The equilibrium constant is \(0.0900\) at \(25^{\circ} \mathrm{C}\) for the reaction $$ \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{Cl}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{HOCl}(g) $$ For which of the following sets of conditions is the system at equilibrium? For those which are not at equilibrium, in which direction will the system shift? a. A \(1.0\) - \(\mathrm{L}\) flask contains \(1.0 \mathrm{~mol} \mathrm{HOCl}, 0.10 \mathrm{~mol} \mathrm{Cl}_{2} \mathrm{O}\), and \(0.10 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}\) b. A 2.0-L flask contains \(0.084 \mathrm{~mol} \mathrm{HOCl}, 0.080 \mathrm{~mol} \mathrm{Cl}_{2} \mathrm{O}\), and \(0.98 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}\) c. A 3.0-L flask contains \(0.25 \mathrm{~mol} \mathrm{HOCl}, 0.0010 \mathrm{~mol} \mathrm{Cl}_{2} \mathrm{O}\), and \(0.56 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}\)

The reaction quotient (Q) is the ratio of the concentrations of the products raised to their stoichiometric coefficients to the concentrations of the reactants raised to their stoichiometric coefficients. For the given reaction, the formula for Q can be written as: \[ Q = \frac{[\mathrm{HOCl}]^2}{[\mathrm{H}_{2}\mathrm{O}] \cdot [\mathrm{Cl}_{2}\mathrm{O}]} \] Note that the brackets denote the concentration of the species in moles per liter (M).

For each set of conditions, we need to calculate the concentration of each species. The concentration can be calculated by dividing the number of moles by the volume of the flask. a. In a 1.0 L flask: [\(\mathrm{HOCl}\)] = \(\frac{1.0 \, \text{mol}}{1.0 \, \text{L}}\) = 1.0 M [\(\mathrm{Cl}_{2}\mathrm{O}\)] = \(\frac{0.10 \, \text{mol}}{1.0 \, \text{L}}\) = 0.10 M [\(\mathrm{H}_{2}\mathrm{O}\)] = \(\frac{0.10 \, \text{mol}}{1.0 \, \text{L}}\) = 0.10 M b. In a 2.0 L flask: [\(\mathrm{HOCl}\)] = \(\frac{0.084 \, \text{mol}}{2.0 \, \text{L}}\) = 0.042 M [\(\mathrm{Cl}_{2}\mathrm{O}\)] = \(\frac{0.080 \, \text{mol}}{2.0 \, \text{L}}\) = 0.040 M [\(\mathrm{H}_{2}\mathrm{O}\)] = \(\frac{0.98 \, \text{mol}}{2.0 \, \text{L}}\) = 0.49 M c. In a 3.0 L flask: [\(\mathrm{HOCl}\)] = \(\frac{0.25 \, \text{mol}}{3.0 \, \text{L}}\) = 0.0833 M [\(\mathrm{Cl}_{2}\mathrm{O}\)] = \(\frac{0.0010 \, \text{mol}}{3.0 \, \text{L}}\) = 0.000333 M [\(\mathrm{H}_{2}\mathrm{O}\)] = \(\frac{0.56 \, \text{mol}}{3.0 \, \text{L}}\) = 0.1867 M

Using the formula for Q from Step 1, we will calculate Q for each set of conditions and compare it to the given K value (0.0900). a. Q = \(\frac{(1.0)^2}{(0.10) \cdot (0.10)}\) = 100. Since Q > K, the system will shift to the left (towards the reactants). b. Q = \(\frac{(0.042)^2}{(0.49) \cdot (0.040)}\) = 0.0900. Since Q = K, the system is at equilibrium. c. Q = \(\frac{(0.0833)^2}{(0.1867) \cdot (0.000333)}\) = 115.46. Since Q > K, the system will shift to the left (towards the reactants).

Based on the comparison of Q and K for each set of conditions: a. The system is not at equilibrium and will shift to the left. b. The system is at equilibrium. c. The system is not at equilibrium and will shift to the left.