Question

Calculate the standard enthalpy change for the reaction between [Fe(H2O)6]2+ and 2Cl- ions to produce [FeCl4]2- complex ion given that the enthalpy of formation for [FeCl4]2- is -392 kJ/mol and the enthalpies of hydration for Fe2+ and Cl- ions are -340 kJ/mol and -381 kJ/mol, respectively.

Answer 1

To calculate the standard enthalpy change for the reaction, we can use the following equation:H_reaction = H_products - H_reactantsThe reaction is as follows:[Fe H2O 6] + 2Cl  [FeCl4] + 6H2OFirst, let's find the enthalpy change for the reactants. Since we are given the enthalpies of hydration for Fe and Cl ions, we can calculate the enthalpy change for the reactants as follows:H_reactants = H_hydration Fe  + 2 * H_hydration Cl H_reactants =  -340 kJ/mol  + 2 *  -381 kJ/mol H_reactants = -340 kJ/mol - 762 kJ/molH_reactants = -1102 kJ/molNow, we are given the enthalpy of formation for [FeCl4], which is -392 kJ/mol. Since the formation of water from its ions is not involved in the reaction, we can consider the enthalpy change for the products as only the enthalpy of formation for [FeCl4]:H_products = H_formation [FeCl4] H_products = -392 kJ/molNow, we can calculate the standard enthalpy change for the reaction:H_reaction = H_products - H_reactantsH_reaction =  -392 kJ/mol  -  -1102 kJ/mol H_reaction = 710 kJ/molTherefore, the standard enthalpy change for the reaction between [Fe H2O 6] and 2Cl ions to produce [FeCl4] complex ion is 710 kJ/mol.