Problem 48 - Chemistry

Question

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

Step-by-Step Solution

Answer

For the mixtures (a-d), the concentration of \(\mathrm{H}_{2}\mathrm{O}\) changes as follows: a) Decrease b) Decrease c) Decrease d) Remain the same For part (e), the concentration of water must be 0.55 M for the mixture to be at equilibrium. In part (f), water is included in the equilibrium expression because it is a product of the reaction in a nonreacting solvent, and its concentration can change affecting the position of equilibrium.

Step 1: Calculate the reaction quotient (Q) for each mixture

For mixture (a): \[ Q = \frac{[0.22] [0.10]}{[0.010] [0.010]} = 22 \] For mixture (b): \[ Q = \frac{[0.22] [0.0020]}{[0.0020] [0.10]} = 22 \] For mixture (c): \[ Q = \frac{[0.88] [0.12]}{[0.044] [6.0]} = 4.0 \] For mixture (d): \[ Q = \frac{[4.4] [4.4]}{[0.88] [10.0]} = 2.2 \]

Step 2: Compare Q with K for each mixture to find the fate of water concentration

Mixture (a): Since \(Q > K\), the reaction will shift to the left (towards reactants), hence the concentration of \(\mathrm{H}_{2}\mathrm{O}\) will decrease. Mixture (b): Since \(Q > K\), the reaction will shift to the left (towards reactants), hence the concentration of \(\mathrm{H}_{2}\mathrm{O}\) will decrease. Mixture (c): Since \(Q > K\), the reaction will shift to the left (towards reactants), hence the concentration of \(\mathrm{H}_{2}\mathrm{O}\) will decrease. Mixture (d): Since \(Q = K\), the reaction is at equilibrium, hence the concentration of \(\mathrm{H}_{2}\mathrm{O}\) will remain the same.

Step 3: Determine the concentration of water for the given mixture in part (e)

For reactants \(\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{C}_{2}\mathrm{H}_{5}\right] = 2.0 M, \left[\mathrm{CH}_{3} \mathrm{CO}_{2}\mathrm{H}\right] = 0.10 M\), and \(\left[\mathrm{C}_{2} \mathrm{H}_{5}\mathrm{OH}\right] = 5.0 M\), we will use the given equilibrium constant (K) to find the concentration of \(\mathrm{H}_{2}\mathrm{O}\): \[ K = \frac{[2.0] [\mathrm{H}_{2}\mathrm{O}]}{[0.10] [5.0]} \] Now, we simply need to find \([\mathrm{H}_{2}\mathrm{O}]\) by solving this equation: \[ [\mathrm{H}_{2}\mathrm{O}] = \frac{K [0.10] [5.0]}{[2.0]} = 0.55 M \] So, the concentration of water must be 0.55 M for the given mixture to be at equilibrium.

Step 4: Explain why water is included in the equilibrium expression

Water is included in the equilibrium expression for this reaction because it is one of the products of the reaction, and its concentration in the nonreacting solvent can change due to the reversibility of the reaction. By including water in the equilibrium expression, it allows accurate prediction and assessment of the position of equilibrium depending on the initial concentrations of the reactants and products. In some reactions, water is excluded from the equilibrium expressions when it is a solvent. However, in this case, the reaction takes place in a non-aqueous (not water) solvent, so water should be considered in the equilibrium expression.

Key Concepts

Reaction Quotient

The reaction quotient (\(Q\)) helps us understand the progress of a chemical reaction at any given moment. It's vital for determining if a reaction is at equilibrium or if it needs to shift in a particular direction. Calculating \(Q\) is similar to how we calculate an equilibrium constant \(K\), but it's done with the current concentrations of reactants and products.
For instance, in our ethyl acetate synthesis, we calculated \(Q\) by dividing the product of the concentrations of the products by the product of the concentrations of the reactants. Comparing \(Q\) to \(K\) tells us which way the reaction will shift to reach equilibrium.

• If \(Q > K\), the reaction will shift towards the reactants.
• If \(Q < K\), it will shift towards the products.
• If \(Q = K\), the system is at equilibrium.

This method is crucial to predict how concentrations will change over time.

Equilibrium Constant

The equilibrium constant (\(K\)) expresses the ratio of concentrations of products to reactants when a reaction is at equilibrium. Each reaction has its own \(K\) value, which can tell us much about the reaction's characteristics and direction at a given temperature. For ethyl acetate synthesis, \(K\) was set at 2.2.
Understanding \(K\):

• A large \(K\) value (greater than 1) indicates that products are favored at equilibrium.
• A small \(K\) value (less than 1) means that reactants are favored.
• The expression for \(K\) only incorporates aqueous and gaseous substances, excluding pure solids and liquids.

In the ethyl acetate problem, calculating and comparing \(K\) helped us predict whether the reaction mixture was at equilibrium or if it needed to be adjusted to achieve it.

Le Chatelier's Principle

Le Chatelier's Principle is an essential concept in chemistry that predicts how an equilibrium system responds to disturbances. If a change in conditions is applied to a system at equilibrium, the system adjusts—shifting in the direction that counteracts the change.
Key influences and adjustments include:

• Concentration changes: Adding or removing reactants or products will shift the equilibrium to restore balance.
• Temperature changes: Increasing temperature favors an endothermic reaction, while decreasing it favors an exothermic one.
• Pressure changes: Increasing pressure favors the side of the reaction with fewer gas molecules.

In the context of ethyl acetate synthesis, when \(Q\) did not match \(K\), Le Chatelier's Principle helped determine the direction of shift to achieve equilibrium, ensuring that the concentration of water and other substances adjusted accordingly.

Reversible Reactions

Reactions that can proceed in both forward and reverse directions are called reversible reactions. They reach a state where both reactants and products exist simultaneously in fixed proportions, known as equilibrium.
Features of reversible reactions include:

• Reversibility allows for the continuous interconversion of reactants and products.
• Equilibrium does not imply equal concentrations but a state where the rates of forward and reverse reactions are equal.
• They are denoted with the double arrow symbol (\(\rightleftharpoons\)).

In the case of ethyl acetate synthesis, understanding the reversible nature of the reaction is key for predicting how changes in concentration, temperature, or other conditions affect the system's position and achieve a balanced state at equilibrium.