Hydrogen sulfide decomposes according to the following reaction,for which Kc = 9.30 multiplied by 10-8 at...

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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 Â Â Â Â Â Â Â Â Â 2H2S Â Â Â Â Â Â Â Â Â Â --- >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 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 Â Â Â Â Â Â Â Â Â Â +2xÂ 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 +x

0.0967-2xÂ 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 2xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 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

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