For the system \[A\left( g \right)+2B\left( g \right)\rightleftharpoons C\left( g \right)\] the equilibrium concentrations are \[(A)~0.06mole/litre~(B)~0.12mole/litre~(C)~0.216mole/litre\] . The \[Keq\] for the reaction is.

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-
\[{{K}_{eq}}=\dfrac{[C]}{[A]{{[B]}^{2}}}\]

Complete Step by Step Solution:
The given reaction is \[A\left( g \right)+2B\left( g \right)\rightleftharpoons C\left( g \right)\].
- Here one mole of reactant $A$ reacts with two moles of reactant $B$ to produce one mole of product $C$.
- Here the given concentration of reactant $A$ is $0.06mole/litre$.
- Here the given concentration of reactant $B$ is $0.12mole/litre$.
- Here the given concentration of product $C$ is $0.216mole/litre$.

The rate constant of the above equilibrium reaction is given as follows-
\[{{K}_{eq}}=\dfrac{[C]}{[A]{{[B]}^{2}}}\]

Putting the values of the concentration of the reactants and products in the above equation we get:
\[{{K}_{eq}}=\dfrac{[0.216]}{[0.06]{{[0.12]}^{2}}}\]
\[=\dfrac{0.216}{0.06\times 0.12\times 0.12}\]
$=250$
Thus the value of the equilibrium constant of the above reaction or \[Keq\]is $250$.

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.