Complete Question
The complete question is shown on the first uploaded image
Answer:
The concentration equilibrium constant is [tex]K_c = 14.39[/tex]
Explanation:
The chemical equation for this decomposition of ammonia is
[tex]2 NH_3[/tex] ↔ [tex]N_2 + 3 H_2[/tex]
The initial concentration of ammonia is mathematically represented a
[tex][NH_3] = \frac{n_1}{V_1} = \frac{29}{75}[/tex]
[tex][NH_3] = 0.387 \ M[/tex]
The initial concentration of nitrogen gas is mathematically represented a
[tex][N_2] = \frac{n_2}{V_2}[/tex]
[tex][N_2] = 0.173 \ M[/tex]
So looking at the equation
Initially (Before reaction)
[tex]NH_3 = 0.387 \ M[/tex]
[tex]N_2 = 0 \ M[/tex]
[tex]H_2 = 0 \ M[/tex]
During reaction(this is gotten from the reaction equation )
[tex]NH_3 = -2 x[/tex](this implies that it losses two moles of concentration )
[tex]N_2 = + x[/tex] (this implies that it gains 1 moles)
[tex]H_2 = +3 x[/tex](this implies that it gains 3 moles)
Note : x denotes concentration
At equilibrium
[tex]NH_3 = 0.387 -2x[/tex]
[tex]N_2 = x[/tex]
[tex]H_2 = 3 x[/tex]
Now since
[tex][NH_3] = 0.387 \ M[/tex]
[tex]x= 0.387 \ M[/tex]
[tex]H_2 = 3 * 0.173[/tex]
[tex]H_2 = 0.519 \ M[/tex]
[tex]NH_3 = 0.387 -2(0.173)[/tex]
[tex]NH_3 = 0.041 \ M[/tex]
Now the equilibrium constant is
[tex]K_c = \frac{[N_2][H_2]^3}{[NH_3]^2}[/tex]
substituting values
[tex]K_c = \frac{(0.173) (0.519)^3}{(0.041)^2}[/tex]
[tex]K_c = 14.39[/tex]
