91Ó°ÊÓ

Enthalpy change of a reaction is equal to the difference of the sum of enthalpy of bonds broken and the enthalpies of bond formed.

$\mathrm{Δ}{{\mathbf{H}}^{\mathbf{∘}}}_{\mathbf{overall}}\mathbf{}\mathbf{=}\underset{}{\overset{}{∑}}\mathrm{Δ}{{\mathbf{H}}^{\mathbf{∘}}}_{\mathbf{bonds}\mathbf{}\mathbf{broken}}\mathbf{}\mathbf{-}\underset{}{\overset{}{∑}}\mathbf{}\mathrm{Δ}{{\mathbf{H}}^{\mathbf{∘}}}_{\mathbf{bonds}\mathbf{}\mathbf{formed}}$

Exothermic reactions are the reactions that have a negative enthalpy change while the endothermic reactions have a positive enthalpy change.

a.

Reaction a

In the given reaction,

The bonds broken are:

${\mathrm{CH}}_3-{\mathrm{CH}}_2-\mathrm{Br}$( $\mathrm{Δ}{\mathrm{H}}^{∘}$= 285 kJ/mol) and H-OH ($\mathrm{Δ}{\mathrm{H}}^{∘}$ = 498 kJ/mol).

The bonds formed are:

role="math" localid="1648188865190" ${\mathrm{CH}}_3-{\mathrm{CH}}_2-\mathrm{OH}$ ($\mathrm{Δ}{\mathrm{H}}^{∘}$ = 393 kJ/mol) and H-Br ($\mathrm{Δ}{\mathrm{H}}^{∘}$ = 368 kJ/mol).

Since the enthalpy of the reaction is positive, the given reaction is endothermic.

b.

Reaction b

In the given reaction,

The bonds broken are:

${\mathrm{CH}}_3-\mathrm{H}$($\mathrm{Δ}{\mathbf{H}}^{∘}$= 435 kJ/mol) and Cl-Cl (= 242 kJ/mol).

The bonds formed are:

${\mathrm{CH}}_3-\mathrm{H}$($\mathrm{Δ}{\mathbf{H}}^{∘}$= 351 kJ/mol) and H-Cl ( = 431 kJ/mol).

$$\begin{gathered} \mathrm{Δ}{{\mathbf{H}}^{\mathbf{∘}}}_{\mathrm{overall}}=\underset{}{\overset{}{∑}}\mathrm{Δ}{{\mathbf{H}}^{\mathbf{∘}}}_{\mathrm{bonds}\mathrm{broken}}-\underset{}{\overset{}{∑}}\mathrm{Δ}{{\mathbf{H}}^{\mathbf{∘}}}_{\mathrm{bonds}\mathrm{formed}} \\ =\left(435+242\right)\mathrm{kJ}/\mathrm{mol}-\left(351+431\right)\mathrm{kJ}/\mathrm{mol} \\ =677\mathrm{kJ}/\mathrm{mol}-782\mathrm{kJ}/\mathrm{mol} \\ =-105\mathrm{kJ}/\mathrm{mol} \end{gathered}$$

Since the enthalpy of the reaction is negative, the given reaction is exothermic.