Answer
In kJ/mol: -34kJ/mol
Or
In kcal/mol: -8kcal/mol
Work Step by Step
In the given reaction, bond breaking and bond formation takes place. Bond breaking is endothermic (positive) whereas bond formation is exothermic.
In order to find $\Delta$ $H^{\circ}$ of the reaction, we add the bond dissociation energies, keeping in mind that where the bond formation takes place, the bond dissociation energy is negative.
Therefore,
In kJ/mol
Bonds broken=
$CH_3CH2-OH=+381$
$H-Br= +368$
Bonds formed=
$CH3CH2-Br=-285$
$H-OH=-498$
$\Delta$$H^{\circ}$= $(+381+368) + (-285) + (-498)$
$\Delta$$H^{\circ}$= $+749+(-783)$
$\Delta$$H^{\circ}$= $+749 -783$
$-34 kJ/mol$
In kcal/mol
Bonds broken=
$CH3CH2-OH= +91$
$H-Br= +88$
Bonds formed=
$CH3CH2-Br=-68$
$H-OH=-119$
$\Delta$$H^{\circ}$=$ (+91+88 )+ (-68)+(-119)$
$\Delta$$H^{\circ}$ =$179 + ( -187)$
$\Delta$$H^{\circ}$ = $179 - 187$
$\Delta$$H^{\circ}$= $-8 kcal/mol$
