Question

The equilibrium concentration of $X,Y$ and $Y{X}_2$ are $4,2$ and $2mol$ respectively for the equilibrium $2X+Y\rightleftharpoons Y{X}_2$. The value of $K_c$ is:
A. $0.625$
B. $0.0625$
C. $6.25$
D. $0.00625$

Answer 1

Hint: We have to remember the law of mass action which gives the relation between the equilibrium constant and concentrations of reactants and products. We need to use an ICE table to solve this problem. ICE table stands for Initial, Change and Equilibrium.

Complete step by step answer:
A chemical system in which the concentration of reactants and products will be constant over time is referred to as chemical equilibrium. When the system is about to attain equilibrium, forward and reverse reactions occur. When it attains equilibrium, both reactions have the same rate.
The chemical reaction is given below:
 $2X+Y\rightleftharpoons Y{X}_2$
It is given that the equilibrium concentrations of $X,Y$ and $Y{X}_2$ are $4,2$ and $2mol$ respectively.
i.e. $4$ $2$ $2$
Equilibrium constant is represented by $K_c$. It can be calculated by dividing the concentration of product rise to the power of its coefficients by concentration of reactant raise to the power of its coefficients.
i.e. $$K_c={\displaystyle \frac{\left[Y{X}_2\right]}{{\left[2X\right]}^2\left[Y\right]}}$$
Thus substituting the values, we get
$$K_c={\displaystyle \frac{\left[2\right]}{{\left[2\times 2\right]}^2\left[2\right]}}$$ since the equilibrium concentrations are given as $4,2,2$ moles.
On simplification, we get
$$K_c={\displaystyle \frac{\left[2\right]}{{\left[4\right]}^2\left[2\right]}}={\displaystyle \frac{2}{16\times 2}}={\displaystyle \frac{2}{32}}=0.0625$$
Thus the value of $K_c$ is $0.0625$.

So, the correct option is B.

Note: When there is a state in which observable changes are not there as time goes by, such state is called equilibrium. If equilibrium constant is greater than one, then it favors products. If equilibrium constant is less than one, then it favors reactants.
We can also use partial pressures of reactants and products for calculating the equilibrium constant. This is because both concentrations and partial pressures are similarly related. When any of the reactants or products has zero, then the reaction will be shifted in that direction.