O2⟶6CO2+3H2O ∆H=−3270kJmol−1 (ii) C(graphite) +O2⟶CO2 ∆H=−394kJmol−1 (iii) H2+
1
2
O2⟶H2O ∆H=−286kJmol−1 Multiplication of Eq. (ii) by 6 and Eq. (iii) by 3 , gives (iv) 6C(s)+6O2⟶6CO2 ∆H=−394×6kJmol−1 =−2364kJmol−1 (v) 3H2+
3
2
O2⟶3H2O ∆H=−286×3kJmol−1 =−858kJmol−1 On inverting Eq, (i), we get (vi) 6CO2+3H2O⟶C6H6+
15
2
O2 ∆H=+3270kJmol−1 Addition of Eq. (iv), (v) and (vi) gives 6C(s)+3H2⟶C6H6 ∆H=+3270+(−2364−858) =+48kJmol−1 Thus, the standard enthalpy of formation of C6H6,(∆fHC6H6)is+48kJmol−1