.........NH4HS(s) ==> NH3(g) + H2S(g)
I........solid.........0........0
C........solid.........x........x
E........solid.........x........x
Kc = (NH3)(H2S)
1.8E-4 = x^2
x = 0.013M. Does this exhaust the 5.00 g NH4HS? 0.13 mols/L x 3L = 0.040 mols H2S and 0.040 mols NH3. How much NH4HS did we have initially. That's 5.00/51.112 = 0.0978 mol; therefore, there is some solid left after this decomposition.
That make the answer to the b part easier. Same answer as in part a. Because the solid is not part of the Kc expression.
Kc for the decomposition of ammonium hydrogen sulfide is 1.8 x 10-4 at 15oC.
a. When 5.00 grams of the pure salt decomposes in a sealed 3.0 L flask at 15oC, what are the equilibrium concentrations of NH3 and H2S?
b. What are the equilibrium concentrations of the products at 15oC if 10.0 grams of the pure salt decomposes in the sealed flask?
1 answer
1 answer