Question

Given,1 mol of XY(g) and 0.2 mol of Y(g) are mixed in 1 L vessel. At equilibrium, 0.6 mol of Y(g) is present. The value of K for the reaction $XY(g)\rightleftharpoons X(g)+Y(g)$ is:
A. $0.04mol{{L}^{-1}}$
B. $0.06mol{{L}^{-1}}$
C. $0.36mol{{L}^{-1}}$
D. $0.40mol{{L}^{-1}}$

Answer 1

Hint: In the question we are asked to find the value of K which is called as the equilibrium constant for the reaction given and the formula to find equilibrium constant of concentration for the reaction is given as $K=\dfrac{[X][Y]}{[XY]}$ where [X] and [Y] is the concentration of product X and Y and [XY] is the concentration of reactant XY.

Complete answer:
From your chemistry lesson you have learned about the equilibrium constant which is denoted by K of a chemical reaction. The equilibrium constant (K) for a chemical reaction gives the relationship between the products and the reactants when the chemical reaction reaches the equilibrium.
Here we have to find the equilibrium constant of the concentration of the chemical reaction. So, the equilibrium constant of concentration for the chemical reaction which is denoted by ${{K}_{c}}$ at equilibrium
Is defined as the ratio of the concentration of the product to the concentration of reactant , each of them raised to their respective stoichiometry.
So, the expression for the equilibrium constant of the reaction given in the question will be:
\[XY(g)\rightleftharpoons X(g)+Y(g)\]
\[{{K}_{c}}=\dfrac{[X][Y]}{[XY]}\]………… (1)
Now, we have to find the concentration of X, Y and XY. You should know that equilibrium constant is measured in moles per litre. So, in the question the initial concentration of XY is given as 1 moles in 1 L and the initial concentration of Y is given as 0.2 in 1 L and for final concentration at equilibrium the moles for Y is given as 0.6 in 1 L.
So,
  \[\begin{align}
  & \underset{\text{at initial concentration}}{\mathop{{}}}\,\underset{\text{0}\text{.1}}{\mathop{\text{XY}}}\,\text{(g)}\rightleftharpoons \underset{\text{0}}{\mathop{\text{X}}}\,\text{(g)+}\underset{\text{0}\text{.2}}{\mathop{\text{Y}}}\,\text{(g)} \\
 & \underset{\text{at eqilibrium}}{\mathop{{}}}\,\text{ }\underset{\text{(0}\text{.1-x)}}{\mathop{\text{ XY}}}\,\text{(g)}\rightleftharpoons \underset{\text{x}}{\mathop{\text{X}}}\,\text{(g)+}\underset{\text{0}\text{.2+x}}{\mathop{\text{Y}}}\,\text{(g)} \\
\end{align}\]
(where x is the dissociation constant)

Now, in the question it is given that at equilibrium the concentration of Y is given as 0.6 mol per liter. Therefore the concentration of Y will be given as:
\[0.2+x=0.6\]
\[\therefore x=0.4\]
So, by knowing the value of x we can also find the concentration of X and XY at equilibrium and it will be given as:
\[[XY]=(1-0.4)=0.6=\dfrac{0.6}{1}=0.6mol{{L}^{-1}}\]
\[[X]=0.4=\dfrac{0.4}{1}=0.4mol{{L}^{-1}}\]
\[[Y]=0.6=\dfrac{0.6}{1}=0.6mol{{L}^{-1}}\]
By putting all the values in equation (1) we will get:
\[{{K}_{c}}=\dfrac{[0.4][0.6]}{[0.6]}=0.4mol{{L}^{-1}}\]
Therefore the value of K will be $0.40mol{{L}^{-1}}$

Thus the correct option will be (D).

Note:
As we know that the expression for concentration is given as $Concentration=\dfrac{moles}{Volume(L)}$. Therefore we have divided the moles of X, Y and XY with 1 L to find the concentration of each of them. Equilibrium constant can also be written in terms of partial pressure if the reaction involves gases.