Solution :-
2H2S --- > 2 H2 +
S2Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
kc = 9.30*10^-8
0.29 mol in 3.0 L container
What is the equilibrium concentration of H2
Lets first calculate the initial concentration of the H2S
Molarity = moles / liter
Molarity of H2S = 0.29 mol / 3.0 L = 0.0967 M
Now lets make the ICE table
         Â
2H2S
          ---
>Â Â Â Â Â Â Â 2
H2Â Â Â Â Â Â Â Â
+Â Â Â Â Â Â Â Â S2
0.0967
MÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
0
-2x
                                       +2x                         Â
+x
0.0967-2x                             Â
2x                       Â
x
Now lets write the Kc equation
Kc= [H2]^2[S2]/[H2S]^2
9.30*10^-8 = [2x]^2[x]/[0.0967-2x]^2
9.30*10^-8 *[0.0967-2x]^2 = [2x]^2[x]
9.30*10^-8 *[0.0967-2x]^2 = 4x^3
Solving this equation to get x
We get
X= 0.00060 M
Therefore the equilibrium concentration of the H2 = 2x =
2*0.0006 M = 1.20 M
