Chemistry - Thermochemistry Question with Solution | TestHub

${\mathrm{H}}_2\left(\mathrm{g}\right)+\mathrm{B}{\mathrm{r}}_2\left(\mathrm{g}\right)\to 2\mathrm{H}\mathrm{B}\mathrm{r}\left(\mathrm{g}\right)$

Enthalpy change for this reaction will equal to the difference in the energy consumed for breaking bonds of reactants and energy released when bonds of products are made

Enthalpy = Sum of bond energies broken - sum of bond energies formed

$∆{\mathrm{H}}^{\mathrm{o}}={\left(\mathrm{B}\mathrm{ }\mathrm{E}\right)}_{\mathrm{r}\mathrm{e}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{t}}-{\left(\mathrm{B}\mathrm{ }\mathrm{E}\right)}_{\mathrm{p}\mathrm{r}\mathrm{o}\mathrm{d}\mathrm{u}\mathrm{c}\mathrm{t}}$

$=\left(433+192\right)-\left(2\times 364\right)$

$=625-728=-\mathrm{103 }\mathrm{k}\mathrm{J}$