Question

Calculate the enthalpy change for the thermite reaction: 2Al(s)+Fe2O3(s)→2Fe(s)+Al2O3(s), ΔH∘rxn=−850 kJ when 12.0 mol of Al undergoes the reaction with a stoichiometrically equivalent amount of Fe2O3. Express your answer to three significant figures and include the appropriate units.

Answer 1

The enthalpy change for the thermite reaction is -9502.4 kJ.

Step-by-step explanation:

The enthalpy change for the thermite reaction can be calculated using Hess's law. The reaction occurs in three distinct steps: the formation of 1 mol of solid aluminum oxide (Al2O3) and 2 mol of liquid iron at its melting point of 1758°C, the conversion of 2 mol of liquid iron at 1758°C to 2 mol of solid iron at 1758°C, and the conversion of 2 mol of solid iron at 1758°C to 2 mol of solid iron at 25°C. The enthalpy changes for these steps are -732.5 kJ/mol of Fe₂O3, -13.8 kJ/mol of Fe, and -45.5 kJ/mol of Fe, respectively.

By Hess's law, the overall enthalpy change for the reaction is the sum of these individual enthalpy changes. Therefore, the enthalpy change for the thermite reaction is (-732.5 kJ/mol of Fe₂O3) + (-13.8 kJ/mol of Fe) + (-45.5 kJ/mol of Fe) = -791.8 kJ/mol of Fe₂O3.

Since 12.0 mol of Al is undergoing the reaction, the enthalpy change is multiplied by the stoichiometric coefficient of Fe₂O3, which is 1 mol. Therefore, the enthalpy change for the thermite reaction is -791.8 kJ/mol of Fe₂O3 × 12.0 mol = -9502.4 kJ.