Question

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 HOCI, 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 HOCI, 0.080 mole of \(\mathrm{Cl}_{2} \mathrm{O},\) and 0.98 mole of \(\mathrm{H}_{2} \mathrm{O}\) c. A 3.0 -L flask contains 0.25 mole of HOCI, 0.0010 mole of \(\mathrm{Cl}_{2} \mathrm{O},\) and 0.56 mole of \(\mathrm{H}_{2} \mathrm{O}\).

Short answer

In summary, for the given conditions: (a) The reaction will shift to the left, as Q > K. (b) The system is at equilibrium, as Q = K. (c) The reaction will shift to the left, as Q > K.

Step-by-step solution

Write the reaction and equilibrium expression

The given reaction is: \[ \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{Cl}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{HOCl}(g) \] The equilibrium constant expression for this reaction is given by: \[ K = \frac{[\mathrm{HOCl}]^{2}}{[\mathrm{H}_{2}\mathrm{O}][\mathrm{Cl}_{2}\mathrm{O}]} \] Step 2: Set up the reaction quotient

Set up the reaction quotient

The reaction quotient, Q, is calculated using the same formula as K, but with initial concentrations instead of equilibrium concentrations. \[ Q = \frac{[\mathrm{HOCl}]^{2}}{[\mathrm{H}_{2}\mathrm{O}][\mathrm{Cl}_{2}\mathrm{O}]} \] Step 3: Calculate Q and compare to K

Calculate Q and compare to K

For each set of conditions (a, b, and c), we will calculate the initial concentrations, and then find Q to compare with K. a) A 1.0-L flask contains 1.0 mole of HOCI, 0.10 mole of \(\mathrm{Cl}_{2}\mathrm{O},\) and 0.10 mole of \(\mathrm{H}_{2}\mathrm{O}\) Initial concentrations: \[ [\mathrm{HOCl}] = \frac{1.0}{1.0} = 1.0 M \] \[ [\mathrm{Cl}_{2} \mathrm{O}] = \frac{0.10}{1.0} = 0.10 M \] \[ [\mathrm{H}_{2} \mathrm{O}] = \frac{0.10}{1.0} = 0.10 M \] Calculate Q: \[ Q = \frac{(1.0)^{2}}{(0.10)(0.10)} = 100 \] Since Q>K (0.0900), the reaction will shift to the left. b) A \(2.0-\) L flask contains 0.084 mole of HOCI, 0.080 mole of \(\mathrm{Cl}_{2}\mathrm{O},\) and 0.98 mole of \(\mathrm{H}_{2}\mathrm{O}\) Initial concentrations: \[ [\mathrm{HOCl}] = \frac{0.084}{2.0} = 0.042 M \] \[ [\mathrm{Cl}_{2} \mathrm{O}] = \frac{0.080}{2.0} = 0.040 M \] \[ [\mathrm{H}_{2} \mathrm{O}] = \frac{0.98}{2.0} = 0.49 M \] Calculate Q: \[ Q = \frac{(0.042)^{2}}{(0.49)(0.040)} = 0.0900 \] Since Q=K, this system is at equilibrium. c) A 3.0-L flask contains 0.25 mole of HOCI, 0.0010 mole of \(\mathrm{Cl}_{2}\mathrm{O},\) and 0.56 mole of \(\mathrm{H}_{2}\mathrm{O}\) Initial concentrations: \[ [\mathrm{HOCl}] = \frac{0.25}{3.0} = 0.0833 M \] \[ [\mathrm{Cl}_{2} \mathrm{O}] = \frac{0.0010}{3.0} = 0.000333 M \] \[ [\mathrm{H}_{2} \mathrm{O}] = \frac{0.56}{3.0} = 0.1867 M \] Calculate Q: \[ Q = \frac{(0.0833)^{2}}{(0.1867)(0.000333)} = 108.6 \] Since Q>K (0.0900), the reaction will shift to the left. In summary, for (a) and (c), the reactions will shift to the left while for (b), the system is at equilibrium.

Key concepts

Chemical Equilibrium

Understanding chemical equilibrium is fundamental to grasping how chemical reactions occur in nature. It represents a state in a reversible chemical reaction where the rate of the forward reaction equals the rate of the reverse reaction. This balance means that the concentrations of the reactants and products remain constant over time, not that the reactants and products are in equal concentration.

For the reaction $$\mathrm{H}_{2} \mathrm{O}(g) + \mathrm{Cl}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{HOCl}(g),$$The equilibrium state would be achieved when the decomposition of water and dichlorine monoxide into hypochlorous acid and the formation of water and dichlorine monoxide from hypochlorous acid happen at the same rate. The equilibrium constant (K) for this reaction is 0.0900 at 25°C, indicating the ratio of product concentrations to reactant concentrations at equilibrium.

Reaction Quotient (Q)

The reaction quotient (Q) is a measure that tells us whether a reaction at a given set of conditions is at equilibrium, and if not, in which direction it will proceed to reach equilibrium. Q is calculated using the same general formula as the equilibrium constant (K), but with initial concentrations rather than equilibrium concentrations.

For example, in part (a) of our problem, Q is calculated as:$$Q = \frac{[\mathrm{HOCl}]^{2}}{[\mathrm{H}_{2}\mathrm{O}][\mathrm{Cl}_{2}\mathrm{O}]}$$With the initial concentrations resulting in $$Q = \frac{(1.0)^{2}}{(0.10)(0.10)} = 100$$which is greater than K. This indicates that the reaction is not at equilibrium and will shift to the left (towards the reactants) to reach equilibrium.

For part (b), the calculated Q is equal to K, pointing to the fact that the system is already at equilibrium, and no shift in the direction of the reaction is required.

Le Chatelier's Principle

Le Chatelier's Principle offers insight into how a system at equilibrium responds to a disturbance or change in conditions. According to this principle, if an external change is imposed on a system at equilibrium, the system will adjust itself to counteract that change and re-establish equilibrium.

In the context of our exercise, if the system for reaction (a) or (c), where Q > K, experiences an increase in the concentration of reactants, the system would respond by shifting the equilibrium position to form more products until K and Q are equal again. Likewise, if there were a decrease in the concentration of products, the reaction would shift to produce more products. This self-adjusting behavior showcases the dynamic nature of chemical equilibrium and allows us to predict the direction of the reaction under various stressors, be it changes in concentration, temperature, or pressure.