Question

2 moles of \[PC{{l}_{5}}\] were heated in a closed vessel of 2 liters capacity. At equilibrium, $40%$ of \[PC{{l}_{5}}\]​ is dissociated into \[PC{{l}_{3}}\] ​ and \[C{{l}_{2}}\] ​. The value of equilibrium constant is-
A. 0.266
B. 0.53
C. 2.66
D 5.3

Answer 1

Hint: To solve this question we have to know about equilibrium constant. In an equilibrium reaction the rate constant of the reaction is given as the ratio of concentration of product to the concentration of reactant. Any coefficient of the reactant or product is used as the power of the concentration.

Formula Used: The formula used in this case is given as-
\[{{K}_{eq}}=\dfrac{[PC{{l}_{3}}][C{{l}_{2}}]}{[PC{{l}_{5}}]}\]

Complete Step by Step Solution:
The reaction can be written as\[PC{{l}_{5}}\left( g \right)\rightleftharpoons PC{{l}_{3}}\left( g \right)+C{{l}_{2}}(g)\].
Initial moles of \[PC{{l}_{5}}\]is 2 moles.
After $40%$ decomposition the moles of \[PC{{l}_{5}}\]left is given as
$ & 2-2\times \dfrac{40}{100} \\ $
$2-0.8=1.2$
Thus the amount of phosphorus trichloride and chlorine gas form is given as $0.8moles$.

Total volume of the vessel is 2 litres.
Thus the molar concentration of \[PC{{l}_{5}}\]is $\dfrac{1.2}{2}=0.6$.
The molar concentration of $PC{{l}_{3}}$ is $\dfrac{0.8}{2}=0.4$.
The molar concentration of $C{{l}_{2}}$ is $\dfrac{0.8}{2}=0.4$
The rate constant of the above equilibrium reaction is given as follows-
\[{{K}_{eq}}=\dfrac{[PC{{l}_{3}}][C{{l}_{2}}]}{[PC{{l}_{5}}]}\]

Putting the values of the concentration of the reactants and products in the above equation we get:
\[{{K}_{eq}}=\dfrac{0.4\times 0.4}{0.6}\]
\[{{K}_{eq}}=0.266\]
Thus the value of the equilibrium constant of the above reaction or \[Keq\] is $0.266$.
Thus the correct option is A.

Note: An equilibrium reaction is that reaction where there is an equilibrium between the reactants and products of the reaction. Here the reaction can proceed in both the forward and backward directions that means reactants react to give product and again product can also break down into the corresponding reactants.