2nd Year Chemistry — MCQs & Practice Questions

Reactivity increases down the group because the outer electron is further from the nucleus and more easily lost.

It floats, melts into a ball, fizzes and moves around, giving off hydrogen: 2Na + 2H₂O → 2NaOH + H₂.

Thermal stability increases down the group. The cations get larger with less charge density, so they polarise/distort the carbonate ion less, making it harder to decompose.

1. Magnesium reacts very slowly with cold water (faster with steam)[1]
2. Calcium reacts readily with cold water, fizzing[1]
3. Ca + 2H₂O → Ca(OH)₂ + H₂[1]
4. balanced equation[1]
5. Reactivity increases down the group[1]
6. because the outer electrons are further from the nucleus and more easily lost[1]

Mg very slow with cold water, Ca reacts readily; Ca + 2H₂O → Ca(OH)₂ + H₂; reactivity rises down the group (electrons lost more easily).

Down the group the cations get larger and have lower charge density, so they polarise (distort) the nitrate ion less. The nitrate is therefore harder to break down, so a higher temperature is needed and thermal stability increases.

1. Reactivity increases down the group[1]
2. the outer electrons are further from the nucleus and more shielded[1]
3. so they are lost more easily, making the metal more reactive[1]
4. Hydroxides become more soluble down the group[1]
5. sulfates become less soluble down the group[1]
6. so barium sulfate is highly insoluble (used in the sulfate test)[1]

Reactivity rises down Group 2 (outer electrons lost more easily); hydroxides get more soluble, sulfates less soluble.

They form ions with more than one oxidation state, form coloured compounds, and act as catalysts (any three; also high melting points/densities).

Iron in the Haber process (or vanadium(V) oxide in the Contact process, or nickel in hydrogenation).

The d-orbitals split into different energy levels; a d-electron absorbs certain wavelengths of visible light to move between them, and the transmitted/reflected light gives the colour.

1. +1 (Cu⁺) and +2 (Cu²⁺)[1]
2. Cu²⁺(aq) is blue (Cu⁺ compounds often white/colourless)[1]
3. A complex ion is a central metal ion[1]
4. surrounded by ligands bonded by coordinate (dative) bonds[1]
5. A pale blue precipitate forms first[1]
6. which dissolves in excess ammonia to give a deep blue solution[1]

Cu⁺ and Cu²⁺ (Cu²⁺ blue); complex = metal ion + ligands (dative bonds); excess NH₃ gives pale blue ppt then deep blue solution.

A pale blue precipitate of copper(II) hydroxide forms first, then dissolves in excess ammonia to give a deep blue solution as water ligands are replaced by ammonia ligands (ligand exchange), changing the colour.

1. Their 3d and 4s sub-shells are close in energy[1]
2. so different numbers of electrons can be lost, giving several stable oxidation states[1]
3. In a complex the d-orbitals split into two energy levels[1]
4. a d-electron absorbs visible light to jump the gap (d–d transition); the transmitted light is the colour[1]
5. They can change oxidation state easily / provide a surface for reaction[1]
6. e.g. iron in the Haber process (or V₂O₅ in the Contact process)[1]

Variable states: close 3d/4s energies; colour: d–d transitions absorbing visible light; catalysis: easy change of oxidation state, e.g. iron in the Haber process.

Compounds with the same molecular formula but different structural arrangements of their atoms.

Propan-1-ol; the functional group is the hydroxyl group, –OH (an alcohol).

In homolytic fission each atom takes one electron, forming two radicals. In heterolytic fission one atom takes both electrons, forming a positive and a negative ion.

1. An electrophile is an electron-pair acceptor (electron-poor species)[1]
2. A nucleophile is an electron-pair donor (electron-rich species)[1]
3. Bromine + ethene is an addition reaction[1]
4. Haloalkane with hot alcoholic KOH is an elimination reaction[1]
5. A functional group is the atom or group of atoms[1]
6. that gives a molecule its characteristic chemical properties[1]

Electrophile accepts an electron pair, nucleophile donates one; Br₂ + ethene = addition, elimination gives the alkene; functional group = the reactive part defining the chemistry.

But-2-ene has a C=C double bond, and there is no rotation about it, so the methyl groups are fixed either on the same side (cis) or opposite sides (trans). Butane has only single bonds, which rotate freely, so no fixed arrangement — and no cis–trans isomers.

1. A nucleophile is an electron-pair donor (electron-rich)[1]
2. an electrophile is an electron-pair acceptor (electron-poor)[1]
3. (i) Bromine acts as an electrophile[1]
4. in an electrophilic addition to the C=C double bond[1]
5. (ii) Hydroxide acts as a nucleophile[1]
6. in a nucleophilic substitution of the haloalkane[1]

Nucleophile = electron-pair donor, electrophile = acceptor; Br₂/ethene = electrophilic addition; OH⁻/bromoethane = nucleophilic substitution.

1. Step 1: hydrate ethene with steam and an acid catalyst to give ethanol[1]
2. (~300 °C, ~60 atm, phosphoric acid)[1]
3. Step 2: oxidise the ethanol with acidified potassium dichromate[1]
4. under reflux, to give ethanoic acid[1]
5. React bromoethane with potassium cyanide in ethanol to give propanenitrile[1]
6. useful because it adds a carbon, lengthening the chain[1]

Ethene → ethanol (steam + acid catalyst) → ethanoic acid (oxidise with acidified dichromate, reflux); bromoethane + KCN → propanenitrile (adds a carbon).

A ring of six carbons with six delocalised electrons spread above and below the ring; all six C–C bonds are identical, between single and double.

Addition would destroy the stable delocalised ring; substitution replaces a hydrogen and preserves the aromatic stability.

Concentrated nitric acid with concentrated sulfuric acid (generating NO₂⁺) gives nitrobenzene, by electrophilic substitution.

1. Benzene's electrons are delocalised around the ring[1]
2. which lowers its energy / makes it more stable than isolated double bonds[1]
3. Chlorine (Cl₂) with a halogen carrier catalyst[1]
4. such as anhydrous AlCl₃ or FeCl₃[1]
5. Benzene has no localised C=C double bond to undergo addition[1]
6. its delocalised, stable ring does not react with bromine water[1]

Delocalisation lowers benzene's energy (extra stability); chlorination needs Cl₂ + AlCl₃/FeCl₃; no localised C=C means no addition, so bromine water is not decolourised.

Concentrated sulfuric acid protonates concentrated nitric acid, which loses water to form the nitronium ion, NO₂⁺. This electrophile substitutes onto the ring, so the reaction is electrophilic substitution.

1. Benzene has a stable delocalised ring of electrons[1]
2. addition would destroy that stability, which is unfavourable[1]
3. substitution replaces an H and restores the stable ring[1]
4. In phenol the –OH lone pair delocalises into the ring[1]
5. making the ring more electron-rich / more reactive to electrophiles[1]
6. with bromine water the orange colour disappears and a white precipitate (2,4,6-tribromophenol) forms[1]

Benzene substitutes to keep its stable ring; phenol is more reactive (–OH lone pair enriches the ring) and decolourises bromine water, forming a white precipitate.

The bond is polar; the halogen is more electronegative, leaving the carbon slightly positive (δ+), which attracts electron-pair-donating nucleophiles.

Ethanol is formed by nucleophilic substitution (OH⁻ replaces Br⁻).

The nitrogen atom has a lone pair of electrons which can accept a proton (H⁺), forming a positively charged ion; so amines are proton acceptors (bases).

1. Hot ethanolic (alcoholic) potassium hydroxide[1]
2. under reflux / heat[1]
3. This is elimination (loss of HBr forms the C=C)[1]
4. Excess ammonia[1]
5. (in a sealed tube / ethanolic solution)[1]
6. This is nucleophilic substitution[1]

→ ethene: hot alcoholic KOH, elimination; → ethylamine: excess ammonia, nucleophilic substitution.

The nitrogen lone pair can accept a proton (H⁺), so amines are bases. Ethylamine + HCl gives the salt ethylammonium chloride (C₂H₅NH₃⁺Cl⁻).

1. Warm aqueous sodium hydroxide (or water)[1]
2. gives ethanol[1]
3. Potassium cyanide (in ethanol) gives propanenitrile[1]
4. useful because it adds a carbon, lengthening the chain[1]
5. The C–I bond is weaker than C–Br[1]
6. so it breaks more easily, making iodoethane react faster[1]

OH⁻ → ethanol; CN⁻ → propanenitrile (adds a carbon); iodoethane is faster because C–I is weaker than C–Br.

1. In SN2 the nucleophile attacks the δ+ carbon as the halogen leaves[1]
2. in a single concerted step[1]
3. favoured by primary haloalkanes (the carbon is easy to approach)[1]
4. In SN1 the halogen leaves first to form a carbocation[1]
5. then the nucleophile attacks the carbocation[1]
6. tertiary haloalkanes form a stable tertiary carbocation, so this route is favoured[1]

SN2: one-step, nucleophile attacks as halide leaves, favoured by primary. SN1: two-step via a carbocation, favoured by tertiary (stable carbocation).

Secondary — the carbon bearing the –OH group is attached to two other carbon atoms.

First an aldehyde, then on further oxidation a carboxylic acid; the dichromate turns from orange to green.

When phenol loses H⁺, the negative charge is delocalised into the benzene ring, stabilising the ion, so H⁺ is released. Ethanol's anion is not stabilised in this way, so it does not release H⁺.

1. Propan-1-ol (primary) → propanal then propanoic acid[1]
2. Propan-2-ol (secondary) → propanone (a ketone)[1]
3. 2-methylpropan-2-ol (tertiary) → no reaction[1]
4. Those that react turn the dichromate from orange to green[1]
5. The tertiary alcohol has no hydrogen on the carbon bearing the –OH[1]
6. so there is nothing for the oxidising agent to remove[1]

1° → aldehyde/acid, 2° → ketone, 3° → no reaction (orange→green for the first two); tertiary resists because there is no H on the OH carbon.

Hydration is a fast, continuous process that gives pure ethanol directly, whereas fermentation is slow, works in batches and produces a dilute impure mixture that must be distilled.

1. When phenol loses H⁺, the negative charge is delocalised into the benzene ring[1]
2. this stabilises the phenoxide ion, so H⁺ is released — weakly acidic[1]
3. ethanol's anion is not stabilised in this way, so it does not release H⁺[1]
4. Add sodium carbonate solution to each[1]
5. the carboxylic acid fizzes, releasing CO₂[1]
6. phenol shows no reaction (too weak an acid)[1]

Phenol is acidic because the ring stabilises the phenoxide ion; ethanol's is not. Sodium carbonate fizzes with a carboxylic acid but not with phenol.

Both contain the carbonyl group C=O; in an aldehyde it is at the end of the chain (–CHO), in a ketone it is in the middle, bonded to two carbon groups.

Warm with Tollens' reagent: an aldehyde forms a silver mirror; a ketone gives no change.

Reduction (e.g. with NaBH₄) gives propan-1-ol; the carbonyl C=O is reduced to a C–OH group (a primary alcohol).

1. Propanal (aldehyde) gives a positive result[1]
2. the blue solution forms a brick-red precipitate[1]
3. Propanone (ketone) gives no change (stays blue)[1]
4. The characteristic reaction is nucleophilic addition[1]
5. The C=O bond is polar with a δ+ carbon[1]
6. so nucleophiles are attracted to and attack the carbonyl carbon[1]

Aldehyde → brick-red precipitate with Fehling's; ketone → stays blue; carbonyl undergoes nucleophilic addition at its δ+ carbon.

Reduction (e.g. with NaBH₄): propanal (an aldehyde) gives propan-1-ol (a primary alcohol); propanone (a ketone) gives propan-2-ol (a secondary alcohol).

1. Add 2,4-DNPH (Brady's reagent) to each[1]
2. both give an orange precipitate, confirming a carbonyl group[1]
3. Warm each with Tollens' reagent (ammoniacal silver nitrate)[1]
4. the aldehyde forms a silver mirror[1]
5. the ketone gives no change[1]
6. (alternatively Fehling's: aldehyde gives a brick-red precipitate, ketone stays blue)[1]

2,4-DNPH gives an orange precipitate with both (carbonyl present); Tollens' gives a silver mirror only with the aldehyde.

The carboxyl group, –COOH; carboxylic acids are weak acids (only partly ionised in water).

Sodium ethanoate, water and carbon dioxide are formed; effervescence (fizzing) is seen.

When a carboxylic acid loses H⁺, the negative charge on the carboxylate ion is delocalised over both oxygen atoms, stabilising it; an alcohol's anion is not stabilised in this way.

1. The reaction is esterification (condensation)[1]
2. The product is ethyl ethanoate (an ester)[1]
3. CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O[1]
4. balanced with water as the other product[1]
5. The sulfuric acid is a catalyst (and dehydrating agent)[1]
6. Esters are used as flavourings / perfumes / solvents (any one)[1]

Esterification → ethyl ethanoate; CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O; H₂SO₄ is the catalyst; esters used as flavourings/perfumes/solvents.

Propyl ethanoate, formed with a concentrated sulfuric acid catalyst (which also removes water).

1. CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O[1]
2. conditions: warm with a concentrated sulfuric acid catalyst[1]
3. the reaction is reversible[1]
4. (i) with dilute acid: ethanoic acid + ethanol[1]
5. (ii) with sodium hydroxide: sodium ethanoate[1]
6. plus ethanol[1]

CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O (conc. H₂SO₄); acid hydrolysis → acid + alcohol; alkaline hydrolysis → sodium ethanoate + ethanol.

The bond is polar; the halogen is more electronegative, leaving the carbon slightly positive (δ+), which attracts electron-pair-donating nucleophiles.

Ethanol is formed by nucleophilic substitution (OH⁻ replaces Br⁻).

The nitrogen atom has a lone pair of electrons which can accept a proton (H⁺), forming a positively charged ion; so amines are proton acceptors (bases).

1. Hot ethanolic (alcoholic) potassium hydroxide[1]
2. under reflux / heat[1]
3. This is elimination (loss of HBr forms the C=C)[1]
4. Excess ammonia[1]
5. (in a sealed tube / ethanolic solution)[1]
6. This is nucleophilic substitution[1]

→ ethene: hot alcoholic KOH, elimination; → ethylamine: excess ammonia, nucleophilic substitution.

The nitrogen lone pair can accept a proton (H⁺), so amines are bases. Ethylamine + HCl gives the salt ethylammonium chloride (C₂H₅NH₃⁺Cl⁻).

1. Warm aqueous sodium hydroxide (or water)[1]
2. gives ethanol[1]
3. Potassium cyanide (in ethanol) gives propanenitrile[1]
4. useful because it adds a carbon, lengthening the chain[1]
5. The C–I bond is weaker than C–Br[1]
6. so it breaks more easily, making iodoethane react faster[1]

OH⁻ → ethanol; CN⁻ → propanenitrile (adds a carbon); iodoethane is faster because C–I is weaker than C–Br.

1. In SN2 the nucleophile attacks the δ+ carbon as the halogen leaves[1]
2. in a single concerted step[1]
3. favoured by primary haloalkanes (the carbon is easy to approach)[1]
4. In SN1 the halogen leaves first to form a carbocation[1]
5. then the nucleophile attacks the carbocation[1]
6. tertiary haloalkanes form a stable tertiary carbocation, so this route is favoured[1]

SN2: one-step, nucleophile attacks as halide leaves, favoured by primary. SN1: two-step via a carbocation, favoured by tertiary (stable carbocation).

Compounds with the same molecular formula but different structural arrangements of their atoms.

Propan-1-ol; the functional group is the hydroxyl group, –OH (an alcohol).

In homolytic fission each atom takes one electron, forming two radicals. In heterolytic fission one atom takes both electrons, forming a positive and a negative ion.

1. An electrophile is an electron-pair acceptor (electron-poor species)[1]
2. A nucleophile is an electron-pair donor (electron-rich species)[1]
3. Bromine + ethene is an addition reaction[1]
4. Haloalkane with hot alcoholic KOH is an elimination reaction[1]
5. A functional group is the atom or group of atoms[1]
6. that gives a molecule its characteristic chemical properties[1]

Electrophile accepts an electron pair, nucleophile donates one; Br₂ + ethene = addition, elimination gives the alkene; functional group = the reactive part defining the chemistry.

But-2-ene has a C=C double bond, and there is no rotation about it, so the methyl groups are fixed either on the same side (cis) or opposite sides (trans). Butane has only single bonds, which rotate freely, so no fixed arrangement — and no cis–trans isomers.

1. A nucleophile is an electron-pair donor (electron-rich)[1]
2. an electrophile is an electron-pair acceptor (electron-poor)[1]
3. (i) Bromine acts as an electrophile[1]
4. in an electrophilic addition to the C=C double bond[1]
5. (ii) Hydroxide acts as a nucleophile[1]
6. in a nucleophilic substitution of the haloalkane[1]

Nucleophile = electron-pair donor, electrophile = acceptor; Br₂/ethene = electrophilic addition; OH⁻/bromoethane = nucleophilic substitution.

1. Step 1: hydrate ethene with steam and an acid catalyst to give ethanol[1]
2. (~300 °C, ~60 atm, phosphoric acid)[1]
3. Step 2: oxidise the ethanol with acidified potassium dichromate[1]
4. under reflux, to give ethanoic acid[1]
5. React bromoethane with potassium cyanide in ethanol to give propanenitrile[1]
6. useful because it adds a carbon, lengthening the chain[1]

Ethene → ethanol (steam + acid catalyst) → ethanoic acid (oxidise with acidified dichromate, reflux); bromoethane + KCN → propanenitrile (adds a carbon).

Carbohydrates (from simple sugars), proteins (from amino acids) and fats/lipids (from fatty acids and glycerol).

Add biuret reagent (sodium hydroxide then copper(II) sulfate solution); a positive result is a colour change from blue to purple/violet.

Starch is broken down (hydrolysed) into simple sugars such as glucose. Enzymes are biological catalysts that speed up this reaction at body temperature, which would otherwise be far too slow.

1. It speeds up a reaction[1]
2. without being used up / by lowering the activation energy[1]
3. As temperature rises up to the optimum, activity increases[1]
4. because particles collide more frequently and with more energy[1]
5. Above the optimum, activity falls sharply[1]
6. because the enzyme is denatured — its shape (active site) changes so substrate no longer fits[1]

Catalyst = speeds a reaction without being used up; activity rises to an optimum then falls as the enzyme denatures (active site changes shape).

Condensation joins two monomers together and releases a small molecule (usually water), building a polymer. Hydrolysis is the reverse: it adds water to split a polymer back into its monomers, as in digestion.

1. Starch: add iodine solution[1]
2. a blue-black colour is a positive result[1]
3. Protein: add biuret solution[1]
4. a purple/violet colour is a positive result[1]
5. A cloudy emulsion shows a lipid (fat) is present[1]
6. dissolve the sample in ethanol then add water — a cloudy white emulsion forms[1]

Starch → iodine blue-black; protein → biuret purple; lipid → cloudy emulsion (ethanol then water).

A zwitterion is an amino acid that has transferred a proton internally from its –COOH to its –NH₂, giving –NH₃⁺ and –COO⁻ ends and no overall charge. It can accept H⁺ (onto –COO⁻) in acid and donate H⁺ (from –NH₃⁺) in alkali, so it resists pH change — acting as a buffer.

Carbohydrates (from simple sugars), proteins (from amino acids) and fats/lipids (from fatty acids and glycerol).

Add biuret reagent (sodium hydroxide then copper(II) sulfate solution); a positive result is a colour change from blue to purple/violet.

Starch is broken down (hydrolysed) into simple sugars such as glucose. Enzymes are biological catalysts that speed up this reaction at body temperature, which would otherwise be far too slow.

1. It speeds up a reaction[1]
2. without being used up / by lowering the activation energy[1]
3. As temperature rises up to the optimum, activity increases[1]
4. because particles collide more frequently and with more energy[1]
5. Above the optimum, activity falls sharply[1]
6. because the enzyme is denatured — its shape (active site) changes so substrate no longer fits[1]

Catalyst = speeds a reaction without being used up; activity rises to an optimum then falls as the enzyme denatures (active site changes shape).

Condensation joins two monomers together and releases a small molecule (usually water), building a polymer. Hydrolysis is the reverse: it adds water to split a polymer back into its monomers, as in digestion.

1. Starch: add iodine solution[1]
2. a blue-black colour is a positive result[1]
3. Protein: add biuret solution[1]
4. a purple/violet colour is a positive result[1]
5. A cloudy emulsion shows a lipid (fat) is present[1]
6. dissolve the sample in ethanol then add water — a cloudy white emulsion forms[1]

Starch → iodine blue-black; protein → biuret purple; lipid → cloudy emulsion (ethanol then water).

A zwitterion is an amino acid that has transferred a proton internally from its –COOH to its –NH₂, giving –NH₃⁺ and –COO⁻ ends and no overall charge. It can accept H⁺ (onto –COO⁻) in acid and donate H⁺ (from –NH₃⁺) in alkali, so it resists pH change — acting as a buffer.

By simple distillation: heat the solution so water boils off, then condense the vapour back to liquid; the salt remains behind.

A pure substance produces a single spot on the chromatogram; a mixture separates into two or more spots.

The Rf value is the ratio of the distance moved by the spot to the distance moved by the solvent front. Rf = distance moved by spot ÷ distance moved by solvent.

1. Add water and stir to dissolve the copper(II) sulfate[1]
2. Filter — sand is the residue in the filter paper[1]
3. Wash and dry the sand[1]
4. Heat the filtrate to evaporate water to the point of crystallisation[1]
5. Leave to cool so crystals form[1]
6. Filter off and dry the crystals[1]

Dissolve, filter (sand = residue), wash and dry it; evaporate the filtrate to crystallisation, cool, filter and dry the crystals.

Dip a clean wire in the salt and hold it in a blue flame: sodium gives a yellow flame, potassium gives a lilac flame.

1. Dissolve the solid and add sodium hydroxide solution[1]
2. a green precipitate forms[1]
3. confirming the iron(II) ion, Fe²⁺[1]
4. To a fresh sample add dilute (hydrochloric or nitric) acid, then barium chloride solution[1]
5. a white precipitate forms[1]
6. confirming the sulfate ion, SO₄²⁻[1]

Fe²⁺: NaOH gives a green precipitate. SO₄²⁻: acid + barium chloride gives a white precipitate.

The mass/charge value of the molecular ion peak (highest m/z, ignoring isotope peaks) gives the relative molecular mass of the compound.

It identifies bonds/functional groups present; for example a strong absorption near 1700 cm⁻¹ indicates a C=O group (or a broad band ~3300 cm⁻¹ indicates O–H).

Number of peaks = number of different hydrogen environments; integration (area) = the ratio of hydrogens in each environment; splitting = the number of hydrogens on neighbouring carbons (n+1 lines for n neighbours).

1. Relative molecular mass = 46[1]
2. The broad ~3300 cm⁻¹ band suggests an O–H group (alcohol)[1]
3. Three peaks means three different hydrogen environments[1]
4. Ethanol, CH₃CH₂OH, has Mr 46[1]
5. It has three H environments (CH₃, CH₂, OH)[1]
6. so X is ethanol[1]

Mr = 46; O–H group present; three H environments (CH₃, CH₂, OH) → X is ethanol, CH₃CH₂OH.

Rf = distance moved by spot ÷ distance moved by solvent = 3.6 ÷ 4.8 = 0.75.

1. Relative molecular mass = 60[1]
2. 1710 cm⁻¹ suggests C=O and the broad 3000 cm⁻¹ suggests the O–H of a carboxylic acid[1]
3. Two peaks means two hydrogen environments[1]
4. A carboxylic acid of Mr 60 is ethanoic acid, CH₃COOH[1]
5. it has two H environments (CH₃ and COOH)[1]
6. and contains both C=O and O–H, matching the IR — so X is ethanoic acid[1]

Mr = 60; C=O + O–H (carboxylic acid); two H environments → X is ethanoic acid, CH₃COOH.

The mass/charge value of the molecular ion peak (highest m/z, ignoring isotope peaks) gives the relative molecular mass of the compound.

It identifies bonds/functional groups present; for example a strong absorption near 1700 cm⁻¹ indicates a C=O group (or a broad band ~3300 cm⁻¹ indicates O–H).

Number of peaks = number of different hydrogen environments; integration (area) = the ratio of hydrogens in each environment; splitting = the number of hydrogens on neighbouring carbons (n+1 lines for n neighbours).

1. Relative molecular mass = 46[1]
2. The broad ~3300 cm⁻¹ band suggests an O–H group (alcohol)[1]
3. Three peaks means three different hydrogen environments[1]
4. Ethanol, CH₃CH₂OH, has Mr 46[1]
5. It has three H environments (CH₃, CH₂, OH)[1]
6. so X is ethanol[1]

Mr = 46; O–H group present; three H environments (CH₃, CH₂, OH) → X is ethanol, CH₃CH₂OH.

Rf = distance moved by spot ÷ distance moved by solvent = 3.6 ÷ 4.8 = 0.75.

1. Relative molecular mass = 60[1]
2. 1710 cm⁻¹ suggests C=O and the broad 3000 cm⁻¹ suggests the O–H of a carboxylic acid[1]
3. Two peaks means two hydrogen environments[1]
4. A carboxylic acid of Mr 60 is ethanoic acid, CH₃COOH[1]
5. it has two H environments (CH₃ and COOH)[1]
6. and contains both C=O and O–H, matching the IR — so X is ethanoic acid[1]

Mr = 60; C=O + O–H (carboxylic acid); two H environments → X is ethanoic acid, CH₃COOH.

The heat energy change for a reaction measured at standard conditions (298 K, 1 atm), with substances in their standard states.

The total enthalpy change of a reaction is independent of the route taken, provided the initial and final conditions are the same. It lets us calculate enthalpy changes that cannot be measured directly.

q = mcΔT, where q = heat energy, m = mass of solution, c = specific heat capacity, ΔT = temperature change.

1. q = mcΔT = 100 × 4.2 × 30[1]
2. = 12600 J = 12.6 kJ[1]
3. Moles ethanol = 0.50 ÷ 46 = 0.0109 mol[1]
4. ΔH = energy ÷ moles = 12.6 ÷ 0.0109[1]
5. = 1156 kJ/mol (magnitude)[1]
6. ΔH is negative (exothermic): ≈ −1160 kJ/mol[1]
7. Heat is lost to the surroundings / incomplete combustion[1]

q = 12.6 kJ; 0.0109 mol; ΔH ≈ −1160 kJ/mol; lower magnitude due to heat loss / incomplete combustion.

Bonds broken = 436 + 242 = 678 kJ. Bonds made = 2 × 431 = 862 kJ. ΔH = 678 − 862 = −184 kJ/mol (exothermic).

1. The energy released when gaseous ions[1]
2. combine to form one mole of solid ionic lattice[1]
3. MgO contains Mg²⁺ and O²⁻ ions with 2+ and 2− charges[1]
4. NaCl contains only 1+ and 1− ions[1]
5. the ions in MgO are also smaller[1]
6. higher charges and smaller radii give stronger electrostatic attraction, so a much larger (more exothermic) lattice energy[1]

Lattice energy = energy released forming a mole of lattice from gaseous ions; MgO's higher charges and smaller ions give much stronger attraction than NaCl.

Nitrogen (from air) and hydrogen (from natural gas / methane): N₂ + 3H₂ ⇌ 2NH₃.

An iron catalyst at about 450 °C (and about 200 atm pressure).

A low temperature would favour a higher yield of ammonia (exothermic forward reaction) but the reaction would be far too slow; a high temperature would be fast but give a low yield. 450 °C is a compromise giving an acceptable yield at a workable rate.

1. To provide nitrogen for plant growth / to increase crop yield[1]
2. The salt is ammonium nitrate[1]
3. NH₃ + HNO₃ → NH₄NO₃ (correct formulae)[1]
4. balanced equation[1]
5. Fertiliser washes into rivers/lakes (leaching)[1]
6. causing eutrophication — algal growth, then oxygen depletion killing aquatic life[1]

Fertilisers supply nitrogen for growth; NH₃ + HNO₃ → NH₄NO₃ (ammonium nitrate); overuse causes eutrophication in waterways.

Some ammonia is oxidised (the Ostwald process) to make nitric acid; the nitric acid is then neutralised with more ammonia: NH₃ + HNO₃ → NH₄NO₃, giving ammonium nitrate.

1. Nitrogen from the air[1]
2. hydrogen from natural gas (methane) / steam[1]
3. N₂ + 3H₂ ⇌ 2NH₃[1]
4. The reaction reaches equilibrium, so only some reactants convert each pass[1]
5. recycling unreacted gases avoids wasting them[1]
6. which raises the overall yield and lowers cost[1]

N₂ (air) + H₂ (natural gas) ⇌ 2NH₃; unreacted gases are recycled because equilibrium limits each pass, improving overall yield and cutting waste.

About 78% nitrogen and about 21% oxygen (the remaining ~1% is mostly argon with a little carbon dioxide).

Add anhydrous copper(II) sulfate — it turns from white to blue in the presence of water (or blue cobalt chloride paper turns pink).

Carbon dioxide is a greenhouse gas; it absorbs and re-emits heat (infrared radiation) that would otherwise escape, so the extra CO₂ traps more heat and raises average global temperatures.

1. Water (moisture)[1]
2. Oxygen (air)[1]
3. Galvanising coats the iron with a layer of zinc[1]
4. The zinc is more reactive than iron, so it corrodes/loses electrons in preference (sacrificial protection)[1]
5. so the iron is protected even where the coating is scratched[1]
6. Painting / oiling / greasing / coating with plastic (any one)[1]

Water + oxygen needed; galvanising = zinc coating that sacrificially protects (zinc more reactive) even if scratched; also painting/oiling.

Sulfur impurities in the coal burn to sulfur dioxide, which dissolves in rainwater to form an acidic solution (sulfurous/sulfuric acid). This acid rain damages plants and forests, acidifies lakes killing fish, or corrodes limestone buildings and metals.

1. Carbon monoxide forms by incomplete combustion[1]
2. when there is not enough oxygen to make CO₂[1]
3. it is toxic because it prevents the blood from carrying oxygen[1]
4. The converter provides a catalyst surface where pollutants react[1]
5. carbon monoxide is oxidised to carbon dioxide[1]
6. 2CO + O₂ → 2CO₂ (and NOₓ is reduced to N₂)[1]

CO from incomplete combustion, toxic (stops blood carrying O₂); catalytic converter oxidises CO to CO₂ (2CO + O₂ → 2CO₂) and reduces NOₓ to N₂.