Given that bond energies of H--Hand Cl--Cl bonds is 430 KJ mol^​- and 240 KJ mol^- respectively. Enthalpy of formation of HCl is -90 KJ mol^-. Bond enthalpy of HCl is

1. 290
2. 425
3. 380
4. 245

Here in the question the answer according to the book is (c) i.e. 380 KJ mol^-. But if we calculate the question for 1 mole of HCl then the answer is coming (b) . I want to ask that while calculating enthalpy, what stoichiometry should we use for the reactants?

Given bond energies of H--H and Cl--Cl bonds which can be represented as follows,
              H_2(g) ——>  2H_(g)                ΔH_(H)  = +430 KJ/mol ----------------(1)
              Cl_2(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,

         $$\begin{gathered} ∆H = \sum _{}^{}∆H_f^0\left(product\right) - \underset{}{\overset{}{\sum ∆H_f^0\left(reac\tan t\right)}} \\ = [∆H_f^0(H) + ∆H_f^0(Cl\left)\right] - [∆H_f^0(HCl\left)\right] \\ = \frac{1}{2}\times 430 + \frac{1}{2}\times 240 - (-90) \\ = 425 KJ/mol \end{gathered}$$

              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.