Answer:
[tex]Kp=5.14[/tex]
Explanation:
Hello,
In this case, the undergoing chemical reaction is:
[tex]N_2+3H_2\rightarrow 2NH_3[/tex]
Thus, the equilibrium expression is written as:
[tex]Kp=\frac{p_{NH_3}^2}{p_{N_2}p_{H_2}^3}[/tex]
And in terms of the reaction extent:
[tex]Kp=\frac{(2x)^2}{(10-x)(10-3x)^3}[/tex]
Thus, from the equilibrium pressure of ammonia we can compute the reaction extent:
[tex]p_{NH_3}=2x=6.0 atm\\\\x=3.0atm[/tex]
Therefore, the equilibrium constant turns out:
[tex]Kp=\frac{(2*3.0)^2}{(10.0-3.0)(10.0-3*3.0)^3}\\\\Kp=5.14[/tex]
Regards.
