For the complete combustion of ethanol, `C_(2)H_(5)OH(l)+3O_(2)(g)rarr2CO_(2)(g)+3H_(2)O(l)` the amount of heat produced as measured in bomb calorimeter is `1364.47KJ mol^(-1)` at `25^(@)C` . Assuming ideality, the enthalpy of combustion, `DeltaH_(C)` , for the reaction will be
`[R=8.314JK^(-1)mol^(-1)]`

Options

A. `-1350.50 kJ mol^(-1)`
B. `- 1366.95kJ mol^(-1)`
C. `-1361.95kJ mol^(-1)`
D. `-1460 .50 kJ mol^(-1)`

Answer

B

Explanation

Given `DeltaU = - 1364.47 kJ mol^(-1)`
`Deltan_(g) = ( n_(p) -n_(r))_("gaseous") = 2-3= -1`
`T = 25 +273= 198 K`
`DeltaH = DeltaU + Deltan_(g) RT`
`= - 1364.47kJ + ( -1)`
`( 8.314 xx 10^(-3) kJ K^(-1) mol^(-1))( 298K)`
`= - 1364 .47 - 1 xx 8.314 xx 10^(-3) xx 298K`
`= - 1364.47 -2477.57 xx 10^(-3)`
`= - 1364.47 - 2.477 = - 1366.95kJ mol^(-1)`