Question

The degree of dissociation $PC{{l}_{5}}$ at 1 atm pressure is 0.2. Calculate the pressure at which $PC{{l}_{5}}$ is dissociated to 50%?
[A] P = 22.19atm
[B] P = 0.128atm
[C] P = 2.165atm
[D] P = 1.7atm

Answer 1

Remember that the relationship between the degree of dissociation ‘$\alpha$’ and the equilibrium constant ‘${{K}_{p}}$’ at pressure ‘P’ is given by-${{K}_{P}}=\dfrac{P\times \alpha }{1-{{\alpha }^{2}}}$ . With this relationship in mind, try to find the answer to the given question.

Complete step by step answer:
Let us first look at the dissociation reaction $PC{{l}_{5}}$ that really takes place before analysing it thoroughly.
Let the number of moles of $PC{{l}_{5}}$ initially be 1 and its degree of dissolution be $\alpha$. This means before dissociation we have 1 mole of $PC{{l}_{5}}$ and zero moles of product (as the reaction has not yet taken place).
Now, as the degree of dissociation is $\alpha$ so after dissociation we will have $\left( 1-\alpha \right)$ moles of the starting material and $\alpha$ moles of each of the dissociated products.
We can write the moles before and after dissociation as-
$PC{{l}_{5}}\rightleftharpoons PC{{l}_{3}}+C{{l}_{2}}$
Then we observe that

| $PC{{l}_{5}}$ | $PC{{l}_{3}}$ | $C{{l}_{2}}$
---|---|---|---
Moles before dissociation:| 1| 0| 0
Moles after dissociation:| $\left( 1-\alpha \right)$ | $\alpha$ | $\alpha$

We observe that due to dissociation, only $\left( 1-\alpha \right)$ moles of reactant remain while an $\alpha$ mole each of the two products are formed.
Now, we know that Pressure = 0.2 atm (Since it has been given)
Calculating the equilibrium constant at given pressure,
$\therefore {{K}_{p}}=\dfrac{{{n}_{PC{{l}_{3}}}}\times {{n}_{C{{l}_{2}}}}}{{{n}_{PC{{l}_{5}}}}}\times {{\left[ \dfrac{P}{\Sigma n} \right]}^{\Delta n}}$
That implies,
${{K}_{P}}=\dfrac{\alpha .\alpha }{(1-\alpha )}[\dfrac{P}{1+\alpha }]=\dfrac{P{{\alpha }^{2}}}{1-{{\alpha }^{2}}}=\dfrac{1\times {{(0.2)}^{2}}}{1-{{(0.2)}^{2}}}$
Thus, we can conclude that the value of equilibrium constant at temperature P = 0.2 atm is given by,${{K}_{p}}$ = 0.0416 atm
Now, we know that the equilibrium constant of a reaction does not change with any change in any factor. So, should the degree of dissociation be now increased to 0.5. Therefore, the pressure can be calculated by- ${{K}_{P}}=\dfrac{P{{\alpha }^{2}}}{1-{{\alpha }^{2}}}$
Plugging in the obtained and already given values into this equation-
$0.0416=\dfrac{P\times {{(0.5)}^{2}}}{1-{{(0.5)}^{2}}}$
Therefore, we can conclude that Pressure ‘P’ is given by P = 0.1248 atm
So, the correct answer is “Option B”.

Note: Dissociation is the separation of ions that occurs when a solid ionic compound dissolves. Non-ionic compounds do not dissociate in water. Be very careful of this distinction so as to ensure you make no silly mistakes when trying to solve questions related to this concept.