Question 14.8: Acetic acid is esterified in the liquid phase with ethanol a...
Acetic acid is esterified in the liquid phase with ethanol at 100°C and atmospheric pressure to produce ethyl acetate and water according to the reaction:
\mathrm{CH}_3 \mathrm{COOH}(l)+\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}(l) \rightarrow \mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_5(l)+\mathrm{H}_2 \mathrm{O}(l)
If initially there is one mole each of acetic acid and ethanol, estimate the mole fraction of ethyl acetate in the reacting mixture at equilibrium.
Data for \Delta H_{f_{298}}^{\circ} \text { and } \Delta G_{f_{298}}^{\circ} are given for liquid acetic acid, ethanol, and water in Table C.4. For liquid ethyl acetate, the corresponding values are:
\Delta H_{f_{298}^{\circ}}^{\circ}=-480,000 \mathrm{~J} \quad \text { and } \quad \Delta G_{f_{298}}^{\circ}=-332,200 \mathrm{~J}<br />
The values of \Delta H_{298}^{\circ} \text { and } \Delta G_{298}^{\circ} for the reaction are therefore:
\Delta H_{298}^{\circ}=-480,000-285,830+484,500+277,690=-3640 \mathrm{~J}
\Delta G_{298}^{\circ}=-332,200-237,130+389,900+174,780=-4650 \mathrm{~J}
By Eq. (14.11b),
\ln K=\frac{-\Delta G^{\circ}}{R T} (14.11b)
\ln K_{298}=\frac{-\Delta G_{298}^{\circ}}{R T}=\frac{4650}{(8.314)(298.15)}=1.8759 \quad \text { or } \quad K_{298}=6.5266For the small temperature change from 298.15 to 373.15 K, Eq. (14.15) is adequate for estimating K. Thus,
\ln \frac{K}{K^{\prime}}=-\frac{\Delta H^{\circ}}{R}\left(\frac{1}{T}-\frac{1}{T^{\prime}}\right) (14.15)
\ln \frac{K_{373}}{K_{298}}=\frac{-\Delta H_{298}^{\circ}}{R}\left(\frac{1}{373.15}-\frac{1}{298.15}\right)
or \text { or } \quad \ln \frac{K_{373}}{6.5266}=\frac{3640}{8.314}\left(\frac{1}{373.15}-\frac{1}{298.15}\right)=-0.2951
and K_{373}=(6.5266)(0.7444)=4.8586
For the given reaction, Eq. (14.5), with x replacing y, yields:
y_i=\frac{n_i}{n}=\frac{n_{i_0}+\nu_i \varepsilon}{n_0+\nu \varepsilon} (14.5)
x_{\mathrm{AcH}}=x_{\mathrm{EtOH}}=\frac{1-\varepsilon_e}{2} \quad x_{\mathrm{EtAc}}=x_{\mathrm{H}_2 \mathrm{O}}=\frac{\varepsilon_e}{2}
Because the pressure is low, Eq. (14.32) is applicable. In the absence of data for the activity coefficients in this complex system, we assume that the reacting species form an ideal solution. In this case Eq. (14.33) is employed, giving:
\prod_i\left(x_i \gamma_i\right)^{\nu_i}=K (14.32)
\prod_i\left(x_i\right)^{\nu_i}=K (14.33)
K=\frac{x_{\mathrm{EtAc}} x_{\mathrm{H}_2 \mathrm{O}}}{x_{\mathrm{AcH}} x_{\mathrm{EtOH}}}
Thus, 4.8586=\left(\frac{\varepsilon_e}{1-\varepsilon_e}\right)^2
Solution yields:
\varepsilon_e=0.6879 \quad \text { and } \quad x_{\mathrm{EtAc}}=0.6879 / 2=0.344
This result is in good agreement with experiment, even though the assumption of ideal solutions may be unrealistic. Carried out in the laboratory, the reaction yields a measured mole fraction of ethyl acetate at equilibrium of about 0.33.