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. `-1366.95KJmol^(-1)`
B. `-1361.95KJmol^(-1)`
C. `-1460.50KJmol^(-1)`
D. `-1350KJmol^(-1)`

Answer

A

Explanation

`C_(2)H_(5)OH(l)+30_(2)rarr2CO_(2)(g)+3H_(2)O(l)`
Amount of heat produced in bomb calorimeter,
`DeltaU=-1364.47KJ mol^(-1)`
Enthalpy of a combustion reaction is
`DeltaH=DeltaU+Delta_(g)RT`
where `DeltaU` =internal energy
`Deltan_(g)` =moles of gas (producets-reactants)
`DeltaR` =Gas constant
`T` =Temperature in `K`
As per euation,
`Deltan_(g)=2-3=-1`
`T=25^(@)C=25+273K`
`implies T=298K`
Thus, `DeltaH=-1364.47+[((-1)xx8.314xx298)/(1000)]`
`DeltaH=-1364.47-2.477=-1366.947KJ mol^(-1)`
Hence, the enthalpy of combustion, `DeltaH_(c)` , for the given reaction will be `-1366.947KJ mol^(-1)` .