Question

Determine the standard enthalpy change for the following reaction: Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g) Given:ΔHf°[Fe2O3(s)] = -824.2 kJ/mol ΔHf°[CO(g)] = -110.5 kJ/mol ΔHf°[Fe(s)] = 0 kJ/mol ΔHf°[CO2(g)] = -393.5 kJ/mol

Answer 1

To determine the standard enthalpy change  H  for the given reaction, we can use the formula:H =  Hf products  -  Hf reactants 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 *  -824.2 kJ/mol  = -824.2 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   -   -824.2 kJ  +  -331.5 kJ  H =  -1180.5 kJ  -  -1155.7 kJ H = -24.8 kJThe standard enthalpy change for the given reaction is -24.8 kJ.