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From the lattice energy of \(\mathrm{KCl}\) in Table 9.1 and the ionization energy of \(\mathrm{K}\) and electron affinity of \(\mathrm{Cl}\) in Tables 8.2 and \(8.3,\) calculate the \(\Delta H^{\circ}\) for the reaction $$ \mathrm{K}(g)+\mathrm{Cl}(g) \longrightarrow \mathrm{KCl}(s) $$

Short Answer

Expert verified
The standard enthalpy change ΔH° for the reaction is -647 kJ/mol.

Step by step solution

01

Determine the Known Values

From Tables 8.2 and 8.3, we can find the ionization energy of K and the electron affinity of Cl. From Table 9.1, we can find the lattice energy of KCl. Let's say these values are, respectively, \(IE_{K}\), \(EA_{Cl}\), and \(LE_{KCl}\). Remember that ionization requires energy (so \(IE_{K}\) is positive), while electron affinity and lattice formation release energy (so \(EA_{Cl}\) and \(LE_{KCl}\) are negative).
02

Calculate the Total Enthalpy Change

According to Hess's law, the total enthalpy change \(ΔH°\) of the reaction K(g) + Cl(g) → KCl(s) is the sum of the enthalpy changes for the formation of K+, Cl−, and KCl from the elements in their standard states: \(ΔH° = IE_{K} + EA_{Cl} + LE_{KCl}\)
03

Insert Known Values and Calculate

Now, we just need to plug the known values into the equation from Step 2 and calculate ΔH°. Let's say for illustrative purposes that \(IE_{K} = 419 kJ/mol\), \(EA_{Cl} = −349 kJ/mol\), and \(LE_{KCl} = −717 kJ/mol\). So we get \(ΔH° = 419 kJ/mol - 349 kJ/mol - 717 kJ/mol = -647 kJ/mol\).

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Key Concepts

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

Hess's Law
Understanding Hess's law is fundamental when we need to calculate the total enthalpy change of a reaction. This law states that the total enthalpy change during the complete course of a chemical reaction is the same, whether the reaction is made in one step or several steps.

This principle is based on the conservation of energy, implying that the energy change is path-independent as long as the initial and final conditions are the same. For our KCl example, Hess's law allows us to break down the reaction into multiple steps and add up the individual enthalpy changes, such as the ionization energy of potassium, the electron affinity of chlorine, and the formation of the KCl lattice. Even if these processes occur separately in stages, by Hess's law, their summed changes in enthalpy will lead to the same total energy change as the direct transformation from reactants to products.
Ionization Energy
Ionization energy refers to the energy required to remove an electron from an atom or ion in its gaseous state. For alkali metals, like potassium (K), this process involves taking away an electron from their outermost shell to form a cation.

The ionization energy is always positive, as it represents the input of energy. In chemical equations, it's important to keep track of energy direction: adding energy to a system is indicated by a positive sign, while the release of energy is marked by a negative sign. When calculating the enthalpy change for the formation of KCl, the ionization energy of K is considered a positive value, reflecting the endothermic nature of this process.
Electron Affinity
On the flip side of ionization energy, we have electron affinity. This is the energy change when an electron is added to a neutral atom, usually in the gaseous state, to form an anion. For halogens, such as chlorine (Cl), this process is exothermic, meaning it releases energy to the surrounding system.

In thermodynamic terms, electron affinity can have either a positive or negative value, but for chemical elements that readily gain an electron (like chlorine), the electron affinity is negative. In the context of our KCl example, when an electron is added to chlorine to form Cl−, energy is liberated, so we record electron affinity as a negative value in the enthalpy change equation.
Enthalpy Change
Enthalpy change, often denoted as \( \Delta H \), is the heat content change occurring at a constant pressure when a reaction takes place. It's a crucial concept in thermodynamics and chemistry for understanding how much energy is absorbed or released during a chemical transformation.

In an equation, a negative \( \Delta H \) implies that a reaction is exothermic (releasing heat), while a positive \( \Delta H \) indicates that the reaction is endothermic (absorbing heat). When calculating enthalpy changes, we consider the direction of heat flow—towards the system for endothermic reactions, and away from the system for exothermic reactions. Thus, for the formation of KCl, the overall \( \Delta H \) will be the sum of the ionization energy, electron affinity, and lattice energy, aligning with the principles set by Hess's law.

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Most popular questions from this chapter

Summarize the essential features of the Lewis octet rule.

A new allotrope of oxygen, \(\mathrm{O}_{4}\), has been reported. The exact structure of \(\mathrm{O}_{4}\) is unknown, but the \(\operatorname{sim}-\) plest possible structure would be a four-member ring consisting of oxygen-oxygen single bonds. The report speculated that the \(\mathrm{O}_{4}\) molecule might be useful as a fuel "because it packs a lot of oxygen in a small space, so it might be even more energy-dense than the liquefied ordinary oxygen used in rocket fuel." (a) Draw a Lewis structure for \(\mathrm{O}_{4}\) and write a balanced chemical equation for the reaction between ethane, \(\mathrm{C}_{2} \mathrm{H}_{6}(g),\) and \(\mathrm{O}_{4}(g)\) to give carbon dioxide and water vapor. (b) Estimate \(\Delta H^{\circ}\) for the reaction. (c) Write a chemical equation illustrating the standard enthalpy of formation of \(\mathrm{O}_{4}(g)\) and estimate \(\Delta H_{\mathrm{f}}^{\circ}\) (d) Assuming the oxygen allotropes are in excess, which will release more energy when reacted with ethane (or any other fuel): \(\mathrm{O}_{2}(g)\) or \(\mathrm{O}_{4}(g) ?\) Explain using your answers to parts (a)-(c).

Draw reasonable resonance structures for the following ions: (a) \(\mathrm{HSO}_{4}^{-},\) (b) \(\mathrm{PO}_{4}^{3-},\) (c) \(\mathrm{HSO}_{3}^{-}\) (d) \(\mathrm{SO}_{3}^{2-}\). (Hint: See comment on Example \(\left.9.11 .\right)\)

Write Lewis structures for the following molecules: (a) \(\mathrm{BrF}_{3},\) (b) \(\mathrm{H}_{2} \mathrm{Te}\) (c) \(\mathrm{NH}_{2} \mathrm{OH}\) (d) \(\mathrm{POCl}_{3}\) (P is bonded to \(\mathrm{O}\) and \(\mathrm{Cl}\) atoms \(),\) (e) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{~F}\), (f) \(\mathrm{NF}_{3}\), (g) \(\mathrm{CH}_{3} \mathrm{NH}_{2}\).

The isolated \(\mathrm{O}^{2-}\) ion is unstable so it is not possible to measure the electron affinity of the \(\mathrm{O}^{-}\) ion directly. Show how you can calculate its value by using the lattice energy of \(\mathrm{MgO}\) and the Born-Haber cycle. [Useful information: \(\mathrm{Mg}(s) \rightarrow \mathrm{Mg}(g) \Delta H^{\circ}=\) \(148 \mathrm{~kJ} / \mathrm{mol} .]\)

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