Problem 70 - General Chemistry: Principles and Modern Applications

Question

Given the following information: $$\frac{1}{2} \mathrm{N}_{2}(\mathrm{g})+\frac{3}{2} \mathrm{H}_{2}(\mathrm{g}) \longrightarrow \mathrm{NH}_{3}(\mathrm{g})\quad\quad\quad\quad\Delta H_{1}^{\circ}$$ $$\mathrm{NH}_{3}(\mathrm{g})+\frac{5}{4} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{NO}(\mathrm{g})+\frac{3}{2} \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \quad \Delta H_{2}^{\circ}$$ $$\mathrm{H}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\quad\quad\quad\Delta H_{3}^{\circ}$$ Determine \(\Delta H^{\circ}\) for the following reaction, expressed in terms of \(\Delta H_{1}^{\circ}, \Delta H_{2}^{\circ},\) and \(\Delta H_{3}^{\circ}\) $$\mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{NO}(\mathrm{g}) \quad \Delta H^{\circ}=?$$

Step-by-Step Solution

Answer

The overall enthalpy change for the desired reaction, in terms of \(\Delta H_{1}^{\circ}, \Delta H_{2}^{\circ},\) and \(\Delta H_{3}^{\circ}\), is \(\Delta H = \Delta H_{1}^{\circ} - 2 \Delta H_{2}^{\circ} + 3 \Delta H_{3}^{\circ}\).

Step 1: Writing the desired equation

Identify the desired equation: \(\mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{NO}(\mathrm{g})\)

Step 2: Modifying the given equations

Transform the given equations into forms that will add in such a way to yield the desired equation. For the first equation, nothing needs to be done. For the second equation, it should be reversed and multiplied by \(2\) to produce \(-2 \mathrm{NH}_{3}(\mathrm{g}) - \frac{5}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow -2 \mathrm{NO}(\mathrm{g}) - 3 \mathrm{H}_{2}\mathrm{O}(\mathrm{l})\). The third equation should be multiplied by \(3\) to yield \(3 \mathrm{H}_{2}(\mathrm{g}) + \frac{3}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow 3 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\)

Step 3: Addition of manipulated equations & enthalpies

Add the three modified equations: \(\frac{1}{2} \mathrm{N}_{2}(\mathrm{g}) + \frac{3}{2} \mathrm{H}_{2}(\mathrm{g}) - 2 \mathrm{NH}_{3}(\mathrm{g}) + 3 \mathrm{H}_{2}(\mathrm{g}) \longrightarrow \mathrm{NH}_{3}(\mathrm{g}) - 2 \mathrm{NO}(\mathrm{g}) - 3 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) + 3 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \). Cancel terms on both sides to obtain the desired equation. Correspondingly, add the enthalpies with modifications. The overall \(\Delta H = \Delta H_{1}^{\circ} - 2 \Delta H_{2}^{\circ} + 3 \Delta H_{3}^{\circ} \).

Key Concepts

Understanding Enthalpy Change

Enthalpy change, denoted as \( \Delta H \), is a crucial concept in chemistry that measures the heat absorbed or evolved during a chemical reaction. It reflects the energy change in the system as reactants convert to products. - If \( \Delta H \) is negative, the reaction is exothermic, meaning it releases heat to the surroundings.- Conversely, if \( \Delta H \) is positive, the reaction is endothermic, absorbing heat from the surroundings.In the context of Hess's Law, which focuses on the enthalpy changes of reactions, we utilize the fact that enthalpy is a state function. This means the overall enthalpy change for a reaction is the same, regardless of the path taken. So, by altering and summing the enthalpy changes of known reactions, we can find the enthalpy change for a reaction that might be difficult to measure directly. In the given exercise, Hess’s Law is applied by manipulating the provided reaction equations. By reversing, multiplying, or canceling parts of these equations, students align them to reflect the new reaction while calculating the overall energy change \( \Delta H = \Delta H_{1}^{\circ} - 2\Delta H_{2}^{\circ} + 3\Delta H_{3}^{\circ} \). This illustrates how enthalpy changes can be strategically combined to achieve the desired result.

Exploring Chemical Reactions

Chemical reactions are transformations where substances, the reactants, convert into different substances, the products. Each reaction is characterized by changes in properties and energy. A balanced chemical equation represents the stoichiometry of the reactants and products involved. Understanding the mechanics of balancing these equations is vital for studying reactions. For students working through Hess's Law exercises, comprehending how to manipulate and combine reactions is essential. In our exercise, three chemical reactions involving nitrogen, hydrogen, oxygen, and water serve as steps toward forming nitrogen monoxide from nitrogen and oxygen. - The initial reactions provide equations with associated enthalpy changes \( \Delta H_1, \Delta H_2, \text{and} \Delta H_3 \).- By adjusting coefficients and reversing reactions, the target reaction is achieved.- Terms are precisely balanced to ensure both mass and energy are conserved.This approach highlights the creativity involved in chemistry, using logical strategies to manipulate known reactions to reveal unknown reaction paths.

Fundamentals of Thermodynamics

Thermodynamics, a branch of physics, deals with energy transformations, particularly the exchange of heat or work between systems and their surroundings. It involves laws and principles that govern these exchanges and explain why reactions occur. The first law, known as the Law of Energy Conservation, is pivotal here. It states that energy cannot be created or destroyed, only transformed. Hence, the total energy content of an isolated system remains constant, which is key when applying Hess’s Law. In practical applications like our exercise:- Energy changes calculated (\( \Delta H \)) are a direct reflection of thermodynamic principles.- The systematic combination of reaction enthalpies shows that the total enthalpy change aligns with the energy conservation law.Thermodynamic stability can also be inferred. Reactions with a significant exothermic enthalpy change \( \Delta H < 0 \) yield more stable products. Grasping these thermodynamic principles enriches our understanding of why certain chemical processes are spontaneous and how energy flows affect reaction feasibility.