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When concentrated sulfuric acid is added to anhydrous potassium chloride and the fumes produced are bubbled into aqueous potassium iodide solution, the observed solution would be colourless solution.

I think the first reaction is: $$\ce{2KCl(aq) + H2SO4(aq) -> K2SO4 + 2HCl(g)}$$

I assumed the second reaction would be: $$\ce{HCl + KI -> KCl + \frac{1}{2}I2 + \frac{1}{2}H2 }$$

However the second reaction is wrong according to the answer book as $\ce{HCl}$ won't react with $\ce{KI}$. From what I know, $\ce{Cl}$ is a stronger oxidising agent than $\ce{I}$, so shouldn't $\ce{I-}$ in $\ce{KI}$ be oxidized to $\ce{I2}$? I am an A-levels student so would appreciate a simpler answer.

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  1. In your first reaction, you added concentrated sulfuric acid to anhydrous potassium chloride. In your equation, you wrote $\ce{KCl (aq)}$, which is incorrect. Under dilute conditions, that reaction would not take place because all reagents and products would be in aqueous ionic state, meaning there is no reaction.

  2. In your second reaction, the equation is completely wrong. It is true that elemental $\ce{Cl2}$ a stronger oxidizing agent than elemental $\ce{Br2}$ or elemental $\ce{I2}$. But in current situation is ionic $\ce{Cl-}$ versus ionic $\ce{I-}$ where both are aqueous. If all reagents are in ionic form and expected products are also ionic, there is no reaction. See your reaction below: $$\ce{H+(aq) + Cl-(aq) +K+(aq) + I-(aq) <=> H+(aq) + I-(aq) +K+(aq) + Cl-(aq)}$$ There is no gas, liquid or solid formed, but all ions. Therefore, there is no reaction.

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As said by @Mathew that the reactants are in aqueous form and hence the solution only contains ions and thus no reaction takes place. But, if the conditions are changed slightly and if some energy is applied, then the reaction should take place.

So, if the reaction takes place in the presence of oxygen at room temperature and if light is applied as a form of energy, then the reaction takes place. Light as energy helps bonds to break and join to form products and oxygen is applied as an additional oxidising agent as hydrochloric acid is not enough to oxidize $\ce{KI}$. Here is the reaction given:

$$\ce{4KI + 4HCl + O2 ->[h\nu] 2I2 + 4KCl + 2H2O}$$

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What happens in the second reaction solution will depend on the concentration of the reagents. When the hydrogen iodide concentration is high, it can form iodine. This is not a reaction of HI with an oxidant. What happens is

2HI ---> H2 + I2

If the HCl concentration in your experiment was very low in the second aqueous solution then I would not expect the reaction to occur. But if you look at an old bottle of concentrated HI, it will always be very dark due to the fact that it contains HI3. I know this as I have used HI solution to convert alcohols into alkyl iodides in lab.

Let us think for a moment about solvent extraction of metals from nitric acid, one of the problems is that the nitrate concentration is not equal to the anayltical nitric acid concentration (by titration) this is due to the fact that the Ka of nitric acid is not infinity. In the same way HI does not have an infinite Ka value.

As the anayltical concentration of HI in a solution increases, as the solvent properties will change and as Ka even in a very dilute solution is not infinity then the concentration of HI (intact HI) will increase. Also the concentration of the HI in the head space above the liquid will also increase.

The thermodynamics of formation of HI from I2 and H2 suggests that it should break down in to its elements. While Max Bondenstein did his work in the gas phase at a much higher temperture, it may be worth looking at.

Sulfuric acid will not form chlorine gas when it reacts with potassium chloride, it will form hydrogen chloride only. I know this as I have done this type of reaction myself.

When I was a teenaged boy I heated sodium chloride and sodium hydrogen sulfate in a test tube, it formed hydrogen chloride gas and not chlorine gas.

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  • $\begingroup$ Thanks. Why does it form H2 and I2 when concentrated but not when dilute? I would appreciate even if you told me what to look for regarding this. $\endgroup$ – libresandals Apr 20 '18 at 9:35

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