Respuesta :
The question is incomplete, the complete question is ;
Ammonia decomposes to form nitrogen and hydrogen, like this:
[tex]2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)[/tex]
Also, a chemist finds that at a certain temperature the equilibrium mixture of ammonia, nitrogen, and hydrogen has the following composition:
Compound Pressure at equilibrium.
[tex]NH_3[/tex] 98.8 atm
[tex]N_2[/tex] 14.8 atm
[tex]H_2[/tex] 77.3 atm
Calculate the value of the equilibrium constant for this reaction.
Answer:
700.30 is the value of the equilibrium constant for this reaction.
Explanation:
[tex]2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)[/tex]
Partial pressure of ammonia gas = [tex]p_1=98.8 atm[/tex]
Partial pressure of nitrogen gas = [tex]p_2=14.8 atm[/tex]
Partial pressure of hydrogen gas = [tex]p_3=77.3 atm[/tex]
The expression of equilibrium constant in terms of partial pressure is given as:
[tex]K_p=\frac{p_2\times (p_3)^3}{(p_1)^2}[/tex]
[tex]K_p=\frac{14.8\times (77.3)^3}{(98.8)^2}=700.30[/tex]
700.30 is the value of the equilibrium constant for this reaction.
