Physical chemistry I: A-level Chemistry AQA revision notes

3.1.1

Atomic structure

Protons, neutrons, electrons and isotopes 3.1.1.1–3.1.1.2

Definitions

Atomic (proton) number, $Z$ : the number of protons in the nucleus; it defines the element.
Mass number, $A$ : the number of protons plus neutrons in the nucleus.
Isotopes: atoms of the same element (same $Z$ ) with different numbers of neutrons, and so different mass numbers.

Key results

Relative charges: proton $+ 1$ , neutron $0$ , electron $- 1$ . Relative masses: proton $1$ , neutron $1$ , electron about $1/1836$ .
Neutrons $= A - Z$ ; for an ion, charge $=$ protons $-$ electrons. So $_{6}^{13} C^{+}$ has 6 protons, 7 neutrons and 5 electrons.

Notes

Almost all of an atom's mass is in a nucleus about $1 0^{- 5}$ times the atom's diameter. The diagram enlarges the nucleus enormously so it can be seen at all.
Electrons are drawn as a cloud because they occupy orbitals (regions of probability), not fixed circular tracks.
Isotopes of an element have the same electron configuration, so they have the same chemical properties. Their physical properties that depend on mass (density, rate of diffusion) differ slightly.
The model of the atom changed as evidence accumulated: Dalton's indivisible spheres, Thomson's 'plum pudding' after the discovery of the electron, Rutherford's nuclear atom after alpha particles were scattered through large angles by gold foil, and Bohr's fixed energy levels to explain line spectra. Each model was kept only as long as it explained the data.
Mass number is a count of particles and is always a whole number; relative isotopic mass is a measured mass relative to carbon-12 and is not exactly whole (e.g. $^{35} Cl$ is 34.969). Exam questions using 'relative isotopic mass' expect the measured quantity.
For an ion, change only the electron count: $^{32} S^{2 -}$ has 16 protons, 16 neutrons and 18 electrons. A common error is to change the proton number, which would make a different element.

Shells, subshells and orbitals 3.1.1.3

Definitions

Orbital: a region around the nucleus that can hold up to two electrons, with opposite spins.
Subshell: a set of orbitals of equal energy within a shell: s (1 orbital, 2 electrons), p (3 orbitals, 6), d (5 orbitals, 10).

Key results

Filling order up to $Z = 36$ : $1 s 2 s 2 p 3 s 3 p 4 s 3 d 4 p$ . Oxygen is $1 s^{2} 2 s^{2} 2 p^{4}$ .
Exceptions: chromium is $[Ar] 3 d^{5} 4 s^{1}$ and copper is $[Ar] 3 d^{10} 4 s^{1}$ .
Transition-metal ions lose their 4s electrons before 3d: Fe is $[Ar] 3 d^{6} 4 s^{2}$ but $Fe^{2 +}$ is $[Ar] 3 d^{6}$ .

Notes

Within a subshell, orbitals fill singly before electrons pair up, because paired electrons in the same orbital repel. This is why oxygen's 2p electrons are drawn as one pair and two single electrons.
4s fills before 3d because, in the neutral atom, 4s is slightly lower in energy. Once 3d is occupied the order of the two levels changes, which is why 4s electrons are the first removed on ionisation.
Chromium and copper adopt a half-filled or full 3d subshell, which has lower energy overall than the 'expected' $3 d^{4} 4 s^{2}$ or $3 d^{9} 4 s^{2}$ .
Each orbital type has a characteristic shape: an s orbital is spherical; the three p orbitals are dumbbells along the x, y and z axes. You are expected to recognise these shapes, not d-orbital shapes, in this topic.
The block of the Periodic Table matches the subshell being filled: Groups 1–2 are the s block, Groups 3–0 (13–18) the p block, and the transition metals the d block. So an element's position gives its outer configuration directly.
Write ions carefully: $Cu^{2 +}$ is $[Ar] 3 d^{9}$ (both the 4s electron and one 3d electron are removed), and $Cr^{3 +}$ is $[Ar] 3 d^{3}$ . Writing $4 s$ electrons in a transition-metal ion is a frequent mark loser.
Ions of main-group elements usually reach a noble-gas configuration: $Na^{+}$ , $Mg^{2 +}$ , $F^{-}$ and $O^{2 -}$ are all $1 s^{2} 2 s^{2} 2 p^{6}$ , isoelectronic with neon.

Ionisation energies reveal electron arrangement 3.1.1.3

Definitions

First ionisation energy: the energy needed to remove one electron from each atom in one mole of gaseous atoms, forming one mole of gaseous $1 +$ ions: $X(g) \to X^{+} (g) + e^{-}$ .
Second ionisation energy: $X^{+} (g) \to X^{2 +} (g) + e^{-}$ , for one mole of ions.

Key results

Successive ionisation energies always increase: each electron is removed from an increasingly positive ion.
A large jump marks a change of shell. For magnesium the jump comes after the second electron ( $1451 \to 7733 kJ mol^{- 1}$ ), showing two outer electrons: Mg is in Group 2.
Across Period 3 first ionisation energy generally rises; down a group it falls.

Notes

Across a period nuclear charge increases while shielding stays similar (electrons are added to the same shell), so outer electrons are held more strongly.
Down a group there are more shells, so the outer electron is further from the nucleus and better shielded; this outweighs the greater nuclear charge.
The dip from Mg to Al: aluminium's outer electron is in a 3p orbital, higher in energy than magnesium's 3s, so it is easier to remove. This is evidence for subshells.
The dip from P to S: sulfur's fourth 3p electron must pair in an occupied orbital, and the repulsion between the paired electrons makes it easier to remove. This is evidence for electron pairing in orbitals.
Three factors control ionisation energy, and every explanation should use them: nuclear charge (proton number), distance of the outer electron from the nucleus (shell or subshell), and shielding by inner electrons. Quoting 'stability of a full shell' alone does not earn credit.
Helium has the highest first ionisation energy of all elements: its electrons are in the first shell, very close to a nucleus of charge +2, with no inner shielding.

Time-of-flight mass spectrometry 3.1.1.2

Key results

Equal kinetic energy: $E_{k} = \frac{1}{2} m v^{2}$ , so $v = 2 E_{k} / m$ and the flight time over drift length $L$ is $t = L m /2 E_{k}$ .
For ions of equal charge, $t \propto m$ : heavier ions take longer.

Method

Ionise: the sample is turned into positive ions (by electron impact or electrospray).
Accelerate: an electric field gives every ion of the same charge the same kinetic energy.
Drift: the ions cross a field-free region; lighter ions travel faster and arrive first.
Detect: ions hit a detector and gain electrons, producing a current proportional to their abundance.

In practice $^{24} Mg^{+}$ ions are accelerated to a kinetic energy of $2.0 \times 1 0^{- 16}$ J and drift 1.5 m. Find the time of flight, and that of $^{25} Mg^{+}$ .

$m = \frac{24 \times 1 0 ^{- 3}}{6.022 \times 1 0 ^{23}} = 3.99 \times 1 0^{- 26} kg$ the mass of one ion
$v = \frac{2 E _{k}}{m} = \frac{2 ( 2.0 \times 1 0 ^{- 16} )}{3.99 \times 1 0 ^{- 26}} = 1.00 \times 1 0^{5} m s^{- 1}$ every ion has the same kinetic energy
$t = \frac{L}{v} = \frac{1.5}{1.00 \times 1 0 ^{5}} = 1.50 \times 1 0^{- 5} s$
$t_{25} = t_{24} \frac{25}{24} = 1.53 \times 1 0^{- 5} s$ heavier ions arrive later: $t \propto m$

Notes

The instrument is kept under a vacuum so ions do not collide with air molecules, which would deflect or slow them.
The output is a spectrum of $m / z$ (mass-to-charge ratio) against relative abundance. For singly charged ions $m / z$ equals the ion's relative mass.
Exam calculations usually require converting relative mass in g mol⁻¹ to the mass of one ion in kg: $m = \frac{M}{1000 N _{A}}$ .
Worked example: an ion of $^{24} Mg^{+}$ ( $m = 24.0/ (1000 \times 6.022 \times 1 0^{23}) = 3.99 \times 1 0^{- 26}$ kg) with kinetic energy $1.00 \times 1 0^{- 16}$ J travels at $v = 2 E_{k} / m = 7.08 \times 1 0^{4} m s^{- 1}$ , crossing a 1.50 m drift region in $2.12 \times 1 0^{- 5}$ s.
Ions of the same mass but different charge are separated too: a $2 +$ ion gains twice the kinetic energy in the same field, so it moves faster and appears at half the $m / z$ of the $1 +$ ion.
Mass spectrometry identifies elements by their isotope pattern, measures $A_{r}$ , and gives the $M_{r}$ of molecules from the molecular-ion peak (the peak at highest $m / z$ , ignoring small isotope peaks).

Electron impact and electrospray ionisation 3.1.1.2

Definitions

Electron impact: high-energy electrons from an electron gun knock an electron off gaseous atoms or molecules: $X(g) + e^{-} \to X^{+} (g) + 2 e^{-}$ .
Electrospray ionisation: the sample, dissolved in a volatile solvent, is sprayed through a high-voltage needle; each molecule gains a proton: $M + H^{+} \to [M+H]^{+}$ .

Key results

Electron impact gives the molecular ion $M^{+}$ , with $m / z = M_{r}$ ; it often causes fragmentation.
Electrospray gives $[M+H]^{+}$ , with $m / z \approx M_{r} + 1$ ; it is a 'soft' method with little fragmentation.

Notes

Electron impact suits elements and small molecules with low $M_{r}$ . Electrospray suits large molecules such as proteins, which would fragment under electron impact.
Electrospray can also add more than one proton, giving multiply charged ions. Their $m / z$ is then about $(M_{r} + z) / z$ , so $z$ must be accounted for.
In electron impact the sample must first be vaporised; in electrospray the sample stays in solution until the solvent evaporates from the charged droplets, which is why it suits involatile and thermally fragile molecules.
Ions are detected because they gain electrons at the detector. The size of the current generated is proportional to the number of ions arriving, which is why peak height measures abundance.
Fragment peaks from electron impact are useful in organic chemistry: the pattern of fragments is a fingerprint of the molecule's structure (see mass spectrometry in organic analysis).

Relative atomic mass from a mass spectrum 3.1.1.2

Definitions

Relative atomic mass, $A_{r}$ : the weighted mean mass of an atom of an element relative to $1/12$ of the mass of an atom of carbon-12. It has no units.

Key results

$A_{r} = \frac{Σ ( isotopic mass \times abundance )}{Σ ( abundance )}$ .
Chlorine: $A_{r} \approx \frac{35 \times 75 + 37 \times 25}{100} = 35.5$ .

Notes

Each peak is one isotope ( $^{35} Cl$ and $^{37} Cl$ have 18 and 20 neutrons); its height is the relative abundance. The mean is weighted, so it lies nearer the more abundant isotope.
Molecular chlorine also gives peaks at $m / z$ 70, 72 and 74 from $Cl_{2}^{+}$ ions, in the ratio $9 : 6 : 1$ ( $0.7 5^{2}$ , $2 \times 0.75 \times 0.25$ , $0.2 5^{2}$ ).
Real isotopic masses are not exact whole numbers, which is why accurate $A_{r}$ values are not simple averages of mass numbers.
Worked example with three isotopes: magnesium has peaks at 24 (78.6%), 25 (10.1%) and 26 (11.3%). $A_{r} = (24 \times 78.6 + 25 \times 10.1 + 26 \times 11.3) /100 = 24.3$ .
Reverse problems are common: if an element has two isotopes, 63 and 65, and $A_{r} = 63.5$ , let the abundance of 63 be $x$ : $63 x + 65 (100 - x) = 6350$ , giving $x = 75%$ .
Relative abundances on a spectrum may be given as heights rather than percentages. Divide by the sum of the heights, not by 100, unless the heights already add up to 100.

Worked example

Explain why the first ionisation energy of aluminium (578 kJ mol⁻¹) is lower than that of magnesium (738 kJ mol⁻¹), despite aluminium having one more proton.

Show worked solution

Magnesium's outer electron configuration is 3s², so its highest-energy electron is removed from a full 3s subshell.

Aluminium's outer configuration is 3s²3p¹, so its highest-energy electron is instead removed from the 3p subshell.

The 3p subshell is at a higher energy than the 3s subshell, and the single 3p electron is additionally shielded from the nucleus by the two 3s electrons beneath it.

Both effects make aluminium's outermost electron easier to remove than magnesium's, and together they outweigh aluminium's one extra proton, so aluminium's first ionisation energy is lower even though its nuclear charge is greater.

This is a specific, subshell-level exception to the general across-period increase in ionisation energy, not a failure of the general trend.

3.1.2

Amount of substance

The mole and molar mass 3.1.2.1–3.1.2.2

Definitions

Mole: the amount of substance containing exactly $6.02214076 \times 1 0^{23}$ specified entities (atoms, molecules, ions or electrons).
Avogadro constant, $N_{A} = 6.022 \times 1 0^{23} mol^{- 1}$ .
Relative molecular mass, $M_{r}$ : the mean mass of a molecule relative to $1/12$ of the mass of a carbon-12 atom. Molar mass, $M$ , is the same number in $g mol^{- 1}$ .

Key results

$n = \frac{m}{M}$ ; $N = n N_{A}$ ; in solution $n = c V$ ; for a gas $n = \frac{p V}{RT}$ .
Water: $M_{r} = 18.0$ (no units); $M = 18.0 g mol^{- 1}$ .

Notes

Always say what the entities are: one mole of $O_{2}$ molecules contains two moles of O atoms.
For ionic compounds use relative formula mass, since they have no discrete molecules.
The amount in moles is the hub that connects mass, number of particles, solution volume and gas volume. Most calculations go mass → moles → ratio → moles → answer.
Worked example: 9.80 g of $H_{2} SO_{4}$ ( $M = 98.1$ ) is 0.0999 mol, containing $0.0999 \times 6.022 \times 1 0^{23} = 6.02 \times 1 0^{22}$ molecules and $4 \times$ that number of oxygen atoms.
Give answers to the number of significant figures in the least precise data value. Keep unrounded values through a multi-step calculation and round only at the end.
Water of crystallisation counts towards the molar mass: $CuSO_{4} \cdot 5 H_{2} O$ has $M = 249.7 g mol^{- 1}$ , not 159.6. Heating to constant mass and comparing masses is how the number of water molecules is found.

Empirical and molecular formulae 3.1.2.4

Definitions

Empirical formula: the simplest whole-number ratio of atoms of each element in a compound.
Molecular formula: the actual number of atoms of each element in one molecule.

Key results

Worked example: C 3.60 g, H 0.600 g, O 4.80 g → 0.300 : 0.600 : 0.300 → $CH_{2} O$ . With $M_{r} = 180$ , $180/30.0 = 6$ , so the molecular formula is $C_{6} H_{12} O_{6}$ .

Method

Write the mass (or percentage) of each element.
Divide each by its $A_{r}$ to get amounts in moles.
Divide every amount by the smallest to get the ratio; scale to whole numbers if needed.
Divide $M_{r}$ by the empirical formula mass, then multiply every subscript by the result.

In practiceA compound is 40.0% C, 6.7% H and 53.3% O by mass, with $M_{r} = 180$ . Find its molecular formula.

$C : \frac{40.0}{12.0} = 3.33, H : \frac{6.7}{1.0} = 6.7, O : \frac{53.3}{16.0} = 3.33$ moles in 100 g
$\div 3.33 : 1 : 2 : 1 ⟹ CH_{2} O$ empirical formula, $M_{r} = 30$
$\frac{180}{30} = 6 ⟹ C_{6} H_{12} O_{6}$

Notes

A ratio such as 1 : 1.5 or 1 : 1.33 must be multiplied (by 2 or 3), not rounded.
Combustion analysis gives the masses indirectly: all the carbon ends up in $CO_{2}$ and all the hydrogen in $H_{2} O$ ; oxygen is found by difference.
Worked example from percentages: 85.7% C and 14.3% H gives $85.7/12.0 = 7.14$ and $14.3/1.0 = 14.3$ , a 1 : 2 ratio, so $CH_{2}$ . With $M_{r} = 56$ , the molecular formula is $C_{4} H_{8}$ .
For a hydrated salt, treat water as a unit: if 2.50 g of hydrated copper(II) sulfate leaves 1.60 g anhydrous solid, $n (CuSO_{4}) = 1.60/159.6 = 0.0100$ and $n (H_{2} O) = 0.90/18.0 = 0.050$ , so the formula is $CuSO_{4} \cdot 5 H_{2} O$ .
Use at least three significant figures in the mole values; rounding too early can turn 1.33 into 1.3 and hide the 3 : 4 ratio.

Balanced equations and limiting reactants 3.1.2.5

Definitions

Limiting reactant: the reactant that is completely used up and so fixes the amount of product formed. The other reactants are in excess.

Key results

Coefficients give mole ratios: $2 H_{2} + O_{2} \to 2 H_{2} O$ means $H_{2} : O_{2} : H_{2} O = 2 : 1 : 2$ .
The reactant with the smallest value of $n / coefficient$ is limiting. 3.00 mol $H_{2}$ with 1.00 mol $O_{2}$ : $O_{2}$ limits, giving 2.00 mol $H_{2} O$ and leaving 1.00 mol $H_{2}$ .

Notes

Balance with coefficients only. Changing a subscript changes the substance.
The particle diagram and the mole calculation are the same reasoning at two scales: 3 molecules of $H_{2}$ with 1 of $O_{2}$ leaves one $H_{2}$ over, exactly as 3 mol with 1 mol leaves 1 mol.
Ionic equations show only the species that change. For precipitation, $Ag^{+} (a q) + Cl^{-} (a q) \to AgCl(s)$ ; spectator ions such as $Na^{+}$ and $NO_{3}^{-}$ are left out. The equation must balance for charge as well as atoms.
Worked example: 5.00 g of magnesium ( $0.206$ mol) with 50.0 cm³ of 2.00 mol dm⁻³ HCl ( $0.100$ mol). Since Mg : HCl is 1 : 2, the acid could react with only 0.0500 mol Mg, so HCl is limiting and 0.0500 mol $H_{2}$ forms.
Industrial processes deliberately use an excess of the cheaper reactant to make sure the more expensive one reacts completely.

Solutions, standard solutions and titrations 3.1.2.5

Definitions

Concentration: amount of solute per unit volume of solution, $c = n / V$ in $mol dm^{- 3}$ with $V$ in $dm^{3}$ .
Titre: the volume delivered from the burette, final reading minus initial reading. Concordant titres agree within $0.10 cm^{3}$ .

Key results

Worked example: mean titre 20.00 cm³ of 0.100 mol dm⁻³ HCl; $n (HCl) = 2.00 \times 1 0^{- 3}$ mol; 1 : 1 with NaOH, so $c (NaOH) = 0.00200/0.02500 = 0.0800 mol dm^{- 3}$ .

Method

Standard solution: weigh the solid, dissolve it in a beaker, transfer quantitatively (with washings) to a volumetric flask, make up to the mark with the bottom of the meniscus on the line, stopper and invert to mix.
Titration: pipette a known volume into a conical flask, add a few drops of indicator, titrate until the indicator just changes colour, and repeat until results are concordant.

In practice25.0 cm $^{3}$ of NaOH(aq) needs a mean concordant titre of 21.60 cm $^{3}$ of 0.100 mol dm $^{- 3}$ HCl. Find the concentration of the NaOH.

$n (HCl) = \frac{21.60}{1000} \times 0.100 = 2.16 \times 1 0^{- 3} mol$
$NaOH + HCl \to NaCl + H_{2} O : n (NaOH) = 2.16 \times 1 0^{- 3} mol$ 1 : 1
$[NaOH] = \frac{2.16 \times 1 0 ^{- 3}}{0.0250} = 0.0864 mol dm^{- 3}$ volume in dm³

Notes

Convert cm³ to dm³ by dividing by 1000 before using $n = c V$ .
The net ionic equation for any strong acid with a strong base is $H^{+} (a q) + OH^{-} (a q) \to H_{2} O (l)$ ; the mole ratio, though, comes from the full equation (e.g. 1 : 2 for $H_{2} SO_{4}$ with NaOH).
Back titrations: when a solid reacts slowly or is insoluble (e.g. a carbonate in an antacid tablet), add a known excess of acid, then titrate the unreacted acid with alkali. Amount reacted = acid added − acid left over.
Uncertainties: a burette reading is ±0.05 cm³, so a titre (two readings) is ±0.10 cm³. Percentage uncertainty = absolute uncertainty / measured value × 100; it is reduced by using larger titres.
Rinse the burette with the solution it will contain and the pipette with the solution it will deliver; the conical flask may contain water, because water does not change the amount of reactant pipetted into it.

The ideal gas equation 3.1.2.3

Definitions

Ideal gas equation: $p V = n RT$ , with $p$ in Pa, $V$ in m³, $T$ in K and $R = 8.31 J mol^{- 1} K^{- 1}$ .

Key results

1.00 mol at 100 kPa and 298 K occupies $V = n RT / p = 0.0248 m^{3} = 24.8 dm^{3}$ .
Conversions: $1 dm^{3} = 1 0^{- 3} m^{3}$ ; $1 cm^{3} = 1 0^{- 6} m^{3}$ ; $1 kPa = 1 0^{3} Pa$ ; $T / K = θ /^{\circ} C + 273$ .

Notes

Unit slips, not algebra, cause most lost marks: always convert to Pa, m³ and K first.
Real gases behave most ideally at low pressure and high temperature, where molecules are far apart and intermolecular forces are negligible.
Molar gas volume is not a constant: it depends on temperature and pressure, which is why 24.8 dm³ applies only at 100 kPa and 298 K.
Worked example: what mass of $CO_{2}$ occupies 500 cm³ at 25 °C and 101 kPa? $n = p V / RT = (101000 \times 5.00 \times 1 0^{- 4}) / (8.31 \times 298) = 0.0204$ mol, so mass $= 0.0204 \times 44.0 = 0.897$ g.
Rearranging gives molar mass from gas measurements: $M = m RT / p V$ . This is how the $M_{r}$ of a volatile liquid is found by vaporising a weighed sample into a gas syringe.
The equation follows from the gas laws (Boyle's $p V$ constant, Charles's $V \propto T$ ) together with Avogadro's principle that equal volumes of gases at the same $T$ and $p$ contain equal numbers of molecules.

Percentage yield and atom economy 3.1.2.5

Definitions

Percentage yield $= \frac{actual yield}{theoretical yield} \times 100$ .
Percentage atom economy $= \frac{mass of desired product}{total mass of reactants} \times 100$ , using the balanced equation.

Key results

Thermal decomposition $CaCO_{3} \to CaO + CO_{2}$ with CaO the desired product: atom economy $= 56.1/100.1 \times 100 = 56.0%$ .
If 44.9 g of CaO is collected where 56.1 g was possible, the yield is $80.0%$ .

Notes

Yield measures practical efficiency (losses, side reactions, incomplete reaction). Atom economy is fixed by the equation: it measures how many reactant atoms end up in the product you want.
A process can have a high yield but low atom economy, producing a lot of waste. Addition reactions have 100% atom economy; substitution and elimination reactions do not.
Higher atom economy means fewer waste products to dispose of and less use of raw materials, which is why it matters for sustainable industrial chemistry.
Yields are reduced by incomplete (equilibrium) reactions, side reactions giving other products, and mechanical losses on transfer, filtration and purification. Recrystallisation, for example, always leaves some product dissolved in the solvent.
Worked example: making ethanol by hydration ( $C_{2} H_{4} + H_{2} O \to C_{2} H_{5} OH$ ) has 100% atom economy; making it by fermentation ( $C_{6} H_{12} O_{6} \to 2 C_{2} H_{5} OH + 2 CO_{2}$ ) has $92.0/180.0 = 51.1%$ .
If a by-product can be sold or used, the effective waste falls even though the atom economy calculation does not change.

3.1.3

Bonding

Ionic and metallic bonding 3.1.3.1, 3.1.3.3

Definitions

Ionic bonding: the electrostatic attraction between oppositely charged ions in a lattice.
Metallic bonding: the attraction between a lattice of positive metal ions and a sea of delocalised electrons.

Key results

Ionic formulae balance charge: $MgCl_{2}$ , $Ca(NO_{3})_{2}$ , $(NH_{4})_{2} SO_{4}$ .
Common polyatomic ions: $NH_{4}^{+}$ , $OH^{-}$ , $NO_{3}^{-}$ , $CO_{3}^{2 -}$ , $SO_{4}^{2 -}$ .

Notes

Ionic solids do not conduct because the ions are fixed in the lattice; when molten or dissolved the ions can move and carry charge.
Metals conduct as solids and liquids because the delocalised electrons can move through the structure. They are malleable because layers of ions can slide while the delocalised electrons keep bonding them.
Both structures have strong attractions acting throughout a giant lattice, so both usually have high melting points. In magnesium, each ion contributes two delocalised electrons, giving stronger bonding than in sodium (one each).
The strength of ionic bonding increases with ionic charge and decreases with ionic size: MgO (2+ and 2−, small ions) melts at about 2850 °C, far above NaCl at 801 °C.
Ionic solids are brittle: a blow shifts layers so that like charges line up, and the repulsion splits the crystal. Compare metals, where layers can slide without this repulsion.
Metallic bonding strength increases with the number of delocalised electrons per ion and with smaller ionic radius, so melting points rise from Na to Mg to Al across Period 3.

Covalent and dative covalent bonds 3.1.3.2

Definitions

Covalent bond: a shared pair of electrons attracted to the nuclei of both bonded atoms.
Co-ordinate (dative covalent) bond: a shared pair in which both electrons come from the same atom.

Key results

$H_{2}$ has one shared pair, $O_{2}$ two (a double bond) and $N_{2}$ three (a triple bond).
Ammonia's lone pair forms a dative bond to $H^{+}$ , giving $NH_{4}^{+}$ : $NH_{3} + H^{+} \to NH_{4}^{+}$ .

Notes

Dots and crosses only show where electrons came from; all electrons are identical.
A dative bond is drawn as an arrow from donor to acceptor. Once formed it is identical to the other bonds: all four N–H bonds in $NH_{4}^{+}$ are equivalent, and the ion is tetrahedral.
Other important dative bonds: $H_{2} O + H^{+} \to H_{3} O^{+}$ , and the bond from N to B in $H_{3} N \to BF_{3}$ , where boron (only six outer electrons) accepts a lone pair. Aluminium chloride dimerises, $Al_{2} Cl_{6}$ , by Cl lone pairs donating to Al.
Every ligand–metal bond in a transition-metal complex is a co-ordinate bond, so this idea returns in 3.2.5.
Multiple bonds are shorter and stronger than single bonds between the same atoms: C–C 154 pm (348 kJ mol⁻¹), C=C 134 pm (612 kJ mol⁻¹).

Shapes of molecules and ions 3.1.3.5

Definitions

Electron-pair repulsion theory: the electron regions around a central atom (bonding pairs and lone pairs) arrange themselves as far apart as possible to minimise repulsion. A multiple bond counts as one region.

Key results

Linear 180° ( $CO_{2}$ ); trigonal planar 120° ( $BF_{3}$ ); tetrahedral 109.5° ( $CH_{4}$ ).
Trigonal pyramidal ≈107° ( $NH_{3}$ ); bent 104.5° ( $H_{2} O$ ); trigonal bipyramidal 90°, 120° and 180° ( $PCl_{5}$ ).
Octahedral 90° ( $SF_{6}$ ); square pyramidal ( $BrF_{5}$ ); square planar ( $XeF_{4}$ ).

Notes

Repulsion order: lone pair–lone pair > lone pair–bonding pair > bonding pair–bonding pair. Each lone pair closes the remaining bond angles by about 2.5°, which is why $NH_{3}$ (one lone pair) is ≈107° and $H_{2} O$ (two) is 104.5°.
The shape names describe where the atoms are, not the electron pairs: water's four electron regions are tetrahedral, but its shape is bent.
Method: count the central atom's outer electrons, add one per bonded atom (for single bonds), adjust for charge, halve to get the number of electron pairs, then subtract bonding pairs to find lone pairs.
Worked example: $ClF_{3}$ . Chlorine has 7 outer electrons, plus 3 from the F atoms, gives 10 electrons = 5 pairs: 3 bonding and 2 lone. Five pairs adopt a trigonal bipyramidal arrangement with the lone pairs equatorial, so the molecule is T-shaped with angles slightly less than 90°.
Ions follow the same method: $NH_{4}^{+}$ (5 + 4 − 1 = 8 electrons, 4 bonding pairs) is tetrahedral, 109.5°; $NH_{2}^{-}$ (5 + 2 + 1 = 8 electrons, 2 bonding and 2 lone pairs) is bent, about 104.5°.
In exam answers state the number of bonding and lone pairs, the principle that pairs repel to be as far apart as possible, and that lone pairs repel more strongly. Then name the shape and give the angle.

Electronegativity and polarity 3.1.3.6

Definitions

Electronegativity: the power of an atom to attract the pair of electrons in a covalent bond.

Key results

A bond between atoms of different electronegativity is polar, with partial charges $δ +$ and $δ -$ .
$CO_{2}$ has polar bonds but is non-polar; $H_{2} O$ is polar.

Notes

Electronegativity increases across a period and up a group; fluorine is the most electronegative element.
Whether a molecule is polar depends on its shape as well as its bonds. In linear $CO_{2}$ the two bond dipoles are equal and opposite and cancel; in bent water they do not, leaving a net dipole.
Symmetrical molecules such as $CCl_{4}$ and $BF_{3}$ are non-polar for the same reason.
Pauling values for reference: F 4.0, O 3.5, N 3.0, Cl 3.0, C 2.5, H 2.1. Bond polarity grows with the difference: C–H (0.4) is nearly non-polar; O–H (1.4) is strongly polar.
Bonding is a continuum: identical atoms share equally (pure covalent), a moderate difference gives a polar covalent bond, and a large difference (above about 1.7–2) gives predominantly ionic bonding.
A polar molecule is deflected by a charged rod held near a thin stream of the liquid; a stream of water bends towards both positive and negative rods, whereas a non-polar liquid such as hexane does not.

Forces between molecules 3.1.3.7

Definitions

London (induced dipole–dipole) forces: attractions between temporary dipoles caused by fluctuating electron distributions, which induce dipoles in neighbouring molecules. They act between all molecules.
Permanent dipole–dipole forces: attractions between polar molecules whose $δ +$ and $δ -$ ends align.
Hydrogen bonding: the attraction between an H atom bonded to N, O or F and a lone pair on an N, O or F atom of a neighbouring molecule.

Key results

Strength (typical): hydrogen bonds > permanent dipole–dipole > London forces, molecule for molecule of similar size.
London forces increase with the number of electrons (molecular size), so boiling points rise down Group 7 and along the alkanes.

Notes

Hydrogen bonding explains why water, ammonia and hydrogen fluoride have much higher boiling points than the other hydrides of their groups.
In ice each water molecule forms four hydrogen bonds (two through its own H atoms, two to its lone pairs), giving an open tetrahedral network. The molecules are further apart than in the liquid, so ice is less dense than water.
Melting or boiling a simple molecular substance overcomes the forces between molecules; the covalent bonds within the molecules remain intact.
London forces can outweigh dipole–dipole forces when molecules differ greatly in size: iodine ( $I_{2}$ , 106 electrons) is a solid at room temperature, whereas HCl (polar, 18 electrons) is a gas.
Shape matters for London forces: long unbranched chains have more surface contact than compact branched isomers, so pentane boils at 36 °C and 2,2-dimethylpropane at 10 °C.
When drawing a hydrogen bond, show the $δ +$ H, the lone pair on the N, O or F of the other molecule, and a dashed line between them, with the O–H···O atoms roughly in a straight line (about 180°).

Structure and physical properties 3.1.3.4

Definitions

The four crystal types: ionic, metallic, macromolecular (giant covalent) and molecular.

Key results

Diamond: each C bonded to four others; very high melting point; no mobile charge carriers, so non-conducting.
Graphite: each C bonded to three others in layers with one delocalised electron per C; conducts along the layers; layers slide because the forces between them are weak.
Iodine: $I_{2}$ molecules held by London forces; low melting point; sublimes on gentle heating.

Notes

To explain a property, name the particles, the forces between them and how strong those forces are. Melting a giant covalent structure breaks covalent bonds; melting a molecular one only overcomes intermolecular forces.
Conductivity needs mobile charged particles: delocalised electrons (metals, graphite) or mobile ions (molten or dissolved ionic compounds).
Energy is absorbed on melting and boiling and released on freezing and condensing.
Silicon(IV) oxide, $SiO_{2}$ , is also macromolecular: each Si is bonded to four O atoms and each O to two Si, giving a very high melting point (about 1700 °C).
Solubility follows the same reasoning: ionic compounds often dissolve in water because the polar water molecules are attracted to the ions; simple non-polar molecules dissolve in non-polar solvents; giant covalent structures dissolve in neither.
Graphite's layers are about 335 pm apart compared with 142 pm within a layer, a direct measure of how weak the interlayer forces are compared with the covalent bonds.

3.1.4

Energetics

Enthalpy change, exothermic and endothermic 3.1.4.1

Definitions

Enthalpy change, $Δ H$ : the heat energy change measured at constant pressure; $Δ H = H (products) - H (reactants)$ .
Exothermic: energy is transferred from the system to the surroundings, so $Δ H < 0$ . Endothermic: energy is taken in from the surroundings, so $Δ H > 0$ .

Notes

On a reaction profile, exothermic products lie below the reactants; endothermic products lie above.
An exothermic reaction warms its surroundings (the thermometer reading rises) because the system loses energy to them.
Examples: combustion, neutralisation and respiration are exothermic; thermal decomposition of calcium carbonate, photosynthesis and dissolving ammonium nitrate are endothermic.
The system is the reacting chemicals; the surroundings are everything else, including the solvent and the container. Energy is conserved overall: what the system loses, the surroundings gain.
Enthalpy, $H$ , cannot be measured absolutely; only changes in enthalpy can. That is why tables list $Δ H$ values relative to agreed reference states.
The reaction-profile diagram on this panel is the starting point for activation energy in kinetics: the peak between reactants and products is the energy barrier, independent of the sign of $Δ H$ .

Standard enthalpy changes 3.1.4.1

Definitions

Standard conditions: 100 kPa and a stated temperature, usually 298 K, with all substances in their standard states (aqueous solutions at 1 mol dm⁻³).
Standard enthalpy of formation, $Δ_{f} H^{\circ}$ : the enthalpy change when one mole of a compound is formed from its elements in their standard states.
Standard enthalpy of combustion, $Δ_{c} H^{\circ}$ : the enthalpy change when one mole of a substance is completely burned in oxygen.

Key results

$Δ_{f} H^{\circ}$ of any element in its standard state is zero.
Formation of methane: $C(s, graphite) + 2 H_{2} (g) \to CH_{4} (g)$ . Combustion: $CH_{4} (g) + 2 O_{2} (g) \to CO_{2} (g) + 2 H_{2} O (l)$ .

Notes

State symbols are part of the definition: forming $H_{2} O (l)$ releases more energy than forming $H_{2} O (g)$ , by the enthalpy of vaporisation.
Fractional coefficients are allowed (and often needed) to keep 'one mole' of the stated substance, e.g. $H_{2} + \frac{1}{2} O_{2} \to H_{2} O$ .
Standard enthalpy of neutralisation: the enthalpy change when one mole of water is formed by the reaction of an acid with a base under standard conditions. For strong acids with strong bases it is about $- 57 kJ mol^{- 1}$ whichever acid is used, because the reaction is always $H^{+} + OH^{-} \to H_{2} O$ .
Standard states are the physical states at 100 kPa and 298 K: carbon is graphite, bromine is $Br_{2} (l)$ , iodine is $I_{2} (s)$ . Using the wrong state changes the value.
The enthalpy of combustion of an element equals the enthalpy of formation of its oxide: $Δ_{c} H^{\circ} (C, graphite) = Δ_{f} H^{\circ} (CO_{2})$ .

Calorimetry 3.1.4.2

Key results

$q = m c Δ T$ , with $m$ the mass of solution or water (g), $c = 4.18 J g^{- 1} K^{- 1}$ and $Δ T$ in K.
$Δ H = - \frac{q}{1000 n}$ in kJ mol⁻¹, where $n$ is the amount that reacted.
Worked example: 50.0 cm³ each of 1.00 mol dm⁻³ HCl and NaOH; corrected $Δ T = 6.80$ K; $q = 100 \times 4.18 \times 6.80 = 2842$ J; $n (H_{2} O) = 0.0500$ mol; $Δ H \approx - 56.8 kJ mol^{- 1}$ .

Method

Use an insulated (polystyrene) cup with a lid; record the temperature of the reactants for a few minutes before mixing.
Mix at a recorded time, stir, and record the temperature at regular intervals afterwards.
Plot temperature against time and extrapolate the cooling line back to the time of mixing to find the corrected $Δ T$ .

In practice50.0 cm $^{3}$ of 1.00 mol dm $^{- 3}$ HCl is mixed with 50.0 cm $^{3}$ of 1.00 mol dm $^{- 3}$ NaOH; the corrected temperature rise is 6.8 K. Find $Δ H_{neut}$ .

$q = m c Δ T = 100 \times 4.18 \times 6.8 = 2842 J$ 100 g of solution, taking $1.00 g cm^{- 3}$
$n (H_{2} O) = 0.0500 mol$
$Δ H = - \frac{2.842 kJ}{0.0500 mol} = - 56.8 kJ mol^{- 1}$ negative: exothermic

Notes

The extrapolation corrects for heat lost while the mixture was warming up, giving the temperature change there would have been with no heat loss.
Assumptions: the solution has the density and specific heat capacity of water, and the cup absorbs no heat. Heat loss and these assumptions make measured values less exothermic than data-book values.
The sign is added at the end: a temperature rise means the reaction released energy, so $Δ H$ is negative.
For combustion of a fuel, heat a known mass of water in a copper can above a spirit burner and weigh the burner before and after. Large heat losses to the air and incomplete combustion give values far smaller in magnitude than data-book values.
Worked example (combustion): burning 0.230 g of ethanol (0.00500 mol) raises 100 g of water by 12.0 K; $q = 100 \times 4.18 \times 12.0 = 5016$ J, so $Δ_{c} H = - 5.016/0.00500 = - 1003 kJ mol^{- 1}$ (data book −1367).
The mass in $q = m c Δ T$ is the mass of what is being heated (the water or solution), not the mass of the fuel or solid reactant.

Hess's law 3.1.4.3

Definitions

Hess's law: the enthalpy change for a reaction is independent of the route taken, provided the initial and final conditions are the same.

Key results

From formation data: $Δ_{r} H^{\circ} = Σ Δ_{f} H^{\circ} (products) - Σ Δ_{f} H^{\circ} (reactants)$ .
From combustion data: $Δ_{r} H^{\circ} = Σ Δ_{c} H^{\circ} (reactants) - Σ Δ_{c} H^{\circ} (products)$ .
Methane combustion from formation data: $[- 393.5 + 2 (- 285.8)] - [- 74.9 + 0] = - 890.2 kJ mol^{- 1}$ .

Notes

The cycles differ in direction: formation arrows point from the elements up to reactants and products, combustion arrows point down to the common combustion products. That is why the subtraction is reversed between the two formulae.
Multiply each value by its coefficient. Reversing an equation reverses the sign of $Δ H$ ; multiplying an equation multiplies $Δ H$ by the same factor.
Hess's law is used for enthalpy changes that cannot be measured directly, such as the formation of methane from carbon and hydrogen.
Worked example with combustion data: for $3 C(s) + 4 H_{2} (g) \to C_{3} H_{8} (g)$ , using $Δ_{c} H$ (C) = −393.5, $Δ_{c} H (H_{2})$ = −285.8 and $Δ_{c} H (C_{3} H_{8})$ = −2220: $Δ_{f} H = 3 (- 393.5) + 4 (- 285.8) - (- 2220) = - 103.7 kJ mol^{- 1}$ .
Hess's law is a consequence of the conservation of energy: if two routes between the same states gave different enthalpy changes, energy could be created by going out one way and back the other.
Experimentally, the enthalpy change for anhydrous copper(II) sulfate becoming hydrated is found by measuring the enthalpies of solution of both the anhydrous and hydrated salts and combining them in a cycle.

Mean bond enthalpies 3.1.4.4

Definitions

Mean bond enthalpy: the energy needed to break one mole of a given covalent bond in gaseous molecules, averaged over a range of compounds. Values are always positive.

Key results

$Δ H \approx Σ (bonds broken) - Σ (bonds formed)$ .
Hydrogenation of ethene: broken $4 (412) + 612 + 436 = 2696$ ; formed $6 (412) + 348 = 2820$ ; $Δ H \approx - 124 kJ mol^{- 1}$ .

Notes

Bond breaking is endothermic and bond making exothermic; a reaction is exothermic when the bonds formed are stronger overall than the bonds broken.
Results are estimates: a mean value ignores how a bond's strength depends on its molecular environment, and the method assumes all species are gases.
The break-everything-then-reform route is a calculation device, not the reaction mechanism.
Values found from mean bond enthalpies differ from those found by Hess's law with formation data, because the specific bonds in a given molecule are not exactly the mean. The Hess's law value is the more accurate.
Worked example: methane combustion. Broken: $4 (C-H) + 2 (O=O) = 4 (412) + 2 (496) = 2640$ . Formed: $2 (C=O) + 4 (O-H) = 2 (743) + 4 (463) = 3338$ . $Δ H \approx - 698 kJ mol^{- 1}$ , much less exothermic than −890 because water is formed as a gas in the bond-enthalpy method and the C=O value in $CO_{2}$ (805) exceeds the mean.
Reverse calculations are common: given $Δ H$ and all but one bond enthalpy, rearrange to find the unknown bond enthalpy.

Worked example

Calculate the standard enthalpy of combustion of ethanol,

C_{2} H_{5} OH (l) + 3 O_{2} (g) \to 2 CO_{2} (g) + 3 H_{2} O (l)

using standard enthalpies of formation:

Δ H_{f} [C_{2} H_{5} OH (l)] = - 277 kJ mol^{- 1}

Δ H_{f} [CO_{2} (g)] = - 394 kJ mol^{- 1}

Δ H_{f} [H_{2} O (l)] = - 286 kJ mol^{- 1}

(and:

Δ H_{f} [O_{2} (g)] = 0

as it is an element in its standard state).

Show worked solution

Δ H_{reaction} = \sum Δ H_{f} (products) - \sum Δ H_{f} (reactants)

Products:

2 \times (- 394) + 3 \times (- 286) = - 788 - 858 = - 1646 kJ mol^{- 1}

Reactants:

(- 277) + 3 \times (0) = - 277 kJ mol^{- 1}

Δ H_{reaction} = - 1646 - (- 277) = - 1646 + 277 = - 1369 kJ mol^{- 1}

This is close to the widely cited data-book value of approximately:

- 1367 kJ mol^{- 1}

for ethanol's enthalpy of combustion, confirming the calculation, and negative as expected for a strongly exothermic combustion reaction.

3.1.5

Kinetics

Collision theory and activation energy 3.1.5.1

Definitions

Rate of reaction: the change in amount or concentration of a reactant or product per unit time.
Activation energy, $E_{a}$ : the minimum energy that particles need to react when they collide.

Notes

A collision leads to reaction only if the particles have at least $E_{a}$ and collide with a suitable orientation; most collisions do neither.
On a reaction profile $E_{a}$ is the height of the barrier from reactants to the top of the curve, whether the reaction is exothermic or endothermic.
Rate can be followed by any property that changes as the reaction proceeds: gas volume, mass lost as a gas escapes, colour (colorimetry), pH, electrical conductivity, or the time to form a fixed amount of precipitate.
The rate of a reaction usually falls as it proceeds, because the reactants are used up and their concentrations fall, so successful collisions become less frequent.
Activation energy exists because bonds must be stretched and partly broken before new ones can form. The top of the barrier corresponds to the unstable arrangement of atoms (the transition state) part-way between reactants and products.
Surface area: for a solid reactant, only particles at the surface can collide, so powdering a solid increases the rate. This is why catalytic converters use a honeycomb coated with a thin layer of metal.

The Maxwell–Boltzmann distribution and temperature 3.1.5.2–3.1.5.3

Definitions

Maxwell–Boltzmann distribution: the spread of molecular energies in a gas at a given temperature.

Key results

The curve starts at the origin (no molecule has zero energy), has a single peak (the most probable energy) and approaches the energy axis without touching it.
The area under the curve equals the total number of molecules; the area beyond $E_{a}$ is the number able to react.

Notes

At a higher temperature the peak moves to higher energy and becomes lower; the area under both curves is the same because the number of molecules is the same.
A small temperature rise greatly increases the area beyond $E_{a}$ , so the rate rises much more than the collision frequency alone would suggest. This, not more frequent collisions, is the main reason rate increases with temperature.
Temperature changes the distribution, not $E_{a}$ .
Exam sketches are marked on detail: the higher-temperature curve must start at the origin, have its peak lower and to the right, cross the original curve once, and lie above it at high energies without meeting the axis.
A rough rule of thumb is that a 10 K rise doubles the rate of many reactions near room temperature. This is because the fraction of molecules with $E \geq E_{a}$ increases exponentially, as quantified later by the Arrhenius equation.
Only gases are described by this exact distribution, but the same idea (a spread of energies, with only the most energetic particles able to react) applies in solutions.

Concentration and pressure 3.1.5.4

Notes

Higher concentration (or higher pressure for a gas) puts more particles in the same volume, so collisions are more frequent and the rate increases.
At constant temperature the fraction of molecules with $E \geq E_{a}$ is unchanged; only the collision frequency rises. Contrast this with heating, which changes the fraction.
Compressing a gas mixture increases the pressure and the concentrations of every gas at once.
In exam answers name the full chain: more particles per unit volume → more frequent collisions → more frequent successful collisions (those with $E \geq E_{a}$ ) → greater rate. 'More collisions' alone is not enough.
Not every reactant concentration affects the rate in the same way. Rate equations (3.1.9) show that some reactants have no effect at all (zero order), because they take part only after the slowest step of the mechanism.
On a graph of product against time, a higher concentration gives a steeper initial gradient. If the extra reactant is also the limiting one, the curve also levels off higher.
Increasing the total pressure by adding an inert gas at constant volume does not change the concentrations of the reactants, and so does not change the rate.

Catalysts 3.1.5.5

Definitions

Catalyst: a substance that increases the rate of a reaction by providing an alternative reaction route of lower activation energy, and is chemically unchanged at the end.

Notes

On the Maxwell–Boltzmann curve a catalyst does not change the distribution; it moves the activation energy line to the left, so more molecules lie beyond it.
On a reaction profile the catalysed route has a lower peak, but the reactant and product enthalpies (and so $Δ H$ ) are the same.
A catalyst takes part in the reaction but is regenerated, so only a small amount is needed.
Heterogeneous catalysts are in a different phase from the reactants (iron in the Haber process; platinum and rhodium in catalytic converters). Reactants adsorb onto active sites on the surface, bonds weaken, reaction occurs and products desorb.
Homogeneous catalysts are in the same phase as the reactants; they work by forming an intermediate species. Examples appear in 3.2.5 (iron ions catalysing peroxodisulfate with iodide).
Catalysts are important for sustainability: by allowing lower temperatures they cut energy use and the associated $CO_{2}$ emissions, and they can make processes with better atom economy practicable.
Catalysts can be poisoned: impurities such as sulfur compounds bind strongly to the active sites and block them, which is why feed gases are purified first.

Measuring rates 3.1.5.1, RP3

Key results

Rate at any moment = gradient of the tangent to the curve; the initial rate is the gradient at $t = 0$ . In the example, $60/25 = 2.4 cm^{3} s^{- 1}$ .
For a fixed endpoint, initial rate $\propto 1/ t$ .

Method

Gas collection: record the volume of gas in a syringe at regular intervals; plot volume against time.
Required practical 3 (sodium thiosulfate with acid): time how long the sulfur precipitate takes to hide a cross viewed from above; repeat at several temperatures, equilibrating both solutions in a water bath before mixing.

In practiceGas volumes of 0, 20, 34, 44 and 50 cm $^{3}$ are recorded at 0, 30, 60, 90 and 120 s. A thiosulfate cross disappears in 50 s at 20 °C and 25 s at 30 °C.

$mean rate = \frac{34 - 0}{60 - 0} = 0.57 cm^{3} s^{- 1}$ over the first minute
The initial rate is the gradient of the tangent at $t = 0$ : steeper than the mean, because the rate falls as reactant is used up.
$rate \propto \frac{1}{t} : \frac{1}{50} = 0.020 s^{- 1} (2 0^{\circ} C), \frac{1}{25} = 0.040 s^{- 1} (3 0^{\circ} C)$ a 10 K rise doubles the rate here

Notes

The curve levels off when the limiting reactant is used up; with excess acid, the amount of calcium carbonate fixes the final volume.
The $1/ t$ method is valid only because the same small amount of sulfur obscures the cross each time, so the reaction has gone the same short way in every run.
Keep concentrations, volumes, cross and viewing depth the same so temperature is the only variable.
Choose the method to match the reaction: a gas syringe or balance for reactions giving a gas, a colorimeter for coloured species (e.g. iodine), a pH probe when $H^{+}$ is consumed or produced.
Draw tangents with a ruler, touching the curve at one point only, and make the triangle used for the gradient as large as possible to reduce the reading error.
Plotting $1/ t$ (a measure of rate) against temperature shows the rate rising steeply with temperature, a link to the Arrhenius treatment in 3.1.9.

3.1.6

Chemical equilibria, Le Chatelier's principle and Kc

Dynamic equilibrium 3.1.6.1

Definitions

Dynamic equilibrium: in a closed system, the forward and reverse reactions occur at equal rates, so the concentrations of reactants and products stay constant.

Notes

Both reactions continue at equilibrium; it only looks static because the rates are equal.
Equal rates do not mean equal concentrations: the equilibrium mixture can be mostly reactants or mostly products.
A closed system exchanges no matter with its surroundings, which is required for equilibrium to be reached.
Equilibrium can be approached from either side: starting with pure $H_{2}$ and $I_{2}$ , or with pure HI, the same equilibrium mixture forms at the same temperature (for the same total amounts of atoms).
Evidence that the equilibrium is dynamic comes from isotopic labelling: adding radioactive iodine as $I_{2}$ to an equilibrium mixture soon puts radioactivity into the HI as well, even though the overall concentrations do not change.
On a concentration–time graph equilibrium is reached when every line becomes flat; on a rate–time graph, when the forward and reverse rate lines meet.
The position of equilibrium describes the relative amounts of reactants and products; it is quantified by the equilibrium constant.

Le Chatelier's principle and concentration 3.1.6.1

Definitions

Le Chatelier's principle: if a factor affecting an equilibrium is changed, the position of equilibrium shifts in the direction that opposes the change.

Notes

Adding A to $A ⇌ B$ makes [A] jump; the forward reaction then runs faster than the reverse until a new equilibrium is reached with more B.
Removing a product, or adding a reactant, shifts the equilibrium to the right; the reverse changes shift it to the left.
At constant temperature $K_{c}$ does not change: the concentrations rearrange to restore the same value.
Worked example: in $Fe^{3 +} + SCN^{-} ⇌ [FeSCN]^{2 +}$ (pale yellow to blood red), adding more $SCN^{-}$ darkens the red colour; adding $Ag^{+}$ , which precipitates $SCN^{-}$ as AgSCN, lightens it.
Another observable example is the chromate–dichromate equilibrium, $2 CrO_{4}^{2 -} + 2 H^{+} ⇌ Cr_{2} O_{7}^{2 -} + H_{2} O$ : adding acid turns yellow to orange; adding alkali removes $H^{+}$ and turns it back.
Le Chatelier's principle predicts the direction of a shift but not how far it goes. Only a $K_{c}$ calculation gives the new concentrations.
Use the principle as a quick prediction, then explain it properly in terms of rates: the change briefly makes one reaction faster than the other.

Pressure and temperature 3.1.6.1

Key results

Increasing pressure favours the side with fewer moles of gas. For $N_{2} + 3 H_{2} ⇌ 2 NH_{3}$ (4 mol → 2 mol), more ammonia forms.
Increasing temperature favours the endothermic direction. With $Δ H < 0$ for ammonia synthesis, heating gives less ammonia and $K_{c}$ decreases.

Notes

Pressure has no effect on the position when both sides have equal moles of gas.
Temperature is the only factor that changes the value of $K_{c}$ . Concentration and pressure change the position but not $K_{c}$ .
Cooling an exothermic equilibrium increases $K_{c}$ and the equilibrium yield, but slows the rate.
Worked example: for $2 SO_{2} (g) + O_{2} (g) ⇌ 2 SO_{3} (g)$ , $Δ H = - 196 kJ mol^{- 1}$ , high pressure favours $SO_{3}$ (3 mol gas → 2) and low temperature favours $SO_{3}$ (exothermic forward reaction).
Pressure changes affect only gases. In $CaCO_{3} (s) ⇌ CaO(s) + CO_{2} (g)$ , only the $CO_{2}$ counts, so increasing the pressure shifts the equilibrium to the left.
In the $2 NO_{2} ⇌ N_{2} O_{4}$ equilibrium (brown to colourless, exothermic forward), a sealed tube goes paler in ice and darker in hot water, a visible demonstration of the temperature effect.

The equilibrium constant, Kc 3.1.6.2

Definitions

For a homogeneous equilibrium $a A + b B ⇌ c C + d D$ : $K_{c} = \frac{[ C ] ^{c} [ D ] ^{d}}{[ A ] ^{a} [ B ] ^{b}}$ , with equilibrium concentrations in mol dm⁻³.
Homogeneous equilibrium: all species in the same phase.

Key results

$K_{c} = \frac{[ NH _{3} ] ^{2}}{[ N _{2} ] [ H _{2} ] ^{3}}$ has units $dm^{6} mol^{- 2}$ .
$K_{c} = \frac{[ HI ] ^{2}}{[ H _{2} ] [ I _{2} ]}$ has no units.

Notes

Work out units by substituting mol dm⁻³ for each concentration and cancelling.
A large $K_{c}$ means the equilibrium lies to the right (mostly products); a small one, to the left.
Only temperature changes $K_{c}$ for a given equation; catalysts do not.
Reversing the equation inverts $K_{c}$ ; doubling all coefficients squares it. For $2 HI ⇌ H_{2} + I_{2}$ , $K_{c} = 1/64.0 = 0.0156$ under the same conditions as the example above.
Pure liquids and solids are omitted from $K_{c}$ expressions for heterogeneous equilibria (not required at AS but needed with $K_{p}$ later). In aqueous reactions where water is a reactant in large excess, its concentration is effectively constant.
Typical errors: using initial instead of equilibrium concentrations, forgetting to raise concentrations to the power of their coefficients, and writing reactants over products.

Calculating Kc from equilibrium amounts 3.1.6.2

Key results

$H_{2} + I_{2} ⇌ 2 HI$ in 2.00 dm³: 0.1000 mol of each reactant, 0.1600 mol HI at equilibrium, so 0.0800 mol of each reactant reacted and 0.0200 mol remains.
$K_{c} = \frac{( 0.0800 ) ^{2}}{0.0100 \times 0.0100} = 64.0$ , with no units.

Method

Write initial amounts, the change (from the stoichiometry) and equilibrium amounts in a table.
Divide each equilibrium amount by the volume to get concentrations.
Substitute into the $K_{c}$ expression and work out the units.

In practice1.00 mol H $_{2}$ and 1.00 mol I $_{2}$ in 2.00 dm $^{3}$ form 1.56 mol HI at equilibrium. Find $K_{c}$ .

$H_{2} + I_{2} ⇌ 2 HI : initial 1.00, 1.00, 0; change - 0.78, - 0.78, + 1.56$ from the stoichiometry
$equilibrium: 0.22, 0.22, 1.56 mol \to 0.11, 0.11, 0.78 mol dm^{- 3}$ divide by 2.00 dm³
$K_{c} = \frac{[ HI ] ^{2}}{[ H _{2} ] [ I _{2} ]} = \frac{0.7 8 ^{2}}{0.11 \times 0.11} = 50.3$ no units: they cancel

Notes

Always use equilibrium amounts, never initial ones.
When the total powers on top and bottom are equal, the volume cancels and amounts can be used directly.
Worked example with volume: for $PCl_{5} ⇌ PCl_{3} + Cl_{2}$ in 2.00 dm³, starting with 0.500 mol $PCl_{5}$ and finding 0.200 mol $Cl_{2}$ at equilibrium: $n (PCl_{5}) = 0.300$ mol. $K_{c} = (0.100 \times 0.100) /0.150 = 0.0667 mol dm^{- 3}$ . Here the volume does not cancel.
Esterification: $CH_{3} COOH + C_{2} H_{5} OH ⇌ CH_{3} COOC_{2} H_{5} + H_{2} O$ . Starting with 1.00 mol of acid and alcohol, 0.667 mol of ester forms, so $K_{c} = (0.667)^{2} / (0.333)^{2} = 4.0$ with no units.
The amount of acid at equilibrium in an esterification can be found by titration with standard NaOH, remembering to subtract the acid catalyst that was added.

Catalysts and industrial compromises 3.1.6.1

Notes

A catalyst speeds up the forward and reverse reactions equally, so equilibrium is reached sooner but the position and yield are unchanged.
In the Haber process a low temperature would give a higher equilibrium yield but too slow a rate; a compromise temperature with an iron catalyst gives a good yield quickly.
High pressure increases yield and rate but costs more in energy and equipment, so a compromise pressure is used.
Ammonia is condensed out and the unreacted nitrogen and hydrogen are recycled, so the overall conversion is high even though each pass converts only a fraction.
Typical Haber conditions are about 450 °C and 200 atm (20 MPa), giving about 15% conversion per pass; recycling raises overall conversion to about 98%.
Ethanol is made industrially by hydrating ethene, $C_{2} H_{4} + H_{2} O ⇌ C_{2} H_{5} OH$ ( $Δ H = - 46 kJ mol^{- 1}$ ), at about 300 °C and 60–70 atm with a phosphoric acid catalyst, another compromise between yield, rate and cost; excess steam pushes the equilibrium to the right.
Methanol is made from synthesis gas, $CO + 2 H_{2} ⇌ CH_{3} OH$ (exothermic, 3 mol → 1 mol gas), so the same reasoning favours high pressure and moderate temperature with a catalyst.
In compromise questions, weigh yield, rate, energy cost, equipment cost and safety, and say which factor each condition is chosen for.

Worked example

1.00 mol of ethanoic acid and 1.00 mol of ethanol were mixed and allowed to reach equilibrium:

CH_{3} COOH + C_{2} H_{5} OH ⇌ CH_{3} COOC_{2} H_{5} + H_{2} O

At equilibrium, two-thirds of the acid and alcohol had been converted to ester and water.

Calculate $K_{c}$ for this reaction.

Show worked solution

At equilibrium: moles of ester $=$ moles of water:

= \frac{2}{3} mol

moles of acid remaining $=$ moles of alcohol remaining:

= 1 - \frac{2}{3} = \frac{1}{3} mol

This reaction has equal numbers of moles on each side (1:1:1:1), so if the mixture occupies a volume $V$ , every concentration term is (moles)/ $V$ , and:

K_{c} = \frac{[ ester ] [ water ]}{[ acid ] [ alcohol ]} = \frac{( n _{ester} / V ) ( n _{water} / V )}{( n _{acid} / V ) ( n _{alcohol} / V )}

and because there are two concentration terms on top and two on the bottom, every factor of $V$ cancels, leaving a ratio of moles only:

K_{c} = \frac{( 2/3 ) \times ( 2/3 )}{( 1/3 ) \times ( 1/3 )} = \frac{4/9}{1/9} = 4

Because the powers of concentration are equal on each side of the expression, the units cancel too, so $K_{c} = 4$ with no units.

3.1.7

Oxidation, reduction and redox equations

Oxidation, reduction, oxidising and reducing agents 3.1.7

Definitions

Oxidation: loss of electrons (oxidation state increases). Reduction: gain of electrons (oxidation state decreases).
Oxidising agent: an electron acceptor, which is itself reduced. Reducing agent: an electron donor, which is itself oxidised.

Key results

$Zn(s) + Cu^{2 +} (a q) \to Zn^{2 +} (a q) + Cu(s)$ : zinc is oxidised (0 → +2) and is the reducing agent; $Cu^{2 +}$ is reduced (+2 → 0) and is the oxidising agent.

Notes

Oxidation and reduction always happen together, and the electrons lost equal the electrons gained.
OIL RIG (oxidation is loss, reduction is gain) applies to electrons.
Displacement reactions are redox: chlorine oxidises bromide ions, $Cl_{2} + 2 Br^{-} \to 2 Cl^{-} + Br_{2}$ ; here chlorine is the oxidising agent and bromide the reducing agent.
Older definitions in terms of oxygen gain or hydrogen loss are special cases of electron transfer: when magnesium burns, Mg loses electrons to O even though no ions are visible in the gas.
A disproportionation reaction is one in which the same element is both oxidised and reduced, e.g. $Cl_{2} + 2 NaOH \to NaCl + NaClO + H_{2} O$ , where chlorine goes from 0 to −1 and +1.
Identify redox by comparing oxidation states before and after; if no element changes oxidation state (as in neutralisation or precipitation), the reaction is not redox.

Assigning oxidation states 3.1.7

Definitions

Oxidation state: a formal count of the electrons an atom has lost or gained if all its bonds were ionic; it is not necessarily a real charge.

Key results

Uncombined elements 0; monatomic ions equal their charge; the states in a species add up to its overall charge.
Group 1 is +1, Group 2 is +2 and fluorine is −1 in compounds. Oxygen is usually −2 (−1 in peroxides, +2 in $OF_{2}$ ); hydrogen is usually +1 (−1 in metal hydrides such as NaH).
S in $H_{2} SO_{4}$ : $2 (+ 1) + x + 4 (- 2) = 0$ , so $+ 6$ . Mn in $MnO_{4}^{-}$ : $x + 4 (- 2) = - 1$ , so $+ 7$ .

Notes

Write oxidation states with the sign first (+2) and ionic charges with the number first (2+).
Roman numerals in names give oxidation states: manganate(VII) contains Mn(+7), iron(II) is $Fe^{2 +}$ .
More worked examples: Cr in $Cr_{2} O_{7}^{2 -}$ is $+ 6$ ( $2 x - 14 = - 2$ ); N in $NH_{4}^{+}$ is $- 3$ ; S in $S_{2} O_{3}^{2 -}$ averages $+ 2$ ; C in $CH_{4}$ is $- 4$ and in $CO_{2}$ is $+ 4$ .
Apply the rules in priority order: F always −1, then Group 1 and 2 metals, then H (+1) and O (−2), and work out the remaining atom last.
An average oxidation state can be fractional when atoms of one element are in different environments: Fe in $Fe_{3} O_{4}$ averages $+ \frac{8}{3}$ because the compound contains both $Fe^{2 +}$ and $Fe^{3 +}$ .

Balancing half-equations in acidic solution 3.1.7

Key results

$MnO_{4}^{-} (a q) + 8 H^{+} (a q) + 5 e^{-} \to Mn^{2 +} (a q) + 4 H_{2} O (l)$ : five electrons gained as Mn goes from +7 to +2.

Method

Balance the element changing oxidation state.
Balance oxygen by adding $H_{2} O$ .
Balance hydrogen by adding $H^{+}$ .
Balance charge by adding electrons to the more positive side.

In practiceWrite the half-equation for $MnO_{4}^{-} \to Mn^{2 +}$ in acidic solution.

$MnO_{4}^{-} \to Mn^{2 +}$ Mn already balanced
$MnO_{4}^{-} \to Mn^{2 +} + 4 H_{2} O$ balance O with water
$MnO_{4}^{-} + 8 H^{+} \to Mn^{2 +} + 4 H_{2} O$ balance H with $H^{+}$
$MnO_{4}^{-} + 8 H^{+} + 5 e^{-} \to Mn^{2 +} + 4 H_{2} O$ charge: $+ 7$ on the left, $+ 2$ on the right

Notes

Check: both sides have 1 Mn, 4 O and 8 H, and total charge $- 1 + 8 - 5 = + 2$ .
The number of electrons equals the change in oxidation state, a useful independent check.
Worked example: dichromate(VI). $Cr_{2} O_{7}^{2 -} \to 2 Cr^{3 +}$ ; add $7 H_{2} O$ on the right; add $14 H^{+}$ on the left; charge on the left $- 2 + 14 = + 12$ , on the right $+ 6$ , so add $6 e^{-}$ on the left: $Cr_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 Cr^{3 +} + 7 H_{2} O$ .
Oxidation half-equations have electrons on the right, e.g. $H_{2} O_{2} \to O_{2} + 2 H^{+} + 2 e^{-}$ and $2 I^{-} \to I_{2} + 2 e^{-}$ .
The four-step method works for any species in acid. For reactions in alkali, AQA questions usually supply the half-equations.

Combining half-equations 3.1.7

Key results

$Fe^{2 +} \to Fe^{3 +} + e^{-}$ (×5) with the manganate(VII) half-equation gives $MnO_{4}^{-} + 8 H^{+} + 5 Fe^{2 +} \to Mn^{2 +} + 4 H_{2} O + 5 Fe^{3 +}$ .

Method

Multiply each half-equation so the electrons lost equal the electrons gained.
Add the half-equations and cancel the electrons (and any species on both sides).

In practiceCombine the manganate(VII) half-equation with $Fe^{2 +} \to Fe^{3 +} + e^{-}$ .

$5 Fe^{2 +} \to 5 Fe^{3 +} + 5 e^{-}$ × 5 to match the five electrons
$MnO_{4}^{-} + 8 H^{+} + 5 Fe^{2 +} \to Mn^{2 +} + 4 H_{2} O + 5 Fe^{3 +}$ add: the electrons cancel
$charge: - 1 + 8 + 10 = 17 = 2 + 15$ check

Notes

Check the overall equation for atoms and total charge: +17 on each side here. No electrons should remain.
This 1 : 5 ratio is the basis of manganate(VII) titrations for iron(II), revisited in transition-metal chemistry.
Worked example: dichromate oxidising iodide. $Cr_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 Cr^{3 +} + 7 H_{2} O$ and $2 I^{-} \to I_{2} + 2 e^{-}$ (×3) combine to $Cr_{2} O_{7}^{2 -} + 14 H^{+} + 6 I^{-} \to 2 Cr^{3 +} + 3 I_{2} + 7 H_{2} O$ .
When the same species ( $H^{+}$ or $H_{2} O$ ) appears on both sides after adding, cancel the smaller number from both sides.
The combined equation gives the reacting ratio used in redox titrations. In iron analysis, for example, 1 mol of $MnO_{4}^{-}$ reacts with 5 mol of $Fe^{2 +}$ , so a titre in moles of manganate is multiplied by 5.