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Chapter 8: Problem 61

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 Cl₂: 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 \(287.5\,\text{kJ/mol}\).

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