Problem 45 - Chemistry

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 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

Answer

In summary, for the given reaction: a. The system is not at equilibrium, and it will shift to the left to reach equilibrium. b. The system is at equilibrium. c. The system is not at equilibrium, and it will shift to the left to reach equilibrium.

Step 1: 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.

Step 2: 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.

Step 3: 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.

Step 4: 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.

Key Concepts

Reaction Quotient

When dealing with reactions, it's essential to know the status of the reaction, whether it's at equilibrium or not. This is where the reaction quotient (Q) comes into play. Q is similar to the equilibrium constant (K), but it can be calculated at any point in the reaction, not just at equilibrium.

By comparing Q and K, we determine the direction in which the reaction needs to shift to reach equilibrium.

• If Q < K, the reaction will proceed in the forward direction to form more products.
• If Q = K, the reaction is at equilibrium, and no shift is needed.
• If Q > K, the reaction will shift in the reverse direction to form more reactants.

For the given exercise, calculating Q for each set of conditions was key to determining whether the system was at equilibrium. Depending on the value of Q compared to K, the reaction could either be balanced or require shifting to reach equilibrium.

Le Chatelier's Principle

Le Chatelier's Principle is a fundamental rule predicting how a system at equilibrium will respond to changes in concentration, temperature, or pressure. It states that if a dynamic equilibrium experiences a disturbance, such as a change in concentration, temperature, or pressure, the equilibrium will shift in a direction that tends to counteract that disturbance.

In the context of the exercise:

• If there's a shift in concentrations, such as more or fewer reactants/products, Le Chatelier's Principle helps us decide the direction of the shift to restore equilibrium.
• For instance, in case (a), Q was greater than K, meaning the product concentration was too high. Hence, the system shifted left to decrease product concentration and increase reactants to minimize the change.
• This principle is crucial when determining how reactions react to changes without needing extensive calculations each time.

Understanding Le Chatelier's Principle provides insights into how chemical systems maintain balance, especially when the current state doesn't meet equilibrium conditions.

Chemical Equilibrium

Chemical equilibrium occurs in a reversible reaction when the rate of the forward reaction equals the rate of the reverse reaction. At this point, the concentrations of reactants and products remain steady, but not necessarily equal.

Key features of chemical equilibrium include:

• The reaction doesn't stop; it continues to proceed in both directions at equal rates.
• It can be affected by changes in concentration, temperature, and pressure.
• Equilibrium constants (K) are vital as they provide the ratio of product and reactant concentrations at equilibrium.

Despite the intricate balance, establishing equilibrium ensures stable concentrations until disrupted. In the exercise, we calculated Q to figure out if the systems were at equilibrium and identified how the system needed to adjust if it wasn't. Recognizing how equilibrium works in real situations is a quintessential part of understanding dynamic processes in chemistry.