Question

Calculate the standard enthalpy change for the following reaction involving solids at 298 K:Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g) Given the following standard enthalpies of formation in kJ/mol: Fe2O3(s) = -825.5 Fe(s) = 0CO2(g) = -393.5 CO(g) = -110.5

Answer 1

To calculate the standard enthalpy change for the reaction, we can use the formula:H =  Hf products  -  Hf reactants where H is the standard enthalpy change, and Hf is the standard enthalpy of formation.For the products:2 mol Fe s : 2  0 kJ/mol = 0 kJ3 mol CO2 g : 3   -393.5 kJ/mol  = -1180.5 kJFor the reactants:1 mol Fe2O3 s : 1   -825.5 kJ/mol  = -825.5 kJ3 mol CO g : 3   -110.5 kJ/mol  = -331.5 kJNow, we can plug these values into the formula:H =  0 kJ +  -1180.5 kJ   -   -825.5 kJ  +  -331.5 kJ  H =  -1180.5 kJ  -  -1157 kJ H = -23.5 kJThe standard enthalpy change for the reaction at 298 K is -23.5 kJ.