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July 1, 2015

Homework Help: chemistry, plz check work

Posted by Anonymous on Monday, March 2, 2009 at 1:26am.

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?

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