Chemical Energetics: Question 9

Worked solution

Part (a): Defining the standard enthalpy change of formation

The standard enthalpy change of formation, $\Delta H_f^{\ominus}$, is the enthalpy change when one mole of a compound is formed from its constituent elements, with all reactants and products in their standard states, under standard conditions (298 K, 100 kPa).

Part (b): Energy cycle and calculation of $\Delta H^{\ominus}$

The cycle links the direct route (the thermite reaction itself) with an indirect route via the elements aluminium, iron and oxygen in their standard states.

Direct route: $2\text{Al(s)} + \text{Fe}_2\text{O}_3\text{(s)} \xrightarrow{\ \Delta H^{\ominus}\ } \text{Al}_2\text{O}_3\text{(s)} + 2\text{Fe(s)}$

Indirect route, both branches starting from the same elements, $2\text{Al(s)} + 2\text{Fe(s)} + \tfrac{3}{2}\text{O}_2\text{(g)}$:

1. The elements form $2\text{Al(s)} + \text{Fe}_2\text{O}_3\text{(s)}$ (i.e. the aluminium stays as the element, and iron combines with oxygen to form $\text{Fe}_2\text{O}_3$), releasing $\Delta H_f^{\ominus}[\text{Fe}_2\text{O}_3]$ (aluminium and iron, as elements, contribute zero).
2. The elements form $\text{Al}_2\text{O}_3\text{(s)} + 2\text{Fe(s)}$ directly (aluminium combines with oxygen; iron stays as the element), releasing $\Delta H_f^{\ominus}[\text{Al}_2\text{O}_3]$.

By Hess’s law, since both branches start at the same elements and finish at the same products, the two routes must give the same total enthalpy change: $\Delta H_f^{\ominus}[\text{Fe}_2\text{O}_3] + \Delta H^{\ominus} = \Delta H_f^{\ominus}[\text{Al}_2\text{O}_3]$

(recall that $\Delta H_f^{\ominus}$ for the elements $\text{Al(s)}$ and $\text{Fe(s)}$ is zero, so they contribute nothing to either side)

Rearranging for the unknown, $\Delta H^{\ominus}$: $\Delta H^{\ominus} = \Delta H_f^{\ominus}[\text{Al}_2\text{O}_3] - \Delta H_f^{\ominus}[\text{Fe}_2\text{O}_3]$

Substituting the values from the table: $\Delta H^{\ominus} = -1676 - (-824)$

$\Delta H^{\ominus} = -1676 + 824 = -852\ \text{kJ mol}^{-1}$

Check (independent recomputation): $-1676-(-824) = -1676+824$. Since $824<1676$, the result stays negative: $1676-824=852$, giving $-852\ \text{kJ mol}^{-1}$, consistent.

Part (c): Exothermic or endothermic?

The reaction is exothermic, because $\Delta H^{\ominus} = -852\ \text{kJ mol}^{-1}$ is negative, meaning energy is released to the surroundings as the products form. (This matches the real thermite reaction, which is famously highly exothermic and produces molten iron.)

Part (d): Why mean bond enthalpies would not be appropriate

Mean bond enthalpies apply to covalent bonds between specific pairs of atoms in discrete molecules, where the same bond type recurs across many different compounds. $\text{Al}_2\text{O}_3$ and $\text{Fe}_2\text{O}_3$, however, are giant ionic lattice structures (and $\text{Al(s)}$ and $\text{Fe(s)}$ are giant metallic lattices), not simple molecules made of discrete, identifiable covalent bonds. There is no single “Al–O bond” or “Fe–O bond” enthalpy that can be measured or looked up in the way there is for, say, a C–H bond, so the bond-enthalpy method (breaking and making a fixed number of individual covalent bonds) cannot be applied to this reaction. Enthalpies of formation, used in (b), are the appropriate route instead.

Final answers

• (a) Enthalpy change when one mole of a compound forms from its elements, all species in their standard states, under standard conditions.
• (b) $\Delta H^{\ominus} = \boxed{-852\ \text{kJ mol}^{-1}}$
• (c) Exothermic ($\Delta H^{\ominus} < 0$).
• (d) $\text{Al}_2\text{O}_3$/$\text{Fe}_2\text{O}_3$ are ionic lattices with no discrete covalent bonds to assign bond enthalpies to; the method only works for covalent molecules.