Acetobacter aceti bacteria convert ethanol to acetic acid (vinegar) under aerobi
ID: 888395 • Letter: A
Question
Acetobacter aceti bacteria convert ethanol to acetic acid (vinegar) under aerobic (i.e. with oxygen) conditions. A continuous fermentation process for acetic acid production is shown in the Figure. The conversion reaction is as follows: The feed stream containing ethanol enters the reactor at a rate of 1.0 kg/hr ethanol. Also, air bubbles into the reactor at a rate of 50.0 L/min (measured at 1 atm, 25 degree C; ). An exit off-gas stream as well as the liquid product stream containing acetic acid and water leave the reactor. Assume reaction goes to completion. What is the limiting reactant? What is the rate of ethanol consumption? What are the fractional conversions of C2H5OH and O2? Determine the outlet mass flow rates of acetic acid and water in the liquid product stream and the volumetric flow rates of oxygen and nitrogen in the offgas stream. Assume all water generated by the reaction ends up in the liquid product stream.Explanation / Answer
Consider 1 hr of fermentation proess.
Mass of ethanol entering the reactor = 1.0 Kg = 1.0 Kg x (1000g / 1 Kg) = 1000 g
Moles of ethanol entering the reactor = 1000 g C2H5OH x (1 mol C2H5OH / 46 g C2H5OH)
= 21.74 mol C2H5OH
Volume of air entering the reactor, V = 50.0 L / min = 50.0L * 60 / hr = 3000 L / hr
Now moles of air entering the reactor can be calculated from ideal gas equation.
n = PV / RT = 1 atm x 3000 L / (0.08205L.atm.mol-1K-1x298K) = 122.7 mol air
Moles of O2 present in 122.7 mol air = 122.7 x 0.21 = 25.8 mol O2
Moles of N2 present in 122.7 mol air = 122.7 x 0.79 = 96.9 mol N2
(a): The chemical reaction occuring during fermentation process is
C2H5OH + O2 ------ > CH3COOH + H2O
1 mol 1 mol 1 mol 1 mol
for the above balanced reaction,
21.74 mol C2H5OH that will react with the moles of O2 = (1 mol O2 / 1 mol C2H5OH) x 21.74 mol C2H5OH
= 21.74 mol O2
Hence moles of O2 remain unreacted = 25.8 mol - 21.74 mol = 4.06 mol O2.
Since C2H5OH is used up completely, it is the limiting reactant (answer)
Fractional conversion of C2H5OH = 21.74 / 21.74 = 1
Fractional conversion of O2 = 21.74 / 25.8 = 0.84
(b) C2H5OH, being the limiting reactant determines the amount of CH3COOH and H2O formed.
Hence moles of CH3COOH formed = (1 mol CH3COOH / 1 mol C2H5OH) x 21.74 mol C2H5OH
= 21.74 mol CH3COOH
mass of CH3COOH formed = 21.74 mol CH3COOH x (60 g CH3COOH / 1 mol CH3COOH) = 1304 g / hr
Hence mass flow rate of acetic acid = 1304 g / hr
moles of H2O formed = (1 mol H2O / 1 mol C2H5OH) x 21.74 mol C2H5OH
= 21.74 mol H2O
mass of H2O formed = 21.74 mol H2O x (18 g H2O / 1 mol H2O) = 391.3 g / hr
Hence mass flow rate of water = 391.3 g / hr
Moles of N2 present in 122.7 mol air = 122.7 x 0.79 = 96.9 mol N2
Since N2 is not used in the reaction, 96.9 mol N2 will flow in the off-gas stream
volume of N2, V = nRT / P = 96.9 mol x (0.08205L.atm.mol-1K-1x298K) / 1 atm = 2369 L/ hr
Hence volumetric flow rate of N2 = 2369 L/hr
moles of O2 remain unreacted = 25.8 - 21.74 = 4.06 mol
volume of O2, V = nRT / P = 4.06 mol x (0.08205L.atm.mol-1K-1x298K) / 1 atm = 2369 L/ hr
Hence volumetric flow rate of O2 = 99.3 L/hr