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 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.

We are given the following reaction: \[ \mathrm{H}_{2}(g) + \mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g) \] In both cases, we need to figure out the equilibrium concentrations of each species, and then compare their final equilibrium concentrations. We will follow the procedures (a) and (b), analyze the changes in moles/reactant ratios, and observe the final equilibrium mixtures.

For Procedure (a): 1. Initially, we have 0.5 moles of H2 and 0.5 moles of I2. 2. In order to reach equilibrium, let's assume that 'x' moles of H2 and I2 react to form 2x moles of HI. At equilibrium, we have: \[ \mathrm{H}_{2}: 0.5-x \, \mathrm{mol} \] \[ \mathrm{I}_{2}: 0.5-x \, \mathrm{mol} \] \[ \mathrm{HI}: 2x \, \mathrm{mol} \]

Now, we add another mole of H2 and let the system reach equilibrium again. Assume that 'y' moles of H2 and I2 react to form 2y moles of HI in this new equilibrium. At this new equilibrium, we have: \[ \mathrm{H}_{2}: (0.5-x) + 1 - y \, \mathrm{mol} \] \[ \mathrm{I}_{2}: (0.5-x) - y \, \mathrm{mol} \] \[ \mathrm{HI}: 2x + 2y \, \mathrm{mol} \]

For Procedure (b): 1. Initially, we have 1.5 moles of H2 and 0.5 moles of I2. 2. In order to reach equilibrium, let's assume that 'z' moles of H2 and I2 react to form 2z moles of HI. At equilibrium, we have: \[ \mathrm{H}_{2}: 1.5-z \, \mathrm{mol} \] \[ \mathrm{I}_{2}: 0.5-z \, \mathrm{mol} \] \[ \mathrm{HI}: 2z \, \mathrm{mol} \]

Now let's compare the final equilibrium mixtures from both the procedures: Procedure (a): \[ \mathrm{H}_{2}: (0.5-x) + 1 - y \, \mathrm{mol} \] \[ \mathrm{I}_{2}: (0.5-x) - y \, \mathrm{mol} \] \[ \mathrm{HI}: 2x + 2y \, \mathrm{mol} \] Procedure (b): \[ \mathrm{H}_{2}: 1.5-z \, \mathrm{mol} \] \[ \mathrm{I}_{2}: 0.5-z \, \mathrm{mol} \] \[ \mathrm{HI}: 2z \, \mathrm{mol} \] Both the cases lead to different equilibrium mixtures. However, the equilibrium constant remains the same under the same experimental conditions. The reaction quotient (Q) will initially be different in both procedures. As the system adjusts the concentration of the species to reach equilibrium, the final Q value will equal the equilibrium constant (K), according to Le Châtelier's principle. In conclusion, the final equilibrium mixtures will indeed be different for the two procedures as they have different initial conditions. However, the equilibrium constant will remain the same for both cases.