An important industrial source of ethanol is the reaction, catalyzed by \(\mathrm{H}_{3} \mathrm{PO}_{4},\) of steam with ethylene derived from oil: $$ \begin{array}{c} \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(g) \\ \Delta H_{\mathrm{rxn}}^{\circ}=-47.8 \mathrm{~kJ} \quad K_{\mathrm{c}}=9 \times 10^{3} \mathrm{at} 600 . \mathrm{K} \end{array} $$ (a) At equilibrium, \(P_{\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}}=200 . \mathrm{atm}\) and \(P_{\mathrm{H}_{0} \mathrm{O}}=400 .\) atm. Cal- culate \(P_{\mathrm{C}_{2} \mathrm{H}_{4}}\). (b) Is the highest yield of ethanol obtained at high or low \(P\) ? High or low \(T ?\) (c) Calculate \(K_{c}\) at \(450 .\) K. (d) In \(\mathrm{NH}_{3}\) manufacture, the yield is increased by condensing the \(\mathrm{NH}_{3}\) to a liquid and removing it. Would condensing the \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) have the same effect in ethanol production? Explain.

For the reaction $$ \mathrm{C}_{2} \mathrm{H}_{4}(g) + \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(g), $$ the equilibrium constant expression for pressure, denoted as \( K_p \), is: \[ K_p = \frac{P_{\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}}}{P_{\mathrm{C}_{2} \mathrm{H}_{4}} \cdot P_{\mathrm{H}_{2} \mathrm{O}} }\]Given: \( K_p = 9 \times 10^3 \).

Insert the given pressures into the equilibrium expression: \[ 9 \times 10^3 = \frac{200}{P_{\mathrm{C}_{2} \mathrm{H}_{4}} \cdot 400}.\]

Rearrange the equilibrium expression to solve for \(P_{\mathrm{C}_{2} \mathrm{H}_{4}}\): \[ P_{\mathrm{C}_{2} \mathrm{H}_{4}} = \frac{200}{9 \times 10^3 \cdot 400}\] \[ P_{\mathrm{C}_{2} \mathrm{H}_{4}} = \frac{200}{3.6 \times 10^6}\] \[ P_{\mathrm{C}_{2} \mathrm{H}_{4}} = 5.56 \times 10^{-5} \mathrm{atm}\]

Since the reaction has a negative enthalpy change \( \Delta H = -47.8 \mathrm{~kJ} \), it is exothermic. Therefore, to maximize ethanol yield, lower temperatures are favored. As for pressure, increasing the pressure will favor the production of fewer gas molecules, in this case, ethanol. Thus, the highest yield of ethanol is obtained at high pressure and low temperature.

Use the van't Hoff equation to calculate \(K_c\) at a different temperature:\[\ln \left( \frac{K_{c2}}{K_{c1}} \right) = \frac{\Delta H_{\text{rxn}}}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right)\]where \( \Delta H_{\text{rxn}} = -47.8 \text{ kJ} \), \( R = 8.314 \text{ J/mol·K} \), \( T_1 = 600 \text{ K} \), \( T_2 = 450 \text{ K} \), and \( K_{c1} = 9 \times 10^3 \). First, convert \( \Delta H_{\text{rxn}} \) to J:\( \Delta H_{\text{rxn}} = -47.8 \times 10^3 \text{ J} \). Now, substitute all values:\[\ln \left( \frac{K_{c2}}{9 \times 10^3} \right) = \frac{-47.8 \times 10^3}{8.314} \left( \frac{1}{600} - \frac{1}{450} \right)\]Calculate the temperature difference term: \[\frac{1}{600} - \frac{1}{450} = -\frac{1}{18000}\]Then, calculate the right side of the van't Hoff equation: \[\frac{-47.8 \times 10^3}{8.314} \left( -\frac{1}{18000} \right) = 0.319\]Now solve for \( K_{c2} \) :\[ \ln \left( \frac{K_{c2}}{9 \times 10^3} \right) = 0.319 \implies \frac{K_{c2}}{9 \times 10^3} \approx e^{0.319} \approx 1.376 \implies K_{c2} \approx 1.376 \times 9 \times 10^3 \approx 1.238 \times 10^4 \].

Condensing \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) to a liquid would remove ethanol from the gas phase. This drives the reaction toward the product side by Le Châtelier's principle as it requires compensating for the ethanol removal, thus increasing the production yield of ethanol.