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Chapter 6: Problem 35

What is the enthalpy change for the preparation of one mole of liquid water from the elements, given the following equations? $$ \begin{aligned} &\mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g) ; \Delta H_{f} \\ &\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g) ; \Delta H_{v a p} \end{aligned} $$

Short Answer

Expert verified
The enthalpy change is \( \Delta H = \Delta H_{f} - \Delta H_{vap} \).

Step by step solution

01

Understanding Given Reactions

We are provided two reactions: 1. The formation of water vapor from hydrogen and oxygen: \( \mathrm{H}_{2}(g) + \frac{1}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{H}_{2} \mathrm{O}(g) \). The enthalpy change for this reaction is \( \Delta H_{f} \) (heat of formation).2. The conversion of liquid water to water vapor: \( \mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{H}_{2} \mathrm{O}(g) \). The enthalpy change here is \( \Delta H_{vap} \) (heat of vaporization).
02

Identifying Target Reaction

The target reaction is the formation of one mole of liquid water from hydrogen gas and oxygen gas: \( \mathrm{H}_{2}(g) + \frac{1}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{H}_{2} \mathrm{O}(l) \).
03

Using Hess's Law

According to Hess's Law, the enthalpy change of the target reaction can be calculated using the given reactions. This involves reversing the second given reaction to represent the condensation process: \( \mathrm{H}_{2} \mathrm{O}(g) \rightarrow \mathrm{H}_{2} \mathrm{O}(l) \), which will have an enthalpy change of \( -\Delta H_{vap} \).
04

Calculating Enthalpy Change for Target Reaction

To find the enthalpy change for the formation of liquid water, \( \Delta H \), apply the following equation: \[ \Delta H = \Delta H_{f} + (-\Delta H_{vap}) \].This simplifies to: \[ \Delta H = \Delta H_{f} - \Delta H_{vap} \].

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Key Concepts

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

Enthalpy Change
Enthalpy change is a crucial concept in chemistry that refers to the heat absorbed or released during a chemical reaction at constant pressure. It is denoted by \( \Delta H \), where a positive \( \Delta H \) indicates an endothermic reaction (heat is absorbed), and a negative \( \Delta H \) demonstrates an exothermic reaction (heat is released). Understanding enthalpy change is key to studying reaction energetics and predicting how different chemical reactions will behave.In the context of Hess's Law, enthalpy change provides a method to determine the heat change of a reaction by combining known reactions. For example, when forming liquid water from hydrogen and oxygen gases, the enthalpy change is calculated by considering the heat of formation of water vapor and adjusting for the required phase change to liquid water. Because Hess's Law states that the total enthalpy change is the same no matter how a reaction proceeds, we can manipulate intermediate reactions to find the desired reaction's enthalpy change.
Heat of Formation
The heat of formation, or enthalpy of formation, \( \Delta H_f \), is the energy change when one mole of a compound is formed from its elements in their standard states. The standard state refers to the most stable form of an element or compound at 1 atm pressure and a specified temperature, usually 25掳C.This concept allows chemists to calculate the energy released or absorbed during the formation of a compound. For example, in the formation of water vapor from hydrogen and oxygen, \( \Delta H_f \) represents the enthalpy change when one mole of water vapor is formed. This reaction usually releases energy, making \( \Delta H_f \) negative, indicating that forming water vapor is an exothermic process that releases heat.The table of standard enthalpies of formation for various compounds helps predict the overall energy change of different reactions by summing the individual heat changes for each component.
Heat of Vaporization
The heat of vaporization, represented as \( \Delta H_{vap} \), is the energy required to convert one mole of a substance from liquid to gas at its boiling point under standard pressure. This concept is integral to understanding phase changes, particularly those involving water, which requires a significant amount of energy to break the hydrogen bonds between molecules.In the given exercise, the heat of vaporization for water is considered when calculating the total enthalpy change for forming liquid water from its gaseous elements. By understanding that \( \Delta H_{vap} \) is positive, indicating energy absorption during vaporization, reversing this reaction (condensation) entails energy release, thus making it negative.Recognizing the role of \( \Delta H_{vap} \) in processes like condensation is essential for manipulating reaction enthalpies using Hess's Law and forming a comprehensive view of energy transformations in chemical reactions.

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Most popular questions from this chapter

A piece of iron was heated to \(95.4^{\circ} \mathrm{C}\) and dropped into a constant-pressure calorimeter containing \(284 \mathrm{~g}\) of water at \(32.2^{\circ} \mathrm{C}\). The final temperature of the water and iron was \(51.9^{\circ} \mathrm{C}\). Assuming that the calorimeter itself absorbs a negligible amount of heat, what was the mass (in grams) of the piece of iron? The specific heat of iron is \(0.449 \mathrm{~J} /\left(\mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\), and the specific heat of water is \(4.18 \mathrm{~J} /\left(\mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\).

With a platinum catalyst, ammonia will burn in oxygen give nitric oxide, NO. $$ \begin{aligned} 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) & \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) \\ \Delta H &=-906 \mathrm{~kJ} \end{aligned} $$ What is the enthalpy change for the following reaction? $$ \mathrm{NO}(g)+\frac{3}{2} \mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{NH}_{3}(g)+\frac{5}{4} \mathrm{O}_{2}(g) $$

When \(15.3 \mathrm{~g}\) of sodium nitrate, \(\mathrm{NaNO}_{3}\), was dissolved in water in a calorimeter, the temperature fell from \(25.00^{\circ} \mathrm{C}\) to \(21.56^{\circ} \mathrm{C}\). If the heat capacity of the solution and the calorimeter is \(1071 \mathrm{~J} /{ }^{\circ} \mathrm{C}\), what is the enthalpy change when \(1 \mathrm{~mol}\) of sodium nitrate dissolves in water? The solution process is $$ \mathrm{NaNO}_{3}(s) \longrightarrow \mathrm{Na}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) ; \Delta H=? $$

A \(10.0-\mathrm{g}\) sample of a mixture of \(\mathrm{CH}_{4}\) and \(\mathrm{C}_{2} \mathrm{H}_{4}\) reacts with oxygen at \(25^{\circ} \mathrm{C}\) and 1 atm to produce \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l) .\) If the reaction produces \(520 \mathrm{~kJ}\) of heat, what is the mass percentage of \(\mathrm{CH}_{4}\) in the mixture?

Nitrous oxide, \(\mathrm{N}_{2} \mathrm{O}\), has been used as a dental anesthetic. The average speed of an \(\mathrm{N}_{2} \mathrm{O}\) molecule at \(25^{\circ} \mathrm{C}\) is \(379 \mathrm{~m} / \mathrm{s}\). Calculate the kinetic energy (in joules) of an \(\mathrm{N}_{2} \mathrm{O}\) molecule traveling at this speed.
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