Selective oxidation of hydrocarbons is a known method to produce alcohols. Howev

Question

Selective oxidation of hydrocarbons is a known method to produce alcohols. However, the alcohols react with the oxygen to produce aldehydes, and the latter react with oxygen to produce organic acids. A 52 mol/s stream consisting of 90.0% ethane and 10.0% nitrogen is mixed with a 37.0 mol/s air stream and fed into a catalytic reactor. The following reactions take place in the reactor:

2 C2H6 + O2 ------> 2 C2H5OH

2 C2H5OH + O2 ------> 2 CH3CHO + 2 H2O

2 CH3CHO + O2 ------> 2 CH3COOH

The oxygen conversion is 82.0%, and the concentration of the ethanol in the product stream is three times that of the aldehyde and four times that of the acetic acid.

(a) What is the fractional conversion of ethane?
(b) What is the production rate of ethanol?

Explanation / Answer

Express the fractional conversion of O2 and the two expressions involving the relationship between different components in the product stream in terms of ?1, ?2, and ?3. You will then have three equations and three unknowns. You can then express the fractional conversion of ethane in terms of ?1 and (nC2H6)1, allowing you to solve for the fractional conversion.

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