Question

The degree of dissociation is $0.4$ at $400K$ and $1atm$ for the gaseous reaction.
$PC{l}_5\rightleftharpoons PC{l}_3+C{l}_2$. Assuming ideal behaviour of all the gases. Calculate the density of the equilibrium mixture at $400K$ and $1$ atm Pressure.

• A. $4.53g/L$
• B. $9.26g/L$
• C. $1.25g/L$
• D. $7.28g/L$

Answer 1

The correct option is A $4.53g/L$

We have, $PC{l}_5\leftrightarrow PC{l}_3+C{l}_2$

Initial mole , $1$ $0$ $0$

Mole at equilibrium $1–0.4$ $0.4$ $0.4$

Total mole at equilibrium = $1–0.4+0.4+0.4=1.4$

$\frac{Normal{M}_{PC{l}_5}}{Observed{M}_{PC{l}_5}}=1+\alpha =1.4$

Therefore, Observed $M_{PC{l}_5}=\frac{208.5}{1.4}$

At equilibrium, $PV=\left(\frac{w}{m}\right)RT$

Or, $\frac{w}{V}=\frac{Pm}{RT}$

density= $\frac{(1\times 208.5)}{(1.4\times 0.0821\times 400)}$

density= $4.53\frac{g}{L}$