Question

The enthalpy change for the reaction, \[\ce{C2H4_{(g)} + H2_{(g)} -> C2H6_{(g)}}\] is −620 J when 100 mL of ethylene and 100 ml of \[\ce{H2}\] react at 1 bar pressure. Calculate the pressure volume type of work and ΔU for the reaction.

Answer 1

Given: 

Enthalpy change (ΔH) = −620 J

Volumes of reactants; \[\ce{C2H4}\] = 100 ml, \[\ce{H2}\] = 100 ml

Pressure (P_ext) = 1 bar

To find:

Pressure volume work (W) and change in internal energy (ΔU) for the given reaction

Formulae: 

1. W = −P_ext ΔV
2. ΔH = ΔU + P_ext ΔV

Calculation:

According to the equation of reaction 1 mole of \[\ce{C2H4}\] reacts with 1 mole of \[\ce{H2}\] to produce 1 mole of \[\ce{C2H6}\]. Hence, 100 ml of \[\ce{C2H4}\] would react with 100 ml of \[\ce{H2}\] to produce 100 ml of \[\ce{C2H6}\].

V₁ = 100 mL + 100 mL = 200 mL = 0.2 dm³

V₂ = 100 mL = 0.1 dm³

From formula (1),

W = −P_ext ΔV

= −1 bar (0.1 dm³ − 0.2 dm³)

= 0.10 dm³ bar

= 0.10 dm³ bar × 100 `"J"/("dm"^3 "bar")` = + 10.00 J

∴ − P_ext = ΔV = 10.00 J

∴ P_ext = ΔV = −10.00 J

From formula (2),

ΔH = ΔU + P_ext ΔV

∴ ΔU = ΔH − P_ext ΔV

= −620 − (−10.00 J) = −610 J

Pressure volume work (W) = +10.00 J and ΔU = −610 J