Question

For the following reaction, the equilibrium constant
KC = 97.0 at 900 K. If the initial concentrations of NH3 and H2S are both 0.20 M, what is the equilibrium concentration of H2S?
H2S (g) + NH3 (g) ---> NH4HS (s)

Answer 1

Make an ICE chart and substitute into Kc and solve.

Answer 2

Could you help me on the ice table..........I could only get the I part :(

I: 0.2 0.2 0

Answer 3

Wait is this right?


I: 0.2 0.2 0
C: -x -x x
E: 0.2-x 0.2-x x

Answer 4

Wait is this right?


I: (0.2) (0.2) (0)
C: (-x) (-x) (x)
E: (0.2-x) (0.2-x) (x)

Answer 5

Now what do I do next once I have the ice table?

Answer 6

Set Kc = (x)(0.2-x)/(0.2-x) .
Solve for x.

0.2-x is H2S equilibrium concentration

Answer 7

Ummmmm I'm waiting to confirm this with DrBob222...Thanks anyways ;D

Answer 8

That looks pretty good.
We can't space on the board; let me show you how to do it. You trick it with periods.
..........NH3(g) + H2S(g) ==> NH4HS(s)
initial....0.2......0.2.........0
change.....-x........-x..........x
equil....0.2-x...0.2-x............

Kc = 97.0 = 1/[(0.2-x)(0.2-x)]
Solve for x. You can do it the long with with the quadratic formula but there is a short way to do it if you notice that the denominator is a perfect square.

Answer 9

initial....0.2......0.2.........0
change.....-x........-x..........x
equil....0.2-x...0.2-x............


equil....0.2-x...0.2-x............ <----what goes here

Did you forget to put an x?????

Answer 10

Note that NH4HS is a solid; therefore, it never goes into the Kc expression. I put a zero and x in the first two lines, although that wasn't necessary, and perhaps should have put one in the last line; however, I knew I didn't intend to use it in Kc and I just let it go.

Answer 11

Okay so why is there a 1 here:

Kc = 97.0 = 1/[(0.2-x)(0.2-x)]


where did you get the 1 from?

Also in this equation:
H2S (g) + NH3 (g) ---> NH4HS (s)

What if there was a 2 in front of the NH4HS(s)...would that make it a 2 in place of the one?

So for example: Kc = 97.0 = 2/[(0.2-x)(0.2-x)]

I know this would change the whole problem but hypothetically speaking would the 2 replace the 1 in this problem even if it is a solid? Or is it a 1 because of the solid.

Answer 12

For the question about the 2, no everything would stay the same. And you can see why with the answer to th second question.
Let's look at it the other way.
NH4HS(s) ==> NH3(g) + H2S(g)
Then Kc = products/reactants and
(NH3)(H2S) = Kc. Since NH4HS is a solid, by definition anything in its normal state = 1 so we could write
Kc = (NH3)(H2S)/1 but no one does that. But you can see that if we take the reverse of that Kc, the new Kc becomes 1/this Kc or 1/[(NH3)(H2S)] = 97.0

Answer 13

Thanks! I understand! You are the best :D