Question

Pure $PC{{l}_{5}}$ is introduced into an evacuated chamber and comes to equilibrium at ${{250}^{\circ }}C$ and 2 atmospheres. The equilibrium mixture contains 40.7% $C{{l}_{2}}$ volume / volume. What is the ${{K}_{p}}$ (atm) (nearest integer)?

Answer 1

The equilibrium reaction of the phosphorus chloride will be: $PC{{l}_{5}}\rightleftharpoons PC{{l}_{3}}+C{{l}_{2}}$, so the equilibrium constant in terms of pressure can be calculated by the formula:
${{K}_{p}}=\dfrac{{{P}_{PC{{l}_{3}}}}\text{ x }{{P}_{C{{l}_{2}}}}}{{{P}_{PC{{l}_{5}}}}}$ .

Complete answer:
The question says that the phosphorus pentachloride is introduced into an evacuated chamber in which it comes in equilibrium with the products, therefore, the reaction will be the dissociation of phosphorus pentachloride into phosphorus trichloride and chlorine gas. The reaction is:
$PC{{l}_{5}}\rightleftharpoons PC{{l}_{3}}+C{{l}_{2}}$
All the compounds in this reaction are in the gas form and one mole of $PC{{l}_{5}}$ will dissociate into one mole of each $PC{{l}_{3}}$ and $C{{l}_{2}}$.
So, the equilibrium constant in terms of pressure can be calculated by the formula:
${{K}_{p}}=\dfrac{{{P}_{PC{{l}_{3}}}}\text{ x }{{P}_{C{{l}_{2}}}}}{{{P}_{PC{{l}_{5}}}}}$
Now, the mixture contains 40.7% of chlorine gas volume/volume, this percentage by volume can be approximated by the mole fraction.
40.7% of chlorine gas will be equal to 0.407 moles. So, in the reaction, the number of moles of $C{{l}_{2}}$ is 0.0407, and therefore, the number of moles of phosphorus trichloride will also be 0.0407.
The total moles in the product side will be = 0.407 + 0.407 = 0.814
Then the number of moles on the reactant side will be = 1 – 0.814 = 0.186
Now, we have to calculate the partial pressure of each component which is equal to the product of mole fraction and total pressure. The total pressure is 2.
The partial pressure of $PC{{l}_{5}}$ = 0.186 x 2 = 0.372
The partial pressure of $PC{{l}_{3}}$ = 0.407 x 2 = 0.814
The partial pressure of $C{{l}_{2}}$ = 0.407 x 2 = 0.814
Putting these values in the formula, we get:
${{K}_{p}}=\dfrac{\text{0}\text{.814 x 0}\text{.814}}{0.372}=1.78\approx 2$
The value of ${{K}_{p}}$ is 2.

Note:
Whenever the pressure is calculated for a gaseous reaction, the number of moles should be considered, and if we have to calculate the equilibrium constant in terms of concentration, then the concentration of the compounds should be considered.