Question

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

2NH3(g) <----> N2(g) + 3H2(g)

When a sufficiently large sample of NH3(g) is introduced into an evacuated vessel at 667 K, the equilibrium concentration of H2(g) is found to be 0.324 M.

Calculate the concentration of NH3 in the equilibrium mixture.

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