MrBostonChemist - 22-10-2014 at 10:54
Last week my professor gave us a bonus question because she LOVES to challenge us and make us think. But she won't give us the answers... 
EVER. My guess is to keep us interested and give us an itch. But this one I swear I
was close too but can't wrap my head around. Please help or advise. 
"The Following equilibrium was studied by analyzing the equilibrium mixture for the amount of H2S produced:
Sb2S3(s) + 3H2(g) <===> 2Sb(s) +3H2S(g)
A vessel whose volume was 2.50L was filled with 0.0100 mol of each of the two reactants. After the mixture came to equilibrium in the closed
vessel at 440(degrees C), the gaseous mixture was removed and the hydrogen sulfide was dissolved in water. Sufficient lead (II) nitrate was added to
react completely with the hydrogen sulfide. If 1.029g of precipitate were collected, what is the value of the equilibrium constant in terms of
concentration at this temperature?"
~MBC
Metacelsus - 22-10-2014 at 11:17
Hint: You won't be able to include the solids in the equation, just the gases.
Magpie - 22-10-2014 at 17:46
Show us how you calculated the moles of H2S that were generated.
blogfast25 - 23-10-2014 at 04:52
Mol of H2S formed = 1.029 g / 239.3 g/mol = 0.0043 mol (1 mol H2S forms one mol PbS on analysis)
Mol H2 left = 0.01 - 0.0043 = 0.0057 mol (1 mol H2S requires 1 mol H2)
Both are in the same volume so no need to calculate actual concentrations, here:
K = [H<sub>2</sub>S]<sup>3</sup>/[H<sub>2</sub>]<sup>3</sup> =
(0.0043/2.5)<sup>3</sup>/(0.0057/2.5)<sup>3</sup> = 0.429
Could also be calculated from ΔG<sub>reaction</sub> and Nernst.
[Edited on 23-10-2014 by blogfast25]
Magpie - 23-10-2014 at 09:11
Blogfast you get an "A" but the OP didn't do any of the work himself.