Chapter U5.109, Problem 6E

Interpretation Introduction

Interpretation: The amount of energy required to extract 6 mol of tin from tin (II) oxide 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 6E

The amount of energy required to extract 6 mol of tin from tin (II) oxide $\text{1714}\text{.8 kJ}$ .

Explanation of Solution

The balanced chemical reaction for the extraction of tin from tin (II) oxide is:

  $\text{2SnO(s) }\to {\text{ 2Sn(s) + O}}_{\text{2}}\text{(g)}$

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

  $\begin{array}{l}\text{SnO(s)}=\text{ -285}\text{.8 kJ/mol}\\ \text{Sn(s) = 0 kJ/mol}\\ {\text{O}}_{\text{2}}\text{(g) = 0 kJ/mol}\end{array}$

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

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

Substituting the values:

  $\begin{array}{l}{\text{ΔH}}_{\text{f}}^{\text{o}}\text{ = }\left[\text{2(0 kJ/mol) + (0 kJ/mol)}\right]\text{ - 2}\Delta {\text{H}}_{\text{f}}^{\text{o}}\text{(- 285}\text{.8 kJ/mol)}\\ {\text{ΔH}}_{\text{f}}^{\text{o}}\text{ = +571}\text{.6 kJ/mol}\end{array}$

Thus, the heat of reaction for the extraction of 2 moles of tin from 2 moles of tin (II) oxide is $\text{+571}\text{.6 kJ}$ .

Since, the amount of energy required to extract 2 moles of tin is $\text{+571}\text{.6 kJ}$ so, the amount of energy required to extract 6 mol of tin is:

  $\text{6 mol}\times \frac{\text{+571}\text{.6 kJ}}{2\text{ mol}}=\text{ 1714}\text{.8 kJ}$

Hence, the amount of energy required to extract 6 mol of tin from tin (II) oxide $\text{1714}\text{.8 kJ}$ .

Conclusion

Thus, $\text{1714}\text{.8 kJ}$ is the amount of energy required to extract 6 mol of tin from tin (II) oxide.