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

every chemical equation adheres to the law of conservation of mass, which states that matter cannot be created or destroyed. therefore, there must be the same number of atoms of each element on each side of a chemical equation.