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Identify the products of fermentation and dehydration of ethanol, and describe the conditions for fermentation and the reagents for reduction and decarboxylation.

Chemistry

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Answer

c

23200 kJ\text{23200 kJ}

Step-by-step solution

3 steps

Part a(i)

Butan-1-ol

CH3CH2CH2CH2OH\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH}

Part a(ii)

Hexanoic acid

CH3CH2CH2CH2CH2COOH\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_2\text{COOH}

Part b(i)

The conditions necessary for fermentation of glucose to take place are

  • Yeast (as a source of enzymes like zymase)
  • Absence of oxygen (anaerobic conditions)
  • Temperature between 25∘C25^\circ\text{C} and 40∘C40^\circ\text{C} (optimum around 37∘C37^\circ\text{C})

Part b(ii)

One reagent that can be used to carry out process S (reduction of ethanoic acid to ethanol) is lithium aluminium hydride (LiAlH4\text{LiAlH}_4).

Part b(iii)

  • Gas P is ethene (CH2=CH2\text{CH}_2=\text{CH}_2). This is formed by the dehydration of ethanol using concentrated sulfuric acid.
  • Gas T is methane (CH4\text{CH}_4). This is formed by the decarboxylation of sodium ethanoate using dry sodium hydroxide (soda lime).

(iv) Sodium hydroxide is kept dry during the reaction by mixing it with calcium oxide (CaO\text{CaO}) to form soda lime. Calcium oxide is a drying agent and also helps prevent the sodium hydroxide from melting.

Part c

23200 kJ\text{23200 kJ}

  1. Calculate the molar mass of ethanol (C2H5OH\text{C}_2\text{H}_5\text{OH}).

    Molar mass of C2H5OH=(2×12.01)+(6×1.008)+(1×15.999)=24.02+6.048+15.999=46.067 g/mol\text{C}_2\text{H}_5\text{OH} = (2 \times 12.01) + (6 \times 1.008) + (1 \times 15.999) = 24.02 + 6.048 + 15.999 = 46.067 \text{ g/mol}.
    Using approximate atomic masses (C=12, H=1, O=16):
    Molar mass of C2H5OH=(2×12)+(6×1)+(1×16)=24+6+16=46 g/mol\text{C}_2\text{H}_5\text{OH} = (2 \times 12) + (6 \times 1) + (1 \times 16) = 24 + 6 + 16 = 46 \text{ g/mol}.

  2. Calculate the number of moles of ethanol in 1 litre.

    Given that 1 litre of ethanol has a mass of 780 g.

    Moles of ethanol=MassMolar mass=780 g46 g/mol≈16.9565 mol\text{Moles of ethanol} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{780 \text{ g}}{46 \text{ g/mol}} \approx 16.9565 \text{ mol}

  3. Calculate the total heat energy released.

    Given that 1 mole of ethanol releases 1370 kJ of heat energy.

    Heat energy released=Moles of ethanol×Heat released per mole\text{Heat energy released} = \text{Moles of ethanol} \times \text{Heat released per mole} Heat energy released=16.9565 mol×1370 kJ/mol\text{Heat energy released} = 16.9565 \text{ mol} \times 1370 \text{ kJ/mol} Heat energy released≈23230.355 kJ\text{Heat energy released} \approx 23230.355 \text{ kJ}

    Rounding to three significant figures:

    Heat energy released≈23200 kJ\text{Heat energy released} \approx \boxed{\text{23200 kJ}}
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