Question

The vapour density of $P{Cl}_5$ at ${200}^{o}C$ and ${252}^{o}C$ are $70.2$ and $57.2$ respectively at one atmosphere. Calculate its value of dissociation constant at these temperatures.

Answer 1

The molar mass $=2\times$vapour density
The vapour density of $PC{l}_5$ at 200 deg C is 70.2
The molar mass $=2\times 70.2=140.4$ g/mol.
$PC{l}_5\rightleftharpoons PC{l}_3+C{l}_2$
The molar mass of undissociated $PC{l}_5$ 208.2 g/mol.
The molar mass of $PC{l}_3$ 137.3 g/mol.
The molar mass of $C{l}_2$ 71 g/mol.
If x is the degree of dissociation at 200 deg C, then
The average molar mass of the mixture $=\frac{208.2(1-x)+137.3x+71x}{1-x+x+x}=140.4$
$\frac{208.2-208.2x+137.3x+71x}{1+x}=140.4$
$208.2+0.1x=140.4+140.4x$
$67.8=140.3x$
$x=0.483$
The number of moles of $PC{l}_3=$ the number of moles of $C{l}_2=x=0.483$
The number of moles of $PC{l}_5=1-0.483=0.517$
The parital pressure of $PC{l}_3=$ the partial pressure of $C{l}_2=0.483\times 1atm=0.483atm$
The partial pressure of $PC{l}_5=0.517\times 1atm=0.517atm$
$K_p=\frac{P_{PC{l}_3}\times P_{C{l}_2}}{P_{PC{l}_5}}$
$K_p=\frac{0.483atm\times 0.483atm}{0.517}$
$K_p=0.452atm$
The molar mass $=2\times$vapour density
The vapour density of $PC{l}_5$ at 252 deg C is 57.2
The molar mass $=2\times 57.2=114.4$ g/mol.
$PC{l}_5\rightleftharpoons PC{l}_3+C{l}_2$
The molar mass of undissociated $PC{l}_5$ 208.2 g/mol.
The molar mass of $PC{l}_3$ 137.3 g/mol.
The molar mass of $C{l}_2$ 71 g/mol.
If x is the degree of dissociation at 252 deg C, then
The average molar mass of the mixture $=\frac{208.2(1-x)+137.3x+71x}{1-x+x+x}=114.4$
$208.2-208.2x+137.3x+71x=114.4+114.4x$
$93.8=114.3x$
$x=0.821$
The number of moles of $PC{l}_3=$ the number of moles of $C{l}_2=x=0.821$
The number of moles of $PC{l}_5=1-0.821=0.179$
The parital pressure of $PC{l}_3=$ the partial pressure of $C{l}_2=0.821\times 1atm=0.821atm$
The partial pressure of $PC{l}_5=0.179\times 1atm=0.179atm$
$K_p=\frac{P_{PC{l}_3}\times P_{C{l}_2}}{P_{PC{l}_5}}$
$K_p=\frac{0.821atm\times 0.821atm}{0.179}$
$K_p=3.75atm$