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.

The standard enthalpies of formation of C_2H_5OH_((l)), CO_(2(g)) and

1.	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.
Answer» SOLUTION :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