Answer:
[tex]p_{NO_2}^{EQ}=0.394atm\\p_{N_2O_4}^{EQ}=1.07atm[/tex]
Explanation:
Hello,
At first, it is convenient to compute Kp, since it remains constant during the proposed situation:
[tex]Kp=\frac{(p_{NO_2}^{EQ})^2}{p_{N_2O_4}^{EQ}}=\frac{(0.269atm)^2}{0.500atm}=0.145[/tex]
Now, since the volume is halved, the pressures of N₂O₄ and NO₂ are doubled, 1.00 atm and 0.538 atm respectively, but now they are considered as the initial pressures, so a change [tex]x[/tex] is now present in order to take the reaction again to the equilibrium. Besides, the reaction changes as the products have more moles as:
[tex]2NO_2(g)\rightleftharpoons N_2O_4(g)[/tex]
Thus, the law of mass action goes (considering the reverse reaction):
[tex]Kp'=\frac{1}{Kp}=\frac{1}{0.145} =\frac{1+x}{(0.538-2x)^2}[/tex]
Hence, solving for [tex]x[/tex] by using solver or the quadratic equation, one obtains:
[tex]x_1=0.072atm\\x_2=0.50atm[/tex]
Nonetheless, the feasible solution is 0.072 atm as 0.50 atm leads to a negative pressure of NO₂, thus, the new equilibrium pressures are:
[tex]p_{NO_2}^{EQ}=0.538-2*0.072=0.394atm\\p_{N_2O_4}^{EQ}=1+0.072=1.07atm[/tex]
Best regards.
