Chapter U5.109, Problem 8E

Interpretation Introduction

Interpretation:

The formation of a compound that will produce more energy should be selected from formation 6 mol of magnesium oxide, $\text{MgO}$ or formation of 2 mol of aluminum oxide, ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}$.

Concept introduction:

The change in enthalpy by the formation of a compound in 1 mole from its constituent elements in their standard state is said to be standard heat of formation. The ${\text{ΔH}}_{\text{f}}^{\text{o}}$ for a chemical reaction can be calculated using the formula

${\text{ΔH}}_{\text{f}}^{\text{o}}{\text{ = ΣnH}}_{\text{f}}^{\text{o}}{\text{(products) - ΣmH}}_{\text{f}}^{\text{o}}\text{(reactants)}$

Where m and n are stoichiometric numbers.

Answer to Problem 8E

The amount of energy released is more during the formation of 6 mol of magnesium oxide.

Explanation of Solution

The balanced chemical reaction for the formation of magnesium oxide, $\text{MgO}$ from magnesium and oxygen, is

${\text{2Mg(s) + O}}_{\text{2}}\text{(g) }\to \text{ 2MgO(s)}$

The value of heat of formation of all the species involved in the reaction is

$\begin{array}{c}\text{MgO(s)}=\text{ -602 kJ/mol}\\ \text{Mg(s) = 0 kJ/mol}\\ {\text{O}}_{\text{2}}\text{(g) = 0 kJ/mol}\end{array}$

The ${\text{ΔH}}_{\text{f}}^{\text{o}}$ for a chemical reaction is calculated as

${\text{ΔH}}_{\text{f}}^{\text{o}}\text{ = 2}\Delta {\text{H}}_{\text{f}}^{\text{o}}\text{(MgO)}-\left[2\Delta {\text{H}}_{\text{f}}^{\text{o}}\text{(Mg) + }\Delta {\text{H}}_{\text{f}}^{\text{o}}{\text{(O}}_{\text{2}}\text{)}\right]$

Substituting the values:

$\begin{array}{l}{\text{ΔH}}_{\text{f}}^{\text{o}}\text{ = 2}\Delta {\text{H}}_{\text{f}}^{\text{o}}\text{(- 602 kJ/mol)}-\left[\text{2(0 kJ/mol) + (0 kJ/mol)}\right]\\ {\text{ΔH}}_{\text{f}}^{\text{o}}\text{ = -1204 kJ/mol}\end{array}$

Thus, the heat of reaction for the formation of 2 moles of magnesium oxide is $\text{-1204 kJ}$.

Since the amount of energy released to form 2 moles of magnesium oxide is $\text{-1204 kJ}$ so, the amount of energy released to form 6 mol of magnesium oxide is

$\text{6 mol}\times \frac{\text{-1204 kJ/mol}}{2\text{ mol}}=\text{ -3612 kJ}$

Thus, the amount of energy released when 6 mol of magnesium oxide formed is $\text{3612 kJ}$.

The balanced chemical reaction for the formation of aluminum oxide, ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}$ from aluminum and oxygen, is

${\text{4Al(s) + 3O}}_{\text{2}}\text{(g) }\to {\text{ 2Al}}_{\text{2}}{\text{O}}_{\text{3}}\text{(s)}$

The value of heat of formation of all the species involved in the reaction is

$\begin{array}{c}{\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}\text{(s) = -1676 kJ/mol}\\ \text{Al(s) = 0 kJ/mol}\\ {\text{O}}_{\text{2}}\text{(g) = 0 kJ/mol}\end{array}$

The ${\text{ΔH}}_{\text{f}}^{\text{o}}$ for a chemical reaction is calculated as

${\text{ΔH}}_{\text{f}}^{\text{o}}\text{ = 2}\Delta {\text{H}}_{\text{f}}^{\text{o}}{\text{(Al}}_{\text{2}}{\text{O}}_{\text{3}}\text{)}-\left[4\Delta {\text{H}}_{\text{f}}^{\text{o}}\text{(Al) + 3}\Delta {\text{H}}_{\text{f}}^{\text{o}}{\text{(O}}_{\text{2}}\text{)}\right]$

Substituting the values,

$\begin{array}{l}{\text{ΔH}}_{\text{f}}^{\text{o}}\text{ = 2}\Delta {\text{H}}_{\text{f}}^{\text{o}}\text{(- 1676 kJ/mol)}-\left[\text{4(0 kJ/mol) + 3(0 kJ/mol)}\right]\\ {\text{ΔH}}_{\text{f}}^{\text{o}}\text{ = -3352 kJ/mol}\end{array}$

Thus, the heat of reaction for the formation of 2 moles of aluminum oxide is $\text{-3352 kJ}$.

Hence, the amount of energy released is more during the formation of 6 mol of magnesium oxide.

Conclusion

The amount of energy released is more during the formation of 6 mol of magnesium oxide.