3.1 Physical chemistry — notes on one page

3.1.1.1 · Fundamental particles

• Atomic models changed when new experimental evidence could not be explained by an earlier model.
• The accepted model has a very small nucleus containing protons and neutrons, with electrons occupying the surrounding space.
• A proton has relative charge $+1$ and relative mass $1$; a neutron has charge $0$ and relative mass $1$; an electron has charge $-1$ and relative mass about $1/1836$.
• A neutral atom has equal numbers of protons and electrons.
• An ion forms only by losing or gaining electrons, so its nucleus and element identity remain unchanged.

Exam tip. For a particle-comparison question, state both relative mass and relative charge using the values in the data table.

3.1.1.2 · Mass number and isotopes

Definition. The mass number A is the total number of protons and neutrons in an atom; the atomic (proton) number Z is the number of protons. Isotopes are atoms of the same element with the same atomic number but different mass numbers.

• You must be able to determine the number of fundamental particles in atoms and ions from the mass number, the atomic number and the charge:
• neutrons are A minus Z, and electrons equal Z adjusted for the charge. The specification then builds the time-of-flight mass spectrometer on top of this, limited to five stages you should be able to name in order: ionisation, acceleration to give all ions the same kinetic energy, ion drift, ion detection, and data analysis.
• Ionisation is by electron impact or by electrospray; electron impact knocks an electron out, which is written as X(g) gives X⁺(g) + e⁻. Because acceleration gives every ion the same kinetic energy, heavier ions move more slowly and take longer to drift the flight tube, so time of flight increases with mass.
• The mass spectrometer gives accurate relative isotopic masses and relative abundances, which lets you calculate a relative atomic mass as a weighted mean: multiply each isotopic mass by its abundance, add, and divide by the total abundance.
• The same instrument identifies elements and determines relative molecular mass from the molecular ion peak.

What the examiner wants. State symbols on ionisation equations and the exact number of decimal places asked for on a relative atomic mass. The scheme routinely specifies '1 decimal place' or 'must be an integer' and marks strictly.

Exam tip. In a calculation from a mass spectrum, show the abundance weighting before giving the final weighted mean.

3.1.1.3 · Electron configuration

• For atoms up to $Z=36$, electrons fill $1s$, $2s$, $2p$, $3s$, $3p$, $4s$, $3d$, then $4p$. Chromium is $\mathrm{[Ar]3d^5\,4s^1}$ and copper is $\mathrm{[Ar]3d^{10}\,4s^1}$; transition-metal ions lose $4s$ electrons before $3d$.
• First ionisation energy removes one electron from each atom in one mole of gaseous atoms.
• Across Period 3, increasing nuclear charge generally raises first ionisation energy, with sub-shell and electron-pairing anomalies.
• Down Group 2, increasing distance and shielding lower it.
• A large successive-ionisation jump shows removal from a shell closer to the nucleus and provides evidence for shell structure.

Exam tip. For a successive-ionisation graph, locate the large jump first and link it explicitly to a change of electron shell.

3.1.2.1 · Relative atomic mass and relative molecular mass

• Relative atomic mass, $A_r$, is the weighted mean mass of an atom of an element relative to one twelfth of the mass of a carbon-12 atom. Relative molecular mass, $M_r$, is the mean mass of a molecule on the same scale.
• Both are ratios and therefore have no units.
• Calculate $M_r$ by multiplying each element's $A_r$ by the number of its atoms and summing.
• For an ionic substance, the equivalent quantity is relative formula mass because the solid contains a lattice rather than separate molecules.
• Brackets and waters of crystallisation must be counted fully.

Exam tip. Expand every bracketed or hydrated group into atom counts before substituting relative atomic masses.

3.1.2.2 · The mole and the Avogadro constant

• One mole contains the Avogadro constant, $N_A$, of specified entities, such as atoms, molecules, ions, electrons or formula units.
• Amount of substance links measurable quantities to particle number: $n=m/M$, $N=nN_A$, and for a solution $n=cV$ when $V$ is in $\mathrm{dm^3}$.
• A secure calculation first converts the given quantity to moles, applies the chemical formula or equation ratio, then converts to the requested quantity.
• The named entity matters: one mole of $\mathrm{MgCl_2}$ formula units contains two moles of chloride ions.

Exam tip. Write the entity beside every mole value so that formula-unit multipliers are not lost.

3.1.2.3 · The ideal gas equation

• The ideal gas equation is $pV=nRT$.
• The gas constant $R$ has the value $8.31$ in units of $\mathrm{J\,K^{-1}\,mol^{-1}}$, so pressure must be in $\mathrm{Pa}$, volume in $\mathrm{m^3}$, temperature in $\mathrm{K}$ and amount in $\mathrm{mol}$.
• Convert $\mathrm{kPa}$ to $\mathrm{Pa}$ by multiplying by $10^3$, $\mathrm{cm^3}$ to $\mathrm{m^3}$ by multiplying by $10^{-6}$, and $\mathrm{dm^3}$ to $\mathrm{m^3}$ by multiplying by $10^{-3}$.
• Convert Celsius using $T/\mathrm{K}=\theta/^{\circ}\mathrm{C}+273$.
• Rearranging the equation before substitution reduces unit and algebra errors in multi-step calculations.

Exam tip. Show every SI conversion on a separate line before substituting into $pV=nRT$.

3.1.2.4 · Empirical and molecular formula

• An empirical formula is the simplest whole-number ratio of atoms of each element; a molecular formula gives the actual numbers of atoms in one molecule.
• From masses or percentages, divide each value by the element's $A_r$ to obtain relative moles, divide every result by the smallest, then multiply all ratios by the same small integer if necessary.
• A ratio near $1.5$ requires doubling, not rounding.
• To obtain a molecular formula, calculate the empirical formula mass and find the integer $M_r/(\text{empirical formula mass})$.
• Combustion data must be converted from product moles to constituent atom moles.

Exam tip. Keep unrounded mole values until the common whole-number multiplier has been identified.

3.1.2.5 · Balanced equations and associated calculations

Definition. A balanced equation gives the mole ratio in which substances react. Percentage atom economy is the molecular mass of the desired product divided by the sum of the molecular masses of all the reactants, multiplied by 100.

• Almost every quantitative question in 7405 routes through this point. You are expected to write balanced equations, full and ionic, for reactions you have studied, and to balance unfamiliar equations when the reactants and products are given.
• From a balanced equation you must be able to calculate masses, volumes of gases, percentage yields, percentage atom economies, and concentrations and volumes for reactions in solution. Percentage yield compares the mass actually obtained with the theoretical maximum from the limiting reactant, so the first step is always to identify which reactant limits the reaction.
• Atom economy, by contrast, is fixed by the equation alone and is unaffected by how well the reaction is carried out; a high atom economy means less waste, better use of raw materials and lower disposal costs, which is the economic, ethical and environmental argument the specification asks for.
• Gas volume questions use pV = nRT with SI units — pressure in Pa, volume in m³, temperature in K — and the commonest lost marks come from failing to convert cm³ to m³ or kPa to Pa.
• Report answers to the number of significant figures the least accurate datum allows.

What the examiner wants. Every step written on its own line with units. These schemes award marks step by step and allow error carried forward throughout, so a wrong intermediate value used correctly still earns the marks that follow it.

Exam tip. For a multi-stage calculation, label the moles of each substance and show the equation ratio explicitly.

3.1.3.1 · Ionic bonding

• Ionic bonding is the strong electrostatic attraction between oppositely charged ions throughout a giant lattice. Electron transfer forms the ions but is not itself the bond.
• Simple-ion charges can be predicted from Periodic Table groups, and the ionic formula uses the smallest whole-number ratio giving zero overall charge.
• Compound ions remain intact; for example, two $\mathrm{Al^{3+}}$ ions balance three $\mathrm{SO_4^{2-}}$ ions to give $\mathrm{Al_2(SO_4)_3}$.
• Brackets are required when more than one compound ion is present.
• Explanations of high melting point must refer to strong attractions acting in many directions through the lattice.

Exam tip. A bonding-definition question requires both electrostatic attraction and oppositely charged ions.

3.1.3.2 · Nature of covalent and dative covalent bonds

• A covalent bond is a shared pair of electrons. Double and triple bonds contain two and three shared pairs.
• In an ordinary covalent bond, each bonded atom contributes one electron to the shared pair. In a co-ordinate, or dative covalent, bond, both electrons in the shared pair are supplied by one atom.
• The displayed-formula arrow starts at a lone pair on the electron-pair donor and points towards the electron-deficient acceptor.
• Once formed, the dative bond behaves like any other covalent bond.
• Dot-and-cross diagrams must show outer-shell electrons, charges and the origin of the bonding pair accurately.

Exam tip. In a dative-bond diagram, place the tail of the arrow exactly on the donating lone pair.

3.1.3.3 · Metallic bonding

• Metallic bonding is the strong electrostatic attraction between a lattice of positive metal ions and delocalised electrons.
• The electrons are not attached to a particular ion and can move through the structure, so a metal conducts electricity when solid or liquid.
• Metallic attraction is non-directional, allowing layers of ions to slide while remaining attracted to the delocalised electrons; this explains malleability and ductility.
• Bond strength generally increases with greater ionic charge, more delocalised electrons and smaller ionic radius because attraction between the ions and electron density is stronger.
• Metals do not consist of separate molecules.

Exam tip. A conductivity explanation must name the charged particles and state that the delocalised electrons are free to move.

3.1.3.4 · Bonding and physical properties

• The four crystal types are ionic, metallic, macromolecular and molecular. Sodium chloride, magnesium, diamond or graphite, and iodine or ice are respective examples.
• Ionic and metallic solids contain giant lattices; macromolecular substances have many covalent bonds; molecular solids contain discrete molecules held by intermolecular forces. Melting a molecular solid overcomes intermolecular attractions, not covalent bonds inside molecules.
• Ionic substances conduct only when molten or dissolved because ions can move. Metals and graphite conduct through delocalised electrons.
• Structure is inferred from combined melting-point, solubility and conductivity evidence.
• A requested diagram must show the specified number and type of particles.

Exam tip. When deducing crystal type, link each observed property to the particles, attractions and available mobile charge carriers.

3.1.3.5 · Shapes of simple molecules and ions

• Electron charge clouds repel and arrange as far apart as possible around a central atom. Count each bonding region, including a multiple bond as one region, and each lone pair.
• Repulsion decreases in the order lone pair–lone pair $>$ lone pair–bond pair $>$ bond pair–bond pair, so lone pairs compress adjacent bond angles.
• Base geometries are linear $180^{\circ}$, trigonal planar $120^{\circ}$, tetrahedral $109.5^{\circ}$, trigonal bipyramidal $90^{\circ}$ and $120^{\circ}$, and octahedral $90^{\circ}$.
• In a trigonal bipyramid, lone pairs occupy equatorial positions first.
• Shape names describe atom positions, not lone pairs.

Exam tip. A full shape answer states the number of bonding and lone pairs, the repulsion argument, the shape and the bond angle.

3.1.3.6 · Bond polarity

• Electronegativity is the power of an atom to attract the bonding pair of electrons in a covalent bond.
• A difference in electronegativity produces an unequal electron distribution: the more electronegative atom is $\delta^-$ and the other is $\delta^+$.
• A bond dipole has both magnitude and direction.
• To decide whether a whole molecule has a permanent dipole, identify every polar bond and use the molecular shape to combine their dipoles.
• Equal bond dipoles arranged symmetrically cancel, so a molecule can contain polar bonds but have no overall permanent dipole.

Exam tip. A polarity explanation must combine electronegativity difference with the molecule's three-dimensional symmetry.

3.1.3.7 · Forces between molecules

• Induced dipole–dipole forces act between all atoms and molecules. Their strength generally increases with electron number and with molecular surface contact.
• Permanent dipole–dipole attractions also act between polar molecules.
• Hydrogen bonding requires hydrogen covalently bonded to nitrogen, oxygen or fluorine and a lone pair on nitrogen, oxygen or fluorine in another molecule.
• Boiling-point comparisons must identify the forces present and compare their overall strength, not simply molecular mass.
• In ice, hydrogen bonds hold water molecules in an open arrangement, making ice less dense than liquid water; the bonds are not described as inherently stronger.

Exam tip. For a boiling-point comparison, name every force present and then identify the strongest difference between the substances.

3.1.4.1 · Enthalpy change

• Enthalpy change, $\Delta H$, is the heat-energy change at constant pressure.
• An exothermic process transfers heat to the surroundings and has $\Delta H<0$; an endothermic process absorbs heat and has $\Delta H>0$.
• A standard enthalpy change uses substances in their standard states at $100\,\mathrm{kPa}$ and a stated temperature.
• Standard enthalpy of combustion is the enthalpy change when one mole of a substance is completely burned in oxygen under standard conditions.
• Standard enthalpy of formation is the change when one mole of a compound forms from its constituent elements in their standard states.

Exam tip. In an enthalpy-definition question, the one-mole condition, standard states and standard conditions are separate marking points.

3.1.4.2 · Calorimetry

• Calorimetry estimates heat transfer using $q=mc\Delta T$, where $m$ is the mass undergoing the temperature change, $c$ is its specific heat capacity and $\Delta T$ is the measured temperature change.
• In solution calorimetry, total solution volume and density commonly estimate mass, while the limiting reagent supplies the reacting amount.
• Convert heat to a molar enthalpy using $\Delta H=-q/n$ when a temperature rise shows that the reaction released heat.
• Required practical 2 measures an enthalpy change, for example by dissolution, neutralisation, displacement or alcohol combustion.
• Heat loss, incomplete combustion, evaporation and heating the apparatus create systematic differences that must be evaluated.

Exam tip. For an uncertainty evaluation, state the direction in which the named heat-transfer error changes the calculated $\Delta H$.

3.1.4.3 · Applications of Hess's law

• Hess's law states that the enthalpy change for a reaction is independent of the route, provided the initial and final states are identical. A Hess cycle replaces an unknown route with known enthalpy changes.
• With formation enthalpies, $\Delta_\mathrm{r}H^\circ=\sum\Delta_\mathrm{f}H^\circ(\text{products})-\sum\Delta_\mathrm{f}H^\circ(\text{reactants})$.
• With combustion enthalpies leading to common products, use the total for reactants minus the total for products.
• Every equation coefficient multiplies its enthalpy value.
• Reversing a chemical equation reverses the sign of $\Delta H$; the algebra must follow the arrows in the cycle.

Exam tip. Write the complete products-minus-reactants or reactants-minus-products expression before entering numerical values.

3.1.4.4 · Bond enthalpies

• Mean bond enthalpy is the mean energy required to break one mole of a specified covalent bond in gaseous molecules.
• Bond breaking is endothermic and bond formation is exothermic, so estimate a gaseous reaction enthalpy using $\Delta H=\sum E(\text{bonds broken})-\sum E(\text{bonds formed})$.
• Count every bond in the displayed structures, include equation coefficients, and cancel unchanged bonds only after the full count is secure.
• The result is approximate because a mean bond enthalpy averages that bond across different molecular environments; it is not the exact energy for one particular molecule.

Exam tip. Draw or inspect every displayed bond and present separate totals for bonds broken and bonds formed.

3.1.5.1 · Collision theory

• Particles must collide before they can react. A collision is successful only when the particles have energy at least equal to the activation energy.
• Activation energy is the minimum energy required for a reaction to occur following a collision between particles. Collision theory separates collision frequency from the fraction of collisions that are successful.
• Concentration and pressure mainly alter collision frequency; temperature strongly alters the fraction above the activation energy.
• Most collisions are unsuccessful because they lack sufficient energy, not because the particles failed to meet.
• Explanations should identify which factor changes and how this affects successful collisions per second.

Exam tip. A rate explanation should finish with the change in successful collisions per second.

3.1.5.2 · Maxwell–Boltzmann distribution

• A Maxwell–Boltzmann distribution for a gas plots number of molecules against molecular energy. The curve starts at the origin, rises to a most-probable-energy peak, then approaches the energy axis without touching it because there is no maximum molecular energy.
• The area under the curve represents the total number of molecules.
• Equal samples at different temperatures therefore have equal total areas.
• At higher temperature the peak becomes lower and moves right, the curve broadens, and a greater area lies beyond a fixed activation energy.
• A sketch must preserve these features and label both axes and $E_\mathrm{a}$.

Exam tip. On a distribution sketch, label the higher-temperature curve and shade or describe the area beyond $E_\mathrm{a}$.

3.1.5.3 · Effect of temperature on reaction rate

• Rate of reaction is the change in concentration of a reactant or product per unit time, although another measured quantity may track reaction progress. Increasing temperature raises mean molecular energy and collision frequency.
• More importantly, the Maxwell–Boltzmann distribution changes so a much greater fraction of molecules has energy at least equal to $E_\mathrm{a}$.
• Consequently, a much larger proportion of collisions is successful and the rate rises.
• For an uncatalysed route, activation energy remains unchanged.
• Required practical 3 investigates how reaction rate changes with temperature, commonly using an initial-rate method with sodium thiosulfate and hydrochloric acid, controlled quantities and repeat measurements.

Exam tip. Use the phrase 'greater proportion of molecules with energy at least equal to $E_\mathrm{a}$' in a temperature-rate explanation.

3.1.5.4 · Effect of concentration and pressure

• Increasing solution concentration places more reactant particles in each unit volume, so particles collide more frequently.
• Increasing the pressure of reacting gases by decreasing their volume likewise raises the number of particles per unit volume and collision frequency.
• Provided temperature is constant, neither change alters the Maxwell–Boltzmann energy distribution, the particles' mean energy or the activation energy.
• The fraction of collisions with sufficient energy therefore remains the same, but more collisions occur each second and hence more successful collisions occur each second.
• Pressure arguments apply to gaseous reactants; changing pressure has little direct effect on the concentration of a solid.

Exam tip. When temperature is fixed, explain concentration or pressure effects through particle density and collision frequency.

3.1.5.5 · Catalysts

• A catalyst increases reaction rate without being changed in chemical composition or amount by the overall reaction. It provides an alternative route with a lower activation energy.
• At fixed temperature, the Maxwell–Boltzmann distribution is unchanged, but the lower $E_\mathrm{a}$ threshold leaves a larger area of molecules able to react, so a greater fraction of collisions is successful.
• A catalyst does not change reactant or product enthalpies, $\Delta H$, the equilibrium constant or the equilibrium position.
• It accelerates both forward and reverse reactions and therefore allows equilibrium to be reached sooner.
• An intermediate may form and later be regenerated.

Exam tip. On a catalysed profile, keep reactant and product levels fixed and lower only the pathway peak.

3.1.6.1 · Chemical equilibria and Le Chatelier's principle

Definition. In a reversible reaction at equilibrium the forward and reverse reactions proceed at equal rates and the concentrations of reactants and products remain constant. Le Chatelier's principle states that when a change is applied to a system at equilibrium, the position of equilibrium shifts so as to oppose that change.

• Applying the principle needs three linked statements every time: name the change, say which way the equilibrium shifts, and say what that shift opposes. Increasing the concentration of a reactant shifts the equilibrium to the right to reduce that concentration.
• Increasing the total pressure shifts the equilibrium towards the side with fewer moles of gas, because that reduces the pressure; if both sides have the same number of gas moles, pressure has no effect on the position.
• Increasing the temperature shifts the equilibrium in the endothermic direction, because that absorbs energy and opposes the temperature rise — so for an exothermic forward reaction, raising the temperature lowers the yield.
• A catalyst does not affect the position of equilibrium at all: it speeds up the forward and reverse reactions equally and only shortens the time taken to reach equilibrium.
• That is why industrial processes use a compromise temperature and pressure: a higher temperature gives a faster rate but a lower yield for an exothermic reaction, and a higher pressure gives a better yield but costs more in energy and stronger vessels, so the conditions chosen are an economic balance rather than the thermodynamic optimum.

What the examiner wants. The word 'opposes' and the reason for the shift, not just its direction. The scheme explicitly ignores 'favours', and a shift stated without what it opposes is capped short of full marks.

Exam tip. For an industrial-condition question, separate the equilibrium-yield effect from the rate, cost and safety compromise.

3.1.6.2 · Equilibrium constant Kc for homogeneous systems

• For $aA+bB\rightleftharpoons cC+dD$, $K_\mathrm{c}=[C]^c[D]^d/([A]^a[B]^b)$ using equilibrium concentrations and equation coefficients as powers.
• Calculate equilibrium amounts through the stoichiometric change, then divide by volume in $\mathrm{dm^3}$ before substitution.
• Derive units from the net concentration powers; they can cancel.
• At fixed temperature, changing concentration, pressure or adding a catalyst does not change $K_\mathrm{c}$, although the equilibrium position may move.
• Temperature changes $K_\mathrm{c}$: heating increases it for an endothermic forward reaction and decreases it for an exothermic forward reaction.

Exam tip. Write the symbolic $K_\mathrm{c}$ expression and its units before substituting equilibrium values.

3.1.7 · Oxidation, reduction and redox equations

Definition. Oxidation is the process of electron loss; oxidising agents are electron acceptors. Reduction is the process of electron gain; reducing agents are electron donors. A change in the oxidation state of an element identifies which element has been oxidised or reduced.

• Assigning oxidation states follows fixed rules: an uncombined element is 0; a simple ion's state equals its charge; the states in a compound sum to zero and in an ion sum to the charge;
• oxygen is normally -2 except in peroxides and with fluorine; hydrogen is normally +1 except in metal hydrides; and fluorine is always -1. Work out the states before and after, then name the element that has gone up as oxidised and the element that has gone down as reduced.
• Half-equations are written separately for the oxidation and the reduction, each balanced for atoms first, then for oxygen using H₂O, then for hydrogen using H⁺, and finally for charge using electrons. To combine them into an overall equation, multiply each half-equation so that the electrons cancel exactly, then add them and cancel any species appearing on both sides.
• Disproportionation, examined under 3.2.3 (Group 7), is the special case where the same element is both oxidised and reduced in one reaction, and the mark scheme wants you to say which species it becomes in each direction, not just to name the term.
• The oxidising agent is the species that is itself reduced, and the reducing agent is the one that is itself oxidised.

What the examiner wants. Both oxidation states quoted with signs and the direction of change stated. Naming the term 'disproportionation' alone is explicitly ignored — the products of each half-process must be identified.

Exam tip. For a 'write the overall redox equation' question, show balanced half-equations and the electron multiplier before cancellation.

3.1.8.1 · Born–Haber cycles (A-level only)

Definition. Lattice enthalpy can be defined either as the enthalpy of lattice dissociation — the enthalpy change when one mole of an ionic solid separates into gaseous ions — or as the enthalpy of lattice formation, the reverse process. A Born-Haber cycle uses Hess's law to calculate it from measurable enthalpy changes.

• The data used in a Born-Haber cycle are the enthalpy of formation of the compound, the enthalpy of atomisation of each element, the ionisation energies of the metal, the electron affinities of the non-metal, and the lattice enthalpy itself. Each definition must be reproduced exactly, including 'per mole' and gaseous state symbols.
• The cycle is built by taking two routes from the elements in their standard states to the ionic solid: the direct route is the enthalpy of formation, and the indirect route atomises both elements, ionises the metal, adds electrons to the non-metal and then forms the lattice.
• Equating the two routes and rearranging gives the unknown.
• Two arithmetic traps account for most lost marks: every atomisation and electron affinity must be multiplied by the number of that ion in the formula, so a chloride of formula MCl₂ needs two atomisations and two electron affinities; and the sign of the final answer must match the definition used, since lattice formation is negative and lattice dissociation positive.
• The same cycles give enthalpies of solution from lattice enthalpy and enthalpies of hydration, and comparing an experimental Born-Haber lattice enthalpy with a perfect-ionic-model value provides evidence for covalent character.

What the examiner wants. The cycle expression written out before any numbers. It carries its own mark, and a sign error then costs one mark instead of the whole question.

Exam tip. Label every gaseous, aqueous and solid state in a Born–Haber cycle because each enthalpy definition depends on state.

3.1.8.2 · Gibbs free-energy change, ΔG, and entropy change, ΔS (A-level only)

Definition. Entropy change, delta S, measures the change in disorder of a system. The Gibbs free-energy change is delta G = delta H - T delta S, and a reaction is feasible when delta G is zero or negative.

• Enthalpy alone is not sufficient to explain feasible change, which is why entropy is introduced: endothermic reactions can still be feasible if they create enough disorder. Entropy increases when a solid becomes a liquid or a gas, when a solid dissolves, and when the number of moles of gas increases; it decreases in the reverse cases.
• Entropy change for a reaction is the sum of the absolute entropies of the products minus the sum for the reactants, each multiplied by its coefficient, and comes out in J K⁻¹ mol⁻¹. That unit mismatch is the central trap:
• delta H is in kJ mol⁻¹, so before substituting into delta G = delta H - T delta S you must either divide delta S by 1000 or multiply delta H by 1000, and the mark scheme awards that conversion its own mark.
• Temperature is always in kelvin. To find the temperature at which a reaction becomes feasible, set delta G to zero, which gives T = delta H / delta S with both in consistent units.
• A plot of delta G against T is a straight line with intercept delta H and gradient -delta S, so both can be read off a graph.

What the examiner wants. The kJ-to-J conversion shown as its own step, temperature in kelvin, and the feasibility conclusion stated in words at the end rather than left implicit in the sign.

Exam tip. In a temperature-threshold calculation, set $\Delta G=0$ explicitly before rearranging for $T$.

3.1.9.1 · Rate equations (A-level only)

Definition. A rate equation has the form rate = k[A]^m[B]ⁿ, where m and n are the orders with respect to A and B and k is the rate constant. Orders are restricted to 0, 1 and 2, and the overall order is the sum of the individual orders.

• The order with respect to a reactant is the power to which its concentration is raised in the rate equation, and it must be found experimentally — it can never be read off the balanced equation.
• The rate constant k depends on temperature through the Arrhenius equation, k = Ae^−Ea/RT, where A is the Arrhenius constant, Ea the activation energy and T the temperature in kelvin; both the equation and the gas constant R are given when required.
• Taking logarithms rearranges this into ln k = -Ea/RT + ln A, so a plot of ln k against 1/T is a straight line of gradient -Ea/R and intercept ln A; multiply the gradient by -R to obtain Ea, then divide by 1000 to quote it in kJ mol⁻¹.
• Raising the temperature increases k, which is why the rate rises.
• The units of k are the part most often dropped: work them out each time from rate in mol dm⁻³ s⁻¹ divided by the concentration terms, so a first-order overall reaction gives s⁻¹, second order gives mol⁻¹ dm³ s⁻¹, and third order gives mol⁻² dm⁶ s⁻¹.

What the examiner wants. Units on the rate constant every time. They carry an independent mark that can be earned even when the numerical value is wrong, and they are the single most commonly omitted mark on this point.

Exam tip. For a rate-constant calculation, show the rearranged equation and derive units from that equation for the method marks.

3.1.9.2 · Determination of rate equation (A-level only)

Definition. The rate equation is an experimentally determined relationship. Orders with respect to reactants are found from rate-concentration data or graphs, and can give information about the rate-determining step of the reaction.

• There are two experimental routes. In the continuous monitoring method you follow one concentration against time and take the gradient of a tangent to the curve to find the rate at that moment;
• a straight line of constant gradient means the rate is independent of concentration, so the reaction is zero order in that species, and the gradient itself is then the rate constant.
• In the initial rate method you draw the tangent at time zero for several starting concentrations and compare: if doubling a concentration doubles the initial rate the order is 1, if it quadruples the rate the order is 2,
• and if the rate is unchanged the order is 0. Plotting rate against concentration gives a straight line through the origin for first order and a curve for second order, and plotting rate against concentration squared straightens a second-order relationship. Once every order is known, write the rate equation and substitute one experiment's data to find k.
• The orders then constrain the mechanism: the species appearing in the rate equation, with their orders, are those involved in or before the rate-determining step, and a species absent from the rate equation takes no part up to that step. Required practical 7 covers both methods.

What the examiner wants. A tangent that touches the curve without crossing it, and the values actually read off it shown in the working. Both are separately credited before the numerical answer.

Exam tip. When deducing order from trials, compare a pair in which only one reactant concentration changes and state both concentration and rate factors.

3.1.10 · Equilibrium constant Kp for homogeneous systems (A-level only)

Definition. Kp is the equilibrium constant calculated from partial pressures for a homogeneous gas-phase system at constant temperature. The partial pressure of a gas is its mole fraction multiplied by the total pressure.

• Build a Kp expression exactly as you would Kc, but with partial pressures and round brackets — square brackets denote concentrations and are penalised here. Products go on top, reactants underneath, each raised to the power of its coefficient in the balanced equation.
• Calculating partial pressures is a three-step routine: find the equilibrium amount in moles of each gas, add them to get the total, divide each by the total to get its mole fraction, then multiply by the total pressure. Mole fractions must sum to 1, which is a useful check.
• The units of Kp follow from the expression:
• cancel the pressure units that appear top and bottom, so a reaction with equal numbers of gas moles on each side gives a dimensionless Kp — and the mark scheme wants that reasoning stated as equal moles of gas on both sides, not merely as 'the units cancel'. Only temperature changes the value of Kp.
• Changing the total pressure or adding a catalyst shifts nothing in the constant itself: a catalyst changes the rate at which equilibrium is reached, not its position or value. For an exothermic forward reaction, raising the temperature decreases Kp.

What the examiner wants. Round brackets for partial pressures, the units of Kp derived from the expression, and only temperature ever allowed to change its value.

Exam tip. In a $K_p$ calculation, write the expression first and keep one pressure unit throughout so the final units follow correctly.

3.1.11.1 · Electrode potentials and cells (A-level only)

Definition. The standard electrode potential, E standard, is the potential of a half-cell measured relative to the standard hydrogen electrode under standard conditions of 298 K, 100 kPa and 1.00 mol dm⁻³ solution of ions.

• Half-equations for electrode reactions are written by the IUPAC convention as reductions, with the electrons on the left, and a more positive E standard means the species on the left is the stronger oxidising agent and the reduction is more likely to occur. To predict a reaction, compare the two E standard values:
• the half-cell with the more positive value runs as written, as a reduction, and the other runs in reverse, as an oxidation.
• The EMF of a cell is the right-hand electrode potential minus the left-hand one, and a spontaneous reaction always gives a positive EMF, so a negative answer signals that the direction has been written backwards.
• The conventional cell representation puts the oxidised and reduced forms of each half-cell on either side of a double vertical line for the salt bridge, with a single line for a phase boundary and a comma between species in the same phase; a platinum electrode is added when neither species is a solid conductor.
• The standard hydrogen electrode itself is hydrogen gas at 100 kPa bubbled over platinum in 1.00 mol dm⁻³ H⁺ at 298 K, and is defined as exactly 0.00 V. Required practical 8 measures a cell EMF with a high-resistance voltmeter and a salt bridge.