Thermodynamics

Thermodynamics

Question

The combustion of benzene (l) gives CO2(g) and H2O(l). Given that heat of combustion of benzene at constant volume is –3263.9 kJ mol–1 at 250C; the heat of combustion (in kJ mol–1) of benzene at constant pressure will be (R = 8.314 JK–1 mol–1)

  • –3267.6

  • 4152.6

  • –452.46

  • 3260

Answer

A.

–3267.6

C6H6 (l) + 152 O2 (g)  6CO2 (g) +3H2O (l)  

Δng = 6 - 7.5
= -1.5 (change in gaseous mole)
ΔU or ΔE = - 3263.9 kJ
ΔH = ΔU + ΔngRT
Δng = - 1.5
R = 8.314 JK-1 mol-1
T = 298 K
So ΔH = -3263.9 + (-1.5) 8.314 x 10-3 x 298
= -3267.6 kJ
ΔH = Heat at constant pressure
ΔU/ΔE = Heat at constant volume
R = gas constant

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