Question

In the system $X+2Y\rightleftharpoons Z$, the equilibrium concentrations are,
$\left[X\right]=0.06mol{L}^{-1}$,$\left[Y\right]=0.12mol{L}^{-1}$, $\left[Z\right]=0.216mol{L}^{-1}$, Find the equilibrium constant of the reaction.
A) $250$
B) $500$
C) $125$
D) $273$

Answer 1

Hint: We know that the equilibrium constant ${\text{K}}_{\text{eq}}$ gives the relationship between products and reactants of a reaction at equilibrium with respect to a specific unit.

Formula used: We can calculate the equilibrium constant by using the given formula,
${\text{K}}_{\text{eq}}\text{ = }{\displaystyle \frac{{\left[\text{C}\right]}^{\text{c}}{\left[\text{D}\right]}^{\text{d}}}{{\left[\text{A}\right]}^{\text{a}}{\left[\text{B}\right]}^{\text{b}}}}$
Where $K_{eq}$ is the equilibrium constant.
The concentration of the reactants and products are denoted as $\left[\text{A}\right]\text{,}\left[\text{B}\right]\text{\& }\left[\text{C}\right]\left[\text{D}\right]$ respectively.

Complete step by step answer:
We also remember that the equilibrium constant of concentration approximates the activities of solutes and gases in dilute solutions with their respective molarities. Anyhow the activities of solids, pure liquids, and solvents are not approximated with their molarities instead these activities are defined to have a value equal to 1 (one).The equilibrium constant expression for the reaction:
The reaction is $X+2Y\rightleftharpoons Z$
Given,
The equilibrium constants of X is $\left[X\right]=0.06mol{L}^{-1}$
The equilibrium constants of Y is $\left[Y\right]=0.12mol{L}^{-1}$
The equilibrium constants of Z is $\left[Z\right]=0.216mol{L}^{-1}$
Now, calculate the equilibrium constant of the reaction as follows,
${\text{K}}_{\text{eq}}\text{ = }{\displaystyle \frac{{\left[\text{C}\right]}^{\text{c}}{\left[\text{D}\right]}^{\text{d}}}{{\left[\text{A}\right]}^{\text{a}}{\left[\text{B}\right]}^{\text{b}}}}$
${\text{K}}_{\text{eq}}\text{ = }{\displaystyle \frac{\left[Z\right]}{\left[X\right]{\left[Y\right]}^2}}$
Substituting the known values in the formula we get,
$\Rightarrow$${\text{K}}_{\text{eq}}\text{ = }{\displaystyle \frac{0.216}{0.06\times {\left(0.12\right)}^2}}$
$\Rightarrow$${\text{K}}_{\text{eq}}\text{ = }{\displaystyle \frac{0.216}{0.000864}}=250$

So, the correct answer is Option A.

Note:
We must know that the concept of the equilibrium constant is applied in various fields of chemistry and pharmacology. In protein-ligand binding the equilibrium constant describes the affinity between a protein and a ligand. A little equilibrium constant indicates a more tightly bound the ligand. Within the case of antibody-antigen binding the inverted equilibrium constant is employed and is named affinity constant.
Using the equilibrium constant we can calculate the $pH$ of the reaction,
Example:
Write the dissociation equation of the reaction.
$HA+H_{2}O\stackrel{}{\to }{H}_3{O}^{+}+A^{-}$
The constant $K_a$ of the solution is$4\times {10}^{-2}$.
The dissociation constant of the reaction $K_a$ is written as,
$K_a={\displaystyle \frac{\left[H_3{O}^{+}\right]\left[A^{-}\right]}{\left[HA\right]}}$
Let us imagine the concentration of $$\left[H_3{O}^{+}\right]\left[A^{-}\right]$$ as x.
$4\times {10}^{-7}={\displaystyle \frac{x^2}{0.08-x}}$
$\Rightarrow$$x^2=4\times {10}^{-7}\times 0.08$
$\Rightarrow$$x=1.78\times {10}^{-4}$
The concentration of Hydrogen is $1.78\times {10}^{-4}$
We can calculate the $pH$ of the solution is,
$pH=-\log \left[H^{+}\right]=3.75$
The $pH$ of the solution is $3.75$.