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Given that bond energies of H-H and Cl-Cl are 430 kJ mol$^{-1}$ and 240 kJ mol$^{-1}$ respectively and $\Delta H_f$ for HCl is $-90$ kJ mol$^{-1}$, bond enthalpy of HCl is
A
380 kJ mol$^{-1}$
B
425 kJ mol$^{-1}$
C
245 kJ mol$^{-1}$
D
290 kJ mol$^{-1}$
Detailed Solution
Formation of one mole of HCl: $\dfrac{1}{2}H_2 + \dfrac{1}{2}Cl_2 \rightarrow HCl$
$\Delta H_{HCl} = \Sigma$ B.E. of reactants $- \Sigma$ B.E. of products
$-90 = \dfrac{1}{2} \times 430 + \dfrac{1}{2} \times 240 -$ B.E. of HCl
$-90 = 215 + 120 -$ B.E. of HCl
$\therefore$ B.E. of HCl $= 215 + 120 + 90$
$= 425$ kJ mol$^{-1}$
$\Delta H_{HCl} = \Sigma$ B.E. of reactants $- \Sigma$ B.E. of products
$-90 = \dfrac{1}{2} \times 430 + \dfrac{1}{2} \times 240 -$ B.E. of HCl
$-90 = 215 + 120 -$ B.E. of HCl
$\therefore$ B.E. of HCl $= 215 + 120 + 90$
$= 425$ kJ mol$^{-1}$