Problem 18 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

The enthalpy of neutralization of a strong acid by a strong base is \(-57.32 \mathrm{~kJ}\) \(\mathrm{mol}^{-1} .\) The enthalpy of formation of water is \(-285.84 \mathrm{~kJ} \mathrm{~mol}^{-1}\). The enthalpy of formation of aqueous hydroxyl ion is (a) \(+228.52 \mathrm{~kJ} / \mathrm{mol}\) (b) \(-114.26 \mathrm{~kJ} / \mathrm{mol}\) (c) \(-228.52 \mathrm{~kJ} / \mathrm{mol}\) (d) \(+114.2 \mathrm{~kJ} / \mathrm{mol}\)

Step-by-Step Solution

Answer

(a) \(+228.52 \mathrm{~kJ} / \mathrm{mol}\)

Step 1: Understanding the Process of Neutralization

When a strong acid reacts with a strong base, the products are typically water and a salt. The enthalpy of neutralization refers to the heat evolved when one equivalent of an acid and one equivalent of a base undergo a neutralization reaction to form water and a salt.

Step 2: Writing the Neutralization Reaction

The general neutralization reaction between a strong acid (H+) and a strong base (OH-) can be written as:\[\text{H+} + \text{OH-} \rightarrow \text{H2O}\]

Step 3: Using Hess's Law

Hess's Law states that the total enthalpy change for a chemical reaction is the same, no matter whether it occurs in one step or in several steps. The enthalpy change for the formation of water from H+ and OH- ions can be related to the enthalpy of neutralization and enthalpies of formation of the reactants and products.

Step 4: Calculating the Enthalpy of Formation of OH-

Based on Hess's Law, the sum of the enthalpy of neutralization and the enthalpy of formation of the products must equal the sum of the enthalpies of formation of the reactants. In this case, we have the enthalpy of formation of water and we need to find the enthalpy of formation of the hydroxyl ion (OH-). The equation representing these relationships is:\[\Delta H_{neutralization} + \Delta H_{f}(\text{H2O}) = \Delta H_{f}(\text{H+}) + \Delta H_{f}(\text{OH-})\]Here, \(\Delta H_{f}(\text{H+})\) is zero because the enthalpy of formation for elements in their standard state, like a proton from a strong acid, is zero. So the equation simplifies to:\[-57.32 \mathrm{~kJ/mol} + (-285.84 \mathrm{~kJ/mol}) = 0 + \Delta H_{f}(\text{OH-})\]Solving for \(\Delta H_{f}(\text{OH-})\) gives us the enthalpy of formation for the hydroxyl ion.

Step 5: Finding the Correct Answer

To find \(\Delta H_{f}(\text{OH-})\), simply rearrange the equation from the previous step:\[\Delta H_{f}(\text{OH-}) = -57.32 \mathrm{~kJ/mol} - (-285.84 \mathrm{~kJ/mol})\]\[\Delta H_{f}(\text{OH-}) = -57.32 \mathrm{~kJ/mol} + 285.84 \mathrm{~kJ/mol}\]\[\Delta H_{f}(\text{OH-}) = +228.52 \mathrm{~kJ/mol}\]Therefore, the correct answer is (a) \(+228.52 \mathrm{~kJ/mol}\).

Key Concepts

Enthalpy of Formation

Enthalpy of formation, denoted by cwhich is the total heat content of a system, provides an understanding of the energy dynamics during chemical reactions. Specifically, it represents the heat change that occurs when one mole of a compound is formed from its elements in their standard states at standard conditions (typically 1 atm pressure and 25°C). For instance, the enthalpy of formation of water, c, reflects the energy released or absorbed when water is formed from hydrogen and oxygen gases.

In educational exercises, you might encounter questions where you need to calculate the enthalpy of formation for a substance not provided in the problem statement. It’s important to gather all known enthalpies of formations relevant to the reaction under consideration. For instance, when a strong acid and a strong base neutralize each other, the formation of water is involved, and its enthalpy of formation is essentially the energy change of this particular reaction step. Understanding this concept is critical in solving problems related to reaction enthalpy.

Hess's Law

Hess's Law is a priceless rule in thermodynamics which asserts that the total enthalpy change for a chemical reaction is constant, irrespective of the pathway the reaction takes. This principle is essential when dealing with chemical reactions, as it allows us to determine the enthalpy change of a complex reaction by breaking it down into simpler steps whose enthalpy changes are known.

From an educational standpoint, students are often asked to apply Hess's Law to calculate unknown enthalpies, such as the enthalpy of formation, by using known enthalpies from multiple steps of a reaction. For example, in neutralization reactions, if the enthalpy of the overall reaction and the enthalpy of formation of the products are known, Hess's Law can enable us to deduce the missing enthalpies of the reactants. This approach is crucial for students to master, as it equips them to tackle problems without direct experimental data for every reaction.

Chemical Reaction Enthalpy

Chemical reaction enthalpy is the heat change associated with a chemical reaction occurring at constant pressure. It’s symbolized by c, and it reflects the energy absorbed or released during the transformation of reactants into products. This heat change can be measured directly in a calorimeter or calculated theoretically using Hess's Law and standard enthalpies of formation.

To assist students in understanding this concept within educational materials, imagine mixing a strong acid with a strong base. The reaction will evolve heat, which is the enthalpy of neutralization. If the experiment is properly insulated, measuring the temperature change of the solution can provide the enthalpy of the reaction. In theoretical problems, students may be required to compute this enthalpy using provided enthalpy values and understanding the stoichiometry of the reaction. Being proficient with chemical reaction enthalpy calculations is fundamental in thermodynamics and for students studying chemistry.

Strong Acid-Base Reaction

A strong acid-base reaction involves the combination of a strong acid with a strong base to form water and a salt in a neutralization reaction. This type of reaction is characterized by the complete dissociation of both the acid and the base in an aqueous solution, which leads to the formation of hydronium ions () and the negative ions of the salt. The reaction typically releases heat, known as the enthalpy of neutralization, and it is generally a large negative value because the formation of water from H+ and OH- ions is exothermic.

When engaging with students on the topic of strong acid-base reactions, it is important to emphasize the concept of complete dissociation, which means that there are no remaining molecules of the original acid or base in solution–only the ions. Also, the understanding of the relationship between the enthalpy of the reaction and the strengths of the acid and base involved, along with the stoichiometry of the reaction, is key to mastering the concept of enthalpy of neutralization.