Question

Help please, What is the minimum mass of NH4HS that must be added to the 5.00-L flask when charged with the 0.350g of pure H2s(g), at 25 ∘C to achieve equilibrium?


Ammonium bisulfide, NH4HS, forms ammonia, NH3, and hydrogen sulfide, H2S, through the reaction
NH4HS(s)⇌NH3(g)+H2S(g)
This reaction has a K value of 0.120 at 25 ∘C.

PNH3, PH2S at equilibrium =
0.322,0.373
bar

mole fraction = 0.537

been trying for hours, any help is greatly appreciated thanks

Answer 1

Is the Kc you list as 0.120 Kp or Kc?

Answer 2

It only says "K" , there is nothing about a Kc or Kp

Answer 3

I found on the web that 0.12 is Kp (not Kc).
mols H2S = 0.35g/34.08 = about 0.0103
Then P = nRT/V. Substitute n, R, T and solve for P which is partial pressure H2S. I get approximately 0.05 atm. You may use bar if you wish. Then

NH4HS ==> NH3 + H2S
solid.....0......0.05
solid.....x.......x
solid.....x.....0.05+x

Kp = pNH3*pH2S
0.12 = x*(0.05+x+
x^2 + 0.05 = 0.12
Solve for x = partial pressure NH3.
x = about 0.07 atm but you need to clean that up. I used atm but you can use bar.
After finding partial pressure NH3, then use PV = nRT and solve for n = # mols NH3. Then since 1 mol H2S = 1 mol NH3 = 1 mol NH4HS, then you see you need enough NH4HS to provide the x pressure ( atm for NH3). Then grams NH4HS = x mols NH3 x molar mass NH4HS. I think that's about 2.5 grams or so but check my work and clean it up for more accuracy. Most of my numbers are estimates.

Answer 4

Note: The oops just before this was worked a different way but I deleted that and reposted the above version. I believe that is right.