[viviaa]

I have a question about the following problem:

H2 + I2 2HI
Calculate the percentage of HI at equilibrium, if HI = 1mol in the beginning and Kc = 25

I tried to solve this problem using the ICE Table:

c[H2] = 0 + x
c[I2] = 0 + x
c[2HI] = 1 - 2x

according to this x=1/3 (…)

However my teacher calculated something different:

c[H2] = 0.5x
c[I2] = 0.5x
c[2HI] = 1-x

-> x≈0.29

What did I do wrong here? Can I not use the ICE Table in this case? It usually works… (To be clear, my teacher does not use the ICE Table but she approved it, so I am allowed to use it. But because she doesn’t use it, her solutions are very confusing because she doesn’t show the intermediate steps.)

Thanks for any answers! (PS english is not my native language so I apologize for any mistakes, especially in the technical language)

[Aldebaran]

Yes you can use an ICE table and this should enable you to get the concentrations which your teacher has. After which it is a straightforward algebra calculation to solve the quadratic
You need to look carefully at which way round the Kc fraction is expressed and also notice that for each mol of HI consumed only half a mol of H₂ and I₂ is produced.

[Aldebaran]

Note the question actually asks for the percentage of HI once you have found x.

[Aldebaran]

Note also that the state symbols are absent from the equation. The calculation is assuming all in the gas phase.

[Babcock_Hall]

Your definition of x is different from your teacher's definition of x.

[Hunter2]

I didn't get 1/3 nor 0.29 for x.
 My result is 0.909.... with teachers approach. Where I am wrong.
The concentration of HI drops from 1 mol to 0.0909 mol means 9% left.  Is that correct?

[Aldebaran]

@ Hunter 2

Although this reaction is often written as a dissociation for the forward direction in this particular question the forward direction is the association thus I reckon Kc (which is 25) is products over reactants and this gives (1- x)^2 /(0.5x)^2. Solving for x gives 0.2857...

Happy to be corrected.

[Hunter2]

Why I have to do that?
The reaction is 2HI => H2 + I2 not H2 + I2 => 2HI
So I have to use products over reactant Kc = 25 = (0,5x)^2/(1-x)^2.
Why I have to use the association and not dissociation of the reaction, what you did?

[Aldebaran]

The equation given in the question is for the synthesis of HI being the forward reaction. My assumption is that the  Kc given is thus representative of the synthesis rather than the dissociation.
Like I said I’m happy to be corrected but I’m pretty sure the constant has to be related to the forward direction of the equation as given.

[Hunter2]

The question  was from original

‘’Calculate the percentage of HI at equilibrium, if HI = 1mol in the beginning and Kc = 25‘’

So if it is the beginning, then it can be only decomposition. There will no new HI.

The Experiment was to put HI im a tube and then get the equilibrium. Not put H2 and I2 in the vessel to get equilibrium.

[Aldebaran]

I take your point but the position at equilibrium is the same from whichever direction it is achieved so to translate the proportions of products and reactants into a constant requires that we use a convention which in my understanding  is normally products over reactants as the equation is given. In this case the product in the given equation is HI.

This link contains a lot of additional stuff but does directly deal with this point:  https://www.sas.upenn.edu/~mcnemar/apchem/CH15.pdf

[Borek]

I would assume Kc is for the reaction as written, not for the reaction as I imagine.