Question

Given the following data: $$\begin{array}{ll}\mathrm{S}(s)+\frac{3}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{SO}_{3}(g) & \Delta H=-395.2 \mathrm{~kJ} \\\2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{SO}_{3}(g) & \Delta H=-198.2\mathrm{~kJ}\end{array}$$ Calculate \(\Delta H\) for the reaction \(\mathrm{S}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{SO}_{2}(g)\)

Short answer

The ΔH for the target reaction, S(s) + O₂(g) → SO₂(g), is 296.1 kJ.

Step-by-step solution

Write down the given reactions and ΔH values

We know the enthalpy changes for the following reactions: 1. S(s) + 3/2 O₂(g) → SO₃(g), ΔH₁ = -395.2 kJ 2. 2 SO₂(g) + O₂(g) → 2 SO₃(g), ΔH₂ = -198.2 kJ Our goal is to find the ΔH value for the reaction: S(s) + O₂(g) → SO₂(g)

Reverse the first reaction

We need to reverse the first reaction to obtain SO₂(g) as a product. To reverse the reaction, we will change the sign of ΔH for that reaction. Reversed reaction 1: SO₃(g) → S(s) + 3/2 O₂(g), -ΔH₁ = 395.2 kJ

Multiply the second reaction by 1/2

We want to obtain only one mole of SO₂(g), so we need to multiply the second reaction by 1/2. Keep in mind that we must also multiply the ΔH value by the same factor. 1/2 (2 SO₂(g) + O₂(g) → 2 SO₃(g)): SO₂(g) + 1/2 O₂(g) → SO₃(g), (1/2)ΔH₂ = -99.1 kJ

Add the modified reactions to obtain the target reaction

Now, add the modified reactions from Step 2 and Step 3: Reversed reaction 1: SO₃(g) → S(s) + 3/2 O₂(g), -ΔH₁ = 395.2 kJ Modified reaction 2: SO₂(g) + 1/2 O₂(g) → SO₃(g), (1/2)ΔH₂ = -99.1 kJ Target reaction: SO₃(g) + SO₂(g) + 1/2 O₂(g) → S(s) + 3/2 O₂(g) + SO₃(g) We see that SO₃(g) is a reactant in reaction 1 and product in reaction 2. They cancel out, and we have the target reaction: S(s) + O₂(g) → SO₂(g)

Calculate ΔH for the target reaction

Now we just need to add the ΔH values for the modified reactions: ΔH (S(s) + O₂(g) → SO₂(g)) = -ΔH₁ + (1/2)ΔH₂ = 395.2 kJ - 99.1 kJ = 296.1 kJ The ΔH for the target reaction, S(s) + O₂(g) → SO₂(g), is 296.1 kJ.

Key concepts

Enthalpy Change

Enthalpy change, denoted as \( \Delta H \), is a measure of the heat absorbed or released during a chemical reaction at constant pressure. It is expressed in kilojoules per mole (kJ/mol). This value helps predict whether a reaction is endothermic (absorbs heat) or exothermic (releases heat). In our exercise, we calculated \( \Delta H \) for the reaction \( \text{S}(s) + \text{O}_2(g) \rightarrow \text{SO}_2(g) \), which turned out to be positive, indicating an endothermic process.
This concept is crucial in understanding energy changes in reactions, aiding in fields like thermochemistry and kinetics. Enthalpy change is often determined using calorimetry or calculated from other reactions, using principles like Hess's Law, which states that the total enthalpy change for a reaction is the same, no matter how many steps the reaction is carried out in.

Thermochemical Equations

Thermochemical equations are balanced chemical equations that include the enthalpy change (\( \Delta H \)) information. These equations help in predicting the energy requirements or releases in chemical processes. The given data in our exercise provided two thermochemical equations, each with its respective \( \Delta H \) value.
Thermochemical equations are used to track the energy profile of reactions, providing critical information for designing industrial processes, where energy efficiency is paramount. They also facilitate calculations involving enthalpy changes using Hess's Law, as they incorporate enthalpy directly as a part of the chemical equation. By integrating \( \Delta H \) into the equation, one can intuitively see whether the process is gaining or losing heat.

Reversing Chemical Reactions

Reversing chemical reactions is an essential concept in solving problems involving Hess's Law. When a reaction is reversed, the sign of the \( \Delta H \) value must also be reversed. This reflects the fact that if a reaction releases energy in one direction, it will absorb the same amount when going in the opposite direction.
In our example, we reversed the given reaction \( \text{SO}_3(g) \rightarrow \text{S}(s) + \frac{3}{2} \text{O}_2(g) \), changing the \( \Delta H \) from \(-395.2 \text{ kJ} \) to \(+395.2 \text{ kJ} \). Reversing reactions is critical in manipulating chemical equations to calculate enthalpy changes for different reactions using Hess's Law.

Combining Chemical Reactions

Combining chemical reactions involves manipulating and summing different reactions to arrive at a target reaction. This often includes altering the coefficients of reactants and products, reversing reactions, or even multiplying reactions by a fraction, as needed.
In the exercise solution, different reactions were combined and manipulated to find the \( \Delta H \) for a target reaction. We modified the second reaction by half to adjust the stoichiometry, then combined it with a reversed reaction, effectively canceling intermediate compounds and obtaining the desired reaction. This method works under the principle of Hess's Law, which allows the addition of reaction enthalpy changes to find the total \( \Delta H \) for the overall reaction. It provides a powerful technique in chemistry for calculating unknown enthalpy changes by using known reactions.