<u>Answer:</u> The concentration of ammonia in the equilibrium mixture is 0.022 M
<u>Explanation:</u>
We are given:
Equilibrium concentration of hydrogen gas = 0.324 M
For the given chemical equation:

<u>Initial:</u> a
<u>At eqllm:</u> a-2x x 3x
Evaluating the value of 'x':

So, equilibrium concentration of nitrogen gas = x = 0.108 M
Equilibrium concentration of ammonia gas = (a - 2x) = [a - 2(0.108)] = (a - 0.216) M
The expression of
for above equation follows:
![K_c=\frac{[N_2][H_2]^3}{[NH_3]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BN_2%5D%5BH_2%5D%5E3%7D%7B%5BNH_3%5D%5E2%7D)
We are given:

Putting values in above expression, we get:

Equilibrium concentration of ammonia gas = (a - 0.216) = [0.238 - 0.216] = 0.022 M
Hence, the concentration of ammonia in the equilibrium mixture is 0.022 M