Answer:
ΔH°rxn = -780kJ
Explanation:
It is possible to find enthalpy change of a reaction by the sum of analogue reactions (Hess's law).
Using the reactions:
1. 2Al(s) + 3/2O2(g) → Al2O3(s) ΔH°rxn = −1601 kJ/mol
2. 2Fe(s) + 3/2O2(g) → Fe2O3(s) ΔH°rxn = −821 kJ/mol
1 - 2:
2Al(s) + Fe2O3(s) → Al2O3(s) + 2Fe(s)
ΔH°rxn = ΔH°rxn(1) - ΔH°rxn(2)
ΔH°rxn = −1601 kJ/mol - (-821 kJ/mol)
ΔH°rxn = -780kJ
Considering the Hess's Law, the enthalpy change for the reaction is -780 kJ.
Hess's Law indicates that the enthalpy change in a chemical reaction will be the same whether it occurs in a single stage or in several stages. That is, the sum of the ∆H of each stage of the reaction will give us a value equal to the ∆H of the reaction when it occurs in a single stage.
In this case you want to calculate the enthalpy change of:
2 Al(s) + Fe₂O₃(s) → 2 Fe(s) + Al₂O₃(s)
which occurs in two stages.
You know the following reactions, with their corresponding enthalpies:
Equation 1: 2 Al(s) + [tex]\frac{3}{2}[/tex] O₂(g) → Al₂O₃(s) ΔH = −1601 kJ/mol
Equation 2: 2 Fe(s) + [tex]\frac{3}{2}[/tex] O₂(g) → Fe₂O₃(s) ΔH = −821 kJ/mol
Because of the way formation reactions are defined, any chemical reaction can be written as a combination of formation reactions, some going forward and some going back.
To obtain the enthalpy of the desired chemical reaction you need 2 moles of Al(s) on reactant side and it is present in first equation so let's write this as such.
Now, 1 mole of Fe₂O₃(s) must be a reactant and is present in the second equation. Since this equation has 1 mole of Fe₂O₃(s) on the product side, it is necessary to locate the Fe₂O₃(s) on the reactant side (invert it). When an equation is inverted, the sign of ΔH also changes.
You know that two equations with their corresponding enthalpies are:
Equation 1: 2 Al(s) + [tex]\frac{3}{2}[/tex] O₂(g) → Al₂O₃(s) ΔH = −1601 kJ/mol
Equation 2: Fe₂O₃(s) → 2 Fe(s) + [tex]\frac{3}{2}[/tex] O₂(g) ΔH = 821 kJ/mol
Adding or canceling the reactants and products as appropriate, and adding the enthalpies algebraically, you obtain:
2 Al(s) + Fe₂O₃(s) → 2 Fe(s) + Al₂O₃(s) ΔH= -780 kJ
Finally, the enthalpy change for the reaction is -780 kJ.
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