The standard enthalpies of combustion ofC_(6)H_(6(l)),C_(("graphite"))and H_(2(g))are respectively -3270 kJmol^(-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 ofC_(6)H_(6(l)),C_(("graphite"))

1.	The standard enthalpies of combustion ofC_(6)H_(6(l)),C_(("graphite"))and H_(2(g))are respectively -3270 kJmol^(-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 :`C_(6)H_(6(L))+(15)/(2)O_(2(g))rarr6CO_(2(g))+3H_(2)O_((l)), DeltaH=-3270 kJ mol^(-1)"...(i)"` `C_((gr))+O_(2(g))rarrCO_(2(g)), DeltaH=-394 kJ mol^(-1)"....(II)"` `H_(2(g))+(1)/(2)O_(2(g))rarrH_(2)O_((l)), DeltaH=-286 kJ mol^(-1)"...(iii)"` Formation of `C_(6)H_(6)` `6C_((gr))+3H_(2(g))rarrC_(6)H_(6(l)),DeltaH=? "...(iv)"` By multiplying eq. (ii) with 6 and eq. (iii) with 3 and adding we get, `6C_((gr))+6O_(2(g))+3H_(2(g))+(3)/(2)O_(2(g))rarr6CO_(2(g))+6H_(2)O_((l)) "...(V)"` `DeltaH=6(-394)+3(-284)=(-2364)+(-858)=-3222 kJ//mol` Now, by substracting eq. (i), from (v) we get `6C_((gr))+3H_(2(g))rarrC_(6)H_(6(l))` `DeltaH=-3222-(-3270)=+48kJ//mol`