Explanation:
Chemical reaction equation for the give decomposition of [tex]NH_{3}[/tex] is as follows:.
[tex]2NH_{3}(g) \rightleftharpoons N_{2}(g) + 3H_{2}(g)[/tex]
And, initially only [tex]NH_{3}[/tex] is present.
The given data is as follows.
[tex]P_{NH_{3}}[/tex] = 2.3 atm at equilibrium
[tex]P_{H_{2}} = 3 \times P_{N_{2}}[/tex] = 0.69 atm
Therefore,
[tex]P_{N_{2}} = \frac{0.69 atm}{3}[/tex]
= 0.23 aatm
So, [tex]P_{NH_{3}}[/tex] = 2.3 - 2(0.23)
= 1.84 atm
Now, expression for [tex]K_{p}[/tex] will be as follows.
[tex]K_{p} = \frac{(P_{N_{2}})(P^{3}_{H_{2}})}{(P^{2}_{NH_{3}})}[/tex]
[tex]K_{p} = \frac{(0.23) \times (0.69)^{3}}{(1.84)^{2}}[/tex]
= [tex]\frac{0.23 \times 0.33}{3.39}[/tex]
= 0.0224
or, [tex]K_{p} = 2.2 \times 10^{-2}[/tex]
Thus, we can conclude that the pressure equilibrium constant for the decomposition of ammonia at the final temperature of the mixture is [tex]2.2 \times 10^{-2}[/tex].