chemistry, plz check work
posted by Anonymous on .
NH3+H2S>NH4HS
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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 2x 2x 2+x
Kc=[NH4HS]/[NH3][H2S]
400=(2+x)/[(2x)(2x)]
400=(2+x)/(44x+x^2)
16001600x+400x^2=2+x
15981599x+400x^2=0
x=0.500 mol
2x=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=12=1)
Kp gives me the pressure, right?

NH3 H2S NH4HS
initial 2 mol 2 mol 2 mol
change x x +x
equilibrium 2x 2x 2+x
I will write this more clearly:
NH3
initial:2 mol
change: x
equil: 2x
H2S
initial: 2 mol
change:x
equil:2x
NH4HS
initial: 2 mol
change:+x
equil:2+x 
by any chance, are you taking chem 125 at calpoly? because my takehome 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.

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.