At standard conditions, if the change in the enthalpy for the following reaction is $-109 {\mathrm{kJmol}}^{-1}$

${\mathrm{H}}_{2\left(\mathrm{g}\right)}+{\mathrm{Br}}_{2\left(\mathrm{g}\right)}\to 2{\mathrm{HBr}}_{\left(\mathrm{g}\right)}$

Given that bond energy of ${\mathrm{H}}_2$ and ${\mathrm{Br}}_2$ is $435 {\mathrm{kJmol}}^{-1}$ and $192 {\mathrm{kJmol}}^{-1}$, respectively, what is the bond energy (in $\mathrm{kJ} {\mathrm{mol}}^{-1}$) of $\mathrm{HBr}?$

• Correct answer
  A

  $368$
• B

  $736$
• C

  $518$
• D

  $259$

Hence, the correct answer is option: 'A'.