WJEC A-Level Chemistry 1410QS
Formulae and Equations
Basic Ideas About Atoms
Chemical Calculations
Bonding
Solid Structures
The Periodic Table
Simple Equilibria and Acid-Base Reactions
Thermochemistry
Rates of Reaction
The Wider Impact of Chemistry
Organic Compounds
Hydrocarbons
Halogenoalkanes
Alcohols and Carboxylic Acids
Instrumental Analysis
Redox and Standard Electrode Potential
Redox Reactions
Chemistry of the P-Block
Chemistry of the D-Block Transition Metals
Chemical Kinetics
Enthalpy Changes for Solids and Solutions
Entropy and Feasibility of Reactions
Equilibrium Constants
Acid-Base Equilibria
Stereoisomerism
Aromaticity
Alcohols and Phenols
Aldehydes and Ketones
Carboxylic Acids and Their Derivatives
Amines
Amino Acids, Peptides and Proteins
Organic Synthesis and Analysis
Practical Skills and Techniques
Planning, Analysis and Evaluation
Core Experiments and Their Chemistry
Mathematical Skills in Chemistry
Alcohols and Carboxylic Acids
Two routes to ethanol: hydration and fermentation — comparing the two routes on rate
- 3 things to know
- AS Unit 2
Overview
Recall questions on hydration and fermentation — comparing the two routes on rate.
What you need to know
Answer each one before you reveal it.
Compare fermentation and the hydration of ethene for their raw materials, speed and purity of product.
Why is bioethanol described as a carbon-neutral fuel?
Questions from this subtopic
Each one with the model answer underneath.
- 1
Calculate the atom economy for the production of ethanol by fermentation of glucose. Use M(C₆H₁₂O₆) = 180.0 g mol⁻¹ and M(C₂H₅OH) = 46.0 g mol⁻¹.
Model answer
The equation is C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂. Mass of desired product per mole of glucose = 2 × 46.0 = 92.0 g. Total mass of products = 180.0 g. Atom economy = (92.0 / 180.0) × 100 = 51.1%.
- 2
90.0 g of glucose is fermented and 36.8 g of ethanol is obtained. Calculate the percentage yield. Use M(C₆H₁₂O₆) = 180.0 g mol⁻¹ and M(C₂H₅OH) = 46.0 g mol⁻¹.
Model answer
n(glucose) = 90.0 / 180.0 = 0.500 mol. Each mole of glucose gives 2 mol of ethanol, so the theoretical yield is 1.00 mol = 1.00 × 46.0 = 46.0 g. Percentage yield = (36.8 / 46.0) × 100 = 80.0%.
- 3
Compare the raw materials used in the two routes to ethanol.
Model answer
Fermentation uses sugars from crops such as sugar cane or sugar beet, which are renewable and can be regrown each season. Hydration uses ethene from crude oil, which is finite and non-renewable.
- 4
Which route to ethanol is faster, and why?
Model answer
Hydration of ethene is much faster. It runs continuously at 300 °C over a catalyst and produces ethanol in seconds of contact time, whereas fermentation is a batch process that takes several days at about 35 °C.
- 5
Compare the purity of the ethanol produced by each route.
Model answer
Hydration of ethene gives ethanol of high purity directly, because ethanol is essentially the only product. Fermentation gives a dilute aqueous mixture containing at most about 15% ethanol together with yeast and other organic material, so it needs fractional distillation before use.
- 6
State one advantage of fermentation over hydration in terms of energy demand.
Model answer
Fermentation runs at about 35 °C and atmospheric pressure, so it needs far less energy for heating and compression than a plant operating at 300 °C and 70 atmospheres. Much of that saving is offset, however, by the energy needed to distil the dilute product.
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