Question

Given that bond energies of $H-H$ and $Cl-Cl$ are $430{\text{kJ mol}}^{-1}$ and $240{\text{kJ mol}}^{-1},$ respectively and $\Delta H_f$ for $HCl$ is $-90{\text{kJ mol}}^{-1},$ bond enthalpy of $HCl$ is:

• A. $380{\text{kJ mol}}^{-1}$
• B. $425{\text{kJ mol}}^{-1}$
• C. $245{\text{kJ mol}}^{-1}$
• D. $290{\text{kJ mol}}^{-1}$

Answer 1

The correct option is B $425{\text{kJ mol}}^{-1}$

$\frac{1}{2}{H}_2+\frac{1}{2}C{l}_2⟶HCl$
$\Delta H=\sum B.E{.}_{\text{(}reactants)}-\sum B.E{.}_{\text{(products)}}$
$\Rightarrow -90=\frac{1}{2}[B.E{.}_{\left(H_2\right)}+B.E{.}_{\left(C{l}_2\right)}]-B.E{.}_{\left(HCl\right)}$
$\Rightarrow -90=\frac{1}{2}(430+240)-B.E{.}_{\left(HCl\right)}$
$\therefore B.E{.}_{\left(HCl\right)}=\frac{1}{2}(430+240)+90=425{\text{kJ mol}}^{-1}$