Question

Ethyl acetate is synthesized in a nonreacting solvent (not water) according to the following reaction: \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} \rightleftharpoons \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{C}_{2} \mathrm{H}_{5}+\mathrm{H}_{2} \mathrm{O} \quad K=2.2\) \(\begin{array}{ll}\text { Acetic acid } & \text { Ethanol } & \text { Ethyl acetate }\end{array}\) For the following mixtures \((\mathrm{a}-\mathrm{d})\), will the concentration of \(\mathrm{H}_{2} \mathrm{O}\) increase, decrease, or remain the same as equilibrium is established? a. \(\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{C}_{2} \mathrm{H}_{5}\right]=0.22 \mathrm{M},\left[\mathrm{H}_{2} \mathrm{O}\right]=0.10 \mathrm{M}\), \(\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right]=0.010 \mathrm{M},\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right]=0.010 \mathrm{M}\) b. \(\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{C}_{2} \mathrm{H}_{5}\right]=0.22 M,\left[\mathrm{H}_{2} \mathrm{O}\right]=0.0020 M\), \(\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right]=0.0020 \mathrm{M},\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right]=0.10 \mathrm{M}\) c. \(\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{C}_{2} \mathrm{H}_{5}\right]=0.88 \mathrm{M},\left[\mathrm{H}_{2} \mathrm{O}\right]=0.12 \mathrm{M}\), \(\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right]=0.044 M,\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right]=6.0 \mathrm{M}\) d. \(\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{C}_{2} \mathrm{H}_{5}\right]=4.4 \mathrm{M},\left[\mathrm{H}_{2} \mathrm{O}\right]=4.4 \mathrm{M}\), \(\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right]=0.88 \mathrm{M},\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right]=10.0 \mathrm{M}\) e. What must the concentration of water be for a mixture with \(\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{C}_{2} \mathrm{H}_{5}\right]=2.0 \mathrm{M},\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right]=0.10 \mathrm{M}\), and \(\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right]\) \(=5.0 M\) to be at equilibrium? f. Why is water included in the equilibrium expression for this reaction?

Short answer

In short, for mixtures (a-d), we calculate the reaction quotient (Q) and compare it with the equilibrium constant (K). If QK, the concentration of water will decrease; if Q=K, the concentration of water will remain the same. For part (e), we set Q=K and solve for the concentration of water, finding that \([\mathrm{H}_{2}\mathrm{O}]=1.1\mathrm{M}\). In part (f), water is included in the equilibrium expression because the solvent used in the reaction is not water, which allows the concentration of water to change and influence the position of equilibrium.

Step-by-step solution

Calculate Q for each mixture

For each given mixture, calculate the reaction quotient (Q) using the concentration of all species involved in the reaction. The formula for Q is: \[Q = \frac{[\mathrm{CH}_{3}\mathrm{CO}_{2}\mathrm{C}_{2}\mathrm{H}_{5}][\mathrm{H}_{2}\mathrm{O}]}{[\mathrm{CH}_{3}\mathrm{CO}_{2}\mathrm{H}][\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}]}\] Calculate Q for each mixture (a-d) and compare it with the equilibrium constant, K (given K=2.2).

Compare Q and K for each mixture (a-d)

Compare the calculated Q value with the provided equilibrium constant K=2.2: - If QK, the reaction shifts in the reverse direction and the concentration of water will decrease. - If Q=K, the reaction is already at equilibrium and the concentration of water will remain the same. Do this for each mixture (a-d).

Calculate the concentration of water for part (e)

For a mixture to be at equilibrium, Q=K. Use this condition to find the concentration of water required for the given mixture in part (e). Plug in the given concentrations for all other species in the Q formula, and set it equal to K=2.2 to find the concentration of water: \[2.2 = \frac{[2.0][\mathrm{H}_{2}\mathrm{O}]}{[0.10][5.0]}\] Solve for the concentration of water, \([\mathrm{H}_{2}\mathrm{O}]\).

Explain why water is included in the equilibrium expression (part f)

Water is included in the equilibrium expression for this reaction because the solvent used in the reaction is not water. Since water is one of the products in the reaction, its concentration can change and influence the position of equilibrium. If water were the solvent, its concentration would be much higher and would not significantly change during the reaction, so it would not be included in the equilibrium expression.

Key concepts

Equilibrium Constant

The equilibrium constant (\( K \)) is a fundamental concept in chemical equilibrium that expresses the ratio of the concentration of the products to the reactants in a reversible chemical reaction, each raised to the power of their stoichiometric coefficients. It is crucial in determining the extent of a reaction and predicting the composition of the reaction mixture when it has reached equilibrium.

For the reaction involving ethyl acetate, acetic acid, and ethanol to produce ethyl acetate and water, the expression for the equilibrium constant is defined as:
\[K = \frac{[\mathrm{CH}_3\mathrm{CO}_2\mathrm{C}_2\mathrm{H}_5][\mathrm{H}_2\mathrm{O}]}{[\mathrm{CH}_3\mathrm{CO}_2\mathrm{H}][\mathrm{C}_2\mathrm{H}_5\mathrm{OH}]}\]
Having a fixed value for a particular reaction at a constant temperature, this constant helps us predict how different changes in the concentrations of reactants and products will affect the state of equilibrium. When the reaction has reached equilibrium, the concentrations of the reactants and products will remain constant. The specific numerical value of the equilibrium constant provides insight into the position of the equilibrium, indicating whether the reactants or products are favored in the reaction mixture at equilibrium.

Reaction Quotient

The reaction quotient (\( Q \)) operates similarly to the equilibrium constant but is used for non-equilibrium conditions. It is calculated using the same formula as the equilibrium constant, but with the initial or current concentrations of the reactants and products, rather than the concentrations at equilibrium.
\[Q = \frac{[\mathrm{CH}_3\mathrm{CO}_2\mathrm{C}_2\mathrm{H}_5][\mathrm{H}_2\mathrm{O}]}{[\mathrm{CH}_3\mathrm{CO}_2\mathrm{H}][\mathrm{C}_2\mathrm{H}_5\mathrm{OH}]}\]
By comparing the reaction quotient (\( Q \)) to the equilibrium constant (\( K \)), we can predict the direction in which the reaction will proceed to reach equilibrium:

• If \( Q < K \), the reaction will proceed in the forward direction, increasing the concentration of products.
• If \( Q > K \), the reaction will move in the reverse direction, increasing the concentration of reactants.
• If \( Q = K \), the reaction mixture is already at equilibrium, and no net change will occur.

The exercise provided required calculating the reaction quotient for the given mixtures and comparing it with the equilibrium constant to predict the changes in water concentration as equilibrium is established. Understanding the relationship between \( Q \) and \( K \) is key to understanding which way the reaction will shift under different scenarios.

Le Chatelier's Principle

Le Chatelier's Principle is a powerful tool in understanding how a system at equilibrium responds to changes in concentration, temperature, or pressure. According to this principle, if an external stress is applied to a reaction mixture at equilibrium, the system will adjust itself to counteract that stress and re-establish equilibrium.

Applying this principle to the reaction from the exercise helps us understand how altering concentrations of reactants or products will affect the equilibrium position:

• Adding more reactants or removing products will shift the equilibrium towards the products to counteract the change.
• Conversely, adding more products or removing reactants will shift the equilibrium towards the reactants.

For instance, if the concentration of \( \mathrm{CH}_3\mathrm{CO}_2\mathrm{C}_2\mathrm{H}_5 \) (ethyl acetate) is increased, Le Chatelier's Principle suggests that the reaction will shift towards the reactants to decrease the concentration of ethyl acetate and re-establish the equilibrium. This shift will affect the concentration of all species involved, including water. By understanding these principles, we can predict the behaviour of a reaction under different conditions and manipulate various parameters to achieve a desired equilibrium state.