Chapter U5.109, Problem 3E

Interpretation Introduction

Interpretation: The amount of energy released when 10 mol of lead (II) oxide formed needs to be determined.

Concept Introduction: The change in enthalpy by 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 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 3E

The amount of energy released when 10 mol of lead (II) oxide formed is $\text{872 kJ}$ .

Explanation of Solution

The balanced chemical reaction for the formation of lead (II) oxide from lead and oxygen is:

  ${\text{2Pb(s) + O}}_{\text{2}}\text{(g) }\to \text{ 2PbO(s)}$

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

  $\begin{array}{l}\text{PbO(s)}=\text{ -218 kJ/mol}\\ \text{Pb(s) = 0 kJ/mol}\\ {\text{O}}_{\text{2}}\text{(g) = 0 kJ/mol}\end{array}$

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

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

Substituting the values:

  $\begin{array}{l}{\text{ΔH}}_{\text{f}}^{\text{o}}\text{ = 2}\Delta {\text{H}}_{\text{f}}^{\text{o}}\text{(- 218 kJ/mol)}-\left[\text{2(0 kJ/mol) + (0 kJ/mol)}\right]\\ {\text{ΔH}}_{\text{f}}^{\text{o}}\text{ = -436 kJ/mol}\end{array}$

Thus, the heat of reaction for the formation of 2 moles of lead (II) oxide is $\text{-436 kJ/mol}$ .

Since, the amount of energy released to form 2 moles of lead (II) oxide is $\text{-436 kJ/mol}$ so, the amount of energy released to form 10 mol of lead (II) oxide is:

  $\text{10 mol}\times \frac{\text{-436 kJ/mol}}{2\text{ mol}}=\text{ -872 kJ}$

Hence, the amount of energy released when 10 mol of lead (II) oxide formed is $\text{872 kJ}$ .

Conclusion

  $\text{872 kJ}$ is the amount of energy released when 10 mol of lead (II) oxide isformed.