Answer:
To calculate the standard enthalpy change for the reaction 2Al(s) + Fe₂O3(s) → 2Fe(s) + Al₂O3(s), we can use the following equation:
ΔHrxn = ΣΔHf (products) - ΣΔHf (reactants)
where ΔHrxn is the standard enthalpy change for the reaction, ΔHf is the standard enthalpy of formation, and the summation symbol indicates that we are summing up the values for all of the products and reactants involved in the reaction.
First, we need to find the standard enthalpy of formation for each of the products and reactants. The standard enthalpies of formation for Al₂O3(s) and Fe₂O3(s) are given in the problem statement as -1669.8 kJ/mol and -822.2 kJ/mol, respectively.
For the reactants Al(s) and Fe(s), we can use the following equations to calculate their standard enthalpies of formation:
ΔHf (Al(s)) = [ΔHf (Al₂O3(s)) - (3/2)ΔHf (O₂(g))]/2
= [-1669.8 kJ/mol - (3/2)(0 kJ/mol)]/2
= -835.0 kJ/mol
ΔHf (Fe(s)) = [ΔHf (Fe₂O3(s)) - ΔHf (O₂(g))]/2
= [-822.2 kJ/mol - 0 kJ/mol]/2
= -411.1 kJ/mol
Now that we have the standard enthalpies of formation for all of the products and reactants, we can use the equation for ΔHrxn to calculate the standard enthalpy change for the reaction:
ΔHrxn = [2(-411.1 kJ/mol) + (-822.2 kJ/mol)] - [2(-835.0 kJ/mol) + (-1669.8 kJ/mol)]
= -1644.4 kJ/mol
Therefore, the standard enthalpy change for the reaction 2Al(s) + Fe₂O3(s) → 2Fe(s) + Al₂O3(s) is -1644.4 kJ/mol.