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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

Respuesta :

Answer:

The equilibrium concentration of NH3 =  0.141 mol/L

Explanation:

The balanced equation for the reaction is :

[tex]NH_{4}HS(s)\rightleftharpoons NH_{3}(g)+H_{2}S(g)[/tex]

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 :

[tex]Kc=[NH_{3}][H_{2}S][/tex]

Now , look at the equation ,

[tex]NH_{4}HS(s)\rightleftharpoons NH_{3}(g)+H_{2}S(g)[/tex]

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 :

[tex]Concentration=\frac{Moles}{Volume(L)}[/tex]

[tex][NH_{3}]=\frac{0.28 +x}{2}[/tex]

Similarly concentration of H2S

[tex][H_{2}S]=\frac{x}{2}[/tex]

Put the value in given equation:

[tex]Kc=[NH_{3}][H_{2}S][/tex]

Kc = 0.00016

[tex]0.00016=\frac{(0.28 +x)(x)}{2\times 2}[/tex]

[tex]0.00016\times 4 =(0.28 +x)(x)[/tex]

[tex]6.4\times 10^{-4}=0.28x +x^{2}[/tex]

[tex]x^{2}+.28x-6.4\times 10^{-4}=0[/tex]

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

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

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