Question

What is the standard enthalpy change for the reaction of [Fe(H2O)6]2+(aq) + 2Cl-(aq) → [FeCl4]2-(aq) + 6H2O(l)? Given the following bond enthalpies: Fe-Cl = 242 kJ/mol, O-H = 463 kJ/mol, and H-Cl = 431 kJ/mol.

Answer 1

To calculate the standard enthalpy change for the reaction, we need to determine the bond enthalpies of the reactants and products and then subtract the sum of the bond enthalpies of the reactants from the sum of the bond enthalpies of the products.Reactants:1. [Fe H2O 6]2+ aq : There are no Fe-Cl or H-Cl bonds in this complex, so we only need to consider the O-H bonds. There are 6 H2O molecules, each with 2 O-H bonds, so there are a total of 12 O-H bonds. The bond enthalpy for O-H is 463 kJ/mol, so the total bond enthalpy for [Fe H2O 6]2+ aq  is 12 * 463 = 5556 kJ/mol.2. 2Cl- aq : There are no bonds in Cl- ions, so the bond enthalpy is 0 kJ/mol.Products:1. [FeCl4]2- aq : There are 4 Fe-Cl bonds in this complex. The bond enthalpy for Fe-Cl is 242 kJ/mol, so the total bond enthalpy for [FeCl4]2- aq  is 4 * 242 = 968 kJ/mol.2. 6H2O l : There are 6 H2O molecules, each with 2 O-H bonds, so there are a total of 12 O-H bonds. The bond enthalpy for O-H is 463 kJ/mol, so the total bond enthalpy for 6H2O l  is 12 * 463 = 5556 kJ/mol.Now, we can calculate the standard enthalpy change for the reaction:H =  Bond enthalpies of products  -  Bond enthalpies of reactants H =  968 + 5556  -  5556 + 0 H = 6524 - 5556H = 968 kJ/molThe standard enthalpy change for the reaction is 968 kJ/mol.