The bond enthalpies of H-H, Cl-Cl, and H-Cl bonds are 437 kJ/mol, 239 kJ/mol, and 433 kJ/mol respectively. Calculate the enthalpy of formation of HCl.

Step-by-step solution

The reaction for the formation of HCl is: 1 2 H ₂(g) + 1 2 Cl ₂(g) HCl (g) The enthalpy of formation H_f is given by the difference between the bond enthalpies of reactants and products: H_f = BE ( reactants ) - BE ( products ) H_f = ( 1 2 BE ( H-H ) + 1 2 BE ( Cl-Cl ) ) - BE ( H-Cl ) Substituting the given values: H_f = ( 1 2 437 + 1 2 239 ) - 433 H_f = (218.5 + 119.5) - 433 H_f = 338 - 433 = -95 kJ/mol Answer: -95 kJ/mol