Question

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

Answer 1

The equilibrium concentration of NH3 = 0.141 mol/L

Step-by-step explanation:

The balanced equation for the reaction is :

`NH_(4)HS(s)\rightleftharpoons NH_(3)(g)+H_(2)S(g)`

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 :

`Kc=[NH_(3)][H_(2)S]`

Now , look at the equation ,

`NH_(4)HS(s)\rightleftharpoons NH_(3)(g)+H_(2)S(g)`

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 :

`Concentration=(Moles)/(Volume(L))`

`[NH_(3)]=(0.28 +x)/(2)`

Similarly concentration of H2S

`[H_(2)S]=(x)/(2)`

Put the value in given equation:

`Kc=[NH_(3)][H_(2)S]`

Kc = 0.00016

`0.00016=((0.28 +x)(x))/(2* 2)`

`0.00016* 4 =(0.28 +x)(x)`

`6.4* 10^(-4)=0.28x +x^(2)`

`x^(2)+.28x-6.4* 10^(-4)=0`

Use the formula of quadratic equation to solve the value of
`x=\frac{-b\pm \sqrt{b^(2)-4(ac)}}{2a}`

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