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Suppose 1 mol of $\ce{KI (aq)}$ is mixed with $\pu{1 mol}$ of $\ce{I_2^*}$, where $\ce{I_2^*}$ represents a di-iodine molecule which has been radioactively labelled, and the substances are allowed to react for an infinite amount of time. Assume radioactive labeling does not affect reactivity of the atom.

i) Predict the identity of all species in the mixture after an infinite amount of time

ii) Predict the number of moles of all species in the mixture after an infinite amount of time.

For part i), my answer was that after an infinite amount of time, all radioactive tracer decays and hence $\ce{I_2^*}$ becomes $\ce{I_2}$. Then following from the triiodide equilibrium equation ($\ce{I_2 + I^- <=> I_3^-}$), we have that the final species are $\ce{K^+, I_2, I^-, I_3^-}$.

For part ii), after an infinite amount of time, equilibrium is achieved for the triiodide equation, hence the rates of the forward and backward reaction are the same.

Now at the beginning we have $\pu{1 mol}$ of $\ce{I_2}$, $\pu{1 mol}$ of $\ce{I^-}$ and $\pu{0 mol}\ \ce{I_3^-}$, and at the end we have $1-x\ \mathrm{mol}\ \ce{I_2}, 1-x\ \mathrm{mol}\ \ce{I^-}$, and $x\ \mathrm{mol}\ \ce{I_3^-}$. The number of mols on both sides of the equation are the same (my understanding of equilibrium) and hence $(1-x) + (1-x) = x$ gives us $x = \frac23$. Hence there are $\pu{1 mol}\ \ce{K^+}, \pu{\frac13 mol}\ \ce{I_2}, \pu{\frac13 mol}\ \ce{I^-}, \pu{\frac23 mol}\ \ce{I_3^-}$.

My question is, is my reasoning for the above two questions correct? (especially the equilibrium part, I'm not too certain about that). If it is not, then could I ask what the correct way to approach the above question would be?

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