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 $$ \mathrm{K}(g)+\mathrm{Cl}(g) \longrightarrow \mathrm{KCl}(s) $$ (a) Draw three resonance structures to represent the ion. (b) Given the following information $$ 2 \mathrm{H}+\mathrm{H}^{+} \longrightarrow \mathrm{H}_{3}^{+} \quad \Delta H^{\circ}=-849 \mathrm{~kJ} / \mathrm{mol} $$ and $$ \mathrm{H}_{2} \longrightarrow 2 \mathrm{H} \quad \Delta H^{\circ}=436.4 \mathrm{~kJ} / \mathrm{mol} $$ calculate \(\Delta H^{\circ}\) for the reaction $$ \mathrm{H}^{+}+\mathrm{H}_{2} \longrightarrow \mathrm{H}_{3}^{+} $$

Step by step solution

01

Part A - Step 1: Initiate the Born-Haber cycle for KCl

The Born-Haber cycle for KCl formation involves several steps. We have the ionization energy for K and electron affinity for Cl, as well as the lattice energy of KCl. Using these, we can write the steps of the cycle as: (1) Atomization of K to K(g), (2) Ionization of K(g) to \(K^+\) (ionization energy = IE1), (3) Atomization of Cl to Cl(g), (4) Addition of an electron to Cl(g) to form \(-Cl^-\) (electron affinity = EA), (5) The formation of KCl from \(K^+\) and \(-Cl^-\) (lattice energy = LE). The overall energy change, \(\Delta H^{\circ}\), is the sum of these steps. It is important to note that we need to remember the signs of the ionization energy and electron affinity.

02

Part A - Step 2: Calculate the enthalpy change

The sum of the steps in the Born-Haber cycle gives us the enthalpy change for the reaction. We get \(\Delta H^{\circ}=\) IE1 - EA + LE. Insert the given values from the tables into the formula and calculate the result accordingly.

03

Part B - Step 1: Analyze the given reactions

The information provided gives us the enthalpy changes for the dissociation of \(H_2\) into 2H and the reaction of 2H with \(H^+\) to form \(H_3^+\). We need to calculate the enthalpy change for the reaction of \(H^+\) with \(H_2\) to form \(H_3^+\). Per Hess's law, we can add or subtract the provided reactions to obtain the desired reaction and its enthalpy change.

04

Part B - Step 2: Apply Hess's law

Subtracting the reaction \(H_2\) to 2H with \(\Delta H^{\circ}=436.4 \mathrm{~kJ} / \mathrm{mol}\) from \(2 \mathrm{H}+\mathrm{H}^{+} to \mathrm{H}_{3}^{+} with \Delta H^{\circ}=-849 \mathrm{~kJ} / \mathrm{mol} \) we obtain the desired reaction \(H^+\)+\(H_{2} \longrightarrow H_{3}^{+}\) . The resulting \(\Delta H^{\circ}\) can be found by subtracting the \(\Delta H^{\circ}\) for the second reaction from the first.

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