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The equation given in the answer key was

$$\ce{ClF + 2KBr → KCl + KF + Br2}$$

However won't the $\ce{KCl}$ also react with $\ce{ClF}$ to form $\ce{KF}$ and $\ce{Cl2}$, resulting in the net reaction being

$$\ce{2ClF + 2KBr → 2KF + Cl2 + Br2}~?$$

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    $\begingroup$ It may depend on molar ratio of reagents. Both equations may be correct at some conditions. $\endgroup$ – Poutnik Apr 12 at 13:41
  • $\begingroup$ Yea poutnik I was thinking the same thing, problem was the question didn't specify. Thanks though, wanted to confirm the second equation could work. $\endgroup$ – madeye moody Apr 12 at 16:54
  • $\begingroup$ Perhaps check ΔG on both? Just suggesting as I haven't checked anything $\endgroup$ – ManRow Apr 13 at 2:22
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Let's subtract the two reactions from each other to simplify things a bit:

$$\ce{2ClF + 2KBr -> 2KF + Cl2 + Br2}\ \ \ \ \ \ \ \ \ \ [1]$$

$$\ce{ClF + 2KBr -> KCl + KF + Br2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [2]$$

$$\ce{ClF + KCl -> KF + Cl2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ [3] = [1]-[2]]$$

Now we can discuss [2] and [3] separately by splitting them up into half reactions.

Reaction 2

$$\ce{ClF + 2KBr -> KCl + KF + Br2}$$ Potassium is a spectator ion. The reduction half reaction is:

$$\ce{ClF + 2e- -> Cl- + F-}$$

The oxidation half reaction is:

$$\ce{2Br- -> Br2 + 2e-}$$

Reaction 3

$$\ce{ClF + KCl -> KF + Cl2}$$

Potassium is a spectator ion. Fluorine does not change oxidation state. Chlorine undergoes comproportionation (oxidation states +1 and -1 to oxidation state of 0). The reduction half reaction is:

$$\ce{ClF + e- -> 1/2 Cl2 + F-}$$

The oxidation half reaction is:

$$\ce{Cl- -> 1/2 Cl2 + e-}$$

Which reaction happens?

That depends on the reduction potentials, the kinetics of the two reactions and the concentration of species present. I'm not sure how you would figure it out without experimental (or computational) data.

Comparing the oxidation half reactions, you could make an argument about electronegativities of bromine vs chlorine, but I have no insight to compare the two reducation half reactions, which are very different.

The Wikipedia article on chlorine monofluoride states:

Many of its properties are intermediate between its parent halogens, $\ce{Cl2}$ and $\ce{F2}$.

If you substitute $\ce{Cl2}$ for $\ce{ClF}$ in reactions 2 and 3, they both look reasonable (reaction 3 becomes a self-exchange, i.e. equilibrium constant = 1, whereas reaction 2 reduces chlorine and oxidizes bromine, products are favored).

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