Question

Consider the reaction:
2Al(s) + Fe2O3(s) → Al2O3(s) + 2Fe(s)

The ΔHf for Fe2O3(s) = -824.3 kJ/mole.
The ΔHf for Al2O3(s) = -1675.7 kJ/mole.
Finish the equation.

ΔHrxn = [(1)( kJ/mole) + (2)( kJ/mole)] - [(1)( kJ/mole) + (2) ( kJ/mole)]

Answer 1

Answer:

ΔHrxn = [(1) x (-1675.7 kJ/mole) + (2) x (0 kJ/mole)] - [(1) x (-824.3 kJ/mole) + (2) x (0 kJ/mole)] = -1675.7 kJ/mole + 824.3 kJ/mole = -851.4 kJ/mole.

Explanation:

• To calculate ΔH of the reaction, we use the mathematical relation:

• ΔHrxn = ∑ΔHproducts - ∑ΔHreactants.

• So, ΔHrxn = [(n x ΔHf for Al₂O₃) + (n x ΔHf for Fe)] - [(n x ΔHf for Fe₂O₃) + (n x ΔHf for Al)]

• where n is the number of moles of each reactant.

• ΔHrxn = [(1) x (-1675.7 kJ/mole) + (2) x (0 kJ/mole)] - [(1) x (-824.3 kJ/mole) + (2) x (0 kJ/mole)] = -1675.7 kJ/mole + 824.3 kJ/mole = -851.4 kJ/mole.

• ΔHf for Al and Fe = 0.0 kJ/mole; since the heat for formation of elements that found in nature is equal 0.0 kJ/mole