Answer
In kJ/mol: -17kJ/mol
In kcal/mol: -4kcal/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_{3})_3$$C-OH$=$+381$
$H-Cl= +431$
Bonds formed=
$(CH_{3})_3$$C-Cl$=$-331$
$H-OH=-498$
$\Delta$$H^{\circ}$= $(+381+431) + (-331) + (-498)$
$\Delta$$H^{\circ}$= $+812+(-829)$
$\Delta$$H^{\circ}$= $+812 -829$
$-17 kJ/mol$
In kcal/mol
Bonds broken=
$(CH_{3})_3$$C-OH$= $+91$
$H-Cl= +103$
Bonds formed=
$(CH_{3})_3$$C-Cl$=$-79$
$H-OH=-119$
$\Delta$$H^{\circ}$=$ (+91+103 )+ (-79)+(-119)$
$\Delta$$H^{\circ}$ =$194 + ( -198)$
$\Delta$$H^{\circ}$ = $194 - 198$
$\Delta$$H^{\circ}$= $-4 kcal/mol$
