Question

A 10 L box contains ${{O}_{3}}$ and ${{O}_{2}}$ at equilibrium at 2000K. ${{K}_{p}}=4\times {{10}^{14}}$ atm for $2{{O}_{3}}(g)\rightleftharpoons 3{{O}_{2}}(g)$ . Assume that ${{P}_{{{O}_{2}}}}>>{{P}_{{{O}_{3}}}}$ and if total pressure is 8 atm, then partial pressure of ${{O}_{3}}$ will be:
(A) $8\times {{10}^{-5}}atm$
(B) $11.3\times {{10}^{-7}}atm$
(C) $9.71\times {{10}^{-8}}atm$
(D) $9.71\times {{10}^{-2}}atm$

Answer 1

Hint:. Chemical equilibrium is the state of a system in which the concentration of the reactant and the concentration of the products are not changing with time and the system does not require any further change in properties.

Complete step by step answer:
Consider a general reversible reaction,
$A+B\rightleftharpoons C+D$
Where in the above-balanced chemical reaction A and B are reactants, C and D are products.
Then equilibrium constant, ${{K}_{c}}=\dfrac{[C][D]}{[A][B]}$ and this expression is called the equilibrium constant expression, and the subscript ‘c’ indicates that equilibrium constant expressed in concentrations of mol/L.

For equilibrium constant in gaseous systems, the equilibrium constant is generally expressed in terms of partial pressure.
Given the gaseous system, $2{{O}_{3}}(g)\rightleftharpoons 3{{O}_{2}}(g)$
The total pressure of the system = 8 atm
Then, ${{P}_{{{O}_{2}}}}+{{P}_{{{O}_{3}}}} = 8atm$ -- (1)

According to the question, the partial pressure of ozone in the gaseous system is very much less than the partial pressure of oxygen. i.e, ${{P}_{{{O}_{2}}}}>>{{P}_{{{O}_{3}}}}$
Therefore, from equation (1) partial pressure of oxygen, ${{P}_{{{O}_{2}}}}$ = 8 atm
the equilibrium constant in terms of partial pressure, given ${{K}_{p}}=4\times {{10}^{14}}atm$

Equilibrium constant expression for given chemical reaction,
${{K}_{p}}=\dfrac{{{({{P}_{{{O}_{2}}}})}^{3}}}{{{({{P}_{{{O}_{3}}}})}^{2}}}$
Substitute the partial pressure of oxygen and ${{K}_{p}}$ values in the above equation,
$\begin{align}
 & 4\times {{10}^{14}}=\dfrac{{{8}^{3}}}{{{({{P}_{{{O}_{3}}}})}^{2}}} \\
 & \Rightarrow {{P}_{{{O}_{3}}}}=11.3\times {{10}^{-7}}atm \\
\end{align}$
Hence, the partial pressure ${{O}_{3}}$ will be $11.3\times {{10}^{-7}}atm$
So, the correct answer is “Option B”.

Note: The product of concentrations of the reaction products raised to the respective stoichiometric coefficient in the balanced chemical equation divided by the product of reactants raised to their stoichiometric coefficients has a constant value at a constant temperature. This law is known as equilibrium law or law of chemical equilibrium.