Question

For the reaction ${N_2}{O_3} \rightleftharpoons NO + N{O_2}$ total pressure = p, degree of dissociation = 50%, then ${K_p}$ would be:
A. $3P$
B. $2P$
C. $\dfrac{P}{3}$
D. $\dfrac{P}{2}$

Answer 1

Hint: We can calculate the equilibrium constant ${K_p}$ using the partial pressures of $N{O_2}$ , $NO$ and ${N_2}{O_3}$ . The partial pressures of $N{O_2}$ , $NO$ and ${N_2}{O_3}$ are calculated using the moles of $N{O_2}$ , $NO$ and ${N_2}{O_3}$ and the total moles. The moles of $N{O_2}$ , $NO$ and ${N_2}{O_3}$ are calculated using the degree of dissociation.

Complete step by step answer:
Given data contains,
Degree of dissociation $\left( \alpha \right)$ is 50% that is equal to 0.50.
Let us take the total pressure P.
We can the equilibrium reaction as,
Reaction :${N_2}{O_3} \rightleftharpoons N{O_2} + NO$
\[
\begin{array}{*{20}{c}}
{{\text{Initial moles}}}&{{\text{ }}1}&{{\text{ }}0}&{{\text{ }}0}
\end{array} \\
\begin{array}{*{20}{c}}
{{\text{At equilibrium}}}&{1 - \alpha }&\alpha &{{\text{ }}\alpha }
\end{array} \\
 \]
 We now calculate the total number of moles. The total number of moles is the sum of $N{O_2}$ , $NO$ and ${N_2}{O_3}$ .
We can now substitute the values of $N{O_2}$ , $NO$ and ${N_2}{O_3}$ to get the total number of moles.
Total number of moles = $1 - \alpha + \alpha + \alpha $
Total number of moles = $1 + \alpha $
We know that the value of $\alpha $ is 0.50. Let us now substitute the value of $\alpha $ in the expression of total number of moles.
Total number of moles = $1 + 0.50$
Total number of moles = $1.5$
The total number of moles is $1.5$.
We can now calculate the moles of $N{O_2}$, $NO$ and ${N_2}{O_3}$ using the value of $\alpha $.
Moles of ${N_2}{O_3}$ = $\left( {1 - \alpha } \right)$
Moles of ${N_2}{O_3}$ = $1 - 0.50$
Moles of ${N_2}{O_3}$ = $0.50$
Moles of ${N_2}{O_3}$ is $0.50$moles.
Moles of $N{O_2}$ = $\alpha $
Moles of $N{O_2}$ = $0.50$
Moles of $N{O_2}$ = $0.50$
Moles of $N{O_2}$ is $0.50$ moles.
Moles of $NO$ = $\alpha $
Moles of $NO$ = $0.50$
Moles of $NO$ = $0.50$
Moles of $NO$ is $0.50$ moles.
Let us now calculate the partial pressures of $N{O_2}$, $NO$ and ${N_2}{O_3}$.
We can calculate the partial pressures of $N{O_2}$, $NO$ and ${N_2}{O_3}$ using the moles of $N{O_2}$, $NO$ and ${N_2}{O_3}$ and the total number of moles.
Moles of ${N_2}{O_3}$ is $0.50$ moles.
Moles of $N{O_2}$ is $0.50$ moles.
Moles of $NO$ is $0.50$ moles.
The total number of moles is $1.5$.
Partial pressure of ${N_2}{O_3}$ = $P\left( {\dfrac{{{\text{Moles of }}{N_2}{O_3}}}{{{\text{Total number of moles}}}}} \right)$
Partial pressure of ${N_2}{O_3}$ = $P\left( {\dfrac{{0.50}}{{1.5}}} \right)$
Partial pressure of ${N_2}{O_3}$ = $\dfrac{P}{3}$
The partial pressure of ${N_2}{O_3}$is $\dfrac{P}{3}$.
Partial pressure of $N{O_2}$ = $P\left( {\dfrac{{{\text{Moles of }}N{O_2}}}{{{\text{Total number of moles}}}}} \right)$
Partial pressure of $N{O_2}$ = $P\left( {\dfrac{{0.50}}{{1.5}}} \right)$
Partial pressure of $N{O_2}$ = $\dfrac{P}{3}$
The partial pressure of $N{O_2}$is $\dfrac{P}{3}$.
Partial pressure of $NO$ = $P\left( {\dfrac{{{\text{Moles of }}NO}}{{{\text{Total number of moles}}}}} \right)$
Partial pressure of $NO$ = $P\left( {\dfrac{{0.50}}{{1.5}}} \right)$
Partial pressure of $NO$ = $\dfrac{P}{3}$
The partial pressure of $NO$is$\dfrac{P}{3}$.
Using the partial pressures of $N{O_2}$ , $NO$ and ${N_2}{O_3}$ , let us calculate the value of ${K_P}$.
The formula to calculate ${K_P}$ is,
${K_P} = \dfrac{{{P_{N{O_2}}} \times {P_{NO}}}}{{{P_{{N_2}{O_3}}}}}$
We know the partial pressures of $N{O_2}$, $NO$ and ${N_2}{O_3}$ are $\dfrac{P}{3}$. We can substitute these values in the expression of ${K_P}$.
${K_P} = \dfrac{{\left( {\dfrac{P}{3}} \right) \times \left( {\dfrac{P}{3}} \right)}}{{\left( {\dfrac{P}{3}} \right)}}$
${K_P} = \dfrac{P}{3}$
The value of ${K_P}$ is $\dfrac{P}{3}$.

Therefore, the option C is correct.

Note: As we know that there is no change in solubility of liquids and solids with pressure changes, whereas in gases, the solubility increases with increase in pressure. An example of solubility of gases with changes in pressure is carbonated beverages. All carbonated beverages are kept under pressure to raise the amount of carbon dioxide dissolved in the solution. When we open the bottle, the pressure present above the solution reduces and it leads to effervescences of the solution, and few amounts of carbon dioxide bubbles.