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Chapter 8: Problem 55

Use the following data to estimate \(\Delta H_{\mathrm{f}}^{\circ}\) for potassium chloride. $$ \mathrm{K}(s)+\frac{1}{2} \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{KCl}(s) $$ $$ \begin{array}{lr} \text { Lattice energy } & -690 . \mathrm{kJ} / \mathrm{mol} \\ \text { Ionization energy for } \mathrm{K} & 419 \mathrm{~kJ} / \mathrm{mol} \\\ \text { Electron affinity of Cl } & -349 \mathrm{~kJ} / \mathrm{mol} \\ \text { Bond energy of } \mathrm{Cl}_{2} & 239 \mathrm{~kJ} / \mathrm{mol} \\ \text { Enthalpy of sublimation for } \mathrm{K} & 64 \mathrm{~kJ} / \mathrm{mol} \end{array} $$

Short Answer

Expert verified
The estimated standard enthalpy of formation for potassium chloride is \(-443 \,\mathrm{kJ/mol}\).

Step by step solution

01

Write down the given data

We start by writing the given data in their respective symbols: - Lattice energy: \(U_{\mathrm{lattice}} = -690 \,\mathrm{kJ/mol}\) - Ionization energy of K: \(IE_{\mathrm{K}} = 419 \,\mathrm{kJ/mol}\) - Electron affinity of Cl: \(EA_{\mathrm{Cl}} = -349 \,\mathrm{kJ/mol}\) - Bond energy of Cl: \(BE_{\mathrm{Cl_2}} = 239 \,\mathrm{kJ/mol}\) - Enthalpy of sublimation for K: \(\Delta H_{\mathrm{sub}}^{\mathrm{K}} = 64 \,\mathrm{kJ/mol}\)
02

Apply Hess's Law

Using Hess's Law, we can determine the enthalpy change for the formation of KCl from its constituent elements by adding up the individual energy changes: $$\Delta H_{\mathrm{f}}^{\circ}(\mathrm{KCl}) = \Delta H_{\mathrm{sub}}^{\mathrm{K}} + \frac{1}{2} \cdot BE_{\mathrm{Cl_2}} + IE_{\mathrm{K}} + EA_{\mathrm{Cl}} + U_{\mathrm{lattice}}$$
03

Calculate the enthalpy change

Now, we can plug in the given values for each term into the equation from step 2: $$\Delta H_{\mathrm{f}}^{\circ}(\mathrm{KCl}) = 64\,\mathrm{kJ/mol} + \frac{1}{2} \cdot 239\,\mathrm{kJ/mol} + 419\,\mathrm{kJ/mol} - 349\,\mathrm{kJ/mol} - 690\,\mathrm{kJ/mol}$$
04

Solve for the enthalpy of formation

Simplifying the equation, we get: $$\Delta H_{\mathrm{f}}^{\circ}(\mathrm{KCl}) = 64 + \frac{1}{2} \cdot 239 + 419 - 349 - 690$$ Calculating the value, we find: $$\Delta H_{\mathrm{f}}^{\circ}(\mathrm{KCl}) = -443\,\mathrm{kJ/mol}$$ So, the estimated standard enthalpy of formation for potassium chloride is \(-443 \,\mathrm{kJ/mol}\).

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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 \(\mathrm{O}-\mathrm{Cl}\)

Some of the important properties of ionic compounds are as follows: i. low electrical conductivity as solids and high conductivity in solution or when molten ii. relatively high melting and boiling points iii. brittleness iv. solubility in polar solvents How does the concept of ionic bonding discussed in this chapter account for these properties?

Which of the following molecules have net dipole moments? For the molecules that are polar, indicate the polarity of each bond and the direction of the net dipole moment of the molecule. a. \(\mathrm{CH}_{2} \mathrm{Cl}_{2}, \mathrm{CHCl}_{3}, \mathrm{CCl}_{4}\) b. \(\mathrm{CO}_{2}, \mathrm{~N}_{2} \mathrm{O}\) c. \(\mathrm{PH}_{3}, \mathrm{NH}_{3}\)

Rationalize the following lattice energy values: $$ \begin{array}{|lc|} \hline & \text { Lattice Energy } \\ \text { Compound } & (\mathrm{kJ} / \mathrm{mol}) \\ \hline \mathrm{CaSe} & -2862 \\ \mathrm{Na}_{2} \mathrm{Se} & -2130 \\ \mathrm{CaTe} & -2721 \\ \mathrm{Na}_{2} \mathrm{Te} & -2095 \\ \hline \end{array} $$

Place the species below in order of the shortest to the longest nitrogen- oxygen bond. \(\begin{array}{lllll}\mathrm{H}_{2} \mathrm{NOH}, & \mathrm{N}_{2} \mathrm{O}, & \mathrm{NO}^{+}, & \mathrm{NO}_{2}^{-}, & \mathrm{NO}_{3}^{-}\end{array}\) \(\left(\mathrm{H}_{2} \mathrm{NOH}\right.\) exists as \(\mathrm{H}_{2} \mathrm{~N}-\mathrm{OH} .\).
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