Unit 3: Inorganic 2 and Physical 2 Flashcards

The enthalpy change when one mole of a solid ionic lattice is completely dissociated into its constituent gaseous ions under standard conditions.
NaCl(s) → Na⁺(g) + Cl⁻(g) (always endothermic, ΔH > 0).

The enthalpy change when one mole of a solid ionic lattice is formed from its constituent gaseous ions under standard conditions.
Na⁺(g) + Cl⁻(g) → NaCl(s) (always exothermic, ΔH < 0).

The enthalpy change when one mole of specified gaseous ions is completely dissolved in water to form one mole of aqueous ions under standard conditions.
Mg²⁺(g) + aq → Mg²⁺(aq) (always exothermic).

Δsol H = (Lattice Dissociation Enthalpy) + Σ(Δhyd H of all ions).
Enthalpy of solution alone does not determine solubility or dissolution rate; overall thermodynamic feasibility is governed by Gibbs free energy (ΔG = ΔH − TΔS), where a positive entropy of solution (ΔS) can drive dissolution even when Δsol H is endothermic. Dissolution rate is governed by kinetic factors (activation energy).

ΔG = ΔH − TΔS.
A reaction is thermodynamically feasible when ΔG ≤ 0.
Unit trap: ΔH is usually in kJ mol⁻¹, but ΔS is in J K⁻¹ mol⁻¹ (must divide ΔS by 1000).

At the threshold of feasibility, ΔG = 0.
Therefore, T = ΔH / ΔS (with ΔH and ΔS expressed in identical energy units, usually J mol⁻¹).

The theoretical model assumes purely spherical, unpolarised ions with point charges. In AgCl, the small Ag⁺ cation polarises the large, electron-rich Cl⁻ anion, introducing significant covalent character that strengthens the lattice beyond pure ionic attraction.

1. Platinum foil electrode coated with porous platinum black.
2. Pure hydrogen gas at a pressure of 100 kPa.
3. Aqueous solution of H⁺ ions at a concentration of 1.00 mol dm⁻³ at 298 K (25 °C). Potential is defined as exactly 0.00 V.

A strip of filter paper soaked in an unreactive electrolyte solution (such as saturated KNO₃ or NH₄NO₃). It completes the electrical circuit by allowing ions to flow between half-cells to maintain electrical neutrality, without allowing bulk solution mixing.

E°cell = E°(Right / Cathode / Reduction) - E°(Left / Anode / Oxidation).
A positive E°cell indicates the cell reaction is thermodynamically feasible under standard conditions.

A proton (H⁺) donor that is only partially dissociated (or ionised) in aqueous solution.
Example: CH₃COOH(aq) ⇌ CH3COO⁻(aq) + H⁺(aq).

Ka = ([H⁺][A⁻]) / [HA].
Units: mol dm⁻³.
A larger Ka value corresponds to a stronger weak acid (lower pKa, where pKa = -log₁₀ Ka).

1. [H⁺] = [A⁻] (all H⁺ ions originate from acid dissociation; water self-ionisation is negligible).
2. [HA]equilibrium = [HA]initial (the extent of acid dissociation is so small that initial concentration is effectively unchanged).
Formula: [H⁺] = √(Ka × [HA]).

A solution containing a weak acid (HA) and its conjugate base salt (A⁻).
• On adding H⁺: A⁻ + H⁺ → HA.
• On adding OH⁻: HA + OH⁻ → A⁻ + H₂O.
Because the ratio [HA]/[A⁻] changes minimally, pH remains virtually constant.

[H⁺] = Ka × ([HA] / [A⁻]) or [H⁺] = Ka × (moles of acid / moles of salt).
Volumes cancel because both components are in the same total solution volume.

The indicator must undergo its complete colour change strictly within the steep, vertical section of the titration curve (the rapid pH transition encompassing the equivalence point).
• Strong acid - strong base: Methyl orange or phenolphthalein.
• Weak acid - strong base: Phenolphthalein (pH 8.3 - 10.0).
• Strong acid - weak base: Methyl orange (pH 3.1 - 4.4).
• Weak acid - weak base: No indicator suitable (use a pH meter).

• Na₂O, MgO: Giant ionic lattices with ionic bonding.
• Al₂O₃: Giant ionic lattice with significant covalent character (polarisation).
• SiO₂: Giant covalent macromolecular structure with strong covalent bonds.
• P₄O₁₀, SO₂, SO₃: Simple molecular structures with covalent bonding and weak intermolecular forces.

Na₂O(s) + H₂O(l) → 2NaOH(aq) (or Na₂O + H₂O → 2Na⁺ + 2OH⁻).
Resulting pH: 14 (strongly alkaline due to complete dissociation of NaOH).

P₄O₁₀(s) + 6H₂O(l) → 4H₃PO₄(aq).
Resulting pH: 0 to 1 (strongly acidic due to formation of phosphoric(V) acid).

(a) With acid: Al₂O₃(s) + 6HCl(aq) → 2AlCl₃(aq) + 3H₂O(l).
(b) With alkali: Al₂O₃(s) + 2NaOH(aq) + 3H₂O(l) → 2Na[Al(OH)₄](aq).

SiO₂ has a giant covalent macromolecular lattice. The many strong silicon-oxygen covalent bonds require too much energy to overcome, which cannot be compensated by weak hydration interactions with water. However, it reacts as an acidic oxide with hot concentrated base: SiO₂(s) + 2NaOH(aq) → Na₂SiO₃(aq) + H₂O(l).

A d-block element that forms at least one stable ion with a partially filled (incomplete) d-subshell.
• Sc forms only Sc³⁺ ([Ar] 3d⁰), having an empty d-subshell.
• Zn forms only Zn²⁺ ([Ar] 3d¹⁰), having a completely filled d-subshell.

1. Ligand coordination causes degenerate 3d orbitals to split into two energy levels separated by energy gap ΔE.
2. Visible light photons are absorbed to promote a d-electron from lower to higher d-orbitals: ΔE = hν = hc / λ.
3. The unabsorbed wavelengths of visible light are transmitted or reflected to produce the complementary observed colour.

A ligand that donates two lone pairs of electrons from different atoms to a central metal ion to form two coordinate (dative) bonds.
• Neutral: 1,2-diaminoethane (ethane-1,2-diamine), H₂NCH₂CH₂NH₂.
• Anionic: Ethanedioate (oxalate) ion, C₂O₄²⁻.

[Co(H₂O)6]2+(aq) + 4Cl⁻(aq) ⇌ [CoCl₄]²⁻(aq) + 6H₂O(l).
• Geometry: Octahedral → Tetrahedral.
• Coordination number: 6 → 4 (due to larger chloride ligands and steric repulsion).
• Colour change: Pink solution → Deep blue solution.

3+ metal ions have a smaller ionic radius and higher charge than 2+ ions, resulting in a much higher charge density. The 3+ ion strongly polarises the coordinated water molecules, pulling electron density away from the oxygen atom and weakening the O−H bonds, facilitating H⁺ release into solution.

• Dropwise: Blue precipitate of copper(II) hydroxide forms: [Cu(H₂O)6]2+ + 2NH₃ → [Cu(H₂O)4(OH)2] + 2NH₄⁺.
• Excess: Precipitate dissolves to form a deep blue solution via ligand substitution: [Cu(H₂O)4(OH)2] + 4NH₃ → [Cu(NH₃)4(H₂O)2]2+ + 2OH⁻ + 2H₂O.

• Dropwise: White precipitate of aluminium hydroxide forms: [Al(H₂O)6]3+ + 3OH⁻ → [Al(H₂O)3(OH)3] + 3H₂O.
• Excess: White precipitate dissolves to form a clear, colourless solution of tetrahydroxidoaluminate: [Al(H₂O)3(OH)3] + OH⁻ → [Al(OH)₄]⁻ + 3H₂O.

• 2+ ions: Form insoluble metal carbonates without gas evolution: M²⁺(aq) + CO₃²⁻(aq) → MCO₃(s) (FeCO₃ green precipitate; CuCO₃ blue-green precipitate).
• 3+ ions: Due to high acidity, carbonate acts as a base; forms metal hydroxide precipitate and CO₂ gas effervescence: 2[M(H₂O)6]3+ + 3CO₃²⁻ → 2[M(H₂O)3(OH)3](s) + 3CO₂(g) + 3H₂O(l).

• Fe²⁺: Forms a green precipitate of iron(II) hydroxide, turning brown at the surface on air exposure (oxidised to Fe(III)): [Fe(H₂O)6]2+ + 2OH⁻ → [Fe(H₂O)4(OH)2](s) + 2H₂O.
• Fe³⁺: Forms an orange-brown precipitate of iron(III) hydroxide immediately: [Fe(H₂O)6]3+ + 3OH⁻ → [Fe(H₂O)3(OH)3](s) + 3H₂O.
Neither precipitate dissolves in excess NaOH.

A proton donor that only partially dissociates in aqueous solution.
HA(aq) ⇌ H⁺(aq) + A⁻(aq).
Ka = [H⁺][A⁻] / [HA] (units: mol dm⁻³).

Water self-ionisation is endothermic: H₂O ⇌ H⁺ + OH⁻. Heating increases Kw to 5.48 × 10^-14. [H⁺] increases, so pH decreases to 6.63. However, [H⁺] strictly equals [OH⁻], so water remains strictly neutral.

Assuming [H⁺] = [A⁻] and [HA]eq ≈ [HA]initial:
[H⁺] = √(Ka × [HA]), then pH = -log₁₀[H⁺] (quote to 2 d.p.).

Added OH⁻ reacts with the reservoir of un-dissociated weak acid HA:
HA + OH⁻ → A⁻ + H₂O.
The ratio [HA]/[A⁻] changes very little, so [H⁺] and pH remain nearly constant.

[H⁺] = Ka × ([HA] / [A⁻]) or [H⁺] = Ka × (moles HA / moles A⁻).
Volumes cancel because both are in the same mixture.

The indicator pH transition range must fall entirely within the steep vertical region of the titration curve.
• Strong acid - strong base: Methyl orange or phenolphthalein.
• Weak acid - strong base: Phenolphthalein (pH 8.3 - 10.0).
• Strong acid - weak base: Methyl orange (pH 3.1 - 4.4).

3+ ions have smaller radius and higher charge, giving higher charge density. They strongly polarise coordinated water molecules, weakening O−H bonds and releasing H⁺: [M(H₂O)6]3+ + H₂O ⇌ [M(H₂O)5(OH)]2+ + H₃O⁺.

• Dropwise: Blue precipitate: [Cu(H₂O)6]2+ + 2NH₃ → [Cu(H₂O)4(OH)2] + 2NH₄⁺.
• Excess: Deep blue solution: [Cu(H₂O)4(OH)2] + 4NH₃ → [Cu(NH₃)4(H₂O)2]2+ + 2OH⁻ + 2H₂O.

• Dropwise: White precipitate: [Al(H₂O)6]3+ + 3OH⁻ → [Al(H₂O)3(OH)3] + 3H₂O.
• Excess: Colourless solution: [Al(H₂O)3(OH)3] + OH⁻ → [Al(OH)₄]⁻ + 3H₂O.

• 2+ ions (Cu²⁺, Fe²⁺): Form insoluble metal carbonate precipitates without effervescence (e.g. CuCO₃, FeCO₃).
• 3+ ions (Fe³⁺, Al³⁺): High acidity causes carbonate to act as a base, forming metal hydroxide precipitate and CO₂ gas effervescence.

• Fe²⁺: Green precipitate of [Fe(H₂O)₄(OH)₂], oxidising to brown Fe(III) on contact with air.
• Fe³⁺: Orange-brown precipitate of [Fe(H₂O)₃(OH)₃] immediately.
Neither precipitate dissolves in excess NaOH.

• Temperature: 298 K (25 °C).
• Pressure: 100 kPa of pure H₂(g).
• Concentration: 1.00 mol dm⁻³ H⁺(aq).
• Electrode: Platinum foil coated with platinum black. Potential = 0.00 V.

Oxidation half-cell (anode, more negative E°) on the left; reduction half-cell (cathode, more positive E°) on the right.
Phase boundaries are vertical lines (|) and the salt bridge is double lines (||).
Example: Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s).

It completes the circuit and maintains electrical neutrality by allowing ions to migrate between half-cells without allowing bulk mixing of electrolyte solutions.

E°cell = E°(Right / Reduction) - E°(Left / Oxidation).
A reaction is feasible under standard conditions if E°cell > 0.

Equilibrium: Cu²⁺(aq) + 2e⁻ ⇌ Cu(s).
Increasing [Cu²⁺] shifts equilibrium to the right, increasing the tendency to gain electrons, so the electrode potential becomes more positive.

Anode: 2H₂ + 4OH⁻ → 4H₂O + 4e⁻.
Cathode: O₂ + 2H₂O + 4e⁻ → 4OH⁻.
Overall: 2H₂ + O₂ → 2H₂O (E° = +1.23 V).

• Na₂O, MgO: Giant ionic lattices.
• Al₂O₃: Giant ionic with covalent character (polarisation).
• SiO₂: Giant covalent macromolecular.
• P₄O₁₀, SO₂, SO₃: Simple molecular with covalent bonds and weak van der Waals forces.

• Na₂O(s) + H₂O(l) → 2NaOH(aq) (pH 14, soluble).
• MgO(s) + H₂O(l) → Mg(OH)2 (pH 9 to 10, sparingly soluble).

• P₄O₁₀ + 6H₂O → 4H₃PO₄ (pH 0 to 1, phosphoric acid).
• SO₂ + H₂O → H₂SO₃ (pH 2 - 3, sulfurous acid, weak).
• SO₃ + H₂O → H₂SO₄ (pH 0 - 1, sulfuric acid, strong).

It reacts with both acids and bases:
• With acid: Al₂O₃ + 6HCl → 2AlCl₃ + 3H₂O.
• With hot alkali: Al₂O₃ + 2NaOH + 3H₂O → 2Na[Al(OH)₄].

SiO₂ has a giant covalent lattice; breaking many strong Si−O bonds requires too much energy. However, it reacts as an acidic oxide with hot concentrated base to form silicate: SiO₂ + 2NaOH → Na₂SiO₃ + H₂O.

The enthalpy change when one mole of a solid ionic lattice is completely broken into its constituent gaseous ions under standard conditions.
NaCl(s) → Na⁺(g) + Cl⁻(g) (ΔH > 0, always endothermic).

The enthalpy change when one mole of a solid ionic lattice is formed from its constituent gaseous ions under standard conditions.
Na⁺(g) + Cl⁻(g) → NaCl(s) (ΔH < 0, always exothermic).

The enthalpy change when one mole of gaseous atoms is formed from an element in its standard state under standard conditions.
½ Cl₂(g) → Cl(g) (ΔH > 0, always endothermic).

The 1st EA is exothermic because the incoming electron is attracted by the positive nuclear charge of the neutral atom.
The 2nd EA is endothermic because energy is required to overcome electrostatic repulsion between the negative ion (e.g. O⁻) and the incoming electron.

The theoretical model assumes purely spherical, unpolarised ions with point charges. In AgCl, the small Ag⁺ ion polarises the large Cl⁻ ion, introducing covalent character that strengthens the lattice beyond pure ionic forces.

ΔG = ΔH − TΔS.
A process is feasible when ΔG ≤ 0.
Unit warning: Convert ΔS from J K⁻¹ mol⁻¹ to kJ K⁻¹ mol⁻¹ by dividing by 1000.

Set ΔG = 0: 0 = ΔH − TΔS ⇒ T = ΔH / ΔS (ensuring ΔH and ΔS are in matching energy units).

A d-block element that forms at least one stable ion with a partially filled (incomplete) d-subshell.
Sc ([Ar] 3d⁰ in Sc³⁺) and Zn ([Ar] 3d¹⁰ in Zn²⁺) are not transition metals.

Ligand coordination splits degenerate 3d orbitals into two energy levels with gap ΔE.
Visible photons are absorbed to promote a d-electron (ΔE = hν = hc / λ).
The complementary, transmitted wavelengths produce the observed colour.

A ligand that donates two lone pairs of electrons from different atoms to form two coordinate bonds to a metal ion.
• Neutral: 1,2-diaminoethane (H₂NCH₂CH₂NH₂).
• Charged: Ethanedioate (C₂O₄²⁻).

Multidentate ligands replace monodentate ligands to form more stable complexes.
Example: [Cu(H₂O)6]2+ + EDTA⁴⁻ → [Cu(EDTA)]²⁻ + 6H₂O.
Number of particles increases (2 → 7), so ΔS > 0. Since ΔH ≈ 0, ΔG becomes more negative.

[Co(H₂O)6]2+(aq) + 4Cl⁻(aq) ⇌ [CoCl₄]²⁻(aq) + 6H₂O(l).
• Pink octahedral complex → Deep blue tetrahedral complex.
• Coordination number changes from 6 to 4.

• Heterogeneous: Catalyst is in a different phase from reactants (e.g. Fe(s) in Haber process).
• Homogeneous: Catalyst is in the same phase as reactants (e.g. Fe²⁺(aq) in persulfate-iodide reaction).

2Na + 2H₂O → 2NaOH + H₂
Vigorous: sodium floats, melts and fizzes. pH about 14.

Cold water: Mg + 2H₂O → Mg(OH)₂ + H₂. Very slow, pH about 10.
Steam: Mg + H₂O(g) → MgO + H₂.

Na: yellow flame, white solid (Na₂O).
Mg: bright white flame (MgO).
P: white flame, white smoke (P₄O₁₀).
S: blue flame (SO₂).

Sulfur burns to SO₂ (S + O₂ → SO₂).
Then 2SO₂ + O₂ ⇌ 2SO₃ with a V₂O₅ catalyst.

2Na + Cl₂ → 2NaCl
Mg + Cl₂ → MgCl₂
2Al + 3Cl₂ → Al₂Cl₆
Si + 2Cl₂ → SiCl₄
2P + 5Cl₂ → 2PCl₅ (excess chlorine)

NaCl: giant ionic lattice, strong attraction between oppositely charged ions.
SiCl₄: simple molecules held by weak van der Waals forces.
Much less energy is needed to separate SiCl₄ molecules.

Covalent. It sublimes at about 180 °C.
In the vapour it exists as Al₂Cl₆ dimers, with two Cl→Al dative (co-ordinate) bonds.

SiCl₄ + 2H₂O → SiO₂ + 4HCl
White solid and steamy fumes. pH about 1 to 2.

PCl₅ + 4H₂O → H₃PO₄ + 5HCl
Steamy fumes. pH about 1.

Ionic chlorides just dissolve.
Covalent chlorides are hydrolysed: water attacks the central atom (Si and P can accept a lone pair), and HCl forms.

[Al(H₂O)₆]³⁺ forms. The small, highly charged Al³⁺ polarises a water ligand, which releases H⁺:
[Al(H₂O)₆]³⁺ + H₂O ⇌ [Al(H₂O)₅(OH)]²⁺ + H₃O⁺

Na₂O 14; MgO 9 to 10; Al₂O₃ and SiO₂ 7 (insoluble); P₄O₁₀ 0 to 1; SO₂ 2 to 3; SO₃ 0 to 1.

Both have P at the centre with one P=O double bond.
H₃PO₄: three P−O−H groups.
PO₄³⁻: three P−O⁻ groups.

H₂SO₄: S with two S=O double bonds and two S−O−H groups.
H₂SO₃: S with one S=O double bond, two S−O−H groups and a lone pair on S.

Al₂O₃ + 6HCl → 2AlCl₃ + 3H₂O
Al₂O₃ + 2NaOH + 3H₂O → 2NaAl(OH)₄

VO₂⁺: +5, yellow
VO²⁺: +4, blue
V³⁺: +3, green
V²⁺: +2, violet

It is a mixture of yellow VO₂⁺ and blue VO²⁺, not V³⁺.

VO₂⁺ + 2H⁺ + e⁻ → VO²⁺ + H₂O

E° for Zn²⁺/Zn is −0.76 V. This is more negative than E° for each vanadium step, so zinc can reduce each one.

In alkaline solution, metal ions are oxidised more easily.
Green Fe(OH)₂ darkens to brown Fe(OH)₃ in air.
[Co(NH₃)₆]²⁺ is oxidised by air to [Co(NH₃)₆]³⁺.

[Ag(NH₃)₂]⁺, a mild oxidising agent.
It oxidises aldehydes but not ketones, and is reduced to silver:
[Ag(NH₃)₂]⁺ + e⁻ → Ag + 2NH₃

MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O
Mole ratio MnO₄⁻ : Fe²⁺ = 1 : 5

2MnO₄⁻ + 16H⁺ + 5C₂O₄²⁻ → 2Mn²⁺ + 10CO₂ + 8H₂O
Warm to about 60 °C. Mole ratio 2 : 5.

The first permanent pale pink colour. The titration is self-indicating.

HCl: MnO₄⁻ oxidises Cl⁻ to Cl₂, so the titre is too large.
Ethanoic acid: weak, so not enough H⁺.
HNO₃: an oxidising agent itself.

A catalyst in a different phase from the reactants; the reaction happens at active sites on its surface.
A ceramic honeycomb support gives a large surface area using little expensive metal.

Impurities block the active sites.
Sulfur poisons the iron catalyst in the Haber process. Lead poisons catalytic converters.
The catalyst must be cleaned or replaced, which costs money.

SO₂ + V₂O₅ → SO₃ + V₂O₄
V₂O₄ + ½O₂ → V₂O₅
V changes between +5 and +4.

A catalyst in the same phase as the reactants. The reaction goes through an intermediate species.

Uncatalysed: two negative ions repel, so it is slow.
S₂O₈²⁻ + 2Fe²⁺ → 2SO₄²⁻ + 2Fe³⁺
2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂
Each step is between oppositely charged ions.

Mn²⁺ (a product) is the catalyst:
4Mn²⁺ + MnO₄⁻ + 8H⁺ → 5Mn³⁺ + 4H₂O
2Mn³⁺ + C₂O₄²⁻ → 2Mn²⁺ + 2CO₂
Rate starts slow, speeds up as Mn²⁺ forms, then slows as reactants run out (S-shaped graph).

[Cu(H₂O)₆]²⁺ + 4NH₃ → [Cu(NH₃)₄(H₂O)₂]²⁺ + 4H₂O
Blue to deep blue solution. Substitution is incomplete; co-ordination number stays 6.

[Co(H₂O)₆]²⁺ + 6NH₃ → [Co(NH₃)₆]²⁺ + 6H₂O
Pink to pale brown (straw) solution, which darkens in air (oxidised to Co³⁺).

Cl⁻ is larger than H₂O, so only four fit around the metal ion. The shape changes from octahedral to tetrahedral.

[Cu(H₂O)₆]²⁺ + 4Cl⁻ → [CuCl₄]²⁻ + 6H₂O: blue to yellow/green
[Co(H₂O)₆]²⁺ + 4Cl⁻ → [CoCl₄]²⁻ + 6H₂O: pink to blue

[Fe(H₂O)₆]³⁺ + 4Cl⁻ → [FeCl₄]⁻ + 6H₂O
The solution turns yellow.

[Cu(H₂O)₆]²⁺ + 3H₂NCH₂CH₂NH₂ → [Cu(H₂NCH₂CH₂NH₂)₃]²⁺ + 6H₂O
Each H₂NCH₂CH₂NH₂ is bidentate.

Haem has Fe(II) held by a multidentate porphyrin ligand. O₂ forms a co-ordinate bond to Fe(II).
CO binds more strongly and does not readily come off, so less O₂ is carried.

[Cu(H₂O)₆]²⁺: octahedral, 90°
[CuCl₄]²⁻: tetrahedral, 109.5°
Cisplatin [Pt(NH₃)₂Cl₂]: square planar, 90°
[Ag(NH₃)₂]⁺: linear, 180°

Cisplatin is the cis isomer of square planar [Pt(NH₃)₂Cl₂]: the two Cl ligands are at 90°.
In the trans isomer they are opposite each other, at 180°.

It is octahedral. Cis: the two Cl ligands are at 90°. Trans: the two Cl ligands are at 180°.

Complexes with three bidentate ligands, for example [Ni(H₂NCH₂CH₂NH₂)₃]²⁺. They form non-superimposable mirror images.

A change in oxidation state, co-ordination number or ligand.

ΔE = hc/λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / (600 × 10⁻⁹) = 3.32 × 10⁻¹⁹ J
× 6.02 × 10²³ = 200 kJ mol⁻¹

1. Choose a filter of the colour that is absorbed.
2. Measure the absorbance of standard solutions and plot a calibration curve.
3. Read off the unknown.
4. If the colour is pale, add a ligand to intensify it.

Mg needs the first AND second ionisation energies. Chlorine needs 2 × the enthalpy of atomisation and 2 × the electron affinity, because there are two Cl⁻ ions.

Sum of steps = 148 + 738 + 1451 + 244 − 698 = +1883. Lattice formation = −642 − 1883 = −2525 kJ mol⁻¹.

The enthalpy change when one mole of a covalent bond is broken, with the molecule and the fragments all in the gaseous state. Example: Cl₂(g) → 2Cl(g), +244 kJ mol⁻¹.

The bonding has some covalent character. Example: AgCl, ionic model −833 kJ mol⁻¹ but Born-Haber −905 kJ mol⁻¹. Ag⁺ polarises the Cl⁻ electron cloud.

ΔsolH = 711 − 322 − 364 = +25 kJ mol⁻¹ (endothermic). The measured value is about +17 kJ mol⁻¹; data sources vary slightly.

ΔS = 386 − 585 = −199 J K⁻¹ mol⁻¹. T = −92 000 / −199 = 462 K. Feasible below 462 K. Remember to convert J to kJ (or kJ to J).

Write each half-equation as a reduction (electrons on the left) with ⇌. The more positive E°, the stronger the oxidising agent on the left.

Yes. Fe³⁺/Fe²⁺ is more positive, so Fe³⁺ is reduced. 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂. EMF = 0.77 − 0.54 = +0.23 V.

Positive electrode: Li⁺ + CoO₂ + e⁻ → Li⁺[CoO₂]⁻. Negative electrode: Li → Li⁺ + e⁻. Recharging reverses these reactions.

Negative: H₂ + 2OH⁻ → 2H₂O + 2e⁻ (E° −0.83 V). Positive: O₂ + 2H₂O + 4e⁻ → 4OH⁻ (E° +0.40 V). Overall 2H₂ + O₂ → 2H₂O, EMF 1.23 V. No recharging: fuel is supplied continuously.

Benefit: light and rechargeable. Risk: the electrolyte is flammable, so damaged or overheated cells can catch fire.

[H⁺] = 1.00 × 10⁻¹⁴ / 0.0500 = 2.00 × 10⁻¹³ mol dm⁻³, so pH = 12.70.

Excess H⁺ = 0.00250 − 0.00200 = 5.00 × 10⁻⁴ mol in 45.0 cm³. [H⁺] = 0.0111 mol dm⁻³, so pH = 1.95.

Weak acid + strong base: phenolphthalein (8.3 to 10.0). Strong acid + weak base: methyl orange (3.1 to 4.4). Strong + strong: either. Weak + weak: none, as there is no sharp vertical section.

Added H⁺: NH₃ + H⁺ → NH₄⁺. Added OH⁻: NH₄⁺ + OH⁻ → NH₃ + H₂O. The ratio of NH₃ to NH₄⁺ hardly changes, so the pH hardly changes.

[HA] and [A⁻] fall by the same factor, so the ratio [HA]/[A⁻] and [H⁺] stay the same. Uses: blood (carbonic acid / hydrogencarbonate), shampoos, food, biological and industrial processes.