$3\mathrm{.}34$ The important chemical intermediate acetaldehyde can be catalytically produced from partial oxidation of ethane. The primary reaction is

$C_2{H}_6+O_2{C}_2{H}_{4}O+H_{2}O$

However, there are a number of side reactions which also occur to a significant degree:

$C_2{H}_6+3O_{2}2C{O}_2+3H_{2}O$

$C_2{H}_6+3O_{2}C{H}_{3}OH+CO+H_{2}O$

Problems

$185$

$C{H}_{3}OH+2O_{2}C{H}_{2}O+H_{2}O$

$2CO+3H_{2}O{C}_2{H}_6+{?}_2{O}_2$

To reduce the formation of these various by$-$products, the reactor must be operated at low $C_2{H}_6$ conversion and high $C_2{H}_6-$to$-O_2$ ratios in the reactor feed. The process must, therefore, involve a high recycle rate and, because air is used as source of oxygen, must have a recycle purge stream to remove the inert $N_2.$ To avoid the loss of valuable ethane in the purge, in the process flowshect shown in Figure P$3\mathrm{.}34,$ the recycle stream is split into two equal parts. One part is subjected to a separation which will preferentially remove a stream of $N_2,CO,$ and $C{O}_2$ for venting. The other half of the recycle stream is sent directly back to the reactor without treatment. Suppose under

$186$

Species Balances in Reacting Systerns

certain operating conditions the vent gas, stream $9,$ is found to contain equal molar amounts of $CO$ and $C{O}_2$; the product stream $($stream $5)$ is found to consist of $33\%C_2{H}_{4}O,331\%C{H}_{3}OH,$ and ${33}_{{?}_{?}}|\%C{H}_{2}O|$; and the reactor outlet stream is found to contain $35\%C_2{H}_6,51\%N_2,1\%C_2{H}_{4}O,$ and $8\mathrm{.}5\%$ combined $CO$ and $C{O}_2.$ Also, it is found that $1mol{H}_{2}O$ is formed in the process $($stream $6)$ per $1$ mol fresh $C_2{H}_6$ fed to the process $($stream $13).$ All compositions and side conditions are in mole units.

$($a$)$ Construct a degree$-$of$-$freedom table and show that the problem is correctly specified.

$($b$).$ Choose the basis location and devise a unit$-$by$-$unit calculation order which ought to be followed if all streams in the flowsheet were to be calculated. Explain your reasoning in detail.

$($c$)$ Calculate the overall yield of $C_2{H}_{4}O$ from $C_2{H}_6.$ The overall yield is defined as

$$

Overall yield $=\frac{\text{o}\text{v}\text{e}\text{r}\text{a}\text{l}\text{l}\text{r}\text{a}\text{t}\text{e}of{C}_2{H}_{4}O\text{p}\text{r}\text{o}\text{d}\text{u}\text{c}\text{t}\text{i}\text{o}\text{n}in\text{t}\text{h}\text{e}\text{p}\text{l}\text{a}\text{n}\text{t}}{\{\begin{array}{ccc}\text{:}[& \text{o}\text{v}\text{e}\text{r}\text{a}\text{l}\text{l}\text{r}\text{a}\text{t}\text{e}of\text{p}\text{r}\text{o}\text{d}\text{u}\text{c}\text{t}\text{i}\text{o}\text{n}of{C}_2{H}_{4}O\text{w}\text{h}\text{i}\text{c}\text{h}\text{c}\text{o}\text{u}\text{l}\text{d}be& ]\\ & \text{o}\text{b}\text{t}\text{a}\text{i}\text{n}\text{e}\text{d}if\text{t}\text{h}\text{e}{C}_2{H}_6\text{c}\text{o}\text{n}\text{v}\text{e}\text{r}\text{t}\text{e}\text{d}\text{w}\text{a}\text{s}\text{a}\text{l}\text{l}\text{u}\text{s}\text{e}\text{d}& to\\ & \text{p}\text{r}\text{o}\text{d}\text{u}\text{c}\text{e}{C}_2{H}_{4}O\text{a}\text{l}\text{o}\text{n}\text{e}& \end{array}}$

For this calculation, use as basis $12mo\frac{l}{h}$ of stream $5.$

$($d$)$ Suppose the reaction

$2CO+O_{2}2C{O}_2$

y

is added to the set given above. Does this reaction have any effect on the balance calculations? If so$,$ what effect? Explain in detail.