Question

Sulfuric acid \(\left(\mathrm{H}_{2} \mathrm{SO}_{4}\right)\), the most widely produced chemical in the world, is made by a two-step oxidation of sulfur to sulfur trioxide, \(\mathrm{SO}_{3}\), followed by reaction with water. Calculate \(\Delta H_{\mathrm{f}}^{\circ}\) for \(\mathrm{SO}_{3}\) in \(\mathrm{kJ} / \mathrm{mol}\), given the following data: $$\begin{array}{ll}\mathrm{S}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{SO}_{2}(g) & \Delta H^{\circ}=-296.8 \mathrm{~kJ} \\ \mathrm{SO}_{2}(g)+1 / 2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{SO}_{3}(g) & \Delta H^{\circ}=-98.9 \mathrm{~kJ} \end{array}$$

Short answer

\( \Delta H_{\mathrm{f}}^{\circ} \) for \( \mathrm{SO}_{3} \) is \(-395.7 \, \text{kJ/mol}\).

Step-by-step solution

Understanding the Problem

We need to calculate the standard enthalpy change of formation, \( \Delta H_{\mathrm{f}}^{\circ} \), for sulfur trioxide \( \mathrm{SO}_{3} \). Using the given enthalpy changes for the formation reactions of \( \mathrm{SO}_{2} \) and \( \mathrm{SO}_{3} \), we can find \( \Delta H_{\mathrm{f}}^{\circ} \) for \( \mathrm{SO}_{3} \).

Recall Hess's Law

Hess's Law states that the total enthalpy change during a chemical reaction is the same, regardless of whether the reaction occurs in one step or multiple steps. Thus, we can add the given enthalpy changes to find the overall enthalpy change for the formation of \( \mathrm{SO}_3 \).

Identify Relevant Equations

The formation of \( \mathrm{SO}_3 \) from sulfur \( \mathrm{S}(s) \) and oxygen \( \mathrm{O}_2(g) \) can be represented as: \( \mathrm{S}(s) + \frac{3}{2}\, \mathrm{O}_2(g) \rightarrow \mathrm{SO}_3(g) \). The given equations are for the sequential production of \( \mathrm{SO}_3 \) via \( \mathrm{SO}_2 \).

Calculate Formation Enthalpy of SO3

The overall enthalpy change \( \Delta H^{\circ} \) for forming \( \mathrm{SO}_3 \) is found by adding the enthalpy changes of the two steps: \[ \Delta H^{\circ} = -296.8 \, \text{kJ/mol} + (-98.9 \, \text{kJ/mol}) = -395.7 \, \text{kJ/mol} \].

Apply the Formation Reaction

The reaction for the formation of \( \mathrm{SO}_3 \) from its elements in their standard states (\( \mathrm{S}(s) + \frac{3}{2} \, \mathrm{O}_2(g) \rightarrow \mathrm{SO}_3(g) \)) allows us to equate \( \Delta H^{\circ} \) to \( \Delta H_{\mathrm{f}}^{\circ}(\mathrm{SO}_3) \). Therefore, \( \Delta H_{\mathrm{f}}^{\circ}(\mathrm{SO}_3) = -395.7 \, \text{kJ/mol} \).

Key concepts

Enthalpy Change

Enthalpy change is a crucial concept in chemistry, marking the energy change during a chemical reaction. It represents the total heat content change of a system as it undergoes a reaction. The symbol for enthalpy change is \( \Delta H \). When a reaction releases heat, it's called exothermic, and \( \Delta H \) is negative. Conversely, for an endothermic reaction, where heat is absorbed, \( \Delta H \) is positive.

Understanding enthalpy is essential to calculating the energy required or released in various chemical processes, such as those involving sulfur trioxide \( \mathrm{SO}_3 \). It allows chemists to predict whether a reaction requires heat input or will release heat during its course.

Hess's Law is often employed in relation to enthalpy, stating that the total change in enthalpy is the same whether a chemical reaction occurs in one step or numerous steps. This law is highly beneficial, as it aids in determining unknown enthalpy changes by using known equations.

Chemical Reactions

Chemical reactions are processes where substances, known as reactants, transform into different substances, known as products. These reactions involve a re-arrangement of atoms and changes in energy levels. Tracking these transformations allows scientists to understand and predict the products and energy outcomes of such reactions.

Chemical reactions can be identified through various characteristics, such as heat release or absorption, color change, gas formation, or the appearance of a precipitate. In our example, the transformation of sulfur \( \mathrm{S}(s) \) and oxygen \( \mathrm{O}_2(g) \) into sulfur trioxide \( \mathrm{SO}_3(g) \) occurs in defined steps with specified energy changes.

By investigating the chemical reaction processes, one can accurately calculate the enthalpy changes using steps and known data. Understanding these procedures is fundamental to succeeding in academic chemical studies and practical applications.

Sulfur Trioxide Formation

Sulfur trioxide \( \mathrm{SO}_3 \) formation is an important chemical process, especially in creating sulfuric acid, a widely produced chemical. The formation occurs in two steps: first, converting sulfur to sulfur dioxide \( \mathrm{SO}_2 \), and then to sulfur trioxide \( \mathrm{SO}_3 \).

1. The first reaction involves sulfur combining with oxygen to form sulfur dioxide:

• \( \mathrm{S}(s) + \mathrm{O}_2(g) \rightarrow \mathrm{SO}_2(g) \)
• Enthalpy change: \( -296.8 \text{ kJ/mol} \)

2. The second step of converting \( \mathrm{SO}_2 \) to \( \mathrm{SO}_3 \) looks like this:

• \( \mathrm{SO}_2(g) + \frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{SO}_3(g) \)
• Enthalpy change: \( -98.9 \text{ kJ/mol} \)

Overall, calculating the enthalpy change for forming \( \mathrm{SO}_3 \) involves adding the enthalpy changes from both steps, resulting in a total \( \Delta H \) of \( -395.7 \text{ kJ/mol} \).

The detailed tracking of these chemical reactions helps achieve accurate enthalpy calculations, crucial for industrial applications and academic comprehension.