What happens when (+) 2-iodobutane is treated with NaI in acetone? After treating the compound with NaI, we should follow SN2 mechanism and the product will follow an optical inversion, giving us (-)2-iodobutane, but the answer is (±)2-iodobutane.

After going through various articles, I am getting reasons, such as there will be an equilibrium created between the reactants and the products, hence we will get both the products (±)2-iodobutane.

And there is the same question where we are treating (R)-2-bromobutane with NaI in acetone, but the answer is (S)-2-iodobutane.

What could be the reason?

I believe we should simply get one product in the question, and there will be no equilibrium be created, as there shouldn't be enough NaI left to form another compound.
There is no equilibrium in the second reaction because the sodium bromide formed by the iodide displacement is not soluble in acetone so there is no Br- to displace iodide. In contrast with the first reaction, NaI is soluble in acetone so there is always excess I- around to continue the reaction. This is called the Finkelstein reaction, further information here finkelstein reaction