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 that are not at equilibrium, in which direction will the system shift? a. A 1.0 -L flask contains 1.0 mole of HOCl, 0.10 mole of $\mathrm{Cl}_{2} \mathrm{O}\( , and 0.10 mole of \)\mathrm{H}_{2} \mathrm{O}$ . b. A 2.0 -L flask contains 0.084 mole of HOCl, 0.080 mole of $\mathrm{Cl}_{2} \mathrm{O}\( , and 0.98 mole of \)\mathrm{H}_{2} \mathrm{O}$ . c. A 3.0 - flask contains 0.25 mole of HOCl, 0.0010 mole of $\mathrm{Cl}_{2} \mathrm{O},\( and 0.56 mole of \)\mathrm{H}_{2} \mathrm{O}$ .

Step by step solution

01

Calculate molar concentrations of each case

For each case, we first need to find the molar concentration of each species in the reaction by dividing the number of moles by the volume of the flask. We can then calculate the reaction quotient Q to determine if the system is at equilibrium or in which direction it will shift.

02

Check case (a) equilibrium conditions

a. A 1.0-L flask contains 1.0 mole of HOCl, 0.10 mole of Cl₂O, and 0.10 mole of H₂O. Concentration of \(\mathrm{HOCl} = \cfrac{1.0 \ \text{mol}}{1.0 \ \text{L}} = 1.0 \ \text{M}\) Concentration of \(\mathrm{Cl}_{2} O= \cfrac{0.10 \ \text{mol}}{1.0 \ \text{L}} = 0.10 \ \text{M}\) Concentration of \(\mathrm{H}_{2}\mathrm{O} = \cfrac{0.10 \ \text{mol}}{1.0 \ \text{L}} = 0.10 \ \text{M}\) Now, calculate Q using the concentrations: Q = \(\frac{[\mathrm{HOCl}]^2}{[\mathrm{H}_{2} \mathrm{O}][\mathrm{Cl}_{2} \mathrm{O}]} = \frac{(1.0)^2}{(0.10)(0.10)} = 100\) Since Q > K (100 > 0.0900), the system is not at equilibrium, and it will shift to the left to reach equilibrium.

03

Check case (b) equilibrium conditions

b. A 2.0-L flask contains 0.084 mole of HOCl, 0.080 mole of Cl₂O, and 0.98 mole of H₂O. Concentration of \(\mathrm{HOCl} = \cfrac{0.084 \ \text{mol}}{2.0 \ \text{L}} = 0.042 \ \text{M}\) Concentration of \(\mathrm{Cl}_{2} O= \cfrac{0.080 \ \text{mol}}{2.0 \ \text{L}} = 0.040 \ \text{M}\) Concentration of \(\mathrm{H}_{2}\mathrm{O} = \cfrac{0.98 \ \text{mol}}{2.0 \ \text{L}} = 0.49 \ \text{M}\) Now, calculate Q using the concentrations: Q = \(\frac{[\mathrm{HOCl}]^2}{[\mathrm{H}_{2} \mathrm{O}][\mathrm{Cl}_{2} \mathrm{O}]} = \frac{(0.042)^2}{(0.49)(0.040)} = 0.0900\) Since Q = K (0.0900 = 0.0900), the system is at equilibrium.

04

Check case (c) equilibrium conditions

c. A 3.0-L flask contains 0.25 mole of HOCl, 0.0010 mole of Cl₂O, and 0.56 mole of H₂O. Concentration of \(\mathrm{HOCl} = \cfrac{0.25 \ \text{mol}}{3.0 \ \text{L}} = 0.083 \ \text{M}\) Concentration of \(\mathrm{Cl}_{2} O= \cfrac{0.0010 \ \text{mol}}{3.0 \ \text{L}} = 0.000333 \ \text{M}\) Concentration of \(\mathrm{H}_{2}\mathrm{O} = \cfrac{0.56 \ \text{mol}}{3.0 \ \text{L}} = 0.19 \ \text{M}\) Now, calculate Q using the concentrations: Q = \(\frac{[\mathrm{HOCl}]^2}{[\mathrm{H}_{2} \mathrm{O}][\mathrm{Cl}_{2} \mathrm{O}]} = \frac{(0.083)^2}{(0.19)(0.000333)} = 110\) Since Q > K (110 > 0.0900), the system is not at equilibrium, and it will shift to the left to reach equilibrium.

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