Answer:
12. is the pressure equilibrium constant for the decomposition of ammonia at the final temperature of the mixture.
Explanation:
[tex]2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)[/tex]
initially
3.0 atm 0 0
At equilibrium
(3.0-2p) p 3p
Equilibrium partial pressure of nitrogen gas = p = 0.90 atm
The expression of a pressure equilibrium constant will be given by :
[tex]K_p=\frac{p_{N_2}\times (p_{H_2})^3}{(p_{NH_3})^2}[/tex]
[tex]K_p=\frac{p\times (3p)^3}{(3.0-2p)^2}[/tex]
[tex]=\frac{0.90 atm\times (3\times 0.90 atm)^3}{(3.0-2\times 0.90 atm)^2}[/tex]
[tex]K_p=12.30\approx 12.[/tex]
12. is the pressure equilibrium constant for the decomposition of ammonia at the final temperature of the mixture.
