91Ó°ÊÓ

Enthalpy changes for the following reactions can be determined experimentally: \(\mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{NH}_{3}(\mathrm{g})\) $$ \Delta_{r} H^{\circ}=-91.8 \mathrm{kJ} / \mathrm{mol}-\mathrm{rxn} $$ \(4 \mathrm{NH}_{3}(\mathrm{g})+5 \mathrm{O}_{2}(\mathrm{g}) \rightarrow 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) $$ \Delta_{\mathrm{r}} H^{\circ}=-906.2 \mathrm{kJ} / \mathrm{mol}-\mathrm{rxn} $$ \(\mathrm{H}_{2}(\mathrm{g})+1 / 2 \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) $$ \Delta_{i} H^{\circ}=-241.8 \mathrm{kJ} / \mathrm{mol}-\mathrm{v} \mathrm{xn} $$ Use these values to determine the enthalpy change for the formation of \(\mathrm{NO}(\mathrm{g})\) from the elements (an enthalpy change that cannot be measured directly because the reaction is reactant-favored). $$ ^{1 / 2} \mathrm{N}_{2}(\mathrm{g})+1 / 2 \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{NO}(\mathrm{g}) \quad \Delta_{r} H^{\circ}=? $$

Step by step solution

01

Write the Target Reaction

The target reaction is \( \frac{1}{2} \mathrm{N}_{2}(\mathrm{g}) + \frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{NO}(\mathrm{g}) \). Our goal is to find \( \Delta_{r} H^{\circ} \) for this reaction using the provided reactions.

02

Analyze the Given Reactions

We have three reactions: 1. \( \mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{NH}_{3}(\mathrm{g}) \) with \( \Delta_{r} H^{\circ} = -91.8 \mathrm{kJ/mol-rxn} \).2. \( 4 \mathrm{NH}_{3}(\mathrm{g})+5 \mathrm{O}_{2}(\mathrm{g}) \rightarrow 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \) with \( \Delta_{r} H^{\circ} = -906.2 \mathrm{kJ/mol-rxn} \).3. \( \mathrm{H}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \) with \( \Delta_{i} H^{\circ} = -241.8 \mathrm{kJ/mol-rev} \).

03

Adjust Given Reactions to Form the Target Reaction

First, note that we need \( \mathrm{NO} \) on the product side. From Reaction 2, we have 4 \( \mathrm{NO} \) formed. We need to adjust this to obtain \( \frac{1}{2} \mathrm{NO} \), so divide the entire Reaction 2 by 8.This gives:\[ \mathrm{NH}_{3}(\mathrm{g}) + \frac{5}{8} \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{NO}(\mathrm{g}) + \frac{3}{2} \mathrm{H}_{2}\mathrm{O}(\mathrm{g}) \]Divide the reaction enthalpy by 8: \(-\frac{906.2}{8} = -113.275 \mathrm{kJ/mol-rxn}\).

04

Add Up Adjusted Reactions to Solve for Target

Now, for each \( \mathrm{NO} \) produced, you need \( \frac{1}{2} \mathrm{N}_{2} \) and \( \frac{1}{2} \mathrm{O}_{2} \). Notice that \( \frac{1}{2} \mathrm{N}_{2} \) is produced from Reaction 1 when 1 mole of \( \mathrm{NH}_{3} \) is used. Therefore, use Reaction 3, which provides \( \mathrm{H}_{2} \mathrm{O} \), independent of the \( \mathrm{NO} \) process and it correlates with forming water.However, by rearranging, based on combined reaction:\[ \frac{1}{2} \mathrm{N}_{2} \rightarrow \mathrm{NH}_{3} \quad \Delta H_{eff1} = -91.8/3 \; (-30.6 \mathrm{kJ/mol-rxn})\]Sum to achieve:\[ \frac{1}{2} \mathrm{N}_{2} + \frac{1}{2} \mathrm{O}_{2} \rightarrow \mathrm{NO} \quad \Delta_{r} H^{\circ} = 30.6 + (-113.275) + \left(3\left(-241.8/2\right)\right) \]Evaluate, knowing the hypothetical, subtraction:\[\quad 90\; (since\ terms\ cancel) \; -113.275 = -21.775\].

05

Solve for Final Enthalpy Change

By adding the adjusted reaction enthalpies, the final enthalpy change for \( \frac{1}{2} \mathrm{N}_{2} + \frac{1}{2} \mathrm{O}_{2} \rightarrow \mathrm{NO} \) gives:\[ \Delta_{r} H^{\circ} = -21.775 \mathrm{kJ/mol} \] this results from net reducing terms and problem solution verification.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Thermochemistry

Thermochemistry is the study of energy transformations during chemical reactions. It's all about understanding how heat is absorbed or released when substances interact. This energy change is usually measured in terms of enthalpy (\( H \)), which represents the total heat content of a system.
In a chemical reaction, the enthalpy change (\( \Delta H \)) tells us whether the reaction is endothermic (absorbing heat) or exothermic (releasing heat).

• **Exothermic reactions** release energy, shown by a negative \( \Delta H \). This means the products have lower energy than the reactants.
• **Endothermic reactions** absorb energy, resulting in a positive \( \Delta H \). Here, the products have higher energy than the reactants.

Understanding these concepts helps in predicting reaction behavior and managing them in industrial processes, where energy exchange is crucial.

Hess's Law

Hess's Law is a principle in chemistry used to calculate enthalpy changes in a way that is practical when direct measurement isn't possible. According to Hess's Law, if a chemical reaction is expressed as a series of steps, the overall enthalpy change is equal to the sum of the enthalpy changes for each step.
This law is extremely useful for calculating enthalpy changes for reactions that are difficult to measure directly, such as the formation of reactive or unstable compounds. For our exercise, we used Hess's Law to find the enthalpy change for the formation of \( \text{NO} \) gas by manipulating given reactions.

• **Key Insight**: The path taken to reach the final reaction does not affect the overall enthalpy change.
• **Application**: Simply add up the enthalpy changes of the individual steps to find the overall change.

Understanding and applying Hess's Law provides flexibility in chemistry, allowing predictions of reaction energetics without direct experimentation.

Chemical Reactions

Chemical reactions involve the reorganization of atoms and are described by chemical equations. The focus of thermochemistry in reactions is to understand how these atomic changes correlate with energy changes.
In our exercise, several reactions were considered:

• Combining nitrogen and hydrogen to form ammonia shows an exothermic process.
• Burning ammonia in oxygen results in the formation of nitrogen monoxide and water.
• Forming water from hydrogen and oxygen is another exothermic process.

Balancing such equations ensures that the law of conservation of mass is obeyed, meaning that atoms are neither created nor destroyed.

• **Reactants and Products**: Reactions start with reactants and end with products, and keeping track of \( \Delta H \) helps in understanding the energy landscape of these transformations.
• **Stoichiometry**: Correct stoichiometric coefficients ensure the right pathway to calculate enthalpy changes with Hess's Law.

By understanding these basics, you can predict not just the quantities of substances involved in reactions but also the energy transformations accompanying these changes.