Question

The equilibrium constant ${K_p}$ for the reaction ${N_2}{O_4} \rightleftharpoons 2N{O_2}$ is $4.5$. What would be the average molar mass ( in $g/mol$) of an equilibrium mixture of ${N_2}{O_4}$ and $N{O_2}$ formed by the dissociation of Pure ${N_2}{O_4}$ at total pressure of 2 atm?
(A) $69$
(B) $57.5$
(C) $80.5$
(D) $85.5$

Answer 1

Hint:At first we will note down the data given in the question. Then we will calculate the total number of moles hypothetically. We will find the partial pressure of each component. Then we will write the equation of equilibrium constant of the reaction. From there we will get the actual mole value. And from moles we can calculate the average molar mass.

Complete step-by-step solution:Step1. We are provided with the following data about the equation ${N_2}{O_4} \rightleftharpoons 2N{O_2}$
Equilibrium constant ${K_p}$ =4.5
The total pressure = 2atm
Step2. Let us assume that there was one mole at the starting of the reaction. Now let us assume that ‘a’ moles of dissociates and they will form ‘’ moles of .
So at equilibrium the total moles will be models.
Step3. Now we will find the partial pressure of $4.5$${N_2}{O_4}$.
The formula of partial pressure is $p = \dfrac{{moles}}{{total moles}} \times \Pr essure$
${p_{{N_2}{O_4}}} = \dfrac{{1 - a}}{{1 + a}} \times P$ ; Here the P is the total pressure.
${p_{N{O_2}}} = \dfrac{{2a}}{{1 + a}} \times P$ ; Here the P is the total pressure.
Step4: We will write the equation of equilibrium constant that will give us the value of ‘a’.
${K_p} = \dfrac{{{{({p_{{N_2}{O_4}}})}^2}}}{{{p_{N{O_2}}}}}$
$\Rightarrow 4.5 = \dfrac{{4{a^2}}}{{1 - {a^2}}} \times 2$ , $2$ is the total pressure.
$\therefore a = 0.6$ .
The moles dissociated will be $0.6$.
Step5. The total moles are $1.6$
The mass of ${N_2}{O_4}$ is $92$ and the mass of $N{O_2}$ is $46$ .
The average molecular mass = $\dfrac{{(1 - 0.6) \times 92}}{{1.6}} + \dfrac{{2 \times 0.6 \times 46}}{{1.6}} = 57.5$

Hence the correct option is (B).

Note:The equilibrium state is the state of the reaction where there is enough time has passed and the reaction has come at halt. It loses its tendency to move forward. The equilibrium constant is the value of the reaction quotient.