Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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 \(\Delta H_{3}^{\circ}\) $$\mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{NO}(\mathrm{g}) \quad \Delta H^{\circ}=?$$

Short Answer

Expert verified
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 by step solution

01

Writing the desired equation

Identify the desired equation: \(\mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{NO}(\mathrm{g})\)
02

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 \(3 \mathrm{H}_{2}(\mathrm{g}) + \frac{3}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow 3 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\)
03

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 \(\Delta H = \Delta H_{1}^{\circ} - 2 \Delta H_{2}^{\circ} + 3 \Delta H_{3}^{\circ} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Use Hess's law to determine \(\Delta H^{\circ}\) for the reaction \(\mathrm{C}_{3} \mathrm{H}_{4}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{g}) \longrightarrow \mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{g}),\) given that $$\mathrm{H}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \quad \Delta H^{\circ}=-285.8 \mathrm{kJ}$$ $$\begin{aligned} \mathrm{C}_{3} \mathrm{H}_{4}(\mathrm{g})+4 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow & 3 \mathrm{CO}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \\ && \Delta H^{\circ}=-1937 \mathrm{kJ} \end{aligned}$$ $$\begin{array}{r} \mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{g})+5 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow 3 \mathrm{CO}_{2}(\mathrm{g})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \\ \Delta H^{\circ}=-2219.1 \mathrm{kJ} \end{array}$$

The combustion of methane gas, the principal constituent of natural gas, is represented by the equation $$\begin{aligned} \mathrm{CH}_{4}(\mathrm{g})+2 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g})+& 2 \mathrm{H}_{2} \mathrm{O}(1) \\ \Delta H^{\circ} &=-890.3 \mathrm{kJ} \end{aligned}$$ (a) What mass of methane, in kilograms, must be burned to liberate \(2.80 \times 10^{7} \mathrm{kJ}\) of heat? (b) What quantity of heat, in kilojoules, is liberated in the complete combustion of \(1.65 \times 10^{4} \mathrm{L}\) of \(\mathrm{CH}_{4}(\mathrm{g})\) measured at \(18.6^{\circ} \mathrm{C}\) and \(768 \mathrm{mmHg} ?\) (c) If the quantity of heat calculated in part (b) could be transferred with \(100 \%\) efficiency to water, what volume of water, in liters, could be heated from 8.8 to \(60.0^{\circ} \mathrm{C}\) as a result?

A 1.22 kg piece of iron at \(126.5^{\circ} \mathrm{C}\) is dropped into \(981 \mathrm{g}\) water at \(22.1^{\circ} \mathrm{C} .\) The temperature rises to \(34.4^{\circ} \mathrm{C} .\) What will be the final temperature if this same piece of iron at \(99.8^{\circ} \mathrm{C}\) is dropped into \(325 \mathrm{mL}\) of glycerol, \(\mathrm{HOCH}_{2} \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{2} \mathrm{OH}(1)\) at \(26.2^{\circ} \mathrm{C} ?\) For glycerol, \(d=1.26 \mathrm{g} / \mathrm{mL} ; C_{n}=219 \mathrm{JK}^{-1} \mathrm{mol}^{-1}\).

Brass has a density of \(8.40 \mathrm{g} / \mathrm{cm}^{3}\) and a specific heat of \(0.385 \mathrm{Jg}^{-1}\) \(^{\circ} \mathrm{C}^{-1} . \mathrm{A} 15.2 \mathrm{cm}^{3}\) piece of brass at an initial temperature of \(163^{\circ} \mathrm{C}\) is dropped into an insulated container with \(150.0 \mathrm{g}\) water initially at \(22.4^{\circ} \mathrm{C}\) What will be the final temperature of the brass-water mixture?

The internal energy of a fixed quantity of an ideal gas depends only on its temperature. A sample of an ideal gas is allowed to expand at a constant temperature (isothermal expansion). (a) Does the gas do work? (b) Does the gas exchange heat with its surroundings? (c) What happens to the temperature of the gas? (d) What is \(\Delta U\) for the gas?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free