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)

Asked by  | 25th Dec, 2011, 06:17: PM

Expert Answer:

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 x–393.5 + 6x–187.6] – [2x49.04] = -5946 kJ

Now since there are two moles of benzene hence for one mole: -5946 kJ /2 = - 2973 kJ/mol C6H6

Answered by  | 26th Dec, 2011, 10:55: AM

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