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

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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.

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Most popular questions from this chapter

The standard molar heats of combustion of C(graphite) and \(\mathrm{CO}(\mathrm{g})\) are -393.5 and \(-283 \mathrm{kJ} / \mathrm{mol}\) respectively. Use those data and that for the following reaction $$\mathrm{CO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow \mathrm{COCl}_{2}(\mathrm{g}) \quad \Delta H^{\circ}=-108 \mathrm{kJ}$$ to calculate the standard molar enthalpy of formation of \(\mathrm{COCl}_{2}(\mathrm{g})\).

Thermite mixtures are used for certain types of welding, and the thermite reaction is highly exothermic. $$\begin{array}{r} \mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{s})+2 \mathrm{Al}(\mathrm{s}) \longrightarrow \mathrm{Al}_{2} \mathrm{O}_{3}(\mathrm{s})+2 \mathrm{Fe}(\mathrm{s}) \\ \Delta H^{\circ}=-852 \mathrm{kJ} \end{array}$$ \(1.00 \mathrm{mol}\) of granular \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) and \(2.00 \mathrm{mol}\) of granular Al are mixed at room temperature \(\left(25^{\circ} \mathrm{C}\right),\) and a reaction is initiated. The liberated heat is retained within the products, whose combined specific heat over a broad temperature range is about \(0.8 \mathrm{Jg}^{-1}\) \(^{\circ} \mathrm{C}^{-1} .\) (The melting point of iron is \(1530^{\circ} \mathrm{C} .\) ) Show that the quantity of heat liberated is more than sufficient to raise the temperature of the products to the melting point of iron.

How much heat, in kilojoules, is evolved in the complete combustion of (a) \(1.325 \mathrm{g} \mathrm{C}_{4} \mathrm{H}_{10}(\mathrm{g})\) at \(25^{\circ} \mathrm{C}\) and \(1 \mathrm{atm} ;\) (b) \(28.4 \mathrm{L} \mathrm{C}_{4} \mathrm{H}_{10}(\mathrm{g})\) at \(\mathrm{STP} ;(\mathrm{c})\) \(12.6 \mathrm{LC}_{4} \mathrm{H}_{10}(\mathrm{g})\) at \(23.6^{\circ} \mathrm{C}\) and \(738 \mathrm{mmHg} ?\) Assume that the enthalpy change for the reaction does not change significantly with temperature or pressure. The complete combustion of butane, \(\mathrm{C}_{4} \mathrm{H}_{10}(\mathrm{g}),\) is represented by the equation $$\begin{array}{r} \mathrm{C}_{4} \mathrm{H}_{10}(\mathrm{g})+\frac{13}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow 4 \mathrm{CO}_{2}(\mathrm{g})+5 \mathrm{H}_{2} \mathrm{O}(1) \\ \Delta H^{\circ}=-2877 \mathrm{kJ} \end{array}$$

Under the entry \(\mathrm{H}_{2} \mathrm{SO}_{4},\) a reference source lists many values for the standard enthalpy of formation. For example, for pure \(\mathrm{H}_{2} \mathrm{SO}_{4}(1), \Delta H_{\mathrm{f}}^{\circ}=-814.0 \mathrm{kJ} / \mathrm{mol}\) for a solution with \(1 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}\) per mole of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) \(-841.8 ;\) with \(10 \mathrm{mol} \mathrm{H}_{2} \mathrm{O},-880.5 ;\) with \(50 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}\) \(-886.8 ;\) with \(100 \mathrm{mol} \mathrm{H}_{2} \mathrm{O},-887.7 ;\) with \(500 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}\) \(-890.5 ;\) with \(1000 \mathrm{mol} \mathrm{H}_{2} \mathrm{O},-892.3 ;\) with \(10,000 \mathrm{mol}\) \(\mathrm{H}_{2} \mathrm{O},-900.8 ;\) and with \(100,000 \mathrm{mol} \mathrm{H}_{2} \mathrm{O},-907.3\) (a) Explain why these values are not all the same. (b) The value of \(\Delta H_{\mathrm{f}}^{\circ}\left[\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})\right]\) in an infinitely dilute solution is \(-909.3 \mathrm{kJ} / \mathrm{mol} .\) What data from this chapter can you cite to confirm this value? Explain. (c) If \(500.0 \mathrm{mL}\) of \(1.00 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})\) is prepared from pure \(\mathrm{H}_{2} \mathrm{SO}_{4}(1),\) what is the approximate change in temperature that should be observed? Assume that the \(\mathrm{H}_{2} \mathrm{SO}_{4}(1)\) and \(\mathrm{H}_{2} \mathrm{O}(1)\) are at the same temperature initially and that the specific heat of the \(\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})\) is about \(4.2 \mathrm{Jg}^{-1}\) \(^{\circ} \mathrm{C}^{-1}\).

The method of Exercise 97 is used in some bomb calorimetry experiments. A 1.148 g sample of benzoic acid is burned in excess \(\mathrm{O}_{2}(\mathrm{g})\) in a bomb immersed in 1181 g of water. The temperature of the water rises from 24.96 to \(30.25^{\circ} \mathrm{C}\). The heat of combustion of benzoic acid is \(-26.42 \mathrm{kJ} / \mathrm{g} .\) In a second experiment, a \(0.895 \mathrm{g}\) powdered coal sample is burned in the same calorimeter assembly. The temperature of \(1162 \mathrm{g}\) of water rises from 24.98 to \(29.81^{\circ} \mathrm{C}\). How many metric tons (1 metric ton \(=1000 \mathrm{kg}\) ) of this coal would have to be burned to release \(2.15 \times 10^{9} \mathrm{kJ}\) of heat?

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