[Rutherford]

Assume that you prepared a solution that corresponds to the equilibrium state of the system by mixing Fe(III)=0.027M, Fe(II)=0.173M, iodide=0.04M and three-iodide(I₃^-)=0.0866M. If at the beginning all of the iodide was radioactive, what is the concentration of the radioactive iodide (I^-) after obtaining real dynamic equilibrium.

I had an answer, but I don't understand it a bit. What is actually happening here?

[Hunter2]

Basicly

Fe³⁺ + e^- => Fe²⁺

3 I^- => I₃^- + 2 e^-

[Rutherford]

2Fe³⁺+3I^- 2Fe²⁺+I₃^-
I have the beginning concentrations now, but I don't have K_eq to calculate the equilibrium concentrations. What now?

[Hunter2]

With your datas you can calculate K_eq.

[Rutherford]

These are the beginning concentrations, how to use them to calculate K?

[Hunter2]

Assume that you prepared a solution that corresponds to the equilibrium state of the system by mixing

In my understanding these are the concentration in equilibrium.

Iodide is radiactive and Tri-iodide is not. The question now is what happend if the iodine/iodide exchange each other.

[Rutherford]

It says that an equilibrium is obtained by mixing the solutions of given concentrations. So, at the beginning I had ions with given concentrations, mixed them, and then the changes occured, [Fe³⁺] and [I^-] decrease, and the other two increase, so at the end equilibrium is obtained. I understood it this way. If you somehow logically get the answer 0.0053mol/dm³ then you are right.

[Vidya]

they should have given Keq value. You are right that concentration mentioned in the question are the starting or initial concentration.This question is not giving us full information. 

[Rutherford]

I will write the answer that I don't understand hoping that someone understands it and explains more clearly:

While establishing the real dynamic equilibrium all iodine atoms (whetever radioactive or not) are distributed between I^- and I₃^-. Every sample of this mixture must contain the properties of the whole mixture, so the iodides contain that much radioactive iodine as the whole mixture contains. The equilibrium mixture contains 0.04M of I^-(radioactive) and 3*0.0866=0.2598M atoms of non-radioactive iodine from three-iodide, then 0.04/(0.04+0.2598)*100=13.34% of radio-active iodide is in the solution. The equilibrium concentration of iodide must contain the same amount of radioactive I, so 0.1334*0.04=0.0053M.

[fledarmus]

Take particular note of the definition of the starting mixture:

a solution that corresponds to the equilibrium state of the system

This means that none of your reactants or starting materials will be changing concentrations. Everytime an I^- is formed, an I₃^- ion disappears and vice versa.

This question is simply trying to point out that this is a dynamic equilibrium. I don't know the mechanism of this particular reaction, but whatever the mechanism, all of the iodine atoms available in the system can participate and will freely interchange. So if you start with a small amount of radioactive iodine atoms in one form and cold iodine atoms in another form, if you wait long enough, the radioactive iodine will be randomly distributed between the two forms. In this case, you have .04M hot iodine atoms and 0.0866M*3 cold iodine atoms, for a total of 0.2998M iodine atoms. The percentage of hot atoms will be .04M/.2998M = 13.36% (I believe you intended + instead of * in your math). Since the hot atoms are eventually randomly distributed throughout all of the iodine atoms in the system, eventually any iodine atom in the system as 13.36% in the form of hot iodine. That means your I^-, which started as 100% hot iodine, will have dropped to 13.36%. Which is, as you point out, 0.0053M.

By measuring the amount of radioactive I^- at several points during the reaction, you can determine how fast the exchange is occurring and directly measure the reaction rates of a reaction at equilibrium.

[Hunter2]

First the given values for the mixture was the equilibrium values. K was not necessary to use here. Now the dynamic equilibrium between Iodide and Iodine (tri iodide) has to be investigated)

Beside the term 0.04/(0.04*0.2598)*100=13.34% everything is correct. It has to be a + in the equation.

0.04/(0.04+0.2598)*100=13.34%

[Rutherford]

That was a typing mistake I corrected now.

[Rutherford]

Hunter, I admit you were right, I didn't understand the sentence well.
I will comment what fledarmus wrote tomorrow, because this need a lot of thinking.

[Rutherford]

Okay, I thought for quite a time about this problem and I came to the following questions:
I had 0.04M of hot iodine atoms, and 0.0866M*3 of cold iodine atoms. I understood that the share of the hot atoms in the solution compared to the sum of both cold and hot atoms is 13.34%. Now,
1.how do the atoms distribute between themselves?
2.If I had 0.04M of hot atoms, at the end I left with 0.0053M, were the other hot atoms decayed?
3."any iodine atom in the system as 13.36% in the form of hot iodine"
Don't understand this.
4.Why was iron introduced? I still think that there is an easier to understand way to calculate all this by using the concentration of iron, because the answer I wrote wasn't made by the original author of this problem.

[fledarmus]

Let me take these in reverse order - I think it will make it easier to follow.

4. Iron is a catalyst in this reaction - it's concentration really doesn't matter. It is only there to lower the activation energy required to move iodine among the various oxidation states in which it exists.

3. You have 4 possibilities for iodine atoms - they can exist in your reaction either as I^- or as any of the three iodine atoms in I₃^-. When you start your reaction, all of the I^- is hot and each of the three iodine atoms in I₃^- is cold. When the reaction is complete, 13.36% of the I^- is hot and 13.36% of each atom in the I₃^- is also iodine. If you are interested in statistics, you could calculate as well how much I₃^- would have 2 or 3 hot iodine atoms present. The 0.04M of hot iodine in the I^- has not disappeared, the hot atoms are just spread evenly amongst the 0.0866M*3 of cold iodine atoms.

2. No, the other hot atoms haven't decayed - one of the implicit assumptions in this sort of experiment is that the reaction rates are much higher than the decay rate of the radioactive compound. Otherwise you have to add an additional factor to track the loss of iodine atoms due to decay. Ther other hot atoms now exist as part of an I₃^- ion rather than as part of an I^- ion.

1. How the atoms distribute would require you to know (or at least guess at) the mechanism of the reaction. Unfortunately, I do not know exactly what the mechanism of this reaction is, but my understanding is that I^- combines with I₂ to form I₃^- in a reversible process.