Problem 62 - Chemistry: The Central Science

Question

From the enthalpies of reaction \(2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(g) \quad \Delta H=-483.6 \mathrm{~kJ}\) \(3 \mathrm{O}_{2}(g) \stackrel{-\cdots}{\longrightarrow} 2 \mathrm{O}_{3}(g) \quad \Delta H=+284.6 \mathrm{~kJ}\) calculate the heat of the reaction $$ 3 \mathrm{H}_{2}(g)+\mathrm{O}_{3}(g) \longrightarrow 3 \mathrm{H}_{2} \mathrm{O}(g) $$

Step-by-Step Solution

Answer

The enthalpy change (heat of the reaction) for the target reaction \( 3 \mathrm{H}_2(g) + \mathrm{O}_3(g) \longrightarrow 3 \mathrm{H}_2\mathrm{O}(g) \) is -440.8 kJ.

Step 1: Start by comparing the given reactions to the target reaction. Identify which species can be directly mapped to those in the target reaction, and which ones require addition or subtraction of reactions to obtain the desired coefficients. #Step 2: Manipulate given reactions to match target reaction coefficients#

To do this, we will adjust each of the given reactions to prepare them for addition or subtraction later on. We need 3 moles of \( \mathrm{H}_2\) in the target reaction, so we will multiply the first reaction (R1) by 1.5. R1: \( 1.5 \times (2 \mathrm{H}_2(g) + \mathrm{O}_2(g) \longrightarrow 2 \mathrm{H}_2\mathrm{O}(g)) \) R1: \( 3 \mathrm{H}_2(g) + 1.5\mathrm{O}_2(g) \longrightarrow 3 \mathrm{H}_2\mathrm{O}(g) \) As for the second reaction (R2) - R2 does not need any modification, so we can use it as it is. R2: \( 3 \mathrm{O}_2(g) \stackrel{-\cdots}{\longrightarrow} 2 \mathrm{O}_3(g) \) Now, the enthalpies for these adjusted reactions must also be adjusted accordingly: \( \Delta H_1 = -483.6 \: \mathrm{kJ} \times 1.5 = -725.4 \: \mathrm{kJ}\) \( \Delta H_2 = 284.6 \: \mathrm{kJ}\), no change, as R2 isn't modified. #Step 3: Combine reactions to match target reaction#

Step 2: Using R1 and R2, we will now combine (add or subtract) them and obtain the target reaction R1: \( 3 \mathrm{H}_2(g) + 1.5\mathrm{O}_2(g) \longrightarrow 3 \mathrm{H}_2\mathrm{O}(g) \) R2: \( 2 \mathrm{O}_3(g) \longrightarrow 3 \mathrm{O}_2(g) \) Sum (R1 + R2): \( 3 \mathrm{H}_2(g) + 1.5\mathrm{O}_2(g) + 2 \mathrm{O}_3(g) \longrightarrow 3 \mathrm{H}_2\mathrm{O}(g) + 3 \mathrm{O}_2(g) \) From this sum, we will eliminate the overlapping species: Target Reaction: \( 3 \mathrm{H}_{2}(g) + \mathrm{O}_{3}(g) \longrightarrow 3 \mathrm{H}_{2}\mathrm{O}(g) \) #Step 4: Combine enthalpies to find target reaction enthalpy#

Now we will combine the enthalpies of the adjusted reactions to find the target reaction enthalpy (\( \Delta H \)). \( \Delta H = \Delta H_1 + \Delta H_2 \) \( \Delta H = -725.4 \: \mathrm{kJ} + 284.6 \: \mathrm{kJ} \) \( \Delta H = -440.8 \: \mathrm{kJ} \) The enthalpy change (heat of the reaction) for the target reaction \( 3 \mathrm{H}_2(g) + \mathrm{O}_3(g) \longrightarrow 3 \mathrm{H}_2\mathrm{O}(g) \) is -440.8 kJ.

Key Concepts

Understanding Enthalpy Change

Enthalpy change is a fundamental concept in chemical thermodynamics, which is pivotal to understanding how energy is absorbed or released during a chemical reaction. It is represented by the symbol \( \Delta H \). An exothermic reaction releases heat, leading to a negative \( \Delta H \), while an endothermic reaction absorbs heat, resulting in a positive \( \Delta H \). In the given exercise, the target reaction involves ozone (\( \mathrm{O}_3 \)) and hydrogen (\( \mathrm{H}_2 \)) converting into water (\( \mathrm{H}_2\mathrm{O} \)), with a calculated enthalpy change of \(-440.8 \ \mathrm{kJ}\). This indicates that the reaction is exothermic, releasing a significant amount of energy as heat. This release of energy is crucial because it not only influences the spontaneity and feasibility of the reaction, but also provides insights into the potential applications of the reaction in energy production and environmental processes. Understanding these changes helps students and chemists predict reaction behavior and plan experiments effectively.

Reaction Manipulation with Hess's Law

Reaction manipulation is often required to calculate the enthalpy change for reactions that are not directly measurable. This is where Hess's Law becomes invaluable. Hess's Law states that the total enthalpy change for a reaction is the same, regardless of whether it takes place in one step or multiple steps. This allows us to use a series of reactions with known enthalpy changes to determine the enthalpy change of the desired reaction.
In the exercise example, reaction manipulation involved multiplying and combining equations to match the target reaction. We multiplied the first reaction by 1.5 so that the coefficients aligned correctly with the target equation. This manipulation maintained the integrity of the thermodynamic properties due to Hess's Law, hence permitting the accurate calculation of the enthalpy change for the final reaction.

Basics of Chemical Thermodynamics

Chemical thermodynamics is a branch of chemistry that studies the relation between heat, work, and energy within chemical reactions and physical transformations. It forms the foundation for concepts like enthalpy change and Hess's Law. By understanding chemical thermodynamics, one can predict how chemical reactions will behave under different conditions.
Key concepts include:

• System vs. Surroundings: The part of the universe under study is the system, while everything else is the surroundings.
• Energy Conservation: In a closed system, the total energy, comprising internal energy, heat, and work, remains constant.
• Spontaneity and Gibbs Free Energy: Spontaneous reactions tend to be thermodynamically favorable, often assessed via Gibbs Free Energy (\( \Delta G \)).

These principles are essential for calculating and predicting the outcomes of chemical reactions like the one mentioned in the exercise. Applying these concepts allows us to determine not only if a reaction is possible, but also gives insight into the energy changes involved, influencing both the reaction's feasibility and its potential applications.