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Ethylene oxide is an important chemical feedstock used many applications, including the production of polyesters and PET plastic bottles. Globally. approximately 19 million metric tons of ethylene oxide is produced each year by the oxidation of ethylene using a silver oxide catalyst on a porous support with the reaction stoichiometry: C_2 H_4(g) + 1/2 O_2(g) rightarrow C_2 H_4 O(g) ethylene ethylene oxide reaction 1 A competing side reaction is the combustion of ethylene: C_2 H_4 (g) + 1/2 O_2 (g) rightarrow CO_2 (g) + H_2O (g) reaction 2 For a given production process, ethylene and oxygen are supplied to a reactor at25^degree C and 35 atm and the flow rate of ethylene to the reactor is 10.0 mol/min The oxygen is supplied in 25% excess based on reaction 1 and the molar flow rate of oxygen in the outlet stream is 5.00. At these conditions the feed stream DOES NOT behave as an ideal gas. Determine the molar flow rate of the oxygen feed (mol/min). Determine the compressibility factor for the gas contained in the feed stream. Determine the volumetric flow rate of the feed stream in units of standard cubic meters per minute (SCM/min). Determine the volumetric flow rate (m^3/min) of the feed.

Explanation / Answer

Step I

Moles of ethylene entering the reactor = 10

Moles of O2 will react based on reaction 1 = (10/2) = 5

25% excess of O2 = 5 + (1.25 x 5) = 11.25

Hence, molar flow rate of the oxygen feed = 11.25 mol/min

Step II

Density of ethylene = 1.18 kg/m3 = 1180 g/cm3

Volumetric flow rate of ethylene = (10 mol/min) / (1180/28 mole/cm3) = 0.2373 cm3/min

Density of O2 = 1.429 g/L = (1.429/16) mol/L = 0.0893 mole/L

Volumetric flow rate of O2 = 11.25/0.0893 = 125.98 L/min

Step III

Volumetric feed rate = 0.2373 + 125980 = 125980.08 cm3/min

Step IV

Compressibility can be easily calculated by -

Z = PVm / RT

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