Alcohols Worked ExamplesIGCSE Chemistry 0620

(a) The water bath keeps the mixture warm and within the 25–35 °C range in which the yeast works best. Below this range fermentation is very slow; well above it the yeast stops working.

(b) Fermentation must take place in the absence of oxygen. With air present, the yeast would not produce ethanol (and oxygen could oxidise the ethanol that forms).

(c) Glucose breaks down into ethanol and carbon dioxide. Balance C: 6 = 2 × 2 + 2 × 1 ✓.

(d) Bubble the gas through limewater: it turns milky.

(a) Products: carbon dioxide and water. C: 2 $\text{CO}_2$. H: 6 H → 3 $\text{H}_2\text{O}$. O on the right: $4 + 3 = 7$; ethanol already has 1 O, so $6$ O atoms are needed from 3 $\text{O}_2$.

(b) O on the right: $2 + 3 = 5$; subtract the 1 O in ethanol → 4 O atoms = 2 $\text{O}_2$.

(c) Ethanol is used as a solvent and as a fuel.

(a) Read up from 4 hours to the curve and across: 180 $\text{cm}^3$.

(b) After 8 hours the line is horizontal — the volume stays constant at 240 $\text{cm}^3$, so no more gas is being produced.

(c) Moles of carbon dioxide (convert $\text{cm}^3$ to $\text{dm}^3$ by dividing by 1000). Ethanol : carbon dioxide = 2 : 2 = 1 : 1. $M_r(\text{C}_2\text{H}_5\text{OH}) = 46$.

(d) Glucose : carbon dioxide = 1 : 2. $M_r(\text{C}_6\text{H}_{12}\text{O}_6) = 180$.

(e) Fractional distillation — ethanol boils at a lower temperature than water, so it distils off first.

C is correct: fermentation takes days, is a batch process, and the product is ethanol dissolved in water with yeast, so it must be separated by fractional distillation.

A describes the addition of steam to ethene, not fermentation. B is wrong: ethene comes from cracking petroleum fractions — a non-renewable resource (sugar for fermentation is the renewable one). D mixes up the conditions: steam + ethene uses 300 °C, 6000 kPa and an acid catalyst; yeast at 25–35 °C is fermentation.

(a) All three substances are gases under the conditions used.

(b) 300 °C, 6000 kPa (60 atm) and an acid catalyst.

(c)(i) Yield increases: there are 2 moles of gas on the left and 1 on the right, so higher pressure moves the equilibrium to the side with fewer gas molecules — the ethanol side.

(ii) Yield decreases: the forward reaction is exothermic, so raising the temperature moves the equilibrium in the endothermic (backward) direction.

(iii) No change in yield: a catalyst speeds up the forward and backward reactions equally, so equilibrium is reached faster but its position does not move.

(a) Country S: fermentation — sugar (glucose) from sugar cane is a renewable raw material it already has, and the process uses low temperature and ordinary pressure, so it needs little energy.

Country T: catalytic addition of steam to ethene — ethene is available from cracking at its refinery; the process is fast, continuous and gives pure ethanol.

(b) Fermentation: slow (batch process) and the ethanol is impure/dilute, so it must be fractionally distilled. Hydration of ethene: ethene comes from non-renewable petroleum, and the high temperature and pressure need a lot of energy.

(a) $M_r(\text{C}_2\text{H}_5\text{OH}) = 24 + 6 + 16 = 46$.

(b) Ethanol : oxygen = 1 : 3.

(c) Ethanol : carbon dioxide = 1 : 2; ethanol : water = 1 : 3. $M_r(\text{H}_2\text{O}) = 18$.

Count the carbons in the chain (3 → propan-, 4 → butan-) and number from the end nearest the OH group to find its position.

(a) 1: OH on carbon 1 of 3 → propan-1-ol. 2: OH on the middle carbon → propan-2-ol. 3: four carbons, OH on carbon 2 → butan-2-ol. 4: four carbons, OH on an end carbon → butan-1-ol (the OH is drawn on the left, but it is still carbon 1).

(b) 3 and 4 both have 4 C, 10 H and 1 O but the OH is in a different position: 3 and 4. (1 and 2 are also isomers of each other, but they are $\text{C}_3\text{H}_8\text{O}$.)

(a) Count each line: C–C 2, C–H 7, C–O 1, O–H 1.

(b) Oxygen has 6 outer electrons: it uses 2 in its two single bonds, leaving 2 lone pairs on the oxygen atom. Carbon and hydrogen have none.

(c) Outer electrons: C has 4, H has 1, O has 6. Check against the bonds: 11 bonding pairs + 2 lone pairs.

(d) Use atomic (proton) numbers: C = 6, H = 1, O = 8.

(a) Mass burned = before − after; temperature rise = final − initial; then divide.

(b) The longer the carbon chain, the greater the temperature rise per gram — more heat is released per gram burned.

(c) For example: the distance between the flame and the can, the same can, or the same starting temperature of the water.

(d) Reason: heat is lost to the surroundings (the air and the can) instead of heating the water — or incomplete combustion (soot on the can). Improvement: use a draught shield / put a lid on the can, and keep the flame close to the can.

(a) The potassium manganate(VII) must be acidified and aqueous.

(b) Purple to colourless — the manganate(VII) is used up as it oxidises the alcohol.

(c) The carbon chain is unchanged (4 C); the $\text{CH}_2\text{OH}$ end becomes COOH: $\text{CH}_3\text{CH}_2\text{CH}_2\text{COOH}$, butanoic acid.

(d) $M_r(\text{C}_4\text{H}_9\text{OH}) = 48 + 10 + 16 = 74$; $M_r(\text{C}_3\text{H}_7\text{COOH}) = 48 + 8 + 32 = 88$.

Percentage yield = actual ÷ theoretical × 100.

(e) Bacterial oxidation of ethanol to ethanoic acid, which produces vinegar.

(a) $\text{C}_2\text{H}_4 + \text{H}_2\text{O} \rightarrow \text{C}_2\text{H}_5\text{OH}$: 1 mol (28 g) of ethene gives 1 mol (46 g) of ethanol, so the mass ratio is 28 : 46 — the units (tonnes) carry through.

(b) The unreacted ethene is separated and recycled back over the catalyst, so almost all of it is eventually converted.

(c) $\text{C}_6\text{H}_{12}\text{O}_6 \rightarrow 2\text{C}_2\text{H}_5\text{OH} + 2\text{CO}_2$: 180 g of glucose gives $2 \times 46 = 92$ g of ethanol.