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
