For the reaction $\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g),$ consider two possibilities: (a) you mix 0.5 mole of each reactant, allow the system to come to equilibrium, and then add another mole of \(\mathrm{H}_{2}\) and allow the system to reach equilibrium again, or \((b)\) you \(\operatorname{mix} 1.5\) moles of \(\mathrm{H}_{2}\) and 0.5 mole of \(\mathrm{I}_{2}\) and allow the system to reach equilibrium. Will the final equilibrium mixture be different for the two procedures? Explain.

First, let's consider the initial equilibrium system where 0.5 moles of H2 and 0.5 moles of I2 are mixed. Let x be the moles of each reactant consumed to reach the initial equilibrium state, and 2x be the moles of HI formed at the equilibrium state. The final moles at the initial equilibrium will be: H2: 0.5 - x I2: 0.5 - x HI: 2x

When an additional mole (1.0 moles) of H2 is added, the moles at the start of the second step are: H2: 1.5 - x I2: 0.5 - x HI: 2x

Let y be the moles of each reactant consumed in the second equilibrium state, and 2y be the moles of HI formed. The final moles at the second equilibrium will be: H2: (1.5 - x) - y I2: (0.5 - x) - y HI: 2x + 2y

Now let's consider the other method where 1.5 moles of H2 and 0.5 moles of I2 are mixed directly without adding H2 in steps. Let z be the moles of each reactant consumed to reach the equilibrium state and 2z be the moles of HI formed at the equilibrium state. The final moles at equilibrium will be: H2: 1.5 - z I2: 0.5 - z HI: 2z

Now we need to compare the equilibrium mixtures between these two methods. According to Le Chatelier's principle, when a system at equilibrium is disturbed by a change in concentration, the system will adjust its rates of reaction in order to counteract the disturbance, and regain a new equilibrium state. In Procedure (a), we have disturbed the equilibrium state of the reaction by adding an extra mole of H2. The second equilibrium state will again follow the same equilibrium constant. Since the final equilibrium state depends on the concentration of reactants/products and the equilibrium constant, it will be the same as that of Procedure (b). In other words, both procedures described above will have the same final equilibrium state, as the equilibrium depends on the initial concentrations of reactants/products and the equilibrium constant. Hence, the final equilibrium mixtures for both procedures will be the same.