/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 61 Use the following data to estima... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Use the following data to estimate \(\Delta H_{f}^{\circ}\) for potassium chloride. $$\mathrm{K}(s)+\frac{1}{2} \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{KCl}(s)$$ \(\begin{array}{l}{\text { Lattice energy }} & {-690 . \mathrm{kJ} / \mathrm{mol}} \\ {\text { Ionization energy for } \mathrm{K}} & \quad{419 \mathrm{kJ} / \mathrm{mol}} \\ {\text { Electron affinity of } \mathrm{Cl}} & {-349 \mathrm{kJ} / \mathrm{mol}}\\\\{\text { Bond energy of } \mathrm{Cl}_{2}} & \quad {239 \mathrm{kJ} / \mathrm{mol}} \\ {\text { Enthalpy of sublimation for } \mathrm{K}} & \quad {90 . \mathrm{kJ} / \mathrm{mol}}\end{array}\)

Short Answer

Expert verified
The estimated standard enthalpy of formation for potassium chloride, \(\Delta H_{f}^{\circ}(\text{KCl})\), is \(287.5\,\text{kJ/mol}\) using the Born-Haber cycle and given data values.

Step by step solution

01

List the Given Data

We are given the following energy values: - Lattice energy: -690 kJ/mol - Ionization energy for K: 419 kJ/mol - Electron affinity of Cl: -349 kJ/mol - Bond energy of Clâ‚‚: 239 kJ/mol - Enthalpy of sublimation for K: 90 kJ/mol
02

Understand the Born-Haber Cycle

The Born-Haber cycle relates the lattice energy, ionization energy, electron affinity, bond energy, and enthalpy of sublimation to the enthalpy of formation of an ionic compound. We will use this cycle to estimate the enthalpy of formation of potassium chloride (KCl) using the given data.
03

Apply the Born-Haber Cycle

According to the Born-Haber cycle, the enthalpy of formation for KCl(s) can be represented as: \(\Delta H_{f}^{\circ}(\text{KCl}) = \Delta H_{\text{sublimation}}^{\circ}(\text{K}) + \frac{1}{2}\Delta H_{\text{bond}}^{\circ}(\text{Cl}_{2}) + \text{IE}_{\text{K}} - \text{EA}_{\text{Cl}} + \Delta H_{\text{lattice}}^{\circ}(\text{KCl})\) Now we can plug in the given values: \(\Delta H_{f}^{\circ}(\text{KCl}) = 90\,\text{kJ/mol} + \frac{1}{2} (239\,\text{kJ/mol}) + 419\,\text{kJ/mol} - (-349\,\text{kJ/mol}) - 690\,\text{kJ/mol}\)
04

Calculate the Enthalpy of Formation

Now we can perform the calculations: \(\Delta H_{f}^{\circ}(\text{KCl}) = 90\,\text{kJ/mol} + 119.5\,\text{kJ/mol} + 419\,\text{kJ/mol} + 349\,\text{kJ/mol} - 690\,\text{kJ/mol}\) \(\Delta H_{f}^{\circ}(\text{KCl}) = 287.5\,\text{kJ/mol}\) The estimated standard enthalpy of formation for potassium chloride, \(\Delta H_{f}^{\circ}(\text{KCl})\), is \(287.5\,\text{kJ/mol}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Enthalpy of Formation
The enthalpy of formation (\( \Delta H_f^{\circ} \)) is a crucial concept in thermodynamics that tells us how much heat energy is absorbed or released when one mole of a compound is formed from its elements in their standard states. It helps us understand the stability of compounds.
- For potassium chloride (\( \text{KCl} \)), the formation involves combining solid potassium (\( \text{K}(s) \)) and chlorine gas (\( \frac{1}{2} \text{Cl}_2(g) \)).- This process can release or absorb energy, indicating how substances interact to form stable or unstable compounds. In our case, we use the Born-Haber cycle—a method involving several energy changes like lattice energy, ionization energy, and electron affinity—to estimate the enthalpy of formation for potassium chloride. Understanding these contributing factors allows us to grasp why a compound is energetically favorable or not.
Lattice Energy
Lattice energy refers to the amount of energy released when ions come together to form a crystalline lattice. It is a measure of the strength of the forces holding the ions together in the solid.
- For potassium chloride (\( \text{KCl} \)), the lattice energy is given as -690 kJ/mol.- This negative value indicates that energy is released when \( \text{K}^+ \) and \( \text{Cl}^- \) ions form a solid lattice from their gaseous state.A higher absolute value of lattice energy signifies a more stable ionic compound since more energy is required to break the ionic lattice apart. In the Born-Haber cycle, this energy is crucial as it compensates for the energy costs involved in converting atoms to ions and separating molecules.
Ionization Energy
Ionization energy is the amount of energy required to remove an electron from an atom or ion in its gaseous state. This is typically measured in kilojoules per mole (kJ/mol).
- For potassium (\( \text{K} \)), the ionization energy is 419 kJ/mol.- It involves turning a neutral potassium atom into a positively charged potassium ion (\( \text{K}^+ \)).In the Born-Haber cycle, ionization energy is an essential contribution to the overall energy changes for forming a compound like potassium chloride. Although it costs energy to remove an electron, this step is vital for creating ionic bonds and is balanced by other energy releases like electron affinity and lattice energy.
Electron Affinity
Electron affinity refers to the energy change that occurs when an electron is added to a neutral atom in the gaseous state, forming a negative ion.
- In the case of chlorine (\( \text{Cl} \)), the electron affinity is -349 kJ/mol.- This negative value indicates that energy is released when chlorine gains an electron to become \( \text{Cl}^- \).Electron affinity is a critical attribute in the Born-Haber cycle. It represents the gain in stability when an atom obtains an extra electron. In the formation of sodium chloride, the release of this energy helps balance the energy inputs needed for ionization and sublimation, contributing to the overall stability of the resulting ionic compound.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Without using Fig. 8.3, predict which bond in each of the following groups will be the most polar. a. \(\mathrm{C}-\mathrm{H}, \mathrm{Si}-\mathrm{H}, \mathrm{Sn}-\mathrm{H}\) b. \(\mathrm{Al}-\mathrm{Br}, \mathrm{Ga}-\mathrm{Br}, \mathrm{In}-\mathrm{Br}, \mathrm{Tl}-\mathrm{Br}\) c.\(\mathrm{C}-\mathrm{O}\) or \(\mathrm{Si}-\mathrm{O}\) d. \(\mathrm{O}-\mathrm{F}\) or \(\mathrm{O}-\mathrm{Cl}\)

Use the following data (in kJ/mol) to estimate \(\Delta H\) for the reaction \(S^{-}(g)+e^{-} \rightarrow S^{2-}(g)\) . Include an estimate of uncertainty. \(\begin{aligned} \mathrm{S}(s) \longrightarrow \mathrm{S}(g) & \Delta H=277 \mathrm{kJ} / \mathrm{mol} \\ \mathrm{S}(g)+\mathrm{e}^{-} \longrightarrow \mathrm{S}^{-}(g) & \Delta H=-200 \mathrm{kJ} / \mathrm{mol} \end{aligned}\) Assume that all values are known to \(\pm 1 \mathrm{kJ} / \mathrm{mol}\)

Given the following information: Heat of sublimation of \(\mathrm{Li}(s)=166 \mathrm{kJ} / \mathrm{mol}\) Bond energy of \(\mathrm{HCl}=427 \mathrm{kJ} / \mathrm{mol}\) Ionization energy of \(\mathrm{Li}(g)=520 . \mathrm{kJ} / \mathrm{mol}\) Electron affinity of \(\mathrm{Cl}(g)=-349 \mathrm{kJ} / \mathrm{mol}\) Lattice energy of LiCl(s) \(=-829 \mathrm{kJ} / \mathrm{mol}\) Bond energy of \(\mathrm{H}_{2}=432 \mathrm{kJ} / \mathrm{mol}\) Calculate the net change in energy for the following reaction: $$2 \mathrm{Li}(s)+2 \mathrm{HCl}(g) \longrightarrow 2 \mathrm{LiCl}(s)+\mathrm{H}_{2}(g)$$

When molten sulfur reacts with chlorine gas, a vile-smelling orange liquid forms that has an empirical formula of SCl. The structure of this compound has a formal charge of zero on all elements in the compound. Draw the Lewis structure for the vile-smelling orange liquid.

For each of the following, write an equation that corresponds to the energy given. a. lattice energy of \(\mathrm{NaCl}\) b. lattice energy of \(\mathrm{NH}_{4} \mathrm{Br}\) c. lattice energy of \(\mathrm{MgS}\) d. \(\mathrm{O}=\mathrm{O}\) double bond energy beginning with \(\mathrm{O}_{2}(g)\) as a reactant

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.