Question 17.65: a) You are given a value of Kc but the amounts of reactant a......
a) You are given a value of K_\text{c} but the amounts of reactant and product is given in units of pressure.
Convert K_\text{c}\text{ to }K_\text{p}.
Substitute the given values into the equilibrium expression and solve for \text{P}_{\text{C}_2\text{H}_4} \\ K_\text{p}=\frac{\text{P}_{\text{C}_2\text{H}_5\text{OH}}}{\text{P}_{\text{C}_2\text{H}_4}\text{ P}_{\text{H}_2\text{O}}} =\frac{200.}{\text{P}_{\text{C}_2\text{H}_4}(400.)} =1.8270\times 10^2 \\ \text{P}_{\text{C}_2\text{H}_4}=2.7367\times 10^{-3}=3\times 10^{-3}\text{ atm} .
b) The forward direction, towards the production of ethanol, produces the least number of moles of gas and is favored by high pressure. A low temperature favors an exothermic reaction.
c) No, condensing the \text{C}_2\text{H}_5\text{OH} would not increase the yield. Ethanol has a lower boiling point (78.5°C) than water ( 100°C). Decreasing the temperature to condense the ethanol would also condense the water, so moles of gas from each side of the reaction are removed. The direction of equilibrium (yield) is unaffected when there is no net change in the number of moles of gas.