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 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?

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.

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

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.

Yes, you can use the equation Kp = Kc(RT)^n to calculate the equilibrium pressure of H2S. In this equation, Kp represents the equilibrium constant in terms of pressure, Kc is the equilibrium constant in terms of concentration, R is the gas constant, T is the temperature in Kelvin, and n is the difference in moles of gas between the products and reactants in the balanced chemical equation.

In the given reaction:
NH3 + H2S ↔ NH4HS

The balanced chemical equation shows that the difference in moles of gas is -1 (1 mol of H2S - 2 mol of NH3), so n is equal to -1.

To solve for the equilibrium pressure of H2S, you will need the equilibrium constant expressed in terms of pressure (Kp), which can be calculated using the equilibrium constant expressed in terms of concentration (Kc) and the ideal gas law.

To convert from Kc to Kp, you can use the equation:
Kp = Kc(RT)^n

Given that Kc = 400 and the temperature (T) is 35 degrees Celsius, you need to convert it to Kelvin by adding 273:
T = 35 + 273 = 308 K

Now you can substitute the values into the equation:
Kp = Kc(RT)^n
Kp = 400(R * 308)^-1

You will also need to know the value of the gas constant R, which is 0.0821 L * atm / (K * mol).

Let's assume the equilibrium pressure of H2S is represented as P. You can solve for P by rearranging the equation:
P = Kp * (RT)^n

Substituting the values:
P = 400 * (0.0821 * 308)^-1

Calculating this expression will give you the equilibrium pressure of H2S.