Answer: The concentration of ammonia in the equilibrium mixture is 0.022 M
Explanation:
We are given:
Equilibrium concentration of hydrogen gas = 0.324 M
For the given chemical equation:
[tex]2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)[/tex]
Initial: a
At eqllm: a-2x x 3x
Evaluating the value of 'x':
[tex]\Rightarrow 3x=0.324\\\\\Rightarrow x=\frac{0.324}{3}=0.108[/tex]
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 [tex]K_c[/tex] for above equation follows:
[tex]K_c=\frac{[N_2][H_2]^3}{[NH_3]^2}[/tex]
We are given:
[tex]K_c=1.55[/tex]
Putting values in above expression, we get:
[tex]1.55=\frac{0.108\times 0.324}{(a-0.216)}\\\\a=0.238[/tex]
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