Question

Calculate the standard enthalpy change and entropy change for the reaction: 2Fe(s) + 3/2O2(g) -> Fe2O3(s) Given the standard enthalpies of formation of Fe(s), O2(g), and Fe2O3(s) are -0.44 kJ/mol, 0 kJ/mol, and -826 kJ/mol, respectively. Also, the standard molar entropy of Fe2O3(s) is 87.4 J/(mol K).

Answer 1

To calculate the standard enthalpy change  H  for the reaction, we can use the formula:H =  [Hf products ] -  [Hf reactants ]where Hf is the standard enthalpy of formation of each substance.For the given reaction:H = [1 * Hf Fe2O3 ] - [2 * Hf Fe  + 3/2 * Hf O2 ]Substituting the given values:H = [1 *  -826 kJ/mol ] - [2 *  -0.44 kJ/mol  + 3/2 *  0 kJ/mol ]H =  -826 kJ/mol  -  -0.88 kJ/mol H = -825.12 kJ/molNow, to calculate the standard entropy change  S  for the reaction, we need the standard molar entropy values for all substances involved in the reaction. We are only given the standard molar entropy for Fe2O3 s , which is not enough information to calculate S for the reaction.If we had the standard molar entropy values for Fe s  and O2 g , we could use the formula:S =  [S products ] -  [S reactants ]However, without the standard molar entropy values for Fe s  and O2 g , we cannot calculate the standard entropy change for the reaction.