Chapter 9, Problem 9.75P

(a)

Interpretation Introduction

Interpretation:

$\Delta H_{\text{f}}^{°}$ of $\text{MgCl}$ is to be calculated.

Concept introduction:

Hess’s law is used to calculate the enthalpy change of an overall reaction that can be derived as a sum of two or more reaction. According to Hess’s law $\Delta H$ of an overall reaction is equal to the sum of the enthalpy change for each individual reaction. $\Delta H_{\text{overall rxn}}=\Delta H_1+\Delta H_2+\mathrm{.......}+\Delta H_{\text{n}}$

Enthalpy is a state function so the value depends upon the initial state and final state not on the path so $\Delta H$ of an overall reaction can be calculated by the addition or subtraction of the individual steps whose $\Delta H$ is known.

Answer to Problem 9.75P

$\Delta H_{\text{f}}^{°}$ of $\text{MgCl}$ is $-125\text{kJ}$.

Explanation of Solution

The enthalpy change of the following reaction is $\Delta H_1^{°}$.

$\text{Mg}\left(s\right)\to \text{Mg}\left(g\right)$ (1)

The enthalpy change of the following reaction is $\Delta H_2^{°}$.

${\text{Cl}}_2\left(g\right)\to 2\text{Cl}\left(g\right)$ (2)

The enthalpy change of the following reaction is $\Delta H_3^{°}$.

$\text{Mg}\left(g\right)\to {\text{Mg}}^{+}\left(g\right)+e^{-}$ (3)

The enthalpy change of the following reaction is $\Delta H_4^{°}$.

$\text{Cl}\left(g\right)+e^{-}\to {\text{Cl}}^{-}\left(g\right)$ (4)

The enthalpy change of the following reaction is $\Delta H_{\text{lattice}}^{\text{°}}$.

$\text{MgCl}\left(s\right)\to {\text{Mg}}^{+}\left(g\right)+{\text{Cl}}^{-}\left(g\right)$ (5)

Reverse the equation (5).

${\text{Mg}}^{+}\left(g\right)+{\text{Cl}}^{-}\left(g\right)\to \text{MgCl}\left(s\right)$ (6)

The enthalpy change for the reaction (6) is calculated as,

$\Delta H_5^{°}=-\Delta H_{\text{lattice}}^{\text{°}}$

Multiply the equation (2) by $\frac{1}{2}$.

$\frac{1}{2}{\text{Cl}}_2\left(g\right)\to \text{Cl}\left(g\right)$ (7)

The enthalpy change for the reaction (7) is calculated as,

$\Delta H_6^{°}=\frac{1}{2}\Delta H_2^{°}$

Add equation (1),(3), (4), (6) and (7).

$\text{Mg}\left(s\right)+\frac{1}{2}{\text{Cl}}_2\left(g\right)\to \text{MgCl}\left(s\right)$ (8)

The enthalpy change of the final reaction (8) is $\Delta H_{\text{f}}^{°}$.

The expression to calculate $\Delta H_{\text{f}}^{°}$ is as follows:

$\Delta H_{\text{f}}^{°}=\Delta H_1^{°}+\Delta H_3^{°}+\Delta H_4^{°}+\Delta H_5^{°}+\Delta H_6^{°}$ (9)

Substitute $148\text{kJ}$ for $\Delta H_1^{°}$ and $121.5\text{kJ}$ for $\Delta H_3^{°}$, $768\text{kJ}$ for $\Delta H_4^{°}$, $-349\text{kJ}$ for $\Delta H_5^{°}$ and $-783.5\text{kJ}$ for $\Delta H_6^{°}$ n the equation (9).

$\begin{array}{c}\Delta H_{\text{f}}^{°}=148\text{kJ}+121.5\text{kJ}+768\text{kJ}+\left(-349\text{kJ}\right)+\left(-783.5\text{kJ}\right)\\ =-125\text{kJ}\end{array}$

Conclusion

$\Delta H_{\text{f}}^{°}$ of $\text{MgCl}$ is $-125\text{kJ}$.

(b)

Interpretation Introduction

Interpretation:

Whether $\text{MgCl}$ is favoured energetically relative to $\text{Mg}$ and ${\text{Cl}}_2$ is to be determined.

Concept introduction:

The standard enthalpy of formation $\left(\Delta H_{\text{f}}^{°}\right)$ is the enthalpy change of the formation of $1\text{mol}$ of compound from its elements and the reactants and products, in this case, is also present in their standard state.

Standard state includes $1\text{atm}$ pressure for gases, $1M$ concentration for substances in the aqueous phase and the most stable form of substance at $1\text{atm}$ pressure and $298\text{K}$ temperature. For example, metal like sodium exists in solid state and molecular elements like halogen exist in a gaseous state.

Answer to Problem 9.75P

$\text{MgCl}$ is favoured energetically relative to $\text{Mg}$ and ${\text{Cl}}_2$ due to the negative value of heat of formation.

Explanation of Solution

The chemical equation for the formation of $\text{MgCl}$ is as follows:

$\text{Mg}\left(s\right)+\frac{1}{2}{\text{Cl}}_2\left(g\right)\to \text{MgCl}\left(s\right)$

The enthalpy of formation $\left(\Delta H_{\text{f}}^{°}\right)$ of $\text{MgCl}$ is $-125\text{kJ}$.

When the enthalpy of formation is negative then the compound formed is relatively more stable as compared to the elements. Therefore, $\text{MgCl}$ is relatively more stable as compared to $\text{Mg}$ and ${\text{Cl}}_2$.

Conclusion

When the enthalpy of formation is negative then the compound formed is relatively more stable as compared to the elements.

(c)

Interpretation Introduction

Interpretation:

$\Delta H^{°}$ for the conversion of $\text{MgCl}$ to ${\text{MgCl}}_2$ and $\text{Mg}$ is to be calculated.

Concept introduction:

The standard enthalpy of reaction is calculated by the summation of standard enthalpy of formation of the product minus the summation of standard enthalpy of formation of product at the standard conditions. The formula to calculate the standard enthalpy of reaction $\left(\Delta H_{\text{rxn}}^{°}\right)$ is as follows:

$\Delta H_{\text{rxn}}^{°}=\sum m\Delta H_{\text{f (products)}}^{°}-\sum m\Delta H_{\text{f (reactants)}}^{°}$

Here, m and n are the stoichiometric coefficients of reactants and product in the balanced chemical equation.

Answer to Problem 9.75P

$\Delta H^{°}$ for the conversion of $\text{MgCl}$ to ${\text{MgCl}}_2$ and $\text{Mg}$ is $-392\text{kJ}$.

Explanation of Solution

The chemical equation for the formation of ${\text{MgCl}}_2$ is as follows:

$\text{2MgCl}\left(s\right)\to {\text{MgCl}}_2\left(s\right)+\text{Mg}\left(s\right)$

The formula to calculate the standard enthalpy of reaction $\left(\Delta H^{°}\right)$ is as follows:

$\Delta H^{°}=\left[\begin{array}{l}\left\{\Delta H_{\text{f}}^{°}\left[{\text{MgCl}}_2\left(s\right)\right]+\Delta H_{\text{f}}^{°}\left[\text{Mg}\left(s\right)\right]\right\}\\ -\left\{2\Delta H_{\text{f}}^{°}\left[\text{MgCl}\left(s\right)\right]\right\}\end{array}\right]$ (10)

Substitute $-641.6\text{kJ/mol}$ for $\Delta H_{\text{f}}^{°}\left[{\text{MgCl}}_2\left(s\right)\right]$, 0 for $\Delta H_{\text{f}}^{°}\left[\text{Mg}\left(s\right)\right]$ and $-125\text{kJ}$ for $\Delta H_{\text{f}}^{°}\left[\text{MgCl}\left(s\right)\right]$ in the equation (10).

$\begin{array}{c}\Delta H^{°}=\left[\begin{array}{l}\left\{\left(1\text{mol}\right)\left(-641.6\text{kJ/mol}\right)+\left(1\text{mol}\right)\left(0\right)\right\}\\ -\left\{\left(2\text{mol}\right)\left(-125\text{kJ}\right)\right\}\end{array}\right]\\ =-391.6\text{kJ}\\ \approx -392\text{kJ}\end{array}$

Conclusion

$\Delta H^{°}$ for the conversion of $\text{MgCl}$ to ${\text{MgCl}}_2$ and $\text{Mg}$ is $-392\text{kJ}$.

(d)

Interpretation Introduction

Interpretation:

Whether $\text{MgCl}$ is favoured energetically relative to ${\text{MgCl}}_2$ is to be determined.

Concept introduction:

The standard enthalpy of formation $\left(\Delta H_{\text{f}}^{°}\right)$ is the enthalpy change of the formation of $1\text{mol}$ of compound from its elements and the reactants and products, in this case, is also present in their standard state.

Standard state includes $1\text{atm}$ pressure for gases, $1M$ concentration for substances in the aqueous phase and the most stable form of substance at $1\text{atm}$ pressure and $298\text{K}$ temperature. For example, metal like sodium exists in solid state and molecular elements like halogen exist in a gaseous state.

Answer to Problem 9.75P

${\text{MgCl}}_2$ is favoured energetically relative to $\text{MgCl}$ due to the more negative value of heat of formation of ${\text{MgCl}}_2$ as compared to $\text{MgCl}$.

Explanation of Solution

The chemical equation for the formation of ${\text{MgCl}}_2$ is as follows:

$\text{2MgCl}\left(s\right)\to {\text{MgCl}}_2\left(s\right)+\text{Mg}\left(s\right)$

The enthalpy of formation $\left(\Delta H_{\text{f}}^{°}\right)$ of $\text{MgCl}$ is $-125\text{kJ}$.

The enthalpy of formation $\left(\Delta H_{\text{f}}^{°}\right)$ of ${\text{MgCl}}_2$ is $-392\text{kJ}$.

The compound with the more negative enthalpy of formation is relatively more stable than the compound with the less negative enthalpy of formation. Therefore, ${\text{MgCl}}_2$ is relatively more stable as compared to $\text{MgCl}$.

Conclusion

The compound with the more negative enthalpy of formation is relatively more stable as compared to the compound with the less negative enthalpy of formation.