Question

Calculate bond energy of HCl if bond enthalpy of H2 and Cl2 are 430 and 242 KJ/Mol respectively and heat of formation of HCl is - 91 KJ/Mol.

Answer 1

I think this will be the answer please check

Answer 2

Given bond energies of H--H and Cl--Cl bonds which can be represented as follows,
              H2(g) ——>  2H(g)                ΔH(H)  = +430 KJ/mol ----------------(1)
              Cl2(g) ——> 2Cl(g)               ΔH(Cl)  = +240 KJ/mol ----------------(2)
              HCl(g) ——> H(g) + Cl(g)     ΔH(HCl)= ?                    ----------------(3)
Bond enthalpy of HCl [ΔH(HCl)] can be found out as follows,

         ∆H = ∑∆H0f(product) − ∑∆H0f(reactant)        = [∆H0f(H) + ∆H0f(Cl)] − [∆H0f(HCl)]        = 12×430 + 12×240 − (−90)        = 425 KJ/mol

              As far as stoichiometry is concerned for the reactants, when an amount of energy is listed for a balanced chemical equation, it relates to the number of moles of the compound as indicated by its coefficient in the chemical reaction. Here,  ΔH(HCl) = 425 KJ/mol which indicates that 1 mole of HCl, 1 mole of H, 1 mole of Cl all are related to 425 KJ/mol.