Calculate the work done terms of dm^3 bar when 200 ml of ethylene...

The balanced chemical equation for the reaction is:

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

According to Avogadro's law, under identical conditions of temperature and pressure, equal volumes of gases contain equal numbers of moles. Thus, the gases react in a $1:1$ volume ratio.

Initial volume of the mixture, $V_i = 200 \text{ ml} + 150 \text{ ml} = 350 \text{ ml}$

Since $HCl$ is the limiting reagent, $150 \text{ ml}$ of $HCl$ will react with $150 \text{ ml}$ of $C_2H_4$ to form $150 \text{ ml}$ of $C_2H_5Cl(g)$.

Volume of unreacted $C_2H_4 = 200 \text{ ml} - 150 \text{ ml} = 50 \text{ ml}$

Final volume of the gaseous mixture, $V_f = \text{Volume of } C_2H_5Cl \text{ formed} + \text{Volume of unreacted } C_2H_4$

$V_f = 150 \text{ ml} + 50 \text{ ml} = 200 \text{ ml}$

Change in volume, $\Delta V = V_f - V_i = 200 \text{ ml} - 350 \text{ ml} = -150 \text{ ml}$

Converting the volume change to $\text{dm}^3$ (since $1 \text{ dm}^3 = 1000 \text{ ml}$):

$\Delta V = -0.15 \text{ dm}^3$

The work done (W$) at constant external pressure is given by:<br><br>$W = -P_{ext} \Delta V$<br><br>Substituting the given values (P_{ext} = 1 \text{ bar}$):

$W = -1 \text{ bar} \times (-0.15 \text{ dm}^3) = 0.15 \text{ dm}^3 \text{ bar}$

Answer: $0.15$