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Question: What is value of PV type of work for following reaction at 1 bar? $C_2H_{4(g)}$ + $HCl_{(g)}$ → $C_...

What is value of PV type of work for following reaction at 1 bar?

C2H4(g)C_2H_{4(g)} + HCl(g)HCl_{(g)} → C2H5Cl(g)C_2H_5Cl_{(g)} V1_1=0.3 0.2(200 mL) 0.15(150 mL) V2_2=0.15 ∴ first prreactant

A

3.5 J

B

4.5 J

C

9.0 J

D

15 J

Answer

15 J

Explanation

Solution

For the reaction

\ceC2H4(g)+HCl(g)−>C2H5Cl(g)\ce{C2H4(g) + HCl(g) -> C2H5Cl(g)}

the data given are:

  • Ethylene: 200 mL (i.e. “0.2”)
  • HCl: 150 mL (i.e. “0.15”)

Since the reaction uses 1 mole of each gas to form 1 mole of product gas, HCl (150 mL) is the limiting reactant. Thus, the reaction consumes 150 mL of ethylene (from the available 200 mL) so that 50 mL of ethylene remains unreacted.

Initial total gas volume
= 200 mL (ethylene) + 150 mL (HCl)
= 350 mL

Final total gas volume
= 150 mL (C₂H₅Cl formed) + 50 mL (unreacted ethylene)
= 200 mL

Therefore, the change in volume is

ΔV=Vfinal−Vinitial=200 mL−350 mL=−150 mL=−0.15 L\Delta V = V_{\text{final}} - V_{\text{initial}} = 200\,\text{mL} - 350\,\text{mL} = -150\,\text{mL} = -0.15\,\text{L}

At constant pressure (1 bar), the work done in a PV work process is given by

w=−P ΔVw = -P\,\Delta V

Substituting,

w=−(1 bar)×(−0.15 L)=0.15 L barw = - (1\,\text{bar}) \times (-0.15\,\text{L}) = 0.15\,\text{L\,bar}

Using the conversion factor 1 L bar≈100 J1\,\text{L\,bar} \approx 100\,\text{J},

w≈0.15×100 J=15 Jw \approx 0.15 \times 100\,\text{J} = 15\,\text{J}