Question

Calculate the standard enthalpy change for the reaction: Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g) Given the following standard enthalpy of formation values: ΔHf° [Fe2O3(s)] = -824 kJ/mol ΔHf° [CO(g)] = -110 kJ/mol ΔHf° [CO2(g)] = -393.5 kJ/mol ΔHf° [Fe(s)] = 0 kJ/mol.

Answer 1

To calculate the standard enthalpy change for the reaction, we can use the following formula:H reaction  =  Hf products  -  Hf reactants For the products:2 mol of Fe s : 2 * Hf[Fe s ] = 2 * 0 kJ/mol = 0 kJ3 mol of CO2 g : 3 * Hf[CO2 g ] = 3 *  -393.5 kJ/mol  = -1180.5 kJFor the reactants:1 mol of Fe2O3 s : 1 * Hf[Fe2O3 s ] = 1 *  -824 kJ/mol  = -824 kJ3 mol of CO g : 3 * Hf[CO g ] = 3 *  -110 kJ/mol  = -330 kJNow, we can plug these values into the formula:H reaction  =  0 kJ +  -1180.5 kJ   -   -824 kJ  +  -330 kJ  H reaction  =  -1180.5 kJ  -  -1154 kJ H reaction  = -26.5 kJThe standard enthalpy change for the reaction is -26.5 kJ.