Question

For the following reaction in equilibrium
PCl5(g) PCl3(g) + Cl2(g)
Vapour density is found to be 100 when 1 mole of PCl5 is taken in a 10 litre flask at 27* C.
Calculate the equilibrium pressure. Also calculate percentage dissociation of PCl5

Answer 1

Percentage dissociation of $PCl_{5}$ is $4.25$% and Total pressure at equilibrium is $2.57atm.$

Explanation:

a) Calculation of percentage dissociation of $PCl_{5}$.

As per the reaction,

$PCl_{5}(g) \rightarrow PCl_{3}(g) + Cl_{2}(g)$

Dissociation constant $\alpha = \frac{(M_{i}- M_{f} ) }{M_{f}(n-1) }$

where, $M_{i}$ is the molar mass before dissociation.

$M_{f}$ is the molar mass after dissociation.

$n$ are the number of moles of product.

$M_{i}= 31+(5\times35.5)\\M_{i}=208.5$

$M_{f}=2\times$ vapor density

$M_{f}= 2\times 100\\M_{f}=200$

$n=2$

therefore, $\alpha$ can be calculated as

$\alpha =\frac{208.5-200}{200(2-1)} \\\alpha = \frac{4.25}{100}$

Hence, percentage dissociation will be

$\alpha \times 100$

= $\frac{4.25}{100}\times 100$

= $4.25%$%.

Percentage dissociation of $PCl_{5}$ is $4.25$%

b) Calculation of equilibrium pressure.

Total moles at equilibrium $= 1+\alpha (n-1)$

$=1+\frac{4.25}{100} (2-1)\\=1.0425$

Total pressure at equilibrium

$PV=nRT\\P=\frac{n}{V} RT$

$P= \frac{1.0425\times0.082\times300}{10} \\P= 2.57 atm$

Total pressure at equilibrium is $2.57atm.$