(Solved): Ethanol is produced commercially by the hydration of ethylene: C2H4(g)+H2O(v)C2H5 ...

Ethanol is produced commercially by the hydration of ethylene: ${\mathrm{C}}_2{\mathrm{H}}_4(\mathrm{g})+{\mathrm{H}}_2\mathrm{O}(\mathrm{v})\rightleftharpoons {\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}(\mathrm{v})$ Some of the product is converted to diethyl ether in the undesired side reaction $2{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}(\mathrm{v})\rightleftharpoons {\left({\mathrm{C}}_2{\mathrm{H}}_5\right)}_2\mathrm{O}(\mathrm{v})+{\mathrm{H}}_2\mathrm{O}(\mathrm{v})$ The combined feed to the reactor contains 53.7 mole $\%{\mathrm{C}}_2{\mathrm{H}}_4,36.7\%{\mathrm{H}}_2\mathrm{O}$ and the balance nitrogen which enters the reactor at $310^{\circ }\mathrm{C}$. The reactor operates isothermally at $310^{\circ }\mathrm{C}$. An ethylene conversion of $5\%$ is achieved, and the yield of ethanol (moles ethanol produced/mole ethylene consumed) is 0.900 . Data for Diethyl Ether $\begin{array}{l}\Delta {\hat{H}}_{\mathrm{f}}^{\circ }=-272.8\mathrm{kJ}/\mathrm{mol}\text{ for the liquid }\\ \Delta {\hat{H}}_{\mathrm{v}}=26.05\mathrm{kJ}/\mathrm{mol}\text{ (assume independent of }T)\\ C_p\left[\mathrm{kJ}/\left(\mathrm{mol}\cdot {}^{\circ }\mathrm{C}\right)\right]=0.08945+40.33\times 10^{-5}T\left({}^{\circ }\mathrm{C}\right)-2.244\times 10^{-7}{T}^2\end{array}$ (a) Calculate the reactor heating or cooling requirement in $\mathrm{kJ}/\mathrm{mol}$ feed. (b) Why would the reactor be designed to yield such a low conversion of ethylene? What processing step (or steps) would probably follow the reactor in a commercial implementation of this process?