At 27oC for the combustion of two mole of benzene, ∆H-∆E is equal to

2C6H6(l)+15O2(g)→ 12CO2(g)+6H2O(l); ∆ng=12-15=-3; ∆H-∆E=∆nRT;                  =-3(8.314×10-3KJmol-1K-1)(300 K) ;                  =-7.48kJ ;