Question

Calculate the enthalpy of reduction for Fe2O3 given that it is reduced to FeO by hydrogen at 298K and constant pressure. Given that the standard enthalpy of formation of Fe2O3 and FeO are -822kJ/mol and -272kJ/mol respectively, the heat of combustion of hydrogen is -286kJ/mol and the enthalpy of formation of H2O is -285.8kJ/mol.

Answer 1

To calculate the enthalpy of reduction for Fe2O3 to FeO by hydrogen, we can use the following reaction:Fe2O3  s  + H2  g  -> 2 FeO  s  + H2O  g We can use Hess's law to find the enthalpy change for this reaction. Hess's law states that the enthalpy change for a reaction is the same whether it occurs in one step or several steps. We can break down the reaction into the following steps:1. Formation of Fe2O3 from its elements:2 Fe  s  + 3/2 O2  g  -> Fe2O3  s  ; Hf Fe2O3  = -822 kJ/mol2. Formation of FeO from its elements:Fe  s  + 1/2 O2  g  -> FeO  s  ; Hf FeO  = -272 kJ/mol3. Combustion of hydrogen:H2  g  + 1/2 O2  g  -> H2O  g  ; Hc H2  = -286 kJ/molNow, we can write the enthalpy change for the overall reaction as:H reaction  = Hf Fe2O3  - 2 * Hf FeO  + Hc H2 Plugging in the given values:H reaction  =  -822 kJ/mol  - 2 *  -272 kJ/mol  +  -286 kJ/mol H reaction  = -822 + 544 - 286H reaction  = -564 kJ/molTherefore, the enthalpy of reduction for Fe2O3 to FeO by hydrogen at 298K and constant pressure is -564 kJ/mol.