The standard enthalpies of combustion of C_(6)H_(6)(l), C("graphite") and H_(2)(g) are respectively -3270 kJ"mol"^(-1), -394 kJ"mol"^(-1) and -286 kJ"mol"^(-1). What is the standard enthalpy of formation of C_(6)H_(6)(l) in kJ"mol"^(-1)?

The standard enthalpies of combustion of C_(6)H_(6)(l), C("graphite")

1.	The standard enthalpies of combustion of C_(6)H_(6)(l), C("graphite") and H_(2)(g) are respectively -3270 kJ"mol"^(-1), -394 kJ"mol"^(-1) and -286 kJ"mol"^(-1). What is the standard enthalpy of formation of C_(6)H_(6)(l) in kJ"mol"^(-1)?
Answer» `-48` `+48` `-480` `+480` Solution :(i) `C_(6)H_(6)(l) + 15/2O_(2)(g) to 6CO_(2)(g) + 3H_(2)O(l)DeltaH = -3270 kJ"mol"^(-1)` (II) `C(gr) + O_(2)(g) to CO_(2)(g) DeltaH = -394 kJ"mol"^(-1)` (iii) `H_(2)(g) + 1/2O_(2)(g) to H_(2)O(l) DeltaH = -286 kJ"mol"^(-1)` The required equation is (iv) `6C(gr) + 3H_(2)(g) to C_(6)H_(6)(l),DeltaH` = ? Multiply eq.(ii) by 6 and eq(ii) by 3 and adding we get : (v) `6C(gr) + 15/2O_(2)(g) + 3H_(2)(g) to 6CO_(2)(g) + 3H_(2)O(l), DeltaH = -3222 kJ` Subtract eq.(i) from eq.(v) `6C(gr) + 3H_(2)(g) to C_(6)H_(6)(l) DeltaH = + 48 kJ"mol"^(-1)`.