For complete combustion of ethaol,
`C_(2)H_(5)OH(l)+3O_(2)(g)to2CO_(2)(g)+3H_(2)O(L)`
the amount of heat produced as measured in bomb calorimeter, is 1364.47 kJ `mol^(-1)` at `25^(@)C`. Assuming ideality the enthalpy of combustion, `Delta_(c)H`, for the reaction will be `(R=8.314kJmol^(-1))`

Options

A. `-1366.95kJ" "mol^(-1)`
B. `-1361.95kJ " "mol^(-1)`
C. `-1460.95kJ " "mol^(-1)`
D. `-1350.50kJ" "mol^(-1)`

Answer

A

Explanation

(a) `C_(2)H_(5)OH(l)+3O_(2)(g)to2CO_(2)(g)+3H_(2)O(l)`
Bomb calorimeter gives `DeltaU` of the reaction
Given, `DeltaU=-1364.47kJmol^(-1)`
`Deltan_(g)=-1`
`DeltaH=DeltaU+Deltan_(g)RT=-1364.47-(1xx8.314xx298)/(1000)`
`=-1366.93kJmol^(-1)`