Determine the heat of reaction for the liquid phase of lactic acid \left(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}_{3}\right) with ethanol \left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right) to form ethyl lactate \left(\mathrm{C}_{5} \mathrm{H}_{10} \mathrm{O}_{3}\right) and liquid water at 25^{\circ} \mathrm{C}. The heat of combustion of ethyl acetate is -2685(\mathrm{~kJ} / \mathrm{mol}) .(-32.7 \mathrm{~kJ} / \mathrm{mol})
The liquid-phase reaction of lactic acid with ethanol is
\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}_{3}+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} \rightarrow \mathrm{C}_{5} \mathrm{H}_{10} \mathrm{O}_{3}+\mathrm{H}_{2} \mathrm{O}
The combustion reaction of ethyl lactate is
\mathrm{C}_{5} \mathrm{H}_{10} \mathrm{O}_{3}+6 \mathrm{O}_{2} \rightarrow 5 \mathrm{CO}_{2}+5 \mathrm{H}_{2} \mathrm{O}_{(1)}
The following table shows the standard heats of formation data:
Compound \Delta \boldsymbol{I}_{\mathbf{*}}^{\mathbf{o}} \mathbf{( k J / m o l )}
\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}_{3} -687.0
\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} -277.6
\mathrm{C}_{5} \mathrm{H}_{10} \mathrm{O}_{3} -
\mathrm{H}_{2} \mathrm{O} (liquid) -285.8
\mathrm{CO}_{2} (gas) -393.5
\mathrm{O}_{2} (gas) 0

Passo 1

O enunciado pede a entalpia de reação da esterificação em fase líquida
\mathrm{C}_{3}\mathrm{H}_{6}\mathrm{O}_{3}(\ell)+\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}(\ell)\to\mathrm{C}_{5}\mathrm{H}_{10}\mathrm{O}_{3}(\ell)+\mathrm{H}_{2}\mathrm{O}(\ell)
à 25^\circ\mathrm{C}. Vamos usar a lei de Hess e os calores de formação padrão \Delta H_{f}^{\circ}, completando o dado faltante para o \Delta H_{f}^{\circ} do éster a partir do seu calor de combustão.

Passo 2

O calor de combustão do éster segue a reação
\mathrm{C}_{5}\mathrm{H}_{10}\mathrm{O}_{3}(\ell)+6\,\mathrm{O}_{2}(g)\to5\,\mathrm{CO}_{2}(g)+5\,\mathrm{H}_{2}\mathrm{O}(\ell)
e vale \Delta H_{\mathrm{comb}}=-2685\\ \mathrm{kJ/mol}. Aplicando
\Delta H_{\mathrm{comb}} =\sum\Delta H_{f}^{\circ}(\text{produtos}) -\sum\Delta H_{f}^{\circ}(\text{reagentes})
temos
-2685 =\bigl[5\,\Delta H_{f}^{\circ}(\mathrm{CO}_{2})+5\,\Delta H_{f}^{\circ}(\mathrm{H}_{2}\mathrm{O})\bigr] -\bigl[\Delta H_{f}^{\circ}(\mathrm{C}_{5}\mathrm{H}_{10}\mathrm{O}_{3})+6\cdot0\bigr].
Substituindo \Delta H_{f}^{\circ}(\mathrm{CO}_{2})=-393{,}5 e \Delta H_{f}^{\circ}(\mathrm{H}_{2}\mathrm{O})=-285{,}8,
-2685 =\bigl[5(-393{,}5)+5(-285{,}8)\bigr] -\Delta H_{f}^{\circ}(\mathrm{C}_{5}\mathrm{H}_{10}\mathrm{O}_{3}).
Calcula-se
5(-393{,}5)+5(-285{,}8)=-3396{,}5\quad\Longrightarrow\quad -2685=-3396{,}5-\Delta H_{f}^{\circ}(\mathrm{C}_{5}\mathrm{H}_{10}\mathrm{O}_{3}),
logo
\Delta H_{f}^{\circ}(\mathrm{C}_{5}\mathrm{H}_{10}\mathrm{O}_{3}) =-3396{,}5+2685=-711{,}5\\ \mathrm{kJ/mol}.

Passo 3

Agora aplicamos a lei de Hess à reação de esterificação, cujo \Delta H_{\mathrm{rxn}} é
\Delta H_{\mathrm{rxn}} =\sum\Delta H_{f}^{\circ}(\text{produtos}) -\sum\Delta H_{f}^{\circ}(\text{reagentes}).
Substituindo
\begin{aligned} \sum\Delta H_{f}^{\circ}(\text{produtos}) &=\Delta H_{f}^{\circ}(\mathrm{C}_{5}\mathrm{H}_{10}\mathrm{O}_{3})+\Delta H_{f}^{\circ}(\mathrm{H}_{2}\mathrm{O}),\\\ \sum\Delta H_{f}^{\circ}(\text{reagentes}) &=\Delta H_{f}^{\circ}(\mathrm{C}_{3}\mathrm{H}_{6}\mathrm{O}_{3})+\Delta H_{f}^{\circ}(\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}), \end{aligned}
usando \Delta H_{f}^{\circ}(\mathrm{C}_{3}\mathrm{H}_{6}\mathrm{O}_{3})=-687{,}0 e \Delta H_{f}^{\circ}(\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH})=-277{,}6.

Passo 4

Calculando numericamente,
\sum\Delta H_{f}^{\circ}(\text{produtos}) =-711{,}5+(-285{,}8)=-997{,}3\\ \mathrm{kJ/mol},
\sum\Delta H_{f}^{\circ}(\text{reagentes}) =-687{,}0+(-277{,}6)=-964{,}6\\ \mathrm{kJ/mol},
portanto
\Delta H_{\mathrm{rxn}} =-997{,}3-(-964{,}6)=-32{,}7\\ \mathrm{kJ/mol}.

Resposta

\Delta H_{\mathrm{rxn}}=-32{,}7\\ \mathrm{kJ/mol}.

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