Question

In a gas phase reaction, the decomposition of ${\mathrm{PCl}}_5$ takes place at $273°\mathrm{C}$ and 1 atmospheric pressure. Its percentage degree of dissociation is 40 per cent. Assuming that all gases in the reaction behave ideally, calculate the density of the equilibrium mixture. [atomic weight of phosphorus $=31.0$ and chlorine $=35.5$ ]

Answer 1

1. Chemical equation: The given reaction is: $$\begin{gathered} {\mathrm{PCl}}_5\left(\mathrm{s}\right)\stackrel{273°\mathrm{C}}{\to }{\mathrm{PCl}}_3\left(\mathrm{s}\right)+{\mathrm{Cl}}_2\left(\mathrm{g}\right) \\ \mathrm{Phosphorus}\mathrm{Phosphorus}\mathrm{Chlorine} \\ \mathrm{pentachloride}\mathrm{trichloride} \end{gathered}$$

2. Let $\alpha$ is the degree of dissociation and 1 mol/L be the initial concentration.

	${\mathrm{PCl}}_5$	${\mathrm{PCl}}_3$	${\mathrm{Cl}}_2$
I (mol)	1	0	0
C (mol)	-$\alpha$	+$\alpha$	+$\alpha$
E (mol)	$1-\alpha$	$\alpha$	$\alpha$

3. Total number of moles: Number of moles $=1-\alpha +\alpha +\alpha$$=1+\alpha$

4. Degree of dissociation $\left(\mathrm{\alpha }\right)=40\%\mathrm{of}1.0\mathrm{mol}=0.4\mathrm{mol}$

5. Thus, expected molar weight of ${\mathrm{PCl}}_5$ is - $$\begin{gathered} \mathrm{Expected}\mathrm{molar}\mathrm{weight}=\frac{\mathrm{Calculated}\mathrm{molar}\mathrm{weight}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{moles}}=\frac{31.0\mathrm{g}/\mathrm{mol}+(5\times 35.5\mathrm{g}/\mathrm{mol})}{1.4\mathrm{moles}} \\ =\frac{208.5\mathrm{g}/\mathrm{mol}}{1.4}=148.9\mathrm{g}/\mathrm{mol} \end{gathered}$$

6. Ideal gas law: $\mathrm{PV}=\mathrm{nRT}$
where $\mathrm{P}$ is pressure, $\mathrm{V}$ is volume, $\mathrm{n}$ is number of moles, $R$ is gas constant and $T$ is temperature. $n=\frac{W}{M}$, where $\mathrm{W}$ is mass of gas, $\mathrm{M}$ is molecular mass.
$PV=\frac{W}{M}\times RT$
As, $d=\frac{W}{V}$
density, $d=\frac{MP}{RT}$
Expected molar weight of ${\mathrm{PCl}}_5=148.9\mathrm{g}/\mathrm{mol}$
$273°\mathrm{C}=546\mathrm{K}\left[\because 0°\mathrm{C}=273\mathrm{K}\right]$
$$\begin{gathered} \mathrm{d}=\frac{148.9\mathrm{g}/\mathrm{mol}\times 1\mathrm{atm}}{0.082\mathrm{L}.\mathrm{atm}/\mathrm{mol}.\mathrm{K}\times 546\mathrm{K}} \\ =3.32\mathrm{g}/\mathrm{L} \end{gathered}$$

7. Hence, the density = 3.32 g/L.