Question

One mole of pure ethyl alcohol was treated with one mole of pure acetic acid at \(25^{\circ} \mathrm{C}\). One-third of the acid changes into ester at equilibrium. The equilibrium constant for the reaction will be : (a) \(\frac{1}{4}\) (b) 2 (c) 3 (d) 4

Short answer

\(\frac{1}{4}\)

Step-by-step solution

Understanding the Reaction

The reaction between ethyl alcohol (ethanol) and acetic acid is an esterification reaction that forms ethyl acetate and water. The chemical equation for the reaction is: \[C_2H_5OH + CH_3COOH \rightleftharpoons CH_3COOC_2H_5 + H_2O\]

Writing the Equilibrium Expression

The equilibrium expression for the reaction is given by the formula: \[K_c = \frac{[CH_3COOC_2H_5][H_2O]}{[C_2H_5OH][CH_3COOH]}\]Here, \([A]\) denotes the concentration of substance A at equilibrium.

Substituting Equilibrium Concentrations

Given that one-third of the acetic acid is converted to ester at equilibrium: \[\text{Initial concentration of acetic acid} = 1 \text{ mole/L}\]\[\text{Concentration of acetic acid at equilibrium} = \frac{2}{3} \text{ mole/L}\]\[\text{Concentration of ethyl acetate and water at equilibrium} = \frac{1}{3} \text{ mole/L each (since 1 mole of acid gives 1 mole of ester and water)}\]Now substitute these values into the equilibrium expression.

Calculating the Equilibrium Constant

Plugging the equilibrium concentrations into the equilibrium expression gives: \[K_c = \frac{\left(\frac{1}{3}\right)\left(\frac{1}{3}\right)}{\left(\frac{2}{3}\right)\left(\frac{2}{3}\right)}\]\[K_c = \frac{1}{4}\]The equilibrium constant for the reaction is \(\frac{1}{4}\).

Key concepts

Esterification Reaction

Esterification is a chemical process in which an alcohol and an acid react to form an ester and water. This reaction is reversible, meaning the products can react to re-form the reactants. The classic reaction involves a carboxylic acid (such as acetic acid) and an alcohol (like ethyl alcohol). In the given problem, the esterification reaction produces ethyl acetate and water:

\[C_2H_5OH + CH_3COOH \rightleftharpoons CH_3COOC_2H_5 + H_2O\]
For a clearer understanding, ethyl alcohol (C2H5OH) reacts with acetic acid (CH3COOH) to form ethyl acetate (CH3COOC2H5) and water (H2O).

Understanding the mechanism of this reaction helps students to visualize the real-time interaction between the molecules and to predict the formation of products. It's also the foundation for grasping the concept of chemical equilibrium and the effect of varying the concentration of reactants and products.

Chemical Equilibrium

Chemical equilibrium refers to a state in a chemical reaction where the rate of the forward reaction equals the rate of the reverse reaction. Consequently, the concentrations of the reactants and products remain constant over time. It's essential to note that equilibrium does not mean the reactants and products are in equal concentrations, but rather that their concentrations do not change.

In the context of the esterification reaction given in the problem, the equilibrium is reached when the amount of ethyl acetate and water being produced in the forward reaction is equal to the amount being converted back to ethyl alcohol and acetic in the reverse reaction.

The calculation of the equilibrium constant (Kc) is crucial as it quantifies the ratio of the concentration of products to the concentration of reactants at equilibrium. A Kc value provides insight into the position of equilibrium — a low value indicates a reaction favoring the reactants, while a high value suggests a reaction favoring the products.

Concentration of Reactants and Products

When dealing with chemical equilibrium, the concentration of reactants and products plays a pivotal role. For the given exercise, it's important to understand how to calculate the equilibrium constant using the concentrations of the chemicals involved at equilibrium.

Initially, both reactants, ethyl alcohol and acetic acid, were present at 1 mole/L. At equilibrium, one-third of the acetic acid has been converted to ethyl acetate and water, resulting in their concentrations being 1/3 mole/L each. The remaining two-thirds of the acetic acid means its concentration is 2/3 mole/L, and the same applies to the concentration of ethyl alcohol, as it reacts in a 1:1 ratio with acetic acid.

The calculation step in the solution applies these equilibrium concentrations to the equilibrium constant expression:
\[K_c = \frac{[CH_3COOC_2H_5][H_2O]}{[C_2H_5OH][CH_3COOH]}\]
Coming full circle, we can see how the equilibrium constant is sensitive to the concentration of reactants and products. The exercise emphasizes the interdependence of these concentrations and their impact on the equilibrium state of a chemical reaction.