Combustion equation of benzene for 2 moles-

2 C6H6 (l) + 15 O2 (g) 12 CO2 (g) +6 H2O , H= ?

H = (Enthalpy of products - Enthalpy of reactants)

=(12* del H(f) CO2(g) +6* del H(f) H2O)   -  (2* del H(f) C6H6(l) + 0)

=(12*  -94.3    + 6*   -68.4)   -  (2*   +11.7)

=    -1565.4 K. Cal

H= qp

E= qv

H= E +nRT

-1565.4= E  +(12  -  15)0.002 * 298

E=   -1563.6