$200$ 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?

The balanced chemical equation for the reaction is:

$C_2H_4(g) + HCl(g) \rightarrow C_2H_5Cl(g)$

Initial volumes:
$V_{C_2H_4} = 200$ mL
$V_{HCl} = 150$ 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, $HCl$ is the limiting reagent.

Volume of $C_2H_4$ reacted = $150$ mL
Volume of $C_2H_4$ remaining = $200 - 150 = 50$ mL
Volume of $C_2H_5Cl$ formed = $150$ mL

Total final volume $V_f = 50 + 150 = 200$ mL.

Change in volume $\Delta V = V_f - V_i = 200 - 350 = -150$ mL = $-0.15$ L.

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

$W = -P \Delta V$

Substituting the values:

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

Since $1 \text{ L bar} = 100 \text{ J}$, we get:

$W = 0.15 \times 100 \text{ J} = 15 \text{ J}$

Answer: $15$ J