100 ml of $\mathrm{H}_{2(\mathrm{~g})}$ and 100 ml of $\mathrm{Cl}_{2(\mathrm{~g})}$ were allowed to react…
100 ml of $\mathrm{H}_{2(\mathrm{~g})}$ and 100 ml of $\mathrm{Cl}_{2(\mathrm{~g})}$ were allowed to react at 1 bar pressure as
$\mathrm{H}_{2(\mathrm{~g})}+\mathrm{Cl}_{2(\mathrm{~g})} \longrightarrow 2 \mathrm{HCl}_{(\mathrm{g})}$
What will be the PV type of work done diuring reaction?
Zero J
+10 J
-10 J
$\quad-100 \mathrm{~J}$
Solution
1 mole of $\mathrm{H}_2$ reacts with 1 mole of $\mathrm{Cl}_2$ to produce 2 moles of HCl.
Hence, 100 mL of $\mathrm{H}_2$ would react with 100 mL of $\mathrm{Cl}_2$ to produce 200 mL of HCl.
$\begin{aligned}
\mathrm{V}_1 & =100+100=200 \mathrm{~mL}=0.2 \mathrm{dm}^3 \\
\mathrm{~V}_2 & =200 \mathrm{~mL}=0.2 \mathrm{dm}^3 \\
\mathrm{~W} & =-\mathrm{P}_{\text {ext }} \Delta \mathrm{V}=-\mathrm{P}_{\mathrm{ext}}\left(\mathrm{~V}_2-\mathrm{V}_1\right) \\
\mathrm{W} & =-1(0.2-0.2) \\
& =\text { zero } \mathrm{J}
\end{aligned} \quad\left(\because 100 \mathrm{~J}=1 \mathrm{dm}^3 \text { bar }\right) .$