Calculate the lattice enthalpy of solid potassium bromide, \(\mathrm{KBr}(\mathrm{s}) \rightarrow \mathrm{K}^{+}(\mathrm{g})+\mathrm{Br}^{-}(\mathrm{g})\), from the following information: $$ \begin{aligned} &\Delta H_{i}{ }^{\circ}\left(\mathrm{KBr}, \text { s) }=-394 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\right. \\ &\Delta H_{i}{ }^{\circ}(\mathrm{K}, \mathrm{g})=+89.2 \mathrm{~kJ} \cdot \mathrm{mol}^{-1} \end{aligned} $$ First ionization energy of \(\mathrm{K}(\mathrm{g})=+425.0 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\) \(\Delta H_{\text {vap }}{ }^{\circ}\left(\mathrm{Br}_{2}, \mathrm{l}\right)=+30.9 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\) Br-Br bond dissociation cuthalpy \(=+192.9 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\) Electron attachment to \(\mathrm{Br}(\mathrm{g})\) : $$ \mathrm{Br}(\mathrm{g})+\mathrm{e}^{-}(\mathrm{g}) \rightarrow \mathrm{Be}^{-}(\mathrm{g}), \quad \Delta H^{*}=-331.0 \mathrm{~kJ} $$

Step by step solution

01

Determine the Enthalpy of Formation of KBr(s)

The enthalpy of formation of KBr(s), \(\Delta H_f\text{{(}}\text{{KBr}}, \text{{s)}\text{{)}}\), is given as -394 kJ/mol.

02

Consider the Sublimation of Solid Potassium

Solid potassium must sublimate to form potassium gas before it can ionize. The sublimation enthalpy of potassium is given as \(\Delta H_{i}^\circ(\text{{K}}, \text{{g}}) = +89.2 \text{{kJ/mol}}\).

03

Account for the First Ionization Energy of Potassium

The first ionization energy of potassium gas is the energy required to remove one electron from each atom to form potassium ions. It is given as \(+425.0 \text{{kJ/mol}}\).

04

Calculate the Enthalpy Change for Vaporization of Bromine

Bromine in the compound is in the liquid state, so it needs to vaporize to form bromine gas before dissociation. The enthalpy change for vaporization of bromine is \(\Delta H_{\text{{vap}}}^\circ(\text{{Br}}_2, \text{{l}}) = +30.9 \text{{kJ/mol}}\).

05

Consider the Bromine-Bromine Bond Dissociation Enthalpy

To form bromine atoms from molecular bromine, the bromine-bromine bond needs to be broken. The bond dissociation enthalpy for this process is \(+192.9 \text{{kJ/mol}}\).

06

Calculate the Electron Attachment Enthalpy for Bromine

An electron is added to each bromine atom to form bromide ions. The enthalpy change for this process is given as \(\Delta H^* = -331.0 \text{{kJ/mol}}\).

07

Sum All Enthalpies to Calculate Lattice Enthalpy

The lattice enthalpy can be calculated by adding all the enthalpies together: \[\Delta H_{\text{{lattice}}} = \Delta H_{\text{{sub}}} + \Delta H_{\text{{ion}}} + \Delta H_{\text{{vap}}} + \frac{1}{2}\Delta H_{\text{{bond}}} + \Delta H_{\text{{ea}}} + \Delta H_{\text{{f}}}\] Remember to divide the bond dissociation enthalpy by 2 because you have two bromine atoms forming from the molecular bromine, and only one is needed to combine with potassium to form KBr.

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Key Concepts

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

Enthalpy of Formation

Enthalpy of formation, symbolized as \(\Delta H_f^\circ\), refers to the heat change that occurs when one mole of a compound is formed from its elements in their standard states at 1 atm pressure and a specified temperature, typically 25°C or 298 K. For example, the enthalpy of formation of potassium bromide (KBr) is given to be -394 kJ/mol. This negative sign indicates that the formation of KBr from its constituent elements is an exothermic process, meaning that heat is released during the reaction. Understanding enthalpy of formation is crucial, as it provides a standard benchmark for the energy involved in creating compounds.

Let's break this down further with bullet points for clarity:

• Standard conditions: 1 atm and 25°C (or 298 K).
• Negative sign: Indicates an exothermic reaction.
• Exothermic process: Energy is released, making the surroundings warmer.

Grasping this concept helps students understand why certain reactions are more favorable than others and how energy is involved in chemical bonding.

Sublimation Enthalpy

Sublimation enthalpy, denoted by \(\Delta H_{sub}^\circ\), is the energy required to convert one mole of a substance from the solid phase directly into the gas phase without passing through the intermediate liquid phase. For substances like potassium (K), which have to be transformed from solid to gas before reacting, the sublimation enthalpy is a key component in lattice enthalpy calculations. In our example, the sublimation enthalpy for potassium is +89.2 kJ/mol. This positive value reflects the endothermic nature of the sublimation process; it requires energy input to overcome the forces holding the solid structure.

To fully comprehend sublimation enthalpy, consider these points:

• Phase change: Solid to gas directly.
• Endothermic process: Energy is absorbed, cooling the surroundings.
• Calculation importance: It's an essential step in determining total energy changes in reactions involving solids becoming gases.

Recognizing the role of sublimation enthalpy is essential for students when analyzing reactions that involve a phase change.

Ionization Energy

Ionization energy is the amount of energy required to remove an electron from an isolated gaseous atom or ion. The first ionization energy is the energy needed to remove the outermost electron from a neutral gaseous atom. In the context of lattice enthalpy, the ionization energy for potassium is a significant figure, specifically +425.0 kJ/mol for K(g). The positive sign indicates that energy must be added to the atom for the electron to be removed, marking it as an endothermic process.

Ionization energy is important for various reasons:

• Element Reactivity: It can give insights into an element's reactivity.
• Trend in the Periodic Table: Ionization energy generally increases across a period and decreases down a group.
• Formation of Cations: Crucial for understanding how cations form from atoms during chemical reactions.

This concept aids students in predicting how atoms will behave during chemical reactions, especially when ions are formed.

Bond Dissociation Enthalpy

Bond dissociation enthalpy reflects the energy required to break a bond in one mole of a gaseous molecule, producing free atoms in the gas phase. It is often used for diatomic molecules, such as \(\mathrm{Br}_2\). For our potassium bromide example, the bond dissociation enthalpy for \(\mathrm{Br}_2\) is +192.9 kJ/mol, which means it takes this amount of energy to break the \(\mathrm{Br}-\mathrm{Br}\) bond. As a positive value, it too is an endothermic process indicating that energy input is necessary.

Here's what students should take note of regarding bond dissociation enthalpy:

• Gaseous Molecules: The process pertains to gaseous species.
• Homolytic Cleavage: It involves breaking the bond to form neutral atoms.
• Stability Indicator: A high bond dissociation enthalpy implies a strong bond and hence a stable molecule.

Understanding this concept helps students assess the stability of molecules and the energy involved in chemical reactions.

Electron Attachment Enthalpy

Electron attachment enthalpy is the energy change that occurs when an extra electron is added to a neutral atom or molecule in the gas phase to form a negative ion, also known as an anion. In our potassium bromide scenario, it’s the energy change when an electron is added to \(\mathrm{Br}(\mathrm{g})\), and it's given as -331.0 kJ/mol. The negative sign here indicates that energy is released in the process, which means it is exothermic. This process is fundamental in the formation of anions.

Essential aspects of electron attachment enthalpy include:

• Formation of Anions: Responsible for the creation of negatively charged ions.
• Exothermic Nature: Usually releases energy, as most atoms in their standard state readily gain electrons.
• Importance in Reactions: Understanding electron affinities can help predict the feasibility and spontaneity of ionic compound formation.

Having a firm grasp of electron attachment enthalpy helps students gauge the energy changes accompanying ionic bond formation.

Vaporization Enthalpy

Vaporization enthalpy, symbolized as \(\Delta H_{vap}^\circ\), is the amount of energy required to convert one mole of a liquid into a gas at constant temperature and pressure. For bromine (\(\mathrm{Br}_2\)), the substance in our exercise, the vaporization enthalpy is +30.9 kJ/mol. As with other enthalpies we've discussed that have positive values, this also indicates an endothermic change; heat must be absorbed for the liquid to transition into the gaseous state.

To understand vaporization enthalpy better, consider these points:

• Phase Change: From liquid to gas.
• Heat Energy: Absorption is necessary for molecules to overcome their intermolecular forces.
• Predicting Boiling Points: A higher vaporization enthalpy usually corresponds to a higher boiling point.

Students must know about vaporization enthalpy to evaluate the energy required in processes like distillation and the physical properties of liquids.