Problem 45 - Chemical Principles: The Quest for Insight

Question

Two successive stages in the industrial manufacture of sulfuric acid are the combustion of sulfur and the oxidation of sulfur dioxide to sulfur trioxide. From the standard reaction enthalpies $$ \begin{gathered} \mathrm{S}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{SO}_{2}(\mathrm{~g}) \\ \Delta H^{\circ}=-296.83 \mathrm{~kJ} \\ 2 \mathrm{~S}(\mathrm{~s})+3 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{SO}_{3}(\mathrm{~g}) \\ \Delta H^{\circ}=-791.44 \mathrm{~kJ} \end{gathered} $$ Calculate the reaction enthalpy for the oxidation of sulfur dioxide to sulfur trioxide in the reaction \(2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{SO}_{3}(\mathrm{~g})\).

Step-by-Step Solution

Answer

\(-197.78 \text{kJ}\right)

Step 1: Write down the given enthalpies of reaction

The enthalpy change for the combustion of sulfur to sulfur dioxide is given as \[\text{S(s)} + \text{O}_2(\text{g}) \rightarrow \text{SO}_2(\text{g}), \quad \Delta H^\circ_1 = -296.83 \text{kJ}\]The enthalpy change for the oxidation of sulfur to sulfur trioxide is given as \[2 \text{S(s)} + 3 \text{O}_2(\text{g}) \rightarrow 2 \text{SO}_3(\text{g}), \quad \Delta H^\circ_2 = -791.44 \text{kJ}\]

Step 2: Write down the target equation

The target reaction, whose enthalpy change we need to calculate, is \[2 \text{SO}_2(\text{g}) + \text{O}_2(\text{g}) \rightarrow 2 \text{SO}_3(\text{g})\]

Step 3: Apply Hess's Law to find the enthalpy change of the target reaction

By Hess's Law, we can add or subtract given reactions to obtain the target reaction. Multiply the first reaction by 2 to match the amount of sulfur in the second reaction:\[2(\text{S(s)} + \text{O}_2(\text{g}) \rightarrow \text{SO}_2(\text{g}))\]This gives us:\[2\text{S(s)} + 2\text{O}_2(\text{g}) \rightarrow 2\text{SO}_2(\text{g}), \quad 2\Delta H^\circ_1 = 2(-296.83 \text{kJ})\]Next, we subtract this from the second reaction:\[(2 \text{S(s)} + 3 \text{O}_2(\text{g}) \rightarrow 2 \text{SO}_3(\text{g})) - (2\text{S(s)} + 2\text{O}_2(\text{g}) \rightarrow 2\text{SO}_2(\text{g}))\]This simplifies to our target equation:\[\text{O}_2(\text{g}) + 2\text{SO}_2(\text{g}) \rightarrow 2\text{SO}_3(\text{g})\]

Step 4: Calculate the enthalpy change of the target reaction

The enthalpy change for the target reaction can now be derived from the adjusted enthalpies of the given reactions:\[\Delta H^\circ_{\text{target}} = \Delta H^\circ_2 - 2\Delta H^\circ_1\]Substitute the given enthalpy values:\[\Delta H^\circ_{\text{target}} = (-791.44 \text{kJ}) - 2(-296.83 \text{kJ})\]\[\Delta H^\circ_{\text{target}} = -791.44 \text{kJ} + 593.66 \text{kJ}\]\[\Delta H^\circ_{\text{target}} = -197.78 \text{kJ}\]

Key Concepts

Hess's Law

Hess's Law is a principle which states that the total enthalpy change during the complete course of a chemical reaction is the same whether the reaction is made in one step or in several steps. This valuable law is particularly useful for calculating the reaction enthalpies that are difficult to obtain experimentally. The law leverages the fact that enthalpy, being a state function, is independent of the path taken from the initial to the final state.

For example, if a chemical reaction can be expressed as the sum of two or more other reactions, we can also sum the enthalpy changes of these reactions to find the overall enthalpy change. By manipulating equations and their associated enthalpies step by step—as done in the exercise—you can determine the enthalpy of a reaction that may not be found directly in standard tables.

Standard Reaction Enthalpies

Standard reaction enthalpies, also known as standard heats of reaction, provide a measure of the heat absorbed or released during a chemical reaction at standard conditions, which are 298 K temperature and 1 bar pressure. The standard reaction enthalpy is denoted by \( \Delta H^\circ \) and is expressed in kilojoules per mole (kJ/mol).

A negative value of \( \Delta H^\circ \) indicates an exothermic reaction, where heat is released to the surroundings, while a positive value indicates an endothermic reaction, where heat is absorbed from the surroundings. These standard enthalpy values are essential for scientists and engineers to predict how much energy is required for a reaction to occur or how much energy can be obtained from a reaction.

Sulfuric Acid Production

Sulfuric acid is one of the most widely produced chemicals and is used in a multitude of industrial processes. Its production typically involves two main stages, as highlighted in the original exercise. The first stage is the combustion of sulfur to form sulfur dioxide. The second stage is the oxidation of sulfur dioxide to form sulfur trioxide, which can then be used to create sulfuric acid through various processes, such as the contact process.

The efficiency and cost-effectiveness of sulfuric acid production depend significantly on the enthalpy changes of these reactions. Understanding and optimizing these enthalpic changes can lead to more energy-efficient production methods, which is not only beneficial from an economic standpoint but also for environmental conservation.

Chemical Thermodynamics

Chemical thermodynamics is the branch of physical chemistry that deals with the energy changes associated with chemical reactions and the direction in which these reactions occur. It is foundational to understanding reaction enthalpies, spontaneity, equilibrium, and how different factors such as temperature and pressure affect the behavior of chemical systems.

Key concepts in chemical thermodynamics include the laws of thermodynamics, Gibbs free energy, and the equilibrium constant. By utilizing these concepts, chemists can predict whether a reaction will proceed and to what extent, which is crucial in both academic research and industrial applications. For instance, understanding the thermodynamics of the reactions involved in sulfuric acid production allows for the optimization of conditions under which the process is both energetically favorable and economically viable.