When solid nh4hs and 0.28 mol nh3(g) were placed in a 2 l vessel at 24◦c, the equilibriumnh4hs(s)⇀↽nh3(g) + h2s(g)for which kc= 0.00016, was reached. what is the equilibrium concentration of nh3? answer in units of mol/l

The equilibrium concentration of NH3 =  0.141 mol/L

Explanation:

The balanced equation for the reaction is :

Here , you have to write the Kc carefully "Don't write SOLID substance in the expression of Kc"

Kc = equilibrium constant for concentration( Solid are not included in the expression)

So, Kc can be written as :

Now , look at the equation ,

Let x moles are dissociated from it at equilibrium

[NH4HS](s)            NH3                H2S

1                          0.28                 0      (Initial moles)

1-x                       x + 0.28        x      (Moles at equilibrium)

Calculate the concentration at equilibrium :

Moles of NH3 = 0.28 mol

The concentration of NH3 is calculated using :

Similarly concentration of H2S

Put the value in given equation:

Kc = 0.00016

Use the formula of quadratic equation to solve the value of

Here, b= 0.28 , c= -.00064 and   a = 1

you get ,

x = 0.002267 mol/L

or

[NH3] =  (x + 0.28)/2 = 0.002267 + 0.28

[NH3] = 0.2822/2 mol/L

[NH3] = 0.141 mol/L