Question

The value of ${{K}_{c}}$ for the reaction $3{{O}_{2}}(g)\to 2{{O}_{3}}(g)$ at 25 degree C is $2\text{ x 1}{{\text{0}}^{-50}}$. If the equilibrium concentration of ${{O}_{2}}$ in air at 25 degree C is $1.6\text{ x 1}{{\text{0}}^{-2}}$. What is the concentration of ${{O}_{3}}$?

Answer 1

Hint: The formula can be used is ${{K}_{c}}=\dfrac{[p]}{[r]}$, where ${{K}_{c}}$ is the equilibrium constant of the reaction, [p] is the concentration of the product, and [r] is the concentration of the reactant.

Complete step by step solution:
For any reaction, the equilibrium constant is always equal to the ratio of the concentration of the product to the concentration of the reactant. And each concentration is raised to some power that is equal to the number of moles in the chemical reaction. For example,
$aA+bB\to cC+dD$
The equilibrium constant for this reaction will be:
${{K}_{c}}=\dfrac{{{[C]}^{c}}{{[D]}^{d}}}{{{[A]}^{a}}{{[B]}^{b}}}$
So for the reaction given in the question, i.e.,
$3{{O}_{2}}(g)\to 2{{O}_{3}}(g)$
The equilibrium constant for this reaction will be:
${{K}_{c}}=\dfrac{{{[{{O}_{3}}]}^{2}}}{{{[{{O}_{2}}]}^{3}}}$
So given in the question that, the value of ${{K}_{c}}$is at ${25^0}$C is $2\text{ x 1}{{\text{0}}^{-50}}$, and we given the concentration of ${{O}_{2}}$ in air at ${25^0}$C is $1.6\text{ x 1}{{\text{0}}^{-2}}$. And we have to find the value of the concentration of ${{O}_{3}}$. So putting the values in the above formula, we get:
$2\text{ x 1}{{\text{0}}^{-50}}=\dfrac{{{[{{O}_{3}}]}^{2}}}{{{[1.6\text{ x 1}{{\text{0}}^{-2}}]}^{3}}}$
So the concentration of ${{[{{O}_{3}}]}^{2}}$ will be:
${{[{{O}_{3}}]}^{2}}={{[1.6\text{ x 1}{{\text{0}}^{-2}}]}^{3}}\text{ x }2\text{ x 1}{{\text{0}}^{-50}}$
${{[{{O}_{3}}]}^{2}}=8.192\text{ x 1}{{\text{0}}^{-56}}$
So this is the value of square of concentration of ${{O}_{3}}$, to get the concentration of ${{O}_{3}}$ we have to take the square root of the value:
$[{{O}_{3}}]=\sqrt{8.192}\text{ x 1}{{\text{0}}^{-28}}$
$[{{O}_{3}}]=2.86\text{ x 1}{{\text{0}}^{-28}}$

So the concentration of ${{O}_{3}}$ at ${25^0}$C is $2.86\text{ x 1}{{\text{0}}^{-28}}$.

Note: For calculating the equilibrium constant all the components in the equilibrium must be at the same temperature. Don't forget to raise the concentration of the molecule to the number of moles in the chemical reaction.