Question

Acetic acid bacteria of the genus Acetobacter are used to made acetic acid, in the form of vinegar. Given sufficient oxygen, these bacteria can produce vinegar from a variety of alcoholic foodstuffs. Commonly used feeds include apple cider, wine, and fermented grain, malt, rice, or potato mashes. The overall chemical reaction facilitated by these bacteria is:

            C₂H₅OH + O₂ → CH₃COOH + H₂O

One of the first modern commercial processes was the "German method", first practised in Germany in 1823. In this process, fermentation takes place in a tower packed with wood shavings (where the bacteria reside). The alcohol-containing feed is trickled into the top of the tower, and fresh air supplied from the bottom. The aqueous solution collected at the bottom is the desired product.

If we start with 1000L of wine (14% wt/vol of ethanol in water), what is the maximum concentration (wt/vol) of acetic acid that we can possibly obtain? What is the minimum volume of air (containing 20% by volume of O₂) at 1atm and 37C that we must we supply to bring about the maximum concentration of acetic acid in the product.  

Please help and show work! I've been sick the past week and I'm very confused!

Answer 1

Molar mass of ethanol = 46 g

Molar mass of CH₃COOH = 60 g = 0.060 kg

amount of ethanol in 1000 L wine =(1000 * 14%) kg = 140 kg = 140/0.046 = 3043.5 moles

Amount of water in wine = (1000*86%) kg = 860 kg = 860 kg/1kg.L^-1 = 860 L

According to the reaction 1 mole of C₂H₅OH gives 1 mole of CH₃COOH and 1 mol of H₂O. So, 3043.5 moles of C₂H₅OH gives 3043.5 moles of CH₃COOH and 3043.5 moles of H₂O.

3043.5 moles of H₂O = (3043.5 * 0.018 ) kg = 54.78 kg = 54.78 L [as density of water = 1kg.L^-1

Total volume of CH₃COOH solution obtained = 860 L + 54.78 L = 914.78 L

Concentration of CH₃COOH = 3043.5 moles/914.78 L = 3.33 moles/L = (3.33 moles * 0.060 kg.mole^-1) /L

= 0.1998 kg/L

Amount of oxygen needed = 3043.5 moles

Volume of oxygen = nRT/P = 3043.5 moles * 0.082 L.atm.mol^-1.K^-1 * 310 K/1 atm = 77365.7 L

Air needed = (77365.7 * 100)/20 = 386828.85 L