Question

If ${K_p}$ for the reaction ${N_2}{O_4} \rightleftharpoons 2N{O_2}$​ is $0.66$ , then what is the equilibrium pressure of ${N_2}{O_4}$ ​? (Total pressure at equilibrium is $0.5\;atm$ )
(A) $0.168$
(B) $0.322$
(C) $0.1$
(D) $0.5$

Answer 1

${K_p}$ is the equilibrium constant at constant pressure. We are given the values of equilibrium constant at constant pressure $({K_p})$ and the total pressure of the reaction at equilibrium conditions. Now, to solve this question we will first find out the relation between these two terms with the help of the reaction given. Then we will find out the equilibrium pressure of ${N_2}{O_4}$ .

Complete answer:
${K_p}$ is the equilibrium constant at constant pressure. Now, first we will find out the relation between these two terms with the help of the reaction given as follows:
${N_2}{O_4} \rightleftharpoons 2N{O_2}$
At $t = 0$ , the individual moles are:
${N_2}{O_4} = 1$
$N{O_2} = 0$
At equilibrium, when $t = {t_{eq}}$ , the individual pressures are:
${N_2}{O_4} = 1 - x$
$N{O_2} = 2x$
Now, we will find out the mole fraction for both:
Mole fraction of ${N_2}{O_4} = \dfrac{{1 - x}}{{1 + x}}$
Mole fraction of $N{O_2} = \dfrac{{2x}}{{1 + x}}$
Now, we are given the total pressure at equilibrium which is equal to $0.5\;atm$ .
Now partial pressures of both is given as:
{P_{{N_2}{O_4}}} = \left( {\dfrac{{1 - x}}{{1 + x}}} \right) \times 0.5\\_\\_\\_\\_\\_\\_(1)
{P_{N{O_2}}} = \left( {\dfrac{{2x}}{{1 + x}}} \right) \times 0.5\\_\\_\\_\\_\\_(2)
We will now write the equilibrium constant equation as:
{K_p} = \dfrac{{{{[{P_{N{O_2}}}]}^2}}}{{{P_{{N_2}{O_4}}}}}\\_\\_\\_\\_\\_\\_(3)
Now, we are given with the value of equilibrium constant as:
${K_p} = 0.66$
Put these values in above equation, we get
$0.66 = \dfrac{{{{\left[ {\left( {\dfrac{{2x}}{{1 + x}}} \right) \times 0.5} \right]}^2}}}{{\left( {\dfrac{{1 - x}}{{1 + x}}} \right) \times 0.5}}$
On simplifying the above equation, we get:
$0.66 = \dfrac{{4 \times {x^2} \times 0.5}}{{1 - {x^2}}}$
$0.66 - 0.66{x^2} = 2{x^2}$
$2.66{x^2} = 0.66$
Therefore, on solving we get the value of x as:
$x = 0.5$
Now, put the value of x in equation first:
${P_{{N_2}{O_4}}} = \left( {\dfrac{{1 - 0.5}}{{1 + 0.5}}} \right) \times 0.5$
${P_{{N_2}{O_4}}} = \left( {\dfrac{{0.5}}{{1.5}}} \right) \times 0.5$
Therefore, the partial pressure of ${N_2}{O_4}$ at equilibrium is:
${P_{{N_2}{O_4}}} = 0.168atm$
Therefore, the correct option is (A) $0.168$ .

Note:
There is one more equilibrium constant which is known as equilibrium constant at constant concentration and it is denoted by ${K_c}$ . It is independent of the actual amount of reactants and products and pressure of inert material or catalysts. Its value changes with the change in temperature. It can also change with the stoichiometry of the reactants and products.