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Question:

The combustion of benzene (l) gives CO2(g) and H2O(l). Given that the heat of combustion of benzene at constant volume is -3267.6 kJ/mol at 25°C, the heat of combustion (in kJ/mol) of benzene at constant pressure will be: (R = 8.314 J K⁻¹ mol⁻¹)

-3267.46

+3260

-3267.6

+4152.6

Solution:

C6H6(l) + 15/2O2(g) → 6CO2(g) + 3H2O(l)
ΔH = ΔU + ΔnRT = -3267.6 + (6 - 15/2) × 8.314 × 298 / 1000 = -3267.6 + (-1.5) × 8.314 × 298 / 1000 = -3267.6 - 3.717 = -3271.317 kJ/mol ≈ -3271.3 kJ/mol
Hence, option B is correct.