To calculate the overall enthalpy change for the reaction, we need to add the enthalpy changes of all the stages.
ΔH_total = ΔH1 + ΔH2 + ΔH3
ΔH1 = -732.5 kJ mol–1 (given)
ΔH2 = -27.6 kJ mol–1 (given)
ΔH3 = -91 kJ mol–1 (given)
Therefore,
ΔH_total = -732.5 kJ mol–1 + (-27.6 kJ mol–1) + (-91 kJ mol–1)
ΔH_total = -851.1 kJ mol–1
The overall enthalpy change for the reaction is -851.1 kJ mol–1. This means that the reaction is exothermic, releasing 851.1 kJ of energy per mole of the reaction. The negative sign indicates that the reaction releases energy, and the magnitude of the enthalpy change indicates the amount of energy released.
You have been asked to calculate the enthalpy change in the following
reaction: 2Al(s) + Fe2O3(s) → Al2O3(s) + 2Fe(s)
The reaction occurs in the following stages:
Chemical Change Assignment 2
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. You must show your workings within your submission, and explain your
reasoning
1 answer
1 answer