Question

If the bond energies of $H-H$, $Br-Br$ and $H-Br$ are $433$, $192$ and $364kJmo{l}^{-1}$ respectively, the $\Delta H^{\circ }$ for the reaction:
$H_2\left(g\right)+B{r}_2\left(g\right)\to 2HBr\left(g\right)$ is:

• A. $-261kJ$
• B. $+103kJ$
• C. $+261kJ$
• D. $-103kJ$

Answer 1

The correct option is D $-103kJ$

$\begin{array}{cccccc}& H-H& +& Br-Br& \to & 2H-Br\\ Bondenergy& 433& & 192& & 2\times 364\\ \left(kJmo{l}^{-1}\right)& & & & & \end{array}$
Bond energy is the energy released when bonds are broken. The reactants require bonds to be broken and the products, for them to be formed. So we add the bond energy of the reactants and subtract the bond energies of the product, keeping in mind the stoichiometric coefficients.
So, $\Delta H^{\circ }=(433+192)-2\times 364$
$=625-728=-103kJ$