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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.
Solution
Given:
Enthalpy change (ΔH) = −620 J
Volumes of reactants; \[\ce{C2H4}\] = 100 ml, \[\ce{H2}\] = 100 ml
Pressure (Pext) = 1 bar
To find:
Pressure volume work (W) and change in internal energy (ΔU) for the given reaction
Formulae:
- W = −Pext ΔV
- ΔH = ΔU + Pext 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 = −Pext 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
∴ − Pext = ΔV = 10.00 J
∴ Pext = ΔV = −10.00 J
From formula (2),
ΔH = ΔU + Pext ΔV
∴ ΔU = ΔH − Pext ΔV
= −620 − (−10.00 J) = −610 J
Pressure volume work (W) = +10.00 J and ΔU = −610 J
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