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 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 \(0.56 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}\)

Short answer

For the given 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.

Step-by-step solution

Recall the formula for Reaction Quotient

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).

Determine the concentration of species in each condition

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

Calculate Q for each condition and compare to K

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).

Summarize the results for each condition

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.

Key concepts

Reaction Quotient

The reaction quotient, symbolized as \( Q \), is a helpful tool in determining the current state of a chemical reaction relative to its equilibrium. This quotient is calculated using the concentrations of the reactants and products of a reaction, raised to the power of their respective coefficients in the balanced chemical equation. Let's consider the reaction: \[ \mathrm{H}_{2} \mathrm{O}(g) + \mathrm{Cl}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{HOCl}(g) \]Here, the formula for the reaction quotient \( Q \) is:\[ Q = \frac{[\mathrm{HOCl}]^2}{[\mathrm{H}_{2} \mathrm{O}] \cdot [\mathrm{Cl}_{2} \mathrm{O}]} \]Calculating \( Q \) at different stages in a reaction allows us to see how the reaction might shift to reach equilibrium:

• If \( Q < K \) (equilibrium constant), the reaction will shift to the right, producing more products.
• If \( Q > K \), the reaction will shift to the left, favoring the formation of reactants.
• If \( Q = K \), the reaction is at equilibrium, meaning there's no shift required since the rates of the forward and reverse reactions are equal.

Equilibrium Constant

The equilibrium constant, represented as \( K \), is a vital part of understanding chemical equilibrium. It is a fixed number for a given reaction at a specific temperature. This constant provides insight into the balance between reactants and products at equilibrium.For the reaction: \[ \mathrm{H}_{2} \mathrm{O}(g) + \mathrm{Cl}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{HOCl}(g) \]The expression for \( K \) is:\[ K = \frac{[\mathrm{HOCl}]^2}{[\mathrm{H}_{2} \mathrm{O}] \cdot [\mathrm{Cl}_{2} \mathrm{O}]} = 0.0900 \] Breaking it down:

• A higher \( K \) value (>1) means the reaction is product-favored at equilibrium.
• A lower \( K \) value (<1), as in our exercise \( K = 0.0900 \), signifies that reactants are favored at equilibrium.

Understanding \( K \) helps predict how a change in conditions can affect a system at equilibrium, and it serves as a reference when calculating the reaction quotient \( Q \). The relationship between \( Q \) and \( K \) determines the direction in which the reaction will shift to reach equilibrium.

Le Chatelier’s Principle

Le Chatelier’s Principle is crucial for predicting how a system at equilibrium will respond to changes in external conditions. This principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium will shift to counteract the change.Key points to consider:

• Changes in Concentration: Adding or removing a reactant or product will shift the equilibrium to restore balance. For instance, adding more \( \mathrm{HOCl} \) in our reaction will push the equilibrium to the left, creating more \( \mathrm{Cl}_{2} \mathrm{O} \) and \( \mathrm{H}_{2} \mathrm{O} \).
• Changes in Pressure: For reactions involving gases, changing the pressure by altering the volume affects the equilibrium state. An increase in pressure will favor the side with fewer moles of gas.
• Temperature Changes: Altering the temperature can shift equilibrium either toward the products or reactants, depending on whether the reaction is exothermic or endothermic.

Applying Le Chatelier’s Principle helps predict how equilibrium will change when one of the system's conditions is modified.