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Chapter 5: Problem 45

State Hess's law. Explain, with one example, the usefulness of Hess's law in thermochemistry.

Short Answer

Expert verified
Hess's Law allows indirect calculation of enthalpy changes by summing known reactions' enthalpies.

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01

Understanding Hess's Law

Hess's Law states that the total enthalpy change during a chemical reaction is the same regardless of the pathway by which the chemical reaction takes place, provided the initial and final states are the same. This law is a consequence of the first law of thermodynamics which implies that energy is conserved.
02

Setting Up a Thermochemical Equation

Consider the conversion of graphite to diamond. For this transformation, the direct enthalpy change is difficult to measure. However, Hess's law allows the enthalpy change to be calculated indirectly by using the following thermochemical equations: \( C_{graphite} + O_2 \rightarrow CO_2 \) with enthalpy change \( \Delta H_1 \) and \( C_{diamond} + O_2 \rightarrow CO_2 \) with enthalpy change \( \Delta H_2 \).
03

Applying Hess's Law

According to Hess's Law, the enthalpy change for converting graphite to diamond, \( C_{graphite} \rightarrow C_{diamond} \), can be derived from the enthalpy changes of the above reactions using the formula: \( \Delta H_{reaction} = \Delta H_2 - \Delta H_1 \). Thus, \( \Delta H_{reaction} \) is the enthalpy change for converting graphite to diamond without direct measurement.
04

Evaluating Hess's Law Usefulness

Hess's Law is particularly useful in thermochemistry for calculating enthalpy changes for reactions where direct measurement is impractical or impossible. It allows chemists to use known values of enthalpy changes from related reactions and obtain the desired value by summing them up appropriately.

Key Concepts

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

Enthalpy Change
Enthalpy change refers to the heat absorbed or released during a chemical reaction at constant pressure. It is denoted by the symbol \( \Delta H \), where \( \Delta \) signifies the change. This concept is vital because it helps in understanding how energy is exchanged in reactions. When bonds break and new ones form, there are transfers of energy that impact the overall heat content of the system.
  • If \( \Delta H \) is negative, the reaction is exothermic and releases heat.
  • If \( \Delta H \) is positive, the reaction is endothermic and absorbs heat.
Enthalpy changes can be classified further into various types, such as formation, combustion, and neutralization. Hess’s Law leverages these changes to calculate unknown enthalpies by summing known ones from related reactions. The beauty of enthalpy change lies in its constancy—the total enthalpy change remains the same regardless of the path taken, as long as initial and final states are consistent. This principle is pivotal in computational chemistry and the broader understanding of reaction energetics.
Thermochemistry
Thermochemistry is the branch of chemistry that explores the heat involved in chemical reactions and changes of state. It connects chemistry with thermal energy, focusing on how energy in the form of heat is absorbed or released during reactions.
To study these aspects, thermochemists use
  • The First Law of Thermodynamics, emphasizing energy conservation.
  • Calorimetry, to measure heat changes in physical and chemical processes.
  • Hess's Law, for indirect calculation of enthalpy changes.
These tools allow predictions about energy changes during reactions, which is essential for both laboratory and industrial applications. In thermochemistry, experiments are often set at constant pressure, mirroring real-world conditions where reactions take place in open systems. Understanding thermochemistry offers insights into why reactions occur and how we can manipulate conditions to control energy output or consumption for desired industrial applications. This field is crucial in designing energy-efficient processes, fuels, and materials.
First Law of Thermodynamics
The First Law of Thermodynamics, also known as the Law of Energy Conservation, is a fundamental concept stating that energy cannot be created or destroyed, only transformed. This principle underlies all thermochemical equations and plays a critical role in Hess’s Law.
According to this law, the total energy change in a closed system is equal to the heat added to the system minus the work done by it, expressed mathematically as: \[ \Delta U = Q - W \]where \( \Delta U \) is the change in internal energy, \( Q \) is the heat absorbed, and \( W \) is the work done by the system. In the context of chemical reactions, this means energy changes must account for both heat exchange and work done. In a constant-pressure setting, the enthalpy change \( \Delta H \) becomes the primary focus since work done is minimized.
Hess's Law is integrally connected to this, allowing chemists to ensure that calculated energy exchanges are consistent with conservation principles. The First Law reinforces that any enthalpy change is part of a broader energy balance, crucial for designing systems that utilize energy efficiently, such as engines or batteries.

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Most popular questions from this chapter

At \(25^{\circ} \mathrm{C}\), the standard enthalpy of formation of \(\mathrm{HF}(a q)\) is \(-320.1 \mathrm{~kJ} / \mathrm{mol} ;\) of \(\mathrm{OH}^{-}(a q),\) it is \(-229.6 \mathrm{~kJ} / \mathrm{mol} ;\) of \(\mathrm{F}^{-}(a q)\) it is \(-329.1 \mathrm{~kJ} / \mathrm{mol} ;\) and of \(\mathrm{H}_{2} \mathrm{O}(l),\) it is \(-285.8 \mathrm{~kJ} / \mathrm{mol}\). (a) Calculate the standard enthalpy of neutralization of \(\mathrm{HF}(a q)\) \(\mathrm{HF}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{F}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) (b) Using the value of \(-56.2 \mathrm{~kJ}\) as the standard enthalpy change for the reaction \(\mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)\) calculate the standard enthalpy change for the reaction \(\mathrm{HF}(a q) \longrightarrow \mathrm{H}^{+}(a q)+\mathrm{F}^{-}(a q)\)

Determine the enthalpy change for the gaseous reaction of sulfur dioxide with ozone to form sulfur trioxide given the following thermochemical data: $$ \begin{aligned} 2 \mathrm{SO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{2}(g) & \Delta H^{\circ}=-602.8 \mathrm{~kJ} / \mathrm{mol} \\ 3 \mathrm{SO}(g)+2 \mathrm{O}_{3}(g) \longrightarrow 3 \mathrm{SO}_{3}(g) & \\\ \Delta H_{\mathrm{rxn}}^{\circ}=-1485.03 \mathrm{~kJ} / \mathrm{mol} \\ \frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{O}_{3}(g) & \Delta H_{\mathrm{rxn}}^{\circ}=142.2 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$

For which of the following reactions does \(\Delta H_{\mathrm{rxn}}^{\circ}=\Delta H_{\mathrm{f}}^{\circ}\) ? (a) \(\mathrm{H}_{2}(g)+\mathrm{S}(\) rhombic \() \longrightarrow \mathrm{H}_{2} \mathrm{~S}(g)\) (b) \(\mathrm{C}(\) diamond \()+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)\) (c) \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{CuO}(s) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Cu}(s)\) (d) \(\mathrm{O}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{O}_{3}(g)\)

A quantity of \(2.00 \times 10^{2} \mathrm{~mL}\) of \(0.862 \mathrm{M} \mathrm{HCl}\) is mixed with \(2.00 \times 10^{2} \mathrm{~mL}\) of \(0.431 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) in a constant-pressure calorimeter of negligible heat capacity. The initial temperature of the \(\mathrm{HCl}\) and \(\mathrm{Ba}(\mathrm{OH})_{2}\) solutions is the same at \(20.48^{\circ} \mathrm{C}\). For the process $$\mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)$$ the heat of neutralization is \(-56.2 \mathrm{~kJ} / \mathrm{mol}\). What is the final temperature of the mixed solution? Assume the specific heat of the solution is the same as that for pure water.

Given the thermochemical data,$$\begin{array}{ll}\mathrm{A}+6 \mathrm{~B} \longrightarrow 4 \mathrm{C} & \Delta H_{1}=-1200 \mathrm{~kJ} / \mathrm{mol} \\\\\mathrm{C}+\mathrm{B} \longrightarrow \mathrm{D} & \Delta H_{1}=-150 \mathrm{~kJ} / \mathrm{mol}\end{array}$$ Determine the enthalpy change for each of the following: a) \(\mathrm{D} \longrightarrow \mathrm{C}+\mathrm{B}\) d) \(2 \mathrm{D} \longrightarrow 2 \mathrm{C}+2 \mathrm{~B}\) b) \(2 \mathrm{C} \longrightarrow \frac{1}{2} \mathrm{~A}+3 \mathrm{~B}\) e) \(6 \mathrm{D}+\mathrm{A} \longrightarrow 10 \mathrm{C}\) c) \(3 \mathrm{D}+\frac{1}{2} \mathrm{~A} \stackrel{\longrightarrow}{\longrightarrow} \mathrm{C}\)
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