Question

Calculate the standard enthalpy change for the following reaction: [Fe(H2O)6]2+(aq) + SO4 2-(aq) → [Fe(H2O)5 SO4]-(aq) + H2O(l) given that the standard enthalpies of formation of [Fe(H2O)6]2+, [Fe(H2O)5SO4]-, and H2O are -360 kJ/mol, -950 kJ/mol, and -286 kJ/mol, respectively.

Answer 1

To calculate the standard enthalpy change for the reaction, we can use the following equation:H =  Hf products  -  Hf reactants where H is the standard enthalpy change, and Hf is the standard enthalpy of formation.For the given reaction:[Fe H2O 6] aq  + SO aq   [Fe H2O SO] aq  + HO l The standard enthalpies of formation are:Hf [Fe H2O 6]  = -360 kJ/molHf [Fe H2O SO]  = -950 kJ/molHf HO  = -286 kJ/molNote that the standard enthalpy of formation for SO is not given, but since it is a reactant and not a product, it will not affect the calculation.Now, we can plug these values into the equation:H =  Hf [Fe H2O SO]  + Hf HO   -  Hf [Fe H2O 6]  H =  -950 kJ/mol +  -286 kJ/mol   -  -360 kJ/mol H =  -1236 kJ/mol  -  -360 kJ/mol H = -876 kJ/molThe standard enthalpy change for the reaction is -876 kJ/mol.