The standard enthalpies of formation of `C_2H_5OH_((l)), CO_(2(g))` and `H_2O_((l))` are -277 , -393.5 and `-285.5 "kJ mol"^(-1)` respectively .
Calculate the standard enthalpy change for the reaction
`C_2H_5OH_((l)) +3O_(2(g)) to 2CO_(2(g)) +3H_2O_((l))`
The enthalpy of formation of `O_(2(g))` in the standard state is zero , by definition.

Explanation

The standard enthalpy change for the combustion of ethanol can be calculated from the standard enthalpies of formation of `C_2H_5OH_((l)), CO_(2(g))` and `H_2O_((l))` The enthalpies of formation are -277,-393.5 and -285.5 kJ `"mol"^(-1)` respectively.
`C_3H_5OH_((l))+3O_(2(g)) to 2CO_(2(g)) +3H_2O_((l))`
`DeltaH_r^0=[(DeltaH_f^0)_"products"-(DeltaH_f^0)_"reactants"]`
`DeltaH_r^0=[2(DeltaH_f^0)_(CO_2)+3(DeltaH_f^0)_(H_2O)]-[1(DeltaH_f^0)_(C_2H_5OH)+3(DeltaH_f^0)_(O_2)]`
`DeltaH_r^0=["2 mol (-393.5) kJ mol"^(-1) + "3 mol (-285.5 ) kJ mol"^(-1)]-["1 mol (-277) kJ mol"^(-1) +"3 mol (0) kJ mol"^(-1)]`
=[-787-856.5]-[-277]
=-1643.5+277
`DeltaH_r^0` =-1366.5 KJ