Mr Cole Chemistry

Making Ethanol Questions

Production of Ethanol

Alcohols Worksheet

1. Fermentation

Ethanol can be produced by the fermentation of sugar.
a) Write the symbol equation for the reaction.

C6H12O6 → 2C2H5OH + 2CO2

b) State the conditions that are required for process to occur.
  • In solution (dissolved in water).
  • Yeast (to provide the zymase enzyme catalyst).
  • Temperature: approx 30–40oC.
  • Anaerobic conditions (no oxygen).
c) State what is meant by the term anaerobic, explain why that condition is required; and write an equation for the reaction that would take place in aerobic conditions.

Anaerobic means “without oxygen”.

This is required because in aerobic conditions (with oxygen), the yeast would respire aerobically, producing carbon dioxide and water instead of ethanol.

Aerobic Equation: C6H12O6 + 6O2 → 6CO2 + 6H2O

d) Draw a graph demonstrating the effect of temperature on the rate of fermentation. Explain the shape of your graph.
Graph showing rate of reaction vs temperature

The rate of reaction increases with temperature initially due to increased kinetic energy of particles, resulting in more frequent collisions and a greater proportion of collisions having the activation energy. However, after the optimum temperature, the rate decreases rapidly as the enzymes in the yeast denature.

2. Hydration of Ethene

Ethanol can be produced by the hydration of ethene.
a) Write a symbol equation for the reaction.

C2H4(g) + H2O(g) ⇌ C2H5OH(g)

b) State the name of the mechanism for this reaction.

Electrophilic Addition

c) State the conditions required for the reaction.
  • Catalyst: Phosphoric Acid (H3PO4).
  • Temperature: High temperature (~300oC).
  • Pressure: High pressure (~60 atm / 6000 kPa).
d) Outline the mechanism for the reaction. You can use H+ to indicate the presence of the acid catalyst.
Mechanism for the hydration of ethene

1. The double bond in ethene attacks the H+ ion (from the acid catalyst), breaking the double bond and forming a carbocation.

2. A lone pair of electrons on the oxygen of a water molecule attacks the positive carbon atom.

3. One of the hydrogens on the attached water molecule loses its electrons to the oxygen (breaking the O-H bond) and leaves as H+, regenerating the catalyst and forming ethanol.

e) Given that the reaction is a reversible reaction and the enthalpy change of this reaction is negative, state one advantage and one disadvantage of using high temperature during this process.
  • Advantage: Increases the rate of reaction (kinetics).
  • Disadvantage: Shifts the position of equilibrium to the left (reactants side) because the forward reaction is exothermic, resulting in a lower percentage yield.

3. Ethanol as Fuel

In some countries, ethanol is used as fuel. There are advantages and disadvantages to using the two methods to produce ethanol for this use. One claimed advantage is that fermentation is a carbon neutral process.
a) State what is meant by the term carbon neutral.

A process where there is no net net release of carbon dioxide into the atmosphere. The amount of CO2 absorbed during the growth of the crop is equal to the amount released during processing and combustion.

b) Use equations to justify the statement that fermentation is carbon neutral.

1. Photosynthesis (Absorbs 6CO2):
6CO2 + 6H2O → C6H12O6 + 6O2

2. Fermentation (Releases 2CO2):
C6H12O6 → 2C2H5OH + 2CO2

3. Combustion (Releases 4CO2):
2C2H5OH + 6O2 → 4CO2 + 6H2O

Total CO2: Absorbed = 6. Released = 2 + 4 = 6. Net change is zero.

c) Explain the reality is that fermentation isn’t actually carbon neutral.

Energy is required for farming machinery, transport of crops, production of fertilizers, and the distillation of the ethanol mixture. This energy typically comes from burning fossil fuels, which releases additional carbon dioxide into the atmosphere.

d) One of the advantages of hydration of ethene is that is produces purer product. Explain how ethanol from fermentation would be purified in a lab, including a diagram in your answer.
Fractional distillation diagram

Ethanol is purified using fractional distillation. The mixture is heated. Ethanol has a boiling point of 78oC, which is lower than water (100oC). The ethanol vapor rises, passes into the condenser where it cools and turns back into a liquid, and is collected.

4. Percentage Yield Calculations

The process for producing ethanol from ethene is not perfectly efficient.
a) Calculate the percentage yield if 150g of ethene produces 190g of ethanol.

Mr(Ethene) = 28.0
Mr(Ethanol) = 46.0

Moles Ethene = 150 / 28.0 = 5.357 mol
Theoretical Mass Ethanol = 5.357 × 46.0 = 246.4g

Yield = (190 / 246.4) × 100 = 77.1%

b) Calculate the percentage yield if 500g of ethene produces 733g of ethanol.

Moles Ethene = 500 / 28.0 = 17.86 mol
Theoretical Mass Ethanol = 17.86 × 46.0 = 821.4g

Yield = (733 / 821.4) × 100 = 89.2%

c) 5 tonnes of ethene are hydrated, and the process have a percentage yield of 68%. What mass of ethanol will you produce?

Moles Ethene = 5,000,000g / 28.0 = 178,571 mol
Theoretical Mass Ethanol = 178,571 × 46.0 = 8,214,286g = 8.21 tonnes

Actual Mass = 0.68 × 8.21 = 5.6 tonnes

d) Calculate the percentage yield if 10.00m³ of ethene produced 496.3kg of ethanol. The conditions for the reaction were the usual temperature for the reaction, and the lower bound the usual pressure for the reaction.

Conditions assumed:
Temperature = 300°C = 573 K
Pressure = 60 atm = 6079.5 kPa (as 1 atm is 101325Pa)

Step 1: Calculate Moles of Ethene (PV = nRT)
P = 6,079,500 Pa; V = 10.00 m³; T = 573 K; R = 8.31
n = PV / RT
n = (6,079,500 × 10) / (8.31 × 573)
n = 12,767 mol

Step 2: Theoretical Mass
Mass = n × Mr = 12,767 × 46.0 = 587,282g = 587.3 kg

Step 3: Yield
Yield = (496.3 / 587.3) × 100 = 84.5%

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