plz show the steps to work out this problem, thanks. Your dad enlists your help
ID: 1064180 • Letter: P
Question
plz show the steps to work out this problem, thanks.
Your dad enlists your help to make ethanol (in the form of wine) from grapes via fermentation in a batch process. After some research, you inform him that yeast is used to convert sugar (specifically glucose - C_6H_12O_6) to ethanol (C_2H_5OH) under anaerobic conditions. Some of the glucose is also converted to acrylic acid (C_2H_3COOH - which gives the wine its acidic flavor) and water. A fermentation tank is initially filled with 50 L of grape juice (containing 11 wt. % glucose in water with an overall density of 1 g/cm^3). During the fermentation process, 35.6 SCF (standard cubic feet) of C02 are produced and 1 kg of glucose remains at the end of the process. You can neglect the yeast in the calculation. Molar masses: glucose = 144 g/mol: ethanol = 46 g/mol, acrylic acid = 72 g/mol, CO_2 = 44 g/mol Balance the reactions above. Draw and clearly label a flow sheet of the fermentation process. Perform a degree of freedom analysis on the process. What is the final molar composition of the fermentation liquid? What is the overall ethanol yield? Your dad tells you that the first batch of wine has the right alcohol content, but is much too acidic. After some careful experiments, you find that the rate equations of the fermentation reactions are second order for reaction (1) and first order for reaction (2) with respect to glucose. What changes would you recommend he make for the next batch?Explanation / Answer
a) 1 Glucose = 2 Ethanol + 2 CO2
1 gluocse = 2 Acrylic acid + 2 Water
c) degree of freedom analysis
f = c + p - 2
c = no of components = 5 (glucose, acrylic acid, ethanol, CO2 & water)
p = no of phases in equilibrium = 2 (gas & liquid)
f = 5 + 2 - 2 = 5
e) 50 L of grape solution has density 1g/cm3
1 L = 1000 cm3
Total mass of grape solution = 50000 g (density x volume)
Total mass of grape solution = 50 kg
Grape solution has 11 % by wt glucose solution
glucose present in grape solution = 11/100 * 50 = 5.5 kg
1 kg glucose remains unreacted. Hence total glucose reacted = 5.5 - 1 = 4.5 kg
CO2 produced = 35.6 Standard cubic feet
At standard cubic feet, Temp = 21 C = 294 K and P = 101325 Pa
1 m3 = 35.6 cubic feet
Volume of gas = 1 m3
Apply ideal gas equation
PV = nRT
101325 *1 = n *8.314 * 294
On solving
n = 41.1 moles of CO2
mass of CO2 produced = 41.1*44 = 1820 g = 1.82 kg
Considering equation
1 Glucose = 1 Ethanol + 2 CO2
Actual ethanol produced = Mass of glucose consumed - Mass of CO2 produced
= 4.5 - 1.82= 2.68 kg
Theoretical yield of ethanol
144 g glucose (1 mol glucose) gives 92 g ethanol (2 moles ethanol)
4500 g glucose gives 92/144* 4500 = 2875 kg ethanol
Overall ethanol yield = Actual yield/ theoretical yield
= 2680/2875 * 100 = 93.16 %
d) Moles of ethanol produced = 2680 / 46 = 58.26
Mass of glucose unreacted = 1 kg
moles of glucose unreacted = 1000/ 144 = 7 moles
Molar composition of fermentation liquid = 7 moles of glucose and 58.26 moles of ethanol
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