Question

If the initial concentrations of NH3(g) and H2S(g) are 2.0 M, what is the equilibrium concentration of NH3(g)?

Answer 1

The question is incomplete, here is the complete question:

$K_c=9.7$ at 900 K for the reaction $NH_3(g)+H_2S(g)\rightleftharpoons NH_4HS(s)$

If the initial concentrations of $NH_3(g)$ and $H_2S(g)$ are 2.0 M, what is the equilibrium  concentration of $NH_3(g)$ ?

Answer: The equilibrium concentration of ammonia is 0.32 M

Explanation:

We are given:

Initial concentration of ammonia = 2.0 M

Initial concentration of hydrogen sulfide = 2.0 M

For the given chemical reaction:

              $NH_3(g)+H_2S(g)\rightleftharpoons NH_4HS(s)$

Initial:        2.0      2.0

At eqllm:    2.0-x     2.0-x            x

The expression of $K_c$ for above equation follows:

$K_c=\frac{1}{[NH_3][H_2S]}$

The concentration of pure solids and pure liquids are taken as 1

We are given:

$K_c=9.7$

Putting values in above equation, we get:

$9.7=\frac{1}{(2.0-x)\times (2.0-x)}\\\\9.7x^2-38.8x+38.8=1\\\\x=2.32,1.68$

Neglecting the value of x = 1.68 because equilibrium concentration cannot be greater than initial concentration

So, equilibrium concentration of ammonia = 2 - x = (2 - 1.68) = 0.32 M

Hence, the equilibrium concentration of ammonia is 0.32 M