Question

Calculate the enthalpy change of reduction for Fe2O3 using the following data:Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)    ΔH°rxn = -26.8 kJ mol^-1Fe2O3(s) + 2Al(s) → Al2O3(s) + 2Fe(l)     ΔH°rxn = +824.2 kJ mol^-1Note: The standard enthalpy of formation for CO2(g) is -393.5 kJ mol^-1 and for Al2O3(s) is -1675.7 kJ mol^-1.

Answer 1

To calculate the enthalpy change of reduction for Fe2O3, we need to find the enthalpy change for the reaction:Fe2O3 s   2Fe s  + 3/2 O2 g We can do this using Hess's Law, which states that the enthalpy change of a reaction is the same whether it occurs in one step or several steps. We can manipulate the given reactions to obtain the desired reaction and then add their enthalpy changes.First, let's manipulate the given reactions:1. Fe2O3 s  + 3CO g   2Fe s  + 3CO2 g     Hrxn = -26.8 kJ mol^-1   Multiply by 2/3 to get 2Fe s  on the right side:    2/3  Fe2O3 s  + 3CO g   2Fe s  + 3CO2 g      2/3 Fe2O3 s  + 2CO g   2Fe s  + 2CO2 g     Hrxn = -17.867 kJ mol^-12. Fe2O3 s  + 2Al s   Al2O3 s  + 2Fe l      Hrxn = +824.2 kJ mol^-1   Reverse the reaction to get 2Fe s  on the left side:   Al2O3 s  + 2Fe l   Fe2O3 s  + 2Al s      Hrxn = -824.2 kJ mol^-1Now, let's add the manipulated reactions:2/3 Fe2O3 s  + 2CO g   2Fe s  + 2CO2 g     Hrxn = -17.867 kJ mol^-1Al2O3 s  + 2Fe l   Fe2O3 s  + 2Al s      Hrxn = -824.2 kJ mol^-1--------------------------------------------------------------2/3 Fe2O3 s  + 2CO g  + Al2O3 s  + 2Fe l   2Fe s  + 2CO2 g  + Fe2O3 s  + 2Al s     Hrxn = -842.067 kJ mol^-1Now, cancel out the terms that appear on both sides of the reaction:2/3 Fe2O3 s   2Fe s  + 1/3 O2 g     Hrxn = -842.067 kJ mol^-1Finally, multiply the reaction by 3/2 to get the enthalpy change of reduction for Fe2O3:Fe2O3 s   2Fe s  + 3/2 O2 g     Hrxn =  -842.067 kJ mol^-1  *  3/2  = -1263.1 kJ mol^-1The enthalpy change of reduction for Fe2O3 is -1263.1 kJ mol^-1.