200 mL of ethylene gas and 150 mL of HCl gas were allowed to react at 1 bar. Pressure to form gaseous ethyl chloride. What is the work done during the reaction?

Question

TestHub · Accepted Answer

The balanced chemical equation for the reaction is:

$C_{2} H_{4} (g) + H Cl (g) \to C_{2} H_{5} Cl (g)$

Initial volumes:
$V_{C_{2} H_{4}} = 200$ mL
$V_{H Cl} = 150$ mL

Total initial volume $V_{i} = 200 + 150 = 350$ mL.

Since the reaction occurs at constant pressure and temperature, the volumes of gases are directly proportional to their number of moles. Here, $H Cl$ is the limiting reagent.

Volume of $C_{2} H_{4}$ reacted = $150$ mL
Volume of $C_{2} H_{4}$ remaining = $200 - 150 = 50$ mL
Volume of $C_{2} H_{5} Cl$ formed = $150$ mL

Total final volume $V_{f} = 50 + 150 = 200$ mL.

Change in volume $Δ V = V_{f} - V_{i} = 200 - 350 = - 150$ mL = $- 0.15$ L.

The work done during the reaction at constant pressure is given by:

$W = - P Δ V$

Substituting the values:

$W = - 1 bar \times (- 0.15 L) = 0.15 L bar$

Since $1 L bar = 100 J$ , we get:

$W = 0.15 \times 100 J = 15 J$

Answer: $15$ J

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