In the reaction \[PC{{l}_{5}}\left( g \right)\rightleftarrows PC{{l}_{3}}\left( g \right)+C{{l}_{2}}\left( g \right)\] the equilibrium concentration of \[PC{{l}_{5}}\]​ and \[PC{{l}_{3}}\] ​ are \[0.4\] and \[0.2\] mole/litre respectively. If the value of \[{{K}_{c}}\]​ is\[0.5\]. What is the concentration of \[C{{l}_{2}}\]​ in moles/litre?
(A) \[2.0\]
(B) \[1.5\]
(C) \[1.0\]
(D) \[0.5\]

Hint: The equilibrium constant (\[{{K}_{eq}}\]) is a formula that illustrates the relationship between the concentrations of the reactants and the products.

Complete Step by Step Solution:
When a chemical process reaches equilibrium, the equilibrium constant (often represented by the letter\[K\]) sheds light on the interaction between the reactants and products. For instance, the ratio of the concentration of the products to the concentration of the reactants, each raised to their respective stoichiometric coefficients, can be used to establish the equilibrium constant of concentration (denoted by\[{{K}_{c}}\]) of a chemical reaction at equilibrium. It is significant to remember that there are several kinds of equilibrium constants that establish relationships between the reactants and products of equilibrium reactions in terms of various units.

The ratio between the amounts of reactant and product for a chemical reaction can be described as the equilibrium constant.
At equilibrium, the rate of the forward reaction = the rate of the backward reaction
That is, \[{{R}_{f}}={{R}_{b}}\] or, \[{{K}_{f}}\times \alpha \times {{\left[ A \right]}^{a}}{{\left[ B \right]}^{b}}={{K}_{b}}\times \alpha \times {{\left[ C \right]}^{c}}{{\left[ D \right]}^{d}}\]

The rate constants are constant at a certain temperature. An equilibrium constant is the ratio of the forward reaction rate constant to the backward reaction rate constant, which must be a constant (\[{{K}_{eq}}\]).
\[{{K}_{eq}}={{K}_{f}}/{{K}_{b}}={{\left[ C \right]}^{c}}{{\left[ D \right]}^{d}}/{{\left[ A \right]}^{a}}{{\left[ B \right]}^{b}}={{K}_{c}}\]
Where\[{{K}_{c}}\] indicates the equilibrium constant measured in moles per litre.

According to our question,
The equilibrium reaction is \[PC{{l}_{5}}\left( g \right)\rightleftarrows PC{{l}_{3}}\left( g \right)+C{{l}_{2}}\left( g \right)\]
Using the formula, \[{{K}_{c}}={{\left[ C \right]}^{c}}{{\left[ D \right]}^{d}}/{{\left[ A \right]}^{a}}{{\left[ B \right]}^{b}}\]
Given, \[PC{{l}_{5}}\]= \[0.4\]
\[PC{{l}_{3}}\]= \[0.2\]
\[{{K}_{c}}\]= \[0.5\]
Substituting the values, \[{{K}_{c}}=\left[ PC{{l}_{3}} \right]\left[ C{{l}_{2}} \right]/\left[ PC{{l}_{5}} \right]\]
\[0.5=0.2\left[ C{{l}_{2}} \right]/0.4\]
\[0.5=\left[ C{{l}_{2}} \right]/0.2\]
Thus, \[\left[ C{{l}_{2}} \right]=1.0\]
The correct option is C.

Note: There is a separate formula for calculating the values for liquids and gases, respectively. When conditions are ideal, temperature and concentration have little effect on the calculation, but when conditions are not ideal, they have a substantial impact.