Question

Calculate the standard enthalpy change (ΔH°) for the following reaction involving the coordination compound [Fe(H2O)6]2+:[Fe(H2O)6]2+(aq) + 2Cl-(aq) → [FeCl2(H2O)4](aq) + 2H2O(l) Given: ΔH°f [FeCl2(H2O)4](aq) = -334 kJ/mol ΔH°f [Fe(H2O)6]2+(aq) = -393 kJ/mol ΔH°f H2O(l) = -286 kJ/mol Note: The superscript ° denotes standard conditions (i.e. 1 atm and 25°C) and the superscript f denotes standard enthalpy of formation.

Answer 1

To calculate the standard enthalpy change  H  for the given reaction, we can use the following equation:H =  Hf products  -  Hf reactants where Hf is the standard enthalpy of formation of each compound.For the given reaction:[Fe H2O 6]2+ aq  + 2Cl- aq   [FeCl2 H2O 4] aq  + 2H2O l H = {[1  Hf [FeCl2 H2O 4] aq ] + [2  Hf H2O l ]} - {[1  Hf [Fe H2O 6]2+ aq ]}H = {[1   -334 kJ/mol ] + [2   -286 kJ/mol ]} - {[1   -393 kJ/mol ]}H =  -334 kJ/mol - 572 kJ/mol  -  -393 kJ/mol H =  -906 kJ/mol  + 393 kJ/molH = -513 kJ/molThe standard enthalpy change  H  for the given reaction is -513 kJ/mol.