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For the reaction: ${N_2}{O_{3(g)}} \rightleftharpoons N{O_{(g)}} + N{O_{2(g)}}$total pressure $ = P$ , degree of dissociation $ = 50\% $ . Then ${K_p}$ would be:
A. $3P$
B. $2P$
C. $\dfrac{P}{3}$
D. $\dfrac{P}{2}$

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Last updated date: 23rd Jun 2024
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Answer
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Hint: ${K_p}$ is equilibrium constant for the gaseous mixture. It is used when the concentration of gases at equilibrium is expressed in atmospheric pressure. The degree of dissociation is the fraction of reactant that is dissociated.

Complete step by step answer:
The equilibrium in the given reaction is homogeneous equilibrium since all the components are in the gaseous phase.
Now in the reaction,
${N_2}{O_{3(g)}} \rightleftharpoons N{O_{(g)}} + N{O_{2(g)}}$
Let $a$moles be taken of ${N_2}{O_3}$ in a closed vessel. Once equilibrium is reached $x$ moles of ${N_2}{O_3}$ dissociate into $NO\,and\,N{O_2}$ . One mole of ${N_2}{O_3}$ gives one mole of each $NO\,and\,N{O_2}$ . Thus $x$ moles of ${N_2}{O_3}$ gives $x$ moles of $NO$ and $x$ moles of $N{O_2}$ .
Remember when $a = 1$ , $x$ becomes a degree of dissociation .
So,
Initial moles: $1$$0$$0$
At equilibrium: $1 - \alpha $$\alpha $$\alpha $

${N_2}{O_{3(g)}} \rightleftharpoons N{O_{(g)}} + N{O_{2(g)}}$

It is given that degree of dissociation is $50\% $ . it means that $\alpha = 0.5$ .
So total moles at equilibrium $ = (1 - \alpha ) + \alpha + \alpha = 1 + \alpha $
$ = 1 + 0.5 = 1.5$
Now in case of gaseous reactants and products for calculating ${K_p}$ we consider partial pressure of gaseous reactants and products.
Partial pressure of any gas is equal to the product of total pressure and mole fraction of that gas.
In our example, the total pressure is $P$ and we know the number of moles of each component.
So, partial pressure of ${N_2}{O_3}$ can be given as-

${p_{{N_2}{O_3}}} = P(\dfrac{{{x_{{N_2}{O_3}}}}}{{total\; moles}}) = P(\dfrac{{0.5}}{{1.5}}) = \dfrac{P}{3}$

Partial pressure of $NO$-

${p_{NO}} = P(\dfrac{{{x_{NO}}}}{{total \;moles}}) = P(\dfrac{{0.5}}{{1.5}}) = \dfrac{P}{3}$

Partial pressure of $N{O_2}$ -

${p_{N{O_2}}} = P(\dfrac{{{x_{N{O_2}}}}}{{total\; moles}}) = P(\dfrac{{0.5}}{{1.5}}) = \dfrac{P}{3}$

Now, we need to calculate ${K_p}$ in terms of partial pressure of reactants and products-

${K_p} = \dfrac{{{p_{N{O_2}}} \times {p_{NO}}}}{{{p_{{N_2}{O_3}}}}}$

${K_p} = \dfrac{{(\dfrac{P}{3})(\dfrac{P}{3})}}{{\dfrac{P}{3}}} = \dfrac{P}{3}$

So ${K_p} = \dfrac{P}{3}$

So, the correct answer is Option C.

Note: ${K_p}$ is calculated when the reactants and products are in gaseous phase and the concentrations are expressed in terms of partial pressure .${K_c}$ is the equilibrium constant of the reaction when the concentrations of reactants and products at equilibrium are expressed in molarity. ${K_p}$ is related to ${K_c}$ by the equation-${K_p} = {K_c}{(RT)^{\Delta n}}$ where $\Delta n$ is the difference between the number of moles of gaseous products and reactants, $R$ is gas constant and $T$ is the temperature.