Question

The difference between heats of reaction at constant pressure and at constanl volume for the reaction

\[\ce{2C6H6_{(l)} + 15O2_{(g)} -> 12CO2_{(g)} + 6H2O_{(l)}}\] at 25°C in kJ

Options

• −7.43

• +3.72

• −3.72

• +7.43

Answer 1

−7.43

Explanation:

∆H − ∆U = ∆n_gRT

= −3 mol × 8.314 J K⁻¹ mol⁻¹ × 298 K

= −7432 J

= −7.43 kJ