Calculate the enthalpy of combustion of ethylene at 300 K at constant pressure, If Its heat of combustion at constant volume (∆U) is – 1406 KJ.
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The complete ethylene combustion reaction can be written as,
C2H4(g) + 3O2(g) → 2CO2(g) + 2H2O(I)
∆U = -1406 kJ
∆n = np(g) – nr(g)
∆n = 2 – 4 = -2
∆H = U + RT ∆n(g)
∆H = -1406 + (8.314 x 10-3 x 300 x (-2))
∆H = – 1410.9 kJ