Calculate the standard enthalpy change of reduction for the reaction: Fe2O3(s) + 3H2(g) -> 2Fe(s) + 3H2O(l) Given: Standard enthalpy of formation (ΔHf°) of Fe2O3(s) = -822 kJ/mol Standard enthalpy of formation (ΔHf°) of H2O(l) = -286 kJ/mol Standard enthalpy of fusion (ΔHfus) of Fe(s) = 13.81 kJ/mol Standard enthalpy of vaporization (ΔHvap) of H2(g) = 0.449 kJ/mol Standard enthalpy of formation (ΔHf°) of H2(g) = 0 kJ/mol

Question

Calculate the standard enthalpy change of reduction for the reaction: Fe2O3(s) + 3H2(g) -> 2Fe(s) + 3H2O(l) Given: Standard enthalpy of formation (ΔHf°) of Fe2O3(s) = -822 kJ/mol Standard enthalpy of formation (ΔHf°) of H2O(l) = -286 kJ/mol Standard enthalpy of fusion (ΔHfus) of Fe(s) = 13.81 kJ/mol Standard enthalpy of vaporization (ΔHvap) of H2(g) = 0.449 kJ/mol Standard enthalpy of formation (

Answer 1

To calculate the standard enthalpy change of reduction for the reaction, we can use the following equation:H =  Hf products  -  Hf reactants For the products:2 mol Fe s  + 3 mol H2O l For the reactants:1 mol Fe2O3 s  + 3 mol H2 g Now, we can substitute the given values into the equation:H = [2 * 0  Hf of Fe  + 3 *  -286   Hf of H2O ] - [1 *  -822   Hf of Fe2O3  + 3 * 0  Hf of H2 ]H = [0 +  -858 ] - [ -822  + 0]H = -858 + 822H = -36 kJ/molThe standard enthalpy change of reduction for the reaction is -36 kJ/mol.