State the relative charge and relative mass of an electron.
Physical chemistry
Notes and three levels of exam-style practice for each registered specification leaf in this section.
Open the printable packFundamental particles
- Knowledge of atomic structure has evolved as new evidence produced improved models. The current model has a tiny nucleus containing protons and neutrons, with electrons occupying the space around it.
- A proton has relative charge and relative mass ; a neutron has charge and relative mass ; an electron has charge and relative mass about .
- For a neutral atom, the numbers of protons and electrons are equal. An ion forms when electrons are lost or gained; the nucleus is unchanged.
- When finding an ion's charge, subtract the number of electrons from the number of protons. A common error is to include neutrons, which carry no charge.
Tier 1 · Easy
Tier 2 · Standard
A particle contains 15 protons, 16 neutrons and 18 electrons. State its overall charge and identify which particles are in its nucleus.
Tier 3 · Hard
The nucleus of an atom has charge . Use the proton charge to determine the number of protons. The atom gains two electrons; state the resulting ion charge and its number of electrons.
Mass number and isotopes
- The atomic number is the number of protons, while the mass number is the total number of protons and neutrons; therefore neutrons .
- Isotopes are atoms of the same element with the same number of protons but different numbers of neutrons. Their chemical properties are similar because they have the same electron configuration.
- In a TOF mass spectrometer, gaseous particles are ionised, accelerated to the same kinetic energy, separated by flight time and detected. For mononuclear ions, peak position gives relative isotopic mass and peak height gives relative abundance.
- Calculate as ; an isotope pattern and its weighted mean can identify an element, while a molecular-ion peak can give . A common error is to use percentage abundances without dividing by .
Tier 1 · Easy
For the ion , give its proton count, neutron count and electron count.
Tier 2 · Standard
An element has two isotopes, with abundance and with abundance . Calculate the relative atomic mass of Q.
Tier 3 · Hard
In a TOF mass spectrometer, singly charged ions take to reach the detector. Another singly charged isotope of X takes . All ions receive the same kinetic energy. Use to deduce the mass number of the second isotope.
Electron configuration
- For atoms up to , fill sub-shells in the order , , , , , , , , while recognising and as exceptions.
- When transition-metal ions form, remove electrons from before . Check that the superscripts total the species' electron count.
- First ionisation energy is the energy required to remove one electron from each atom in one mole of gaseous atoms. A successive equation has the form .
- A large jump between successive ionisation energies shows that removal has begun from a shell closer to the nucleus. A common error is to explain a sub-shell anomaly using nuclear charge alone while ignoring distance, shielding and sub-shell energy.
Tier 1 · Easy
Write the electron configuration, in sub-shell notation, of . Sulfur has atomic number 16.
Tier 2 · Standard
Define first ionisation energy and write an equation, including state symbols, for the first ionisation of magnesium.
Tier 3 · Hard
Element X is in Period 3. Its first five ionisation energies, in , are 780, 1580, 3260, 4380 and 16050. Deduce the group and identity of X, write its outer electron configuration, and write an equation for its fifth ionisation.
Relative atomic mass and relative molecular mass
- Relative atomic mass, , is the weighted mean mass of an atom of an element relative to of the mass of a carbon-12 atom.
- Relative molecular mass, , is the mean mass of a molecule relative to of the mass of a carbon-12 atom; both relative quantities have no units.
- Find by multiplying each element's by its subscript and summing. Use the term relative formula mass for ionic compounds because they do not contain discrete molecules.
- Include every atom inside brackets, waters of crystallisation and repeated groups. A common error is to attach to rather than to molar mass.
Tier 1 · Easy
Calculate the relative molecular mass of urea, . Use : H , C , N , O .
Tier 2 · Standard
Calculate the relative formula mass of . Use : H , O , Mg , S .
Tier 3 · Hard
An ionic oxide has formula and relative formula mass 144.1. Element X contains only isotopes and . Calculate and the percentage abundance of . Use .
The mole and the Avogadro constant
- One mole contains the Avogadro constant, , of specified entities; the entity may be an atom, molecule, ion, electron or formula unit.
- Use for mass, for particle count, and for solutions when is in .
- A reliable route is to convert the given quantity to moles, apply any particle or formula-unit multiplier, then convert to the requested quantity.
- State the entity being counted and convert to by dividing by . A common error is to multiply the concentration by a volume still in .
Tier 1 · Easy
Calculate the amount, in moles, in of . Use .
Tier 2 · Standard
A sample contains of oxygen molecules. Calculate the number of molecules. Use .
Tier 3 · Hard
of aluminium sulfate solution contains fully dissociated . Calculate the number of sulfate ions present. Use .
The ideal gas equation
- The ideal gas equation is , with pressure in , volume in , temperature in and amount in .
- Convert to by multiplying by , to by multiplying by , and to by multiplying by .
- Rearrange symbolically before substituting; for example and when using gas data to find molar mass.
- Use absolute temperature, . A common error is to substitute a Celsius temperature or mix litres with SI pressure.
Tier 1 · Easy
of gas occupies at . Calculate its pressure. Take .
Tier 2 · Standard
Calculate the volume occupied by of an ideal gas at and . Give the answer in . Take .
Tier 3 · Hard
A sample of a volatile liquid forms of vapour at and . Calculate its molar mass. Take .
Empirical and molecular formula
- An empirical formula gives the simplest whole-number ratio of atoms of each element, whereas a molecular formula gives the actual number of each type of atom in a molecule.
- Convert each mass or percentage to moles by dividing by , divide all mole values by the smallest, then scale to whole numbers when necessary.
- Calculate the empirical formula mass and use to find the integer multiplier for the molecular formula.
- Do not round a ratio such as directly to ; multiply every ratio by the same small integer. For combustion data, remember that one mole of water contains two moles of hydrogen atoms.
Tier 1 · Easy
A compound contains of aluminium and of oxygen. Determine its empirical formula. Use : Al , O .
Tier 2 · Standard
A compound is carbon, hydrogen and oxygen by mass. Its is 88. Determine its empirical and molecular formulae. Use : H , C , O .
Tier 3 · Hard
Complete combustion of of a compound containing only carbon, hydrogen and oxygen produces of and of . The compound has . Determine its molecular formula. Use : H , C , O .
Balanced equations and associated calculations
- Balance full and ionic equations by conserving atoms and total charge, changing coefficients only and leaving chemical formulae unchanged.
- For reacting quantities, convert the known amount to moles, use the balanced-equation ratio, then convert the required amount to mass, gas volume, concentration or solution volume.
- Percentage yield is , while atom economy is using coefficients. High atom economy reduces waste and raw-material demand, bringing economic, environmental and ethical advantages.
- Check for a limiting reactant before calculating theoretical yield. A common error is to use a mass ratio directly instead of the stoichiometric mole ratio.
Tier 1 · Easy
Balance the equation using the smallest whole-number coefficients.
Tier 2 · Standard
of hydrochloric acid reacts with excess calcium carbonate: . Calculate the maximum mass of calcium carbonate that can react. Use .
Tier 3 · Hard
Urea is made by . A process uses of ammonia and of carbon dioxide. Determine the limiting reactant, the maximum mass of urea, the percentage atom economy for urea and the percentage yield if is obtained. Use molar masses in : , , urea , .
Ionic bonding
- Ionic bonding is the electrostatic attraction between oppositely charged ions throughout a giant lattice.
- Predict simple-ion charges from Periodic Table groups, then combine ions in the smallest ratio that makes the total charge zero.
- Keep a compound ion intact when balancing charge; for example two ions require three ions, giving .
- Brackets are required around more than one compound ion. A common error is to describe ionic bonding as electron transfer; transfer forms the ions, while attraction between ions is the bond.
Tier 1 · Easy
Write the formula of aluminium sulfate from and ions.
Tier 2 · Standard
Explain the nature of ionic bonding in magnesium oxide.
Tier 3 · Hard
An element X forms the ionic compound . Deduce the charge on the X ion, then write the formulae of its nitrate and hydroxide. Nitrate is and hydroxide is .
Nature of covalent and dative covalent bonds
- A single covalent bond is a shared pair of electrons; double and triple bonds contain two and three shared pairs respectively.
- Represent an ordinary covalent bond with a line between atoms, showing one shared pair without implying that either atom supplied both electrons.
- In a co-ordinate or dative covalent bond, both electrons in the shared pair come from one atom; draw an arrow from the lone-pair donor to the electron-pair acceptor.
- After a dative bond forms it behaves as a covalent bond. A common error is to point the arrow towards the donor rather than away from it.
Tier 1 · Easy
State what is meant by a single covalent bond.
Tier 2 · Standard
accepts a lone pair from . Represent the dative covalent bond in the product and state which atom donates the electron pair.
Tier 3 · Hard
Ammonia reacts with a proton to form ammonium. Write an equation for the reaction, show the direction of dative-bond formation, and explain why all four N-H bonds in the ammonium ion are equivalent after formation.
Metallic bonding
- Metallic bonding is the electrostatic attraction between positive ions arranged in a lattice and delocalised electrons.
- The delocalised electrons move through the structure, so metals conduct electricity in both solid and liquid states.
- Metallic bonding is non-directional: layers of ions can slide while remaining attracted to the electron sea, making many metals malleable and ductile.
- Stronger metallic bonding generally follows greater ionic charge, smaller ion size or more delocalised electrons per atom. A common error is to describe discrete metal molecules or electron pairs between particular atoms.
Tier 1 · Easy
Complete the definition of metallic bonding.
Tier 2 · Standard
Explain why a metal conducts electricity when solid and remains conductive when molten.
Tier 3 · Hard
Magnesium has a higher melting point than sodium. Explain this difference using metallic bonding and the electron configurations and .
Bonding and physical properties
- The four crystal types are ionic, metallic, macromolecular (giant covalent) and molecular. When drawing one, show the specified number and type of particles in a representative repeating arrangement, including charges for ions.
- Melting a substance overcomes attractions between particles but does not normally break covalent bonds within simple molecules. Stronger or more extensive attractions require more energy.
- Ionic substances conduct only when ions can move; metals and graphite conduct through delocalised electrons; simple molecular substances usually lack mobile charged particles.
- Use evidence such as melting point and conductivity together before assigning a structure. A common error is to say that graphite conducts because whole carbon atoms move.
Tier 1 · Easy
Iodine forms crystals containing discrete molecules. State the type of crystal structure and the attractions overcome when iodine melts.
Tier 2 · Standard
Graphite has a high melting point and conducts electricity parallel to its layers. Explain both properties in terms of its structure and bonding.
Tier 3 · Hard
Three crystalline solids have these properties. A: high melting point, does not conduct when solid, conducts when molten. B: low melting point, never conducts. C: conducts when solid and is malleable. Deduce the structure type of A, B and C and justify each choice.
Shapes of simple molecules and ions
- Bonding pairs and lone pairs are electron charge clouds that repel and arrange as far apart as possible around the central atom.
- Count all electron pairs, minimise repulsion, then name the shape from atom positions. With five pairs, lone pairs prefer equatorial sites: one, two and three lone pairs give seesaw, T-shaped and linear derivatives respectively.
- Repulsion decreases in the order lone pair-lone pair lone pair-bond pair bond pair-bond pair, so lone pairs compress adjacent bond angles.
- State both shape and bond angle: linear , trigonal planar , tetrahedral , trigonal bipyramidal , and , and octahedral and are the starting geometries. A common error is to count a multiple bond as several charge clouds.
Tier 1 · Easy
State the shape and bond angle of around the boron atom.
Tier 2 · Standard
Use electron-pair repulsion theory to explain the shape of and its H-N-H bond angle of .
Tier 3 · Hard
The ion has four Br-F bonding pairs. Deduce the number of lone pairs on Br, the arrangement of all electron pairs, the molecular shape and the F-Br-F bond angles.
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 unsymmetrical electron distribution: the more electronegative atom is and the other atom is .
- To decide whether a molecule has a permanent dipole, identify its polar bonds and use the molecular shape to determine whether their dipoles cancel.
- A common error is to assume that every molecule containing polar bonds is polar; symmetrically arranged bond dipoles can have a zero resultant.
Tier 1 · Easy
The electronegativities of hydrogen and chlorine are and , respectively. Add the correct partial charge to each atom in an bond.
Tier 2 · Standard
Both and contain polar bonds. Explain why only one of these molecules has a permanent dipole.
Tier 3 · Hard
Fluorine is more electronegative than carbon. Compare the polarity of tetrahedral and tetrahedral , referring to bond dipoles and molecular shape.
Forces between molecules
- Induced dipole–dipole forces act between all atoms and molecules; their strength generally increases with the number of electrons and molecular surface contact.
- Permanent dipole–dipole attractions occur between polar molecules, while hydrogen bonding requires hydrogen bonded to nitrogen, oxygen or fluorine and a lone pair on such an atom in another molecule.
- When comparing melting or boiling points, identify all intermolecular forces and compare their overall strength; stronger attractions require more energy to overcome.
- Hydrogen bonds hold water molecules in an open lattice in ice. A common error is to say that hydrogen bonds are stronger in ice; the lower density is caused by the more open arrangement.
Tier 1 · Easy
State the strongest type of intermolecular force between molecules.
Tier 2 · Standard
Explain why ethane, , has a higher boiling point than methane, .
Tier 3 · Hard
Explain why water has a much higher boiling point than and why solid water is less dense than liquid water.
Enthalpy change
- Enthalpy change, , is the heat energy change at constant pressure; an exothermic change has and an endothermic change has .
- A standard enthalpy change applies at and a stated temperature, with each substance in its standard state.
- 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 for forming one mole of a compound from its elements in their standard states; a common error is to define it for forming any stated amount.
Tier 1 · Easy
A reaction transfers of heat from the reacting chemicals to the surroundings. State whether the reaction is exothermic or endothermic and give the sign of .
Tier 2 · Standard
Define standard enthalpy of formation and standard enthalpy of combustion. Include the amount of substance and the required conditions in each definition.
Tier 3 · Hard
Complete combustion of of a liquid releases under standard conditions. Calculate its standard enthalpy of combustion.
Calorimetry
- Use , with the mass of the substance undergoing the temperature change, its specific heat capacity and a consistently signed temperature change.
- Convert the measured heat change into a molar enthalpy using when a temperature rise shows that the reaction released heat to the surroundings.
- For solution calorimetry, the mass is usually estimated from total solution volume and density; the reacting amount is found from the limiting reagent.
- Heat loss, incomplete combustion and heating the apparatus commonly make an experimental combustion enthalpy less exothermic than the accepted value; do not invent a correction without evidence.
Tier 1 · Easy
A sample of water warms by . Use to calculate the heat gained by the water.
Tier 2 · Standard
Equal portions of hydrochloric acid and sodium hydroxide are combined. Each solution has concentration , and the temperature rises by . Assume density and . Calculate the enthalpy change per mole of water formed.
Tier 3 · Hard
Burning of propan-1-ol, , heats of water by . Use and . Calculate the experimental enthalpy of combustion and suggest one reason its magnitude is lower than the accepted value.
Applications of Hess's law
- Hess's law states that the enthalpy change for a reaction is independent of the route taken, provided the initial and final states are the same.
- Using formation enthalpies, calculate , including equation coefficients.
- Using combustion enthalpies to a common set of products, calculate the total for the reactants minus the total for the products.
- Reversing an equation reverses the sign of its enthalpy change, and multiplying an equation multiplies its enthalpy change; a common error is to alter one without the other.
Tier 1 · Easy
For the changes and , the enthalpy changes are and . Calculate the enthalpy change for .
Tier 2 · Standard
Use and to calculate for .
Tier 3 · Hard
The standard enthalpies of combustion of , and are , and , respectively. Use Hess's law to calculate the enthalpy change for .
Bond enthalpies
- Mean bond enthalpy is the mean energy required to break one mole of a specified covalent bond in gaseous molecules.
- Estimate a gaseous reaction enthalpy with ; breaking bonds is endothermic and forming bonds is exothermic.
- A reliable method is to count every bond on both sides, multiply by the equation coefficients and cancel unchanged bonds only after the count is correct.
- Values are approximate because a tabulated mean averages the same bond across different molecular environments; it is not the exact bond enthalpy in a particular molecule.
Tier 1 · Easy
Define the term mean bond enthalpy.
Tier 2 · Standard
Use the mean bond enthalpies , and to estimate for .
Tier 3 · Hard
Estimate the enthalpy change for using , , and . Explain why a value obtained from standard formation enthalpies may differ.
Collision theory
- Particles must collide before they can react, and a collision leads to reaction only if it has sufficient energy and a suitable orientation where required.
- Activation energy is the minimum energy required for a reaction to occur following a collision between particles.
- Use collision theory by separating collision frequency from the fraction of collisions that are successful; both can affect rate.
- Most collisions do not cause reaction because their energy is below the activation energy; a common error is to claim that particles did not collide at all.
Tier 1 · Easy
Define activation energy.
Tier 2 · Standard
Explain why most collisions between reactant particles do not produce a reaction.
Tier 3 · Hard
A reacting mixture undergoes molecular collisions each second. A fraction have at least the activation energy, and one quarter of these have a suitable orientation. Calculate the number of successful collisions per second and explain the role of the activation energy.
Maxwell–Boltzmann distribution
- A Maxwell–Boltzmann curve for a gas plots number of molecules against molecular energy; it starts at the origin, rises to a maximum and approaches the energy axis without meeting it.
- The area under the curve represents the total number of molecules, so equal samples at different temperatures have equal areas.
- At higher temperature the peak is lower and farther right, the distribution is broader, and a greater area lies beyond a fixed activation energy.
- A common sketching error is to draw a maximum molecular energy or let the curve cross the energy axis; the distribution has a long high-energy tail.
Tier 1 · Easy
State the quantity represented on each axis of a Maxwell–Boltzmann distribution for a gas.
Tier 2 · Standard
Describe how to add a higher-temperature curve to a Maxwell–Boltzmann distribution while keeping the number of gas molecules constant.
Tier 3 · Hard
Two equal samples of the same gas are at temperatures and , where . A fixed energy lies in the high-energy tail, beyond the point where the two Maxwell–Boltzmann curves cross. Explain why the curves must cross and compare the fractions of molecules with energy greater than .
Effect of temperature on reaction rate
- Rate of reaction is the change in concentration of a reactant or product per unit time; another measured quantity may be used when it tracks reaction progress.
- Raising temperature increases particle speeds and collision frequency, but this is not the main reason for the often large rate increase.
- On a Maxwell–Boltzmann distribution, higher temperature gives a much larger area beyond , so a greater fraction of collisions can react.
- A common error is to say that temperature lowers activation energy; without a catalyst, is unchanged and the energy distribution changes.
Tier 1 · Easy
State the effect of increasing temperature on the rate of a reaction, with all other conditions unchanged.
Tier 2 · Standard
Use a Maxwell–Boltzmann distribution to explain why a small temperature increase can cause a large increase in reaction rate.
Tier 3 · Hard
In the first of a reaction, of gas forms at and forms at . Calculate the factor by which the average rate over this interval increases, then explain the change using molecular energies.
Effect of concentration and pressure
- Increasing the concentration of a solution places more reactant particles in a given volume, so collisions occur more frequently.
- Increasing the pressure of reacting gases by decreasing their volume also raises the number of particles per unit volume and the collision frequency.
- At constant temperature, a concentration or pressure change does not alter the Maxwell–Boltzmann energy distribution or the activation energy.
- A common error is to claim that higher pressure makes each gas particle move faster; temperature controls the energy distribution, while pressure here changes particle spacing.
Tier 1 · Easy
State why increasing the concentration of a dissolved reactant usually increases reaction rate.
Tier 2 · Standard
A gaseous reacting mixture is compressed to half its original volume at constant temperature. Explain why its reaction rate increases.
Tier 3 · Hard
The concentration of one aqueous reactant is increased while temperature and all other concentrations are fixed. Explain why the rate changes, distinguishing the effect on collision frequency from any effect on activation energy or particle energy.
Catalysts
- A catalyst increases reaction rate without being changed in chemical composition or amount by the overall reaction.
- It provides an alternative reaction route with a lower activation energy; it does not lower the energy of every molecule.
- At fixed temperature the Maxwell–Boltzmann curve is unchanged, but moving the threshold left increases the area representing molecules able to react.
- A catalyst does not change or the equilibrium position; it speeds both forward and reverse reactions and allows equilibrium to be reached sooner.
Tier 1 · Easy
Define a catalyst.
Tier 2 · Standard
Use a Maxwell–Boltzmann distribution to explain how a catalyst increases the rate of a gas-phase reaction at constant temperature.
Tier 3 · Hard
An exothermic reaction has and an uncatalysed forward activation energy of . A catalyst lowers the forward activation energy to . Calculate the reverse activation energy for each route and state why is unchanged.
Chemical equilibria and Le Chatelier's principle
- At dynamic equilibrium in a closed system, forward and reverse reactions continue at equal rates and reactant and product concentrations remain constant.
- Le Chatelier's principle predicts that an equilibrium shifts in the direction that opposes a change in concentration, pressure or temperature.
- Higher pressure favours the side with fewer moles of gas, while higher temperature favours the endothermic direction; pressure has no positional effect when gaseous mole totals are equal.
- A catalyst does not change equilibrium position. Industrial conditions often compromise between equilibrium yield, reaction rate, safety and cost rather than maximising one factor alone.
Tier 1 · Easy
State two features of a reversible reaction at dynamic equilibrium.
Tier 2 · Standard
For , the forward reaction is exothermic. Predict the effect on the equilibrium yield of ammonia of increasing pressure, increasing temperature and adding a catalyst.
Tier 3 · Hard
Sulfur trioxide is made by the exothermic equilibrium . Explain why an industrial process may use a moderate temperature, a pressure that is not extremely high, and a catalyst.
Equilibrium constant Kc for homogeneous systems
- For , write using equilibrium concentrations in .
- Calculate equilibrium concentrations before substituting, using stoichiometry to relate the changes and dividing equilibrium amounts by the stated volume.
- Derive the units of from the powers in the expression; they may cancel completely for some equations.
- At a fixed temperature, changing concentration or adding a catalyst does not change . Temperature can change it: heating increases for an endothermic forward reaction and decreases it for an exothermic one.
Tier 1 · Easy
Write the expression for for .
Tier 2 · Standard
At equilibrium, , and for . Calculate and state its units.
Tier 3 · Hard
For the exothermic equilibrium , a vessel initially contains of A, of B and no C. At equilibrium it contains of C. Calculate and predict the effect of increasing temperature on its value.
Oxidation, reduction and redox equations
- Oxidation is loss of electrons and reduction is gain of electrons; an oxidising agent accepts electrons and is reduced, while a reducing agent donates electrons and is oxidised.
- For a neutral compound, oxidation states sum to zero; for an ion, they sum to the ionic charge. Uncombined elements have oxidation state zero.
- Build a half-equation by balancing the changing species and charge with electrons; in acidic solution, use and to balance oxygen and hydrogen where needed.
- Before adding half-equations, multiply them so that the numbers of electrons are equal. A common error is to cancel electrons without first balancing both charge and atoms.
Tier 1 · Easy
Determine the oxidation state of manganese in .
Tier 2 · Standard
Combine the half-equations and to give the overall redox equation.
Tier 3 · Hard
In acidic solution, dichromate(VI) ions oxidise ions to . Construct the overall ionic equation.
Born–Haber cycles (A-level only)
- Lattice enthalpy of formation is the enthalpy change when one mole of an ionic solid forms from its gaseous ions; lattice dissociation has the same magnitude and the opposite sign.
- A Born–Haber cycle links enthalpy of formation to atomisation, bond dissociation, ionisation energy, electron affinity and lattice enthalpy using Hess's law.
- For solution cycles, lattice dissociation separates the solid into gaseous ions and hydration enthalpies then convert those ions into aqueous ions; follow every stoichiometric multiplier in the formula unit.
- A substantial difference between a Born–Haber lattice enthalpy and a perfect-ionic-model value is evidence of covalent character. Common calculation errors are reversing the lattice sign or omitting a repeated ion term.
Tier 1 · Easy
Define the lattice enthalpy of formation of .
Tier 2 · Standard
For , . The atomisation enthalpy of sodium is , half the chlorine bond enthalpy is , the first ionisation energy of sodium is and the first electron affinity of chlorine is . Calculate the lattice enthalpy of formation of .
Tier 3 · Hard
The lattice enthalpy of formation of is . The hydration enthalpies of and are and respectively. Calculate the enthalpy of solution of .
Gibbs free-energy change, ΔG, and entropy change, ΔS (A-level only)
- Entropy measures the dispersal of energy and matter; calculate a reaction entropy using , including coefficients.
- Use with temperature in kelvin and consistent energy units; entropy values in usually need converting to .
- Under the stated conditions a change is thermodynamically feasible when is zero or negative, but feasibility does not imply that the reaction is fast.
- On a graph of against , the intercept is and the gradient is . A common error is to give the gradient as .
Tier 1 · Easy
For a reaction, the total standard entropy of the products is and that of the reactants is . Calculate .
Tier 2 · Standard
A reaction has and . Calculate at and state whether the reaction is feasible at this temperature.
Tier 3 · Hard
For a reaction, at and at . Assume and are constant. Determine , and the temperature at which the reaction first becomes feasible.
Rate equations (A-level only)
- A rate equation has the form ; each order is determined experimentally and is restricted here to , or .
- The overall order is the sum of the individual orders. Derive the units of by rearranging the specific rate equation rather than memorising one set of units.
- Increasing temperature increases . The Arrhenius equation is , and gives a straight line against .
- Orders are powers in the experimental rate equation, not balancing numbers from the overall chemical equation; copying stoichiometric coefficients is a common error.
Tier 1 · Easy
A reaction has rate equation . State the factor by which the rate changes when is halved and is doubled at constant temperature.
Tier 2 · Standard
For , the rate is when and . Calculate and give its units.
Tier 3 · Hard
For this calculation use . A reaction has rate constants at and at . Calculate in from .
Determination of rate equation (A-level only)
- Compare experiments in which only one initial concentration changes: an unchanged rate indicates zero order, a proportional change first order, and a squared change second order.
- A concentration–time curve gives an instantaneous rate from the gradient of a tangent; initial-rate methods use the tangent at the start or a clock time whose reciprocal is proportional to rate.
- For a first-order reactant, successive half-lives are constant. A zero-order concentration–time graph is a straight line whose negative gradient has magnitude .
- The experimental rate equation constrains the rate-determining step and any preceding fast equilibrium, but agreement with a proposed mechanism supports rather than proves that mechanism.
Tier 1 · Easy
At constant temperature, doubling while all other concentrations remain unchanged doubles the initial rate. Deduce the order with respect to .
Tier 2 · Standard
Three invented kinetic trials gave these values. Trial 1: , , rate ; trial 2: , , rate ; trial 3: , , rate . Concentrations are in and rates in . Deduce the rate equation and calculate with units.
Tier 3 · Hard
The experimental rate equation for a reaction is . Mechanism I has a single slow step . Mechanism II has a fast equilibrium followed by the slow step . Determine which mechanism is consistent with the rate equation and explain your choice.
Equilibrium constant Kp for homogeneous systems (A-level only)
- For a gas mixture, , where the mole fraction is ; use equilibrium amounts to find equilibrium partial pressures.
- Construct from gaseous products over gaseous reactants, with each partial pressure raised to its balancing number. All species in this homogeneous system are gases.
- At a fixed temperature, changing pressure may change the equilibrium composition but does not change ; only a temperature change changes its value.
- A catalyst speeds attainment of equilibrium but changes neither the equilibrium position nor . Keep pressure units consistent because the units of follow from the expression.
Tier 1 · Easy
A gas mixture contains of and of at a total pressure of . Calculate the partial pressure of each gas.
Tier 2 · Standard
At equilibrium for , there are of and of at a total pressure of . Calculate , including units.
Tier 3 · Hard
Initially, a vessel contains of and of for . At equilibrium there are of and the total pressure is . Calculate . State the effect of increasing the total pressure at constant temperature on the equilibrium position and on .
Electrode potentials and cells (A-level only)
- Standard electrode potentials are measured relative to the standard hydrogen electrode at , and ion concentrations of .
- Tables write electrode half-equations as reductions. The more positive potential is the reduction at the positive electrode; reverse the less positive half-equation for oxidation.
- Calculate . A positive value supports thermodynamic feasibility under standard conditions, but it gives no information about reaction rate.
- In conventional cell notation, put the oxidation half-cell on the left and reduction half-cell on the right, use a single line for a phase boundary and a double line for the salt bridge; include platinum when no conducting solid is present.
Tier 1 · Easy
Given and , calculate and identify the species reduced.
Tier 2 · Standard
The standard potentials are and . Calculate the EMF, write the overall reaction and give the conventional cell representation.
Tier 3 · Hard
Use and to determine whether ions can disproportionate under standard conditions. Give the equation and calculate .
Commercial applications of electrochemical cells (A-level only)
- Commercial cells may be non-rechargeable, rechargeable or fuel cells. Rechargeable cells use an external potential to drive the electrode reactions in reverse.
- In the simplified lithium cell, the negative reaction is and the positive reaction is .
- An alkaline hydrogen–oxygen fuel cell is supplied continuously with reactants; its overall reaction is , so it is refuelled rather than electrically recharged.
- Evaluate benefits and risks across the full system: operating emissions, energy density and continuous supply must be balanced against manufacture, fuel production, storage, safety, lifetime and disposal.
Tier 1 · Easy
State one difference between a rechargeable cell and a hydrogen–oxygen fuel cell.
Tier 2 · Standard
The alkaline fuel-cell half-equations are and . Deduce the overall equation and explain how the reactions generate a current.
Tier 3 · Hard
An alkaline hydrogen–oxygen fuel cell uses couples with electrode potentials and . Calculate its EMF. Then give one benefit and two risks or limitations of using hydrogen fuel cells in vehicles.
Brønsted–Lowry acid–base equilibria in aqueous solution (A-level only)
- A Brønsted–Lowry acid donates a proton and a Brønsted–Lowry base accepts a proton; acid–base equilibria are proton-transfer reactions.
- A conjugate acid–base pair differs by one ; removing a proton from an acid gives its conjugate base, while adding one to a base gives its conjugate acid.
- Some species are amphoteric and can donate or accept a proton depending on the reaction partner, so identify roles from the equation rather than from a fixed label.
- Show charge as well as atoms when writing proton-transfer equations. A common error is to name a conjugate pair whose formulas differ by more than one proton.
Tier 1 · Easy
In , identify the acid and the base on the left-hand side.
Tier 2 · Standard
For , identify both conjugate acid–base pairs.
Tier 3 · Hard
Use two equations with water to show that is amphoteric. Identify its role in each equation.
Definition and determination of pH (A-level only)
- The pH scale is logarithmic: , with in .
- Reverse the logarithm with . A change of one pH unit represents a factor of ten in hydrogen-ion concentration.
- For a strong monoprotic acid, complete dissociation makes equal to the acid concentration after any dilution or mixing calculation.
- Retain calculator precision until the end. Conventionally, the number of decimal places in pH matches the number of significant figures in .
Tier 1 · Easy
Calculate the pH of a solution for which .
Tier 2 · Standard
A solution has pH . Calculate in .
Tier 3 · Hard
of hydrochloric acid is mixed with of nitric acid. Calculate the pH of the mixture. Assume volumes are additive.
The ionic product of water, Kw (A-level only)
- Water dissociates slightly and its ionic product is ; the value of depends on temperature.
- For a strong base, first use its formula and complete dissociation to find , then calculate and hence pH.
- In a neutral solution . Neutral pH is only at a temperature where .
- Do not take the negative logarithm of and report it directly as pH; that value is pOH unless it is converted using the appropriate .
Tier 1 · Easy
At , . Calculate the pH of sodium hydroxide.
Tier 2 · Standard
At , calculate the pH of barium hydroxide. Use and assume complete dissociation.
Tier 3 · Hard
At a higher temperature, . Calculate the pH of neutral water and the pH of potassium hydroxide at this temperature.
Weak acids and bases Ka for weak acids (A-level only)
- A weak acid dissociates only slightly: , with .
- For a weak monoprotic acid of initial concentration , the approximation is valid only when dissociation is small compared with .
- Acid strength may be expressed as ; a larger and smaller mean a stronger acid.
- When pH is given, use , take , and subtract the dissociated amount from the equilibrium if an exact is required.
Tier 1 · Easy
A weak acid has . Calculate its .
Tier 2 · Standard
Calculate the pH of a solution of a weak monoprotic acid with . Use the small-dissociation approximation.
Tier 3 · Hard
A solution of a weak monoprotic acid has pH . Calculate without assuming that the equilibrium acid concentration equals its initial concentration, and hence calculate .
pH curves, titrations and indicators (A-level only)
- Know the characteristic curves for strong acid–strong base, strong acid–weak base, weak acid–strong base and weak acid–weak base titrations of monoprotic species.
- At equivalence, the reacting acid and base amounts are stoichiometrically equal; the equivalence pH is about only for a strong acid–strong base titration.
- Choose an indicator whose transition range lies within the steep section around the equivalence volume. An indicator is unsuitable when its range falls outside that rapid pH change.
- In a weak acid–strong base titration, at half-equivalence. Do not confuse half-equivalence with the equivalence point.
Tier 1 · Easy
A strong acid–weak base titration has a steep pH change from to . Methyl orange changes colour from pH to , while phenolphthalein changes from pH to . Select the suitable indicator and explain your choice.
Tier 2 · Standard
At , a flask holds of hydrochloric acid. A burette supplies sodium hydroxide. Calculate the equivalence volume and state the approximate pH at equivalence.
Tier 3 · Hard
A flask contains of weak monoprotic acid with . Sodium hydroxide of concentration is added from a burette. Calculate the volume at half-equivalence and the pH at that point. State whether the equivalence pH is below, equal to or above .
Buffer action (A-level only)
- An acidic buffer contains a weak acid and its salt, while a basic buffer contains a weak base and its salt; each pair provides both a proton donor and a proton acceptor.
- In an acidic buffer, removes added and removes added ; in a basic buffer, the weak base removes and its conjugate acid removes .
- For an acidic buffer, , equivalently .
- Before calculating pH after acid or base is added, adjust the amounts of both buffer components stoichiometrically. Dilution alone leaves their ratio, and hence pH approximately, unchanged.
Tier 1 · Easy
An acidic buffer contains equal concentrations of a weak acid and its conjugate base. The acid has . Calculate the pH.
Tier 2 · Standard
An acidic buffer contains and . Explain, using equations, how it resists small additions of acid and base.
Tier 3 · Hard
of a buffer contains of and of . The acid has . Calculate the pH after adding of hydrochloric acid. Assume the volume is unchanged.