Chapter 17.4, Problem 38LC

Interpretation Introduction

Interpretation: The enthalpy change of the chemical reaction is to be calculated using the heats of combustion.

Concept Introduction: Hess’s law states that the overall change in enthalpy for the solution is the sum of all changes independent of the different phases or steps of a reaction.

Answer to Problem 38LC

Thus, the enthalpy change in kJ for the reaction ${\text{2Al(s)+Fe}}_{\text{2}}{\text{O}}_{\text{3}}\text{(s)}\to {\text{2Fe(s)+Al}}_{\text{2}}{\text{O}}_{\text{3}}\text{(s)}$ is $\text{853}\text{.9 kJ}$

Explanation of Solution

The chemical reaction for which enthalpy change is to be calculated

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

  $\text{(1)}$

Enthalpy changes for the combustion of aluminum and iron are

  $\begin{array}{l}\text{2Al(s)+}\frac{\text{3}}{\text{2}}{\text{O}}_{\text{2}}\text{(g)}\to {\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}\text{(s) }\Delta \text{H=-1676}\text{.0 kJ (2)}\\ \text{2Fe(s)+}\frac{\text{3}}{\text{2}}{\text{O}}_{\text{2}}\text{(g)}\to {\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\text{(s) }\Delta \text{H=-822}\text{.1 kJ (3)}\end{array}$

To get the given chemical reaction, we have to reverse the reaction $\text{(3)}$ which will change the sign of enthalpy.

  ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\text{(s)}\to \text{2Fe(s)+}\frac{\text{3}}{\text{2}}{\text{O}}_{\text{2}}\text{(g) }\Delta \text{H=822}\text{.1 kJ (4)}$

Now add the two reactions $\text{(2)}$ and $\text{(4)}$ , we get $\text{(1)}$ and the enthalpies also get added

  $\begin{array}{l}\text{2Al(s)+}\frac{\text{3}}{\text{2}}{\text{O}}_{\text{2}}\text{(g)}\to {\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}\text{(s) }\Delta \text{H=-1676}\text{.0 kJ (2)}\\ {\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\text{(s)}\to \text{2Fe(s)+}\frac{\text{3}}{\text{2}}{\text{O}}_{\text{2}}\text{(g) }\Delta \text{H=822}\text{.1 kJ (4)}\\ \overline{\underset{¯}{{\text{2Al(s)+Fe}}_{\text{2}}{\text{O}}_{\text{3}}\text{(s)}\to {\text{2Fe(s)+Al}}_{\text{2}}{\text{O}}_{\text{3}}\text{(s) }\Delta \text{H= 853}\text{.9}}}\text{ kJ }\end{array}$