Question

Determine the concentration at equilibrium if you start with 4.65 grams of Hydrogen and 200 grams of Iodine in a 4.55 liter container. If you now add an extra 1.25M of I2 after equilibrium, calculate Qc. Recalculate now what the concentrations should be at equilibrium.

H2(g) + I2(g) <----> 2HI(g) Kc = 56


Where do I begin with this question?

Answer 1

You should begin by calculating the original, unreacted concentrations of H2, I2, and HI. From there you can set up a Kc equation and solve for the change in concentrations, which can be easily used to find the concentrations after equilibrium is achieved.

Answer 2

Actually, I think Enrique is talking about the first equilibrium. After that, you will need to set up a second one with the added 1.25 M I2 and run through a second set of calculations. Post your work as far as you can get and explain what you don't understand if you get stuck. We can help you through it.

Answer 3

H2=-0.0037
I2=-0.0037
Hi=0.0276

(0.0276)^2/(0.0037)(0.0037)=56

2.00-x +2.00-x==> 2x

Am I heading in the right direction?

Answer 4

I don't think so.
4.65 g H2 x (1 mol/2g) x (1/4.55L) = not 0.0037 or perhaps I've misinterpreted your work. And what's with the - signs?

Answer 5

I'm not sure of what you did, I got different numbers. Do expect me to be wrong though, I haven't done any chemistry since I took the AP chem test in May. The initial, unreacted concentrations I got were .506M H2, .173M I2, and since there is no HI formed yet that is 0M. AFter that i did my set up as
[HI+2x]^2
56=_________
[H2-x][I2-x]
In which the reactants represent their unreacted concentrations, and x is the change in concentration after the reaction. For HI this value would be 2x, due to its coefficient.

Answer 6

Sorry, 56=[HI+2x]^2/([H2-x][I2-x])

Answer 7

I did the same thing as Enrique with initial (H2) = 0.510; (I2) = 0.173; (HI) = 0, did the set up and my final concentrations are not close to any numbers listed. You know the negative numbers can't be right.