What is the enthalpy of formation of HF if bond enthalpies of H-H, F-F and H-F bonds are 434 kJ/mol, 158 kJ/mol and 565 kJ/mol respectively. ?

Step-by-step solution

The reaction for the formation of HF is: 1 2 H₂(g) + 1 2 F₂(g) HF(g) The enthalpy of formation is given by: _f H = Bond enthalpies of reactants - Bond enthalpies of products _f H = [ 1 2 BE(H-H) + 1 2 BE(F-F) ] - BE(H-F) Substituting the given values: _f H = [ 1 2 (434) + 1 2 (158) ] - 565 _f H = (217 + 79) - 565 _f H = 296 - 565 = -269 kJ/mol Answer: -269 kJ/mol