2 C6H6 (l) + 15 O2 (g) ? 12 CO2 (g) + 6H2O(l)
The difference between heats of reaction at constant pressure and constant volume for the reaction 2 C6H6 (l) + 15 O2 (g) ? 12 CO2 (g) + 6H2O(l)
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water.
2C6H6 (l) + 15O2 (g) -> 12CO2 (g) + 6H2O (l)
Heat released per mole of benzene can be calculated as:
The standard enthalpy of formation of benzene is 49.04 kJ/mol.
?H0 = ? n?H0f (products) - ? m?H0f (reactants)
?H0 = [12?H0 (CO2) + 6?H0 (H2O)] - [ 2 ?H0 (C6H6)]
?H0 = [12 x393.5 + 6x187.6] [2x49.04] = -5946 kJ
Now since there are two moles of benzene hence for one mole: -5946 kJ /2 = - 2973 kJ/mol C6H6