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The enthalpy change for the reaction, CA2HA4A(g)+HA2A(g)⟶CA2HA6A(g) is −620 J when 100 mL of ethylene and 100 ml of HA2 react at 1 bar pressure. Calculate the pressure volume type of work and ΔU for

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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.

Numerical
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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: 

  1. W = −Pext ΔV
  2. 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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Chapter 4: Chemical Thermodynamics - Exercises [Page 89]

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Balbharati Chemistry [English] Standard 12 Maharashtra State Board
Chapter 4 Chemical Thermodynamics
Exercises | Q 4.19 | Page 89

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