Question

Even though so much energy is required to form a metal cation with a \(2+\) charge, the alkaline earth metals form halides with the general formula \(\mathrm{MX}_{2}\), rather than \(\mathrm{MX}\). (a) Use the following data to calculate \(\Delta H_{i}^{\circ}\) of \(\mathrm{MgCl}\) : \(\begin{array}{lr}\mathrm{Mg}(s) \longrightarrow \mathrm{Mg}(g) & \Delta H^{\circ}=148 \mathrm{~kJ} \\ \mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{Cl}(g) & \Delta H^{\circ}=243 \mathrm{~kJ} \\ \mathrm{Mg}(g) \longrightarrow \mathrm{Mg}^{+}(g)+\mathrm{e}^{-} & \Delta H^{\circ}=738 \mathrm{~kJ} \\ \mathrm{Cl}(g)+\mathrm{e}^{-} \longrightarrow \mathrm{Cl}^{-}(g) & \Delta H^{\circ}=-349 \mathrm{~kJ} \\ & \Delta H_{\text {lattice }}^{\circ} \text { of } \mathrm{MgCl}= & 783.5 \mathrm{~kJ} / \mathrm{mol}\end{array}\) (b) Is \(\mathrm{MgCl}\) favored energetically relative to \(\mathrm{Mg}\) and \(\mathrm{Cl}_{2} ?\) Explain. (c) Use Hess's law to calculate \(\Delta H^{\circ}\) for the conversion of \(\mathrm{MgCl}\) to \(\mathrm{MgCl}_{2}\) and \(\mathrm{Mg}\left(\Delta H_{\mathrm{f}}^{\circ}\right.\) of \(\left.\mathrm{MgCl}_{2}=-641.6 \mathrm{~kJ} / \mathrm{mol}\right)\) (d) Is \(\mathrm{MgCl}\) favored energetically relative to \(\mathrm{MgCl}_{2}\) ? Explain.

Short answer

a: ΔH_i^{0} of MgCl = -3.5 kJ/mol. b: MgCl is not favored. c: ΔH = -766.1 kJ/mol. d: MgCl2 is favored.

Step-by-step solution

Sublimation of Magnesium

The first step is to convert magnesium from solid to gas. This energy change is represented by \[\text{Mg}(s) \rightarrow \text{Mg}(g) \ \text{Δ}H^{\text{0}} = 148 \text{ kJ/mol} \]

Bond Dissociation of Chlorine

The next step is the dissociation of one mole of \(\text{Cl}_2\) into two moles of gaseous chlorine atoms.\[\text{Cl}_2(g) \rightarrow 2 \text{Cl}(g) \ \text{Δ}H^{\text{0}} = 243 \text{ kJ/mol} \]

First Ionization Energy of Magnesium

Magnesium gas forms a cation by losing one electron. This step is given by:\[\text{Mg}(g) \rightarrow \text{Mg}^{+}(g) + \text{e}^{-} \ \text{Δ}H^{\text{0}} = 738 \text{ kJ/mol} \]

Electron Affinity of Chlorine

Gaseous chlorine atoms form chloride ions by gaining one electron. This process releases energy:\[\text{Cl}(g) + \text{e}^{-} \rightarrow \text{Cl}^{-}(g) \ \text{Δ}H^{\text{0}} = -349 \text{ kJ/mol} \]

Lattice Energy of MgCl

Combine the magnesium cation and chloride anion to form the MgCl crystal lattice:\[\text{Mg}^{+}(g) + \text{Cl}^{-}(g) \rightarrow \text{MgCl}(s) \ \text{Δ}H^{\text{0}} = -783.5 \text{ kJ/mol} \]

Final Calculation for ΔH_i^{0} of MgCl

Sum all enthalpy changes calculated:\[ \text{Δ} H_i^{\text{0}} (\text{MgCl}) = 148 + 243 + 738 + (-349) + (-783.5) = -3.5 \text{ kJ/mol} \]

Addressing Part (b)

MgCl is not energetically favored relative to Mg and Cl2 because its formation enthalpy is slightly positive (3.5 kJ/mol), indicating its instability.

Using Hess's Law for MgCl to MgCl2

Use Hess's law: \[ \text{MgCl} + \text{Cl} \rightarrow \text{MgCl}_2 \]The data for MgCl2 formation: \[ \text{Δ} H_f^{\text{0}}(\text{MgCl}_2) = -641.6 \text{ kJ/mol} \]

Computing ΔH for Conversion from MgCl to MgCl2

Calculate the ∆H for the conversion:\[ \text{Δ} H = \text{Δ} H_f(\text{MgCl}_2) - \text{Δ} H_i(\text{MgCl}) - \text{BDE of Cl2} \]\[ \text{Δ} H = -641.6 \text{ kJ/mol} - 3.5 - (1/2 * 243) \]\[ \text{Δ} H = -766.1 \text{ kJ/mol }\]

Addressing Part (d)

MgCl2 is more energetically favored relative to MgCl because its formation enthalpy is much more negative (-641.6 kJ/mol for MgCl2 vs. -3.5 kJ/mol for MgCl).

Key concepts

enthalpy change

Enthalpy change is a core concept in chemistry that describes the heat change that occurs in a system at constant pressure. It is denoted by ΔH. When we discuss reactions, we look at the total enthalpy change to determine whether a reaction is exothermic (releases heat) or endothermic (absorbs heat). The enthalpy change can be calculated using various steps, involving breaking bonds and forming new ones.
For instance, in the given exercise, multiple steps like sublimation of magnesium and the dissociation of chlorine contribute to the overall ΔH of the formation of MgCl. Each of these steps has a specific enthalpy change value, and summing them gives us the total enthalpy change for the reaction. This helps us understand if the final compound is stable and if the reaction is energetically favorable.

lattice energy

Lattice energy is an important concept when it comes to ionic compounds. It's the energy released when ions in the gas phase come together to form a solid lattice. The magnitude of lattice energy indicates the strength of the bonds in the ionic compound.
In the exercise, the lattice energy of MgCl is given as -783.5 kJ/mol. This negative value shows that a lot of energy is released when the Mg⁺ and Cl^- ions form the solid MgCl, indicating a very stable ionic lattice. High lattice energy typically correlates with high melting points and stability of the ionic compound.

Hess's Law

Hess's Law is a principle that allows us to calculate the enthalpy change of a reaction, even if it happens in multiple steps. According to Hess’s Law, the total enthalpy change of a reaction is the sum of the enthalpy changes for the individual steps of the reaction. The law relies on the concept that enthalpy is a state function, meaning it depends only on the initial and final states, not the path taken.
In the context of the exercise, Hess's Law is used to calculate the enthalpy change for the conversion of MgCl to MgCl₂. By breaking the overall process into manageable steps, we can use the known enthalpies of formation and bond dissociation values to find the total enthalpy change for the conversion. This involves subtracting and adding the enthalpy values in a logical sequence to get the final result.

stoichiometry

Stoichiometry is the study of the quantitative relationships between the amounts of reactants and products in a chemical reaction. It involves using balanced chemical equations to calculate the quantities of substances needed or produced.
In the exercise, stoichiometry helps in determining the proportions of magnesium and chlorine needed to form compounds like MgCl and MgCl₂. By understanding the stoichiometric ratios, we can precisely calculate the enthalpy changes and use principles like Hess's Law to analyze the energy favorability of the reactions. For example, knowing that one mole of Mg reacts with two moles of Cl₂ to form MgCl₂ allows us to do precise enthalpy and energy calculations.