Question

Use the values in Table 6.2 to calculatefor each reaction. Classify each reaction as endothermic or exothermic.

a.

b.

Short answer

Answer

a. The reaction is endothermic as the $\Delta {\mathrm{H}}^{\circ }$of the reaction is positive.

b. The reaction is exothermic as the ${\mathbf{\Delta H}}^{\mathbf{\circ }}$of the reaction is positive.

Step-by-step solution

Step-by-Step SolutionStep 1: Enthalpy change (△H∘)   in a reaction

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

$\Delta {{\mathbf{H}}^{\mathbf{\circ }}}_{\mathbf{overall}}\mathbf{}\mathbf{=}\sum _{}^{}\mathrm{\Delta }{{\mathbf{H}}^{\mathbf{\circ }}}_{\mathbf{bonds}\mathbf{}\mathbf{broken}}\mathbf{}\mathbf{-}\sum _{}^{}\mathbf{}\mathrm{\Delta }{{\mathbf{H}}^{\mathbf{\circ }}}_{\mathbf{bonds}\mathbf{}\mathbf{formed}}$

Endothermic and exothermic reactions

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

Calculating ΔH∘for the given reactions

a.

Reaction a

In the given reaction,

The bonds broken are:

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

The bonds formed are:

role="math" localid="1648188865190" ${\mathrm{CH}}_3-{\mathrm{CH}}_2-\mathrm{OH}$ ($\Delta {\mathrm{H}}^{\circ }$ = 393 kJ/mol) and H-Br ($\Delta {\mathrm{H}}^{\circ }$ = 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:

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

The bonds formed are:

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

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

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