Question

One mole of ethanol is treated with one mole of ethanoic acid at \(25^{\circ} \mathrm{C}\). Onefourth of the acid changes into ester at equilibrium. The equilibrium constant for the reaction will be (a) \(1 / 9\) (b) \(4 / 9\) (c) 9 (d) \(9 / 4\)

Short answer

The equilibrium constant (K) for the reaction is 1 / 9.

Step-by-step solution

Note given values and the general form of the equilibrium equation

The reaction between ethanol and ethanoic acid to form ester and water is generally represented as: \(\text{Ethanol} + \text{Ethanoic Acid} \rightleftharpoons \text{Ester} + \text{Water}\). Since one mole of ethanoic acid is used, and one-fourth of it is converted into ester, 0.25 moles of ethanoic acid will form 0.25 moles of ester and 0.25 moles of water. At equilibrium, the concentration of the ethanoic acid will be 0.75 moles and ethanol will also be 0.75 moles since the stoichiometry is 1:1. The equilibrium constant (K) is given by the expression \[K = \frac{[\text{Ester}] [\text{Water}]}{[\text{Ethanol}] [\text{Ethanoic Acid}]}\]

Set up the equilibrium constant expression

Place the moles of each component in the equilibrium expression to find K. Since the reaction container's volume isn't specified, we assume it to be the same for all components, so we can directly use mole values for calculating the equilibrium constant: \[K = \frac{(0.25)(0.25)}{(0.75)(0.75)}\]

Calculate the equilibrium constant

Now, compute the value of K using given values from the previous step: \[K = \frac{0.0625}{0.5625} \approx \frac{1}{9}\] The equilibrium constant for the reaction is \(1 / 9\), which corresponds to option (a).

Key concepts

Chemical Equilibrium

Understanding chemical equilibrium is fundamental to grasping how reactions proceed and how they can be controlled. At its core, a chemical equilibrium occurs when the rates of the forward and reverse reactions in a chemical system are equal, leading to no net change in the concentration of reactants and products over time. This does not mean the reaction has stopped; rather, it continues with the exchange of substances at a rate that maintains a constant ratio.

For instance, when ethanol and ethanoic acid react to form ester and water, the reaction reaches a point where the concentration of all substances involved does not appear to change. This state is indicative of chemical equilibrium being established. The equilibrium constant, denoted as 'K', is a numerical value that expresses the ratio of the concentration of products to the concentration of reactants at equilibrium. It is crucial because it tells us the extent to which a reaction proceeds before reaching equilibrium. A higher 'K' value suggests a greater concentration of products at equilibrium, meaning the reaction favors the product side. Conversely, a lower 'K' value indicates a reaction that favors the reactants.

Esterification Reaction

The esterification reaction is a classic example of an organic chemistry process where an alcohol reacts with an acid to form an ester and water. This type of reaction is also a model case for studying chemical equilibria since it often reaches a balance between reactants and products.

The reaction between ethanol and ethanoic acid, as described in the exercise, is a good example of an esterification reaction. A unique feature of esterification is that it's reversible, meaning the formed ester and water can react again to form the original alcohol and acid. This reversibility is crucial when considering the dynamics of chemical equilibrium.

To emphasize the components involved, let's reiterate: the reactants are ethanol and ethanoic acid, and the products are ester and water. At equilibrium, the amount of each has settled into a consistent ratio, unless disturbed by changing conditions or the addition of more reactants or products. This balance is quantitatively expressed by the 'K' value. In educational contexts, understanding esterification helps students grasp key concepts related to both chemical equilibrium and organic reaction mechanisms.

Physical Chemistry

Physical chemistry is the branch of chemistry that deals with the physical properties and behaviors of chemicals, including understanding the principles that govern reactions like esterification. It involves the application of physics concepts to chemical problems, which allows chemists to predict how substances will respond to different environments and how to control reactions for desired outcomes.

One of the cornerstones of physical chemistry is the study of reaction kinetics and dynamics, which tell us how fast reactions occur and the mechanisms by which they proceed. The equilibrium constant 'K' covered in the esterification example is a crucial part of this field as it is derived from the laws of thermodynamics and is a measure of the position of equilibrium for reversible reactions. Concepts from physical chemistry, such as thermochemistry, quantum chemistry, and the structure of atoms and molecules, provide the fundamental knowledge needed to explore and understand complex chemical processes in depth.

In the context of the given exercise, physical chemistry provides the theoretical framework to calculate the equilibrium constant and to predict the behavior of the esterification reaction under different conditions. By exploring these topics, students acquire a powerful toolkit for deciphering and predicting the behavior of chemical systems.