Question

NH3+H2S-->NH4HS
<--
Kc=400 at 35 Celsius
what mass of NH4HS will be present at equilbrium?

My work
NH3 H2S NH4HS
initial 2 mol 2 mol 2 mol
change -x -x +x
equilibrium 2-x 2-x 2+x

Kc=[NH4HS]/[NH3][H2S]
400=(2+x)/[(2-x)(2-x)]
400=(2+x)/(4-4x+x^2)
1600-1600x+400x^2=2+x
1598-1599x+400x^2=0
x=0.500 mol
2-x=1.5
2+x=2.5
What is the mass of NH4HS at equilibrium?
2.50 mol NH4HS(52.0205g/mol)=130 grams of NH4HS

Now, the given Kc=400 but when I solved for Kc by hand by using numbers I got 1.11
Kc=[NH4HS]/[NH3][H2S]
=2.5/[(1.50(1.5)]
=1.111
So is answer for mass correct even if I keep getting a different Kc.

How do I solve for press of H2S at equilibrium.

Can i use the equation Kp=Kc(RT)^n (n is moles=1-2=-1)

Kp gives me the pressure, right?

Answer 1

NH3 H2S NH4HS
initial 2 mol 2 mol 2 mol
change -x -x +x
equilibrium 2-x 2-x 2+x

I will write this more clearly:

NH3
initial:2 mol
change: -x
equil: 2-x

H2S
initial: 2 mol
change:-x
equil:2-x

NH4HS
initial: 2 mol
change:+x
equil:2+x

Answer 2

by any chance, are you taking chem 125 at calpoly? because my take-home quiz has the same exact question on it. anyways, the problem is that you need to use the molarity and not the moles. instead of having 2 mol for your initial concentrations, it needs to be 0.4M (2.00 mol/5.00 L) >>this is assuming that the volume of the container is 5.00 L. you should get the right Kc value if you replace 2 with 0.4. i hope this helps.

Answer 3

Another note here.
You made a math error two places. First the equation should be
400X^2 -1601X + 1598 = 0 (which won't make a large error at all).
And when you solve for X it should be 1.90 and not 0.5. All of that assumes, of course, that the volume is 1 L and from another post that appears not to be the case. In answer to your question, yes, the Kc should be 400 if you substitute the solution to the problem.