Question

At ${{227}^{0}}C$, 60% of 2 moles of $PC{{l}_{5}}$ gets dissociated in a two-litre. The value of ${{K}_{p}}$ will be
(A) 450R
(B) 400R
(C) 50R
(D) 100R

Answer 1

Hint: Start by writing the decomposition reaction of phosphorus pentachloride. Now the number of moles of $PC{{l}_{5}}$ is given. So, we can calculate the number of moles of products and also ${{K}_{c}}$ that is, equilibrium constant in terms of concentration. ${{K}_{p}}$ is the equilibrium constant in terms of partial pressure. Use the formula, ${{K}_{p}} = {{K}_{c}}{{(RT)}^{\Delta n}}$ and calculate the value of ${{K}_{p}}$.

Complete Solution :
In the lower classes of chemistry, we have come across the concepts of finding the equilibrium constant for the reaction when there is change in the concentration at different intervals of time.
- Now, let us see the reaction that takes place when phosphorus pentachloride is being dissociated and also their respective change in concentration with change in time.
The reaction is as shown below:
\[PC{{l}_{5}}\to PC{{l}_{3}}+C{{l}_{2}}\]

Initial conc.	2	0	0
Conc. at${{227}^{0}}C$	2(1-x)	2x	2x

- Now, at ${{227}^{0}}C$, 60% of 2 moles is getting dissociated and forming products having ‘x’ moles. Therefore,$x = \dfrac{2\times 60}{100} = \dfrac{3}{5} moles$.
- Now, we obtained x = $\dfrac{3}{5}$ moles. Equilibrium constant is given as,
\[{{K}_{c}}=\dfrac{[PC{{l}_{3}}][PC{{l}_{2}}]}{[PC{{l}_{5}}]}=\dfrac{4{{x}^{2}}}{2(1-x)}\times \dfrac{1}{2}={{\dfrac{4\times ({3}/{5{{)}^{2}}}\;}{4\times \left( 1-\dfrac{3}{5} \right)}}^{{}}}\]
\[\Rightarrow {{K}_{c}} = \dfrac{9}{10}\]
Therefore, the value of${{K}_{c}}$ is \[\dfrac{9}{10}\].
- We know that, ${{K}_{p}}={{K}_{c}}{{(RT)}^{\Delta n}}$ where R is the universal gas constant, T is the temperature in Kelvin and$\Delta n$ is the change in number of moles.
- $x =\dfrac{3}{5} = 0.6$ moles and so, $2(1-x)=2\left( 1-\dfrac{3}{5} \right)=\dfrac{8}{5} = 1.6moles$
Therefore, $\Delta n$ is equal to 1.
- Substituting the values in the equation${{K}_{p}}={{K}_{c}}{{(RT)}^{\Delta n}}$ we get,
\[{{K}_{p}} = \dfrac{9}{10}(500R) = 450R\]
Therefore, the value of ${{K}_{p}}$ is $450R$.
So, the correct answer is “Option A”.

Note: Remember ${{K}_{p}}={{K}_{c}}{{(RT)}^{\Delta n}}$ where ${{K}_{c}}$ is the equilibrium constant in terms of concentration and ${{K}_{p}}$ is the equilibrium constant in terms of partial pressures which is used in case the reactants are in gaseous state. R is the universal gas constant and T is the temperature in Kelvin and $\Delta n$ is the change in number of moles.