Question

Calculate the standard enthalpy and standard entropy change at 298 K for the following reaction: 2Fe(s) + 3/2O2(g) → Fe2O3(s) Given that the standard enthalpies of formation for Fe(s), O2(g), and Fe2O3(s) are 0 kJ/mol, 0 kJ/mol, and -826 kJ/mol respectively. Also, the standard molar entropy of Fe2O3(s) is 87.4 J/K/mol.

Answer 1

To calculate the standard enthalpy change  H  for the reaction, we can use the following equation:H =   Hf of products  -   Hf of reactants where Hf is the standard enthalpy of formation of each substance.For the given reaction:2Fe s  + 3/2O2 g   Fe2O3 s H = [1 *  -826 kJ/mol ] - [2 * 0 kJ/mol +  3/2  * 0 kJ/mol]H = -826 kJ/molNow, to calculate the standard entropy change  S  for the reaction, we need the standard molar entropy of Fe s  and O2 g . Unfortunately, these values are not provided in the problem. However, we can still set up the equation for S:S =   S of products  -   S of reactants where S is the standard molar entropy of each substance.For the given reaction:2Fe s  + 3/2O2 g   Fe2O3 s S = [1 * 87.4 J/K/mol] - [2 * S Fe  +  3/2  * S O2 ]Without the values for S Fe  and S O2 , we cannot calculate S for the reaction.