H_(5)OH(l)+3O_(2)(g)rarr2CO_(2)(g)+3H_(2)O(l)`, the amount of heta produced as measured in bomb calorimeter, is 1364.47 kJ `mol^(-1)` at `25^(@)`. Assuming ideality the Enthalpy of combustion. `Delta_(C)H`, for the reaction will be `(R=8.314 kJ mol^(-1))`A. `-1366. 95 kJ mol^(-1)`B. `-1361.95 kJ mol^(-1)`C. `-1460.50 kJ mol^(-1)`D. `-1350.50 kJ mol^(-1)`

For complete combustion of ethanol, `C_(2)H_(5)OH(l)+3O_(2)(g)rarr2CO_

1.	For complete combustion of ethanol, `C_(2)H_(5)OH(l)+3O_(2)(g)rarr2CO_(2)(g)+3H_(2)O(l)`, the amount of heta produced as measured in bomb calorimeter, is 1364.47 kJ `mol^(-1)` at `25^(@)`. Assuming ideality the Enthalpy of combustion. `Delta_(C)H`, for the reaction will be `(R=8.314 kJ mol^(-1))`A. `-1366. 95 kJ mol^(-1)`B. `-1361.95 kJ mol^(-1)`C. `-1460.50 kJ mol^(-1)`D. `-1350.50 kJ mol^(-1)`
Answer» Correct Answer - A `C_(2)H_(5)OH(l)+3O_(2)(g)rarr2CO_(2)(g)+3H_(2)O(l)` `DeltaU=-1364.47 KJ//mol` `DeltaH=DeltaU+Deltan_(g)RT` `Deltan_(g)=-1` `DeltaH=-1364.47+[(-1xx8.314xx298)/(1000)]` `=-1364.47-2.4776` `=-1366.94 KJ//mol`.

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