You have been asked to calculate the enthalpy change in the following

Question

reaction: 2Al(s) + Fe2O3(s) → Al2O3(s) + 2Fe(s)
The reaction occurs in the following stages:

1. 2Al(s) + Fe2O3(s) → Al2O3(s) + 2Fe(l) ΔH = -732.5kJ mol–1
2. Al2O3(s) + 2Fe(l) → Al2O3(s) + 2Fe(s) ΔH = -27.6kJ mol–1
3. Al2O3(s) + 2Fe(l) → Al2O3(s) + 2Fe(s) ΔH = -91kJ mol–1
Note stages 2 and 3 look the same, but the reaction is cooling from 1700oC to 25oC
and hence the enthalpy change there. Calculate the overall enthalpy change for the
reaction.

Answer 1

The overall enthalpy change for the reaction is the sum of the enthalpy changes for each stage. Therefore:

ΔHoverall = ΔH1 + ΔH2 + ΔH3

ΔH1 = -732.5 kJ mol-1 (given in the problem statement)

ΔH2 = -27.6 kJ mol-1 (given in the problem statement)

ΔH3 = -91 kJ mol-1 (also given in the problem statement)

ΔHoverall = -732.5 kJ mol-1 + (-27.6 kJ mol-1) + (-91 kJ mol-1)

ΔHoverall = -851.1 kJ mol-1

Therefore, the overall enthalpy change for the reaction is -851.1 kJ mol-1.