Question

The equilibrium constant, Kc, for the following reaction is 1.55 at 667 K.

Answer 1

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:

                     $2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)$

Initial:                  a

At eqllm:           a-2x           x          3x

Evaluating the value of 'x':

$\Rightarrow 3x=0.324\\\\\Rightarrow x=\frac{0.324}{3}=0.108$

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 $K_c$ for above equation follows:

$K_c=\frac{[N_2][H_2]^3}{[NH_3]^2}$

We are given:

$K_c=1.55$

Putting values in above expression, we get:

$1.55=\frac{0.108\times 0.324}{(a-0.216)}\\\\a=0.238$

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