Chemistry Question- chemical equilibrium?

Question

For the reaction, H2(g) + I2(g) (===) 2HI(g), consider two possibilities: (a) you mix 0.5 mol of each reactant, allow the system to come to equilibrium, and then add another mole of H2 and allow the system to reach equilibrium again, or (b) you mix 1.5 mol of H2 and 0.5 mol I2 and allow the system to reach equilibrium. Will the final equilibrium mixture be different for the two procedures? Explain.

Answer 1

No, they will be the same. It is all controlled by the equilibrium constant K = [HI]² / [H2] [I2].

I find it's easier to demonstrate with a simpler example, A ↔ B, with K = 1.

If 1 M of A is and 0 M of B is added to start, the equilibrium mixture is 0.5 M of A and 0.5 M of B. Add another 1M of A, and the equilibrium mixture becomes 1 M of A, and 1 M of B. Start instead with 2M of A and 0 M of B, and the same equilibrium mixture of 1 M of A and 1 M of B results. You can do the same with the given reaction and an assumed K, but the result will be the same.