Question

The chemistry of nitrogen oxides is very versatile. Given the following reactions and their standard enthalpy changes,(1) \(\mathrm{NO}(g)+\mathrm{NO}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{3}(g) \quad \Delta H_{\mathrm{rxn}}^{\circ}=-39.8 \mathrm{~kJ}\) (2) \(\mathrm{NO}(g)+\mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{5}(g) \Delta H_{\mathrm{rxn}}^{0}=-112.5 \mathrm{~kJ}\) (3) \(2 \mathrm{NO}_{2}(g) \rightarrow \mathrm{N}_{2} \mathrm{O}_{4}(g)\) \(\Delta H_{\mathrm{ran}}^{0}=-57.2 \mathrm{~kJ}\) (4) \(2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)\) \(\Delta H_{\mathrm{rxn}}^{0}=-114.2 \mathrm{~kJ}\) (5) \(\mathrm{N}_{2} \mathrm{O}_{5}(s) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{5}(g)\) $$\Delta H_{\mathrm{ran}}^{\circ}=54.1 \mathrm{~kJ}$$calculate the standard enthalpy of reaction for$$\mathrm{N}_{2} \mathrm{O}_{3}(g)+\mathrm{N}_{2} \mathrm{O}_{5}(s) \longrightarrow 2 \mathrm{~N}_{2} \mathrm{O}_{4}(g) $$

Short answer

149.2 kJ

Step-by-step solution

Identify the target reaction

The target reaction is: \(\mathrm{N}_{2} \mathrm{O}_{3}(g)+\mathrm{N}_{2} \mathrm{O}_{5}(s) \longrightarrow 2 \mathrm{~N}_{2} \mathrm{O}_{4}(g)\)

Analyze the given reactions

The following reactions and their standard enthalpy changes are given:1. \(\mathrm{NO}(g)+\mathrm{NO}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{3}(g) \quad \Delta H_{\mathrm{rxn}}^{\circ}=-39.8 \mathrm{~kJ}\)2. \(\mathrm{NO}(g)+\mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{5}(g) \quad \Delta H_{\mathrm{rxn}}^{\circ}=-112.5 \mathrm{~kJ}\)3. \(2 \mathrm{NO}_{2}(g) \rightarrow \mathrm{N}_{2} \mathrm{O}_{4}(g) \quad \Delta H_{\mathrm{rxn}}^{\circ}=-57.2 \mathrm{~kJ}\)4. \(2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) \quad \Delta H_{\mathrm{rxn}}^{\circ}=-114.2 \mathrm{~kJ}\)5. \(\mathrm{N}_{2} \mathrm{O}_{5}(s) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{5}(g) \quad \Delta H_{\mathrm{rxn}}^{\circ}=54.1 \mathrm{~kJ}\)

Reverse reaction 1

Reverse reaction 1 to get \(\mathrm{N}_{2} \mathrm{O}_{3}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{NO}_{2}(g)\) with an enthalpy change of \(\Delta H = 39.8 \mathrm{kJ}\).

Reverse reaction 2

Reverse reaction 2 to get \(\mathrm{N}_{2} \mathrm{O}_{5}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g)\) with an enthalpy change of \(\Delta H = 112.5 \mathrm{kJ}\).

Consider reaction 3 and 5

Use reaction 3: \(2 \mathrm{NO}_{2}(g) \rightarrow \mathrm{N}_{2} \mathrm{O}_{4}(g) \quad \Delta H_{\mathrm{rxn}}^{\circ}=-57.2 \mathrm{~kJ}\)Use reaction 5: \(\mathrm{N}_{2} \mathrm{O}_{5}(s) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{5}(g) \quad \Delta H_{\mathrm{rxn}}^{\circ}=54.1 \mathrm{~kJ}\)

Combine the reactions

Combine the manipulated reactions as follows:[\(\mathrm{N}_{2} \mathrm{O}_{3}(g) \)] + [\(\mathrm{N}_{2} \mathrm{O}_{5}(s) \rightarrow \mathrm{N}_{2} \mathrm{O}_{5}(g)\)] + [\(\mathrm{N}_{2} \mathrm{O}_{5}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g)\)] + 2 * [\(\mathrm{NO}(g)+\mathrm{NO}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{3}(g)\)] + 2* [\(2 \mathrm{NO}_{2}(g) \rightarrow \mathrm{N}_{2} \mathrm{O}_{4}(g)\)]Cancel common species (\(\mathrm{NO}(g)\), \(\mathrm{NO}_{2}(g)\), \(\mathrm{O}_{2}(g)\)) to form the target reaction.

Calculate the total enthalpy change

Sum up the enthalpy changes:\(\Delta H_{\mathrm{rxn}} = 39.8 \mathrm{kJ} + 54.1 \mathrm{kJ} + 112.5 \mathrm{kJ} - 57.2 \mathrm{kJ}\)\(\Delta H_{\mathrm{rxn}} = 149.2 \mathrm{kJ}\)

Key concepts

Standard Enthalpy Change

The standard enthalpy change of a reaction, denoted as \( \Delta H_{\text{rxn}}^{\text{°}} \), is the heat absorbed or released under standard conditions (1 atm pressure and 298.15 K). It is a fundamental concept in thermochemistry because it allows us to understand and quantify the energy changes in chemical reactions. In the exercise provided, different reactions involving nitrogen oxides are given with their respective standard enthalpy changes.
Standard enthalpy changes help us predict the feasibility and spontaneity of reactions. If \( \Delta H_{\text{rxn}}^{\text{°}} \) is negative, the reaction is exothermic (releases heat). If positive, the reaction is endothermic (absorbs heat). For instance, the reaction where \( \mathrm{NO} \) and \( \mathrm{NO}_2 \) form \( \mathrm{N}_2\text{O}_5 \) is given by \( \Delta H_{\text{rxn}}^{\text{°}} = -112.5 \ \text{kJ} \), indicating it releases heat.
Knowing the standard enthalpy changes of individual reactions allows us to calculate the overall enthalpy change for complex reactions, as seen in the exercise.

Thermochemistry

Thermochemistry is the branch of chemistry that deals with the study of heat changes during chemical reactions. It involves understanding how energy is absorbed or released and how it affects matter. One of the core principles of thermochemistry is the conservation of energy: energy cannot be created or destroyed, only transformed.
In practical terms, thermochemistry helps chemists measure and predict the enthalpy changes of reactions. Techniques like calorimetry are used to measure heat changes, which help calculate enthalpy values. The exercise shows various reactions and their enthalpy changes, demonstrating thermochemical calculations.
Thermochemistry relies on standard conditions and specific heat capacities of substances. This way, we can compare different reactions and evaluate their energy profiles accurately. For example, understanding the enthalpy changes in the given nitrogen oxide reactions helps predict how energy-efficient these processes are.
Thermochemistry also ties into real-world applications like industrial processes, where knowing energy changes can optimize efficiency and cost-effectiveness.

Hess's Law

Hess's Law is a crucial principle in thermochemistry that states the total enthalpy change in a chemical reaction is the same, regardless of the number of steps in the reaction. This allows us to determine the enthalpy change of complex reactions by breaking them down into simpler steps with known enthalpy changes. This principle is utilized in the exercise to calculate the standard enthalpy change for the complex reaction.
To apply Hess's Law, we manipulate individual reactions (reversing or multiplying them) and their enthalpy changes to match the target reaction. The reactions given in the exercise are combined and adjusted to derive the enthalpy change for \( \mathrm{N}_2\text{O}_3(g) + \mathrm{N}_2\text{O}_5(s) \rightarrow 2 \mathrm{N}_2\text{O}_4(g) \).
Here's a step-by-step summary of the use of Hess's Law in the exercise:

• Identify the target reaction.
• Analyze given reactions and their enthalpy changes.
• Reverse and/or adjust individual reactions to form the target reaction.
• Combine reactions and sum their enthalpy changes.

By adhering to Hess's Law, we calculated the total enthalpy change of the target reaction as \( \Delta H_{\text{rxn}}^{\text{°}} = 149.2 \ \text{kJ} \).

Chemical Reactions

Chemical reactions are processes where reactants transform into products, involving the breaking and forming of chemical bonds. These reactions can be categorized based on the type of change they involve, such as synthesis, decomposition, single replacement, double replacement, and combustion.
In the exercise, we deal with various reactions involving nitrogen oxides. Each reaction has specific reactants and products, with distinct enthalpy changes. For example, the reaction of \( \mathrm{NO} \) and \( \mathrm{NO}_2 \) forming \( \mathrm{N}_2\text{O}_3 \) and the reaction forming \( \mathrm{N}_2\text{O}_5 \).
Chemical reactions can be influenced by many factors, including temperature, pressure, concentration, and catalysts. Understanding the reactions' enthalpy changes helps predict their behavior under different conditions.
In the provided exercise, combining and manipulating these reactions using Hess's Law allows us to determine the total enthalpy change for a target reaction. This highlights how interconnected chemical reactions and energy changes are in practical scenarios.
Overall, understanding chemical reactions is pivotal for predicting product formation, energy requirements, and the feasibility of chemical processes, both in laboratories and industrial applications.