Calculate the enthalpy of formation of HCl if bond enthalpies of H - H , Cl - Cl and H - Cl bonds are 434 kJ mol ⁻¹ , 242 kJ mol ⁻¹ and 431 kJ mol ⁻¹ , respectively.

Step-by-step solution

The chemical equation for the formation of HCl is: 1 2 H ₂( g ) + 1 2 Cl ₂( g ) HCl ( g ) The enthalpy of formation is given by: _f H = Bond enthalpies of reactants - Bond enthalpies of products _f H = [ 1 2 H_ H-H + 1 2 H_ Cl-Cl ] - H_ H-Cl Substituting the given values: _f H = [ 1 2 (434) + 1 2 (242) ] - 431 _f H = (217 + 121) - 431 _f H = 338 - 431 = -93 kJ mol ⁻¹ Answer: -93 kJ mol ⁻¹