Consider the decomposition of ammonium hydrogen sulfide:

NH4Hs(s) <--> NH3(g) + H2S(g)
In a sealed flask at 25*C are 10.0 g NH4HS, ammonia with a partial pressure of .692 atm, and H2S with a partial pressure of .0532 atm. When equilibrium is established, it is found that the partial pressure of ammonia has increased by 12.4%. Calculate K for the decomp. of NH4HS at 25*C.

I'm not sure exactly how to do this, but my thoughts so far are these:
Finding K= P(NH3) * P(H2S)/1 (since NH4HS is a pure solid)
Since ammonia increased by 12.4%, I would guess that H2S would increase as well. Now, would H2S increase by the same percent as NH3(12.4 %), or by the same number of atm (.08580)?

Since 1 mol NH4HS decomposes to give 1 mol NH3 + 1 mol H2S, I think you convert the initial partial pressure of NH3 and H2S to mols, then add 12.4% to mols NH3 and recalculate partial pressure of NH3. Then subtract new mols NH3 from old mols NH3 to find the mols NH4HS that decomposed. That number of mols added to the initial mols H2S should be the mols H2S that are present at equilibrium. Then convert from mols to partial pressure. (I assumed a volume of 1 L for the PV = nRT but any volume should do as long as it doesn't change.)
Check my thinking.
(I used PV=nRT to calculate mols NH3 and added 12.4% to mols and reconverted to partial pressure NH3 OR just adding 12.4% to partial pressure of NH3 provided the same number for partial pressure NH3 at equilibrium. But for H2S, I think the key is to add the same number of mols to H2S that were added to NH3, then go backward and calculate partial pressure for H2S knowing nRT. Again, check my thinking.)

1 answer

  1. 1.37 g

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