Question

H2S decomposes at 438k ,according to the following equilibrium:
H2S(g)=H2(g)+S(s)
Knowing that Kc=1.1*10-2,determine the gram amount of sulfur formed at the equilibrium coditions in a 5L flask ,where the initial concentration of H2S was 0.7M.
Also,calculate Kp.

Answer 1

......H2S ==> H2 + S(s)
I.....0.7.....0.....0
C.....-x......x.....-
E...0.7-x.....x.....-

NOTE that S is a solid and isn't in the Kc expression: Kc = (H2)/(H2S)
Substitute the E line into Kc expression and solve for x = (H2) in mols/L.
The problem asks for grams S and that isn't anywhere in Kc. But you know that 1 mols S is formed for every 1 mol H2 produced. Therefore, convert M H2 to mols in a 5L container.
mols H2 = M H2 x 5L = ? and convert mols H2 = mols S.
Then grams S = mols S x molar mass S = ?

Answer 2

Find Kp by Kp = Kc(RT)^delta n.
Remember S is a solid and not a gas.