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

Step-by-step 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=(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=(1)/((2.0-x)* (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