(i) C(s)+O2(g)⟶CO2(g),∆H1=−393.5kJ∕mol (ii) H2(g)+
1
2
O2(g)⟶H2O(l),∆H2=−286.0kJ∕mol (iii) C6H6+
15
2
O2(g)⟶6CO2+3H2O+∆H3=−3267.0kJ∕mol (iv) 6C(s)+3H2(g)⟶C6H6∆H4= ?? We can get the required eqn by (i) ×6+ (ii) ×3− (iii) So, ∆H4=6×(−393.5)+3×(−286.0)−(−3267.0)=48kJ∕mol