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Notes/Chemistry/Thermodynamics, rate equations and further equilibria (A-level)
Notes · ChemistryUK · A-Levels

Thermodynamics, rate equations and further equilibria (A-level)

This large A-level-only topic deepens physical chemistry into its most quantitative form. It builds Born-Haber cycles and the ideas of entropy and free energy that decide feasibility, develops rate equations, orders of reaction and the Arrhenius equation, extends equilibrium to Kp for gases, treats electrode potentials and electrochemical cells, and finishes with the quantitative chemistry of acids, bases, pH, Ka and buffers.

7 sections·~24 min reading time·3 competencies·Level Advanced 7

T·141414 / 18
Exam profile
AO1 · Define lattice enthalpy, entropy, order of reaction, standard electrode potential, Ka, Kw and buffer actionAO2 · Calculate Born-Haber terms, free energy, orders and rate constants, Kp, cell EMF and pH valuesAO3 · Interpret rate data, Arrhenius plots, titration curves and electrode-potential data and evaluate feasibility
Operators:definecalculatedeterminededuceexplainevaluate

basic level

This entire topic is A-level only (A2); it is not assessed at AS.

higher level

The full A-Level treats Born-Haber cycles, entropy and free energy, rate equations and the Arrhenius equation, Kp, electrode potentials and electrochemical cells, and the pH of strong and weak acids, bases and buffers, with quantitative rigour throughout.

Depth

Reading depth: In depth

Text

Text size: Standard

Contents · 7 sections▾
  1. Thermodynamics, rate equations and further equilibria (A-level)
    • 01Born-Haber cycles and enthalpies of solution and hydration●
    • 02Entropy and Gibbs free energy●
    • 03Rate equations, orders and the Arrhenius equation●
    • 04Kp for homogeneous gaseous equilibria●
    • 05Electrode potentials and electrochemical cells●
    • 06pH, Ka and the ionic product of water●
    • 07Buffers, titration curves and indicators●
§ 01

Born-Haber cycles and enthalpies of solution and hydration#

●●●AdvancedLPAQA 7405 3.1.8.1LPDfE GCE Chemistry - thermodynamics

Born-Haber cycle for sodium chloride

Born-Haber cycle for NaClenergy level diagram, 6 levels, 5 transitions, Data: Na(s) + 1/2Cl2(g), Na(g) + 1/2Cl2(g), Na+(g) + e- + 1/2Cl2(g), Na+(g) + e- + Cl(g), Na+(g) + Cl-(g), NaCl(s)enthalpy / kJ mol−1Na(s) + 1/2Cl2(g)Na(g) + 1/2Cl2(g)Na+(g) + e- + 1/2Cl2(g)Na+(g) + e- + Cl(g)Na+(g) + Cl-(g)NaCl(s)atomise Na +1071st IE +496atomise Cl +122electron affinity −349lattice enthalpy −787
Fig. 1Each measurable step (atomisation, ionisation, electron affinity, formation) fixes the lattice enthalpy of NaCl as -787 kJ mol^-1.

Key points

The lattice enthalpy of formation is the enthalpy change when one mole of a solid ionic compound is formed from its gaseous ions - a strongly exothermic quantity that measures the strength of the ionic bonding. It cannot be measured directly, so it is found using a Born-Haber cycle, an application of Hess's law that links it to measurable enthalpy changes: the enthalpy of formation of the compound, the enthalpies of atomisation of the elements, the ionisation energies of the metal and the electron affinities of the non-metal.
The cycle is built by going from the elements to the solid by two routes. The direct route is the enthalpy of formation. The indirect route atomises each element (endothermic), ionises the metal (endothermic first ionisation energy, and further ones for higher charges), adds electrons to the non-metal (the first electron affinity is exothermic; a second, adding an electron to an already negative ion, is endothermic), and finally brings the gaseous ions together into the lattice (the lattice enthalpy). Setting the two routes equal and rearranging gives the unknown lattice enthalpy.
Lattice enthalpies also predict physical behaviour. A more exothermic lattice enthalpy (stronger ionic bonding) results from ions of higher charge and smaller radius, because the electrostatic attraction ∝q+q−r++r−\propto \dfrac{q_+ q_-}{r_+ + r_-}∝r+​+r−​q+​q−​​ is greater. Comparing the theoretical lattice enthalpy (calculated from a perfectly ionic model) with the experimental value from the Born-Haber cycle reveals covalent character: a large discrepancy means the bonding has significant covalent character, as in silver halides, where the cation polarises the anion.
Two further cycles handle dissolving. The enthalpy of solution is the enthalpy change when one mole of solid dissolves in enough water to give an infinitely dilute solution; it equals the lattice enthalpy of dissociation (endothermic, breaking the lattice) plus the enthalpies of hydration of the ions (exothermic, water surrounding each ion). The enthalpy of hydration is more exothermic for smaller, more highly charged ions. Whether a salt dissolves endothermically or exothermically depends on the balance between breaking the lattice and hydrating the ions.
ΔHlatt=ΔHf−(ΔHatom+IE+ΔHatom+EA)\Delta H_{\text{latt}} = \Delta H_f - \left(\Delta H_{\text{atom}} + \text{IE} + \Delta H_{\text{atom}} + \text{EA}\right)ΔHlatt​=ΔHf​−(ΔHatom​+IE+ΔHatom​+EA)

Lattice enthalpy from a Born-Haber cycle

Rearranged Hess's law: the lattice enthalpy equals the enthalpy of formation minus the sum of atomisation, ionisation and electron-affinity terms.

ΔHsol=−ΔHlatt+∑ΔHhyd\Delta H_{\text{sol}} = -\Delta H_{\text{latt}} + \sum \Delta H_{\text{hyd}}ΔHsol​=−ΔHlatt​+∑ΔHhyd​

Enthalpy of solution

Breaking the lattice (endothermic) plus hydrating the ions (exothermic) gives the enthalpy of solution.

Worked example

Lattice enthalpy of sodium chloride

Calculate the lattice enthalpy of formation of NaCl given (kJ mol^-1): enthalpy of formation -411, atomisation of Na +107, first ionisation energy of Na +496, atomisation of Cl +122, first electron affinity of Cl -349.

  1. 01Sum the indirect steps to the gaseous ions

    107+496+122−349=+376 kJ mol−1107 + 496 + 122 - 349 = +376\ \text{kJ mol}^{-1}107+496+122−349=+376 kJ mol−1 to reach Na+(g) + Cl-(g).

  2. 02Apply Hess's law

    Formation (direct) = to the gaseous ions + lattice enthalpy, so −411=376+ΔHlatt-411 = 376 + \Delta H_{\text{latt}}−411=376+ΔHlatt​.

  3. 03Solve

    ΔHlatt=−411−376=−787 kJ mol−1\Delta H_{\text{latt}} = -411 - 376 = -787\ \text{kJ mol}^{-1}ΔHlatt​=−411−376=−787 kJ mol−1.

    ΔHlatt=−411−376=−787 kJ mol−1\Delta H_{\text{latt}} = -411 - 376 = -787\ \text{kJ mol}^{-1}ΔHlatt​=−411−376=−787 kJ mol−1

Result: The lattice enthalpy of formation of NaCl is -787 kJ mol^-1.

Exam focus

  • Construct a Born-Haber cycle and calculate the lattice enthalpy (or a missing term), taking care with signs and with second ionisation energies or electron affinities.
  • Explain a difference between theoretical and experimental lattice enthalpies in terms of covalent character.

Typical mistakes

  • Getting a sign wrong, especially the second electron affinity (endothermic) or reversing the lattice-enthalpy direction.
  • Forgetting to double an atomisation enthalpy for a diatomic element or to include both ionisation energies for a 2+ ion.

Active revision

Using a Born-Haber cycle, calculate the lattice enthalpy of formation of potassium chloride given: enthalpy of formation -437, atomisation of K +89, first ionisation energy of K +419, atomisation of Cl +122, electron affinity of Cl -349 (all kJ mol^-1).

Active recall

Recall the key points — then reveal.

Sources: GCE AS and A level subject content for the sciences (Department for Education) · AQA A-level Chemistry 7405 specification (AQA)

§ 02

Entropy and Gibbs free energy#

●●●AdvancedLPAQA 7405 3.1.8.2LPDfE GCE Chemistry - entropy and free energy

Free energy against temperature

Function graph, delta G = delta H - T(delta S) = 178 - 0.161*x, 1 marked pointsGraph of delta G = delta H - T(delta S), roots at x = 1105.59, y-intercept at y = 178, decreasing, on the interval x from 0 to 16002004006008001000120014001600−5050100feasible aboveheredelta G = 0delta G = deltaH − T(delta S)delta G / kJ mol−1temperature / K
Fig. 2For an endothermic reaction with positive delta S, delta G falls with temperature and becomes negative (feasible) above T = delta H / delta S.

Key points

Entropy, SSS, is a measure of the disorder or the number of ways of arranging the particles and their energy in a system; the greater the disorder, the higher the entropy. Gases have much higher entropies than liquids, which are higher than solids, because their particles are more randomly arranged and their energy more spread out. So a reaction that produces more moles of gas, or converts a solid to a liquid or gas, has a positive entropy change ΔS\Delta SΔS (an increase in disorder).
Whether a reaction is feasible (thermodynamically spontaneous) is decided not by enthalpy alone but by the Gibbs free-energy change, ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta SΔG=ΔH−TΔS. A reaction is feasible when ΔG≤0\Delta G \le 0ΔG≤0. This combines the enthalpy factor ΔH\Delta HΔH (systems tend towards lower energy) with the entropy factor TΔST\Delta STΔS (systems tend towards greater disorder), weighted by the temperature TTT in kelvin. Note the unit trap: ΔH\Delta HΔH is usually in kJ mol−1\text{kJ mol}^{-1}kJ mol−1 but ΔS\Delta SΔS in J K−1mol−1\text{J K}^{-1}\text{mol}^{-1}J K−1mol−1, so one must be converted before they are combined.
The sign of ΔS\Delta SΔS determines how temperature affects feasibility. An endothermic reaction (ΔH>0\Delta H > 0ΔH>0) with a positive entropy change becomes feasible only above a certain temperature, because the −TΔS-T\Delta S−TΔS term must grow large enough to overcome the positive ΔH\Delta HΔH. The temperature at which the reaction just becomes feasible is found by setting ΔG=0\Delta G = 0ΔG=0, giving T=ΔH/ΔST = \Delta H / \Delta ST=ΔH/ΔS. This explains why limestone must be heated strongly before it decomposes.
Reading the four sign combinations is examinable. If ΔH\Delta HΔH is negative and ΔS\Delta SΔS positive, ΔG\Delta GΔG is always negative and the reaction is feasible at all temperatures. If ΔH\Delta HΔH is positive and ΔS\Delta SΔS negative, it is never feasible. The other two cases are temperature-dependent: exothermic with a negative ΔS\Delta SΔS is feasible only at low temperature, while endothermic with a positive ΔS\Delta SΔS is feasible only at high temperature. A useful check is that a feasible reaction may still be too slow to observe - feasibility is about thermodynamics, not rate.
ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta SΔG=ΔH−TΔS

Gibbs free-energy change

Feasible when delta G <= 0; convert delta S from J to kJ (or delta H to J) so the units match, with T in kelvin.

T=ΔHΔS(at ΔG=0)T = \dfrac{\Delta H}{\Delta S}\quad(\text{at } \Delta G = 0)T=ΔSΔH​(at ΔG=0)

Temperature of feasibility

Setting delta G = 0 gives the temperature at which the reaction just becomes feasible.

Worked example

Temperature at which decomposition becomes feasible

For CaCO3(s) -> CaO(s) + CO2(g), delta H = +178 kJ mol^-1 and delta S = +161 J K^-1 mol^-1. Find the minimum temperature at which the decomposition is feasible.

  1. 01Set delta G = 0

    At the feasibility temperature delta G = 0, so T = delta H / delta S.

  2. 02Match the units

    Convert delta S to kJ: 161 J K^-1 mol^-1 = 0.161 kJ K^-1 mol^-1.

  3. 03Calculate T

    T=178/0.161=1106 KT = 178 / 0.161 = 1106\ \text{K}T=178/0.161=1106 K (about 833 C).

    T=1780.161=1106 KT = \dfrac{178}{0.161} = 1106\ \text{K}T=0.161178​=1106 K

Result: The decomposition becomes feasible above about 1106 K (833 C), which is why limestone must be strongly heated.

Exam focus

  • Predict the sign of delta S for a reaction and calculate delta G, watching the kJ/J unit mismatch between delta H and delta S.
  • Find the temperature at which a reaction becomes feasible from T = delta H / delta S, and state that feasibility does not guarantee an observable rate.

Typical mistakes

  • Combining delta H in kJ with delta S in J without converting - one must be changed so the units match.
  • Concluding that a feasible reaction (delta G < 0) must be fast - it may have a high activation energy and be very slow.

Active revision

For a reaction ΔH=+58 kJ mol−1\Delta H = +58\ \text{kJ mol}^{-1}ΔH=+58 kJ mol−1 and ΔS=+176 J K−1mol−1\Delta S = +176\ \text{J K}^{-1}\text{mol}^{-1}ΔS=+176 J K−1mol−1. Calculate ΔG\Delta GΔG at 298 K and state whether the reaction is feasible at this temperature.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Chemistry 7405 specification (AQA)

§ 03

Rate equations, orders and the Arrhenius equation#

●●●AdvancedLPAQA 7405 3.1.9.1LPDfE GCE Chemistry - rate equations

First-order decay has a constant half-life

Function graph, [A] = 100*exp(-0.0347*x), 3 marked pointsGraph of [A], y-intercept at y = 100, decreasing, on the interval x from 0 to 80102030405060708020406080100502512.5[A]concentration / %time / s
Fig. 3A constant half-life (here 20 s) is the fingerprint of a first-order reaction: 100 to 50 to 25 to 12.5 in equal times.

Key points

The rate equation relates the rate of a reaction to the concentrations of the reactants: rate=k[A]m[B]n\text{rate} = k[\text{A}]^m[\text{B}]^nrate=k[A]m[B]n, where kkk is the rate constant, and mmm and nnn are the orders of reaction with respect to A\text{A}A and B\text{B}B. The orders are found only by experiment - they cannot be read off the balanced equation. The overall order is m+nm + nm+n. The order tells you how the rate responds to concentration: for a first-order reactant, doubling its concentration doubles the rate; for second order, doubling it quadruples the rate; for zero order, changing it has no effect.
Orders are usually found by the initial-rates method: several experiments are run in which the concentration of one reactant is changed while the others are held constant, and the effect on the initial rate is observed. If doubling [A][\text{A}][A] doubles the rate, the reaction is first order in A\text{A}A; if it quadruples the rate, second order; if the rate is unchanged, zero order. Alternatively, a concentration-time graph reveals a first-order reactant by its constant half-life - the time for the concentration to halve is the same throughout, a signature no other order shares.
The rate constant kkk has units that depend on the overall order, worked out by rearranging the rate equation. For a first-order reaction kkk has units s−1\text{s}^{-1}s−1; for second order, mol−1dm3s−1\text{mol}^{-1}\text{dm}^{3}\text{s}^{-1}mol−1dm3s−1; for third order, mol−2dm6s−1\text{mol}^{-2}\text{dm}^{6}\text{s}^{-1}mol−2dm6s−1. The rate equation also gives clues to the mechanism: the rate-determining step is the slowest step, and only the species involved in it (and steps before it) appear in the rate equation, with their orders reflecting how many are involved.
The rate constant increases with temperature according to the Arrhenius equation k=Ae−Ea/RTk = A e^{-E_a/RT}k=Ae−Ea​/RT, where AAA is the pre-exponential (frequency) factor and EaE_aEa​ the activation energy. Taking natural logarithms gives ln⁡k=ln⁡A−EaRT\ln k = \ln A - \dfrac{E_a}{RT}lnk=lnA−RTEa​​, so a plot of ln⁡k\ln klnk against 1/T1/T1/T is a straight line of gradient −Ea/R-E_a/R−Ea​/R and intercept ln⁡A\ln AlnA. This is the standard way of finding an activation energy from rate data measured at several temperatures.
rate=k[A]m[B]n\text{rate} = k[\text{A}]^m[\text{B}]^nrate=k[A]m[B]n

Rate equation

Orders m and n are found by experiment, not from the equation; overall order is m + n.

ln⁡k=ln⁡A−EaRT\ln k = \ln A - \dfrac{E_a}{RT}lnk=lnA−RTEa​​

Arrhenius equation (logarithmic form)

A plot of ln k against 1/T is a straight line of gradient -Ea/R and intercept ln A.

Arrhenius plot

Function graph, ln k = ln A - Ea/RT = 30 - 6.017*x, 1 marked pointsGraph of ln k = ln A - Ea/RT, decreasing, on the interval x from 2.4 to 3.62.42.62.833.23.43.66810121416gradient = −Ea/Rln k = ln A -Ea/RTln k1/T / 10−3 K−1
Fig. 4ln k against 1/T is a straight line of gradient -Ea/R; here 1/T is in units of 10^-3 K^-1.
Worked example

Finding orders from initial rates

For A + B -> products: Exp 1 [A]=0.10, [B]=0.10, rate=2.0x10^-3; Exp 2 [A]=0.20, [B]=0.10, rate=4.0x10^-3; Exp 3 [A]=0.10, [B]=0.20, rate=8.0x10^-3 (all mol dm^-3 and mol dm^-3 s^-1). Find the rate equation and k.

  1. 01Order in A

    From Exp 1 to 2, [A] doubles and the rate doubles, so the reaction is first order in A.

  2. 02Order in B

    From Exp 1 to 3, [B] doubles and the rate quadruples (x4 = 2^2), so it is second order in B.

  3. 03Rate equation and k

    rate = k[A][B]^2; using Exp 1, k=2.0×10−30.10×(0.10)2=2.0 mol−2dm6s−1k = \dfrac{2.0\times10^{-3}}{0.10 \times (0.10)^2} = 2.0\ \text{mol}^{-2}\text{dm}^{6}\text{s}^{-1}k=0.10×(0.10)22.0×10−3​=2.0 mol−2dm6s−1.

    k=2.0×10−30.10×(0.10)2=2.0 mol−2 dm6 s−1k = \dfrac{2.0\times10^{-3}}{0.10 \times (0.10)^2} = 2.0\ \text{mol}^{-2}\,\text{dm}^{6}\,\text{s}^{-1}k=0.10×(0.10)22.0×10−3​=2.0 mol−2dm6s−1

Result: rate = k[A][B]^2 (overall third order), with k = 2.0 mol^-2 dm^6 s^-1.

Exam focus

  • Determine the order with respect to each reactant from initial-rates data, then write the rate equation and calculate k with its units.
  • Use an Arrhenius plot (ln k against 1/T) to find the activation energy from the gradient -Ea/R.

Typical mistakes

  • Reading orders straight from the balanced equation instead of from experimental data.
  • Quoting the rate constant without deducing its correct units from the overall order.

Active revision

For a reaction the rate is rate=k[A][B]2\text{rate} = k[\text{A}][\text{B}]^2rate=k[A][B]2. When [A]=[B]=0.10 mol dm−3[\text{A}] = [\text{B}] = 0.10\ \text{mol dm}^{-3}[A]=[B]=0.10 mol dm−3 the rate is 4.0×10−4 mol dm−3s−14.0 \times 10^{-4}\ \text{mol dm}^{-3}\text{s}^{-1}4.0×10−4 mol dm−3s−1. Calculate kkk with its units.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Chemistry 7405 specification (AQA)

§ 04

Kp for homogeneous gaseous equilibria#

●●●AdvancedLPAQA 7405 3.1.10.1LPDfE GCE Chemistry - Kp

From moles to partial pressures

N2 + 3H2 = 2NH3 at 200 kPa totalTable with 4 columns and 4 rows, Data: Gas · moles · mole fraction · partial pressure / kPa; N2 · 0.20 · 0.167 · 33.3; H2 · 0.60 · 0.500 · 100.0; NH3 · 0.40 · 0.333 · 66.7; total · 1.20 · 1.000 · 200.0GASMOLESMOLE FRACTIONPARTIAL PRESSURE / KPAN20.200.16733.3H20.600.500100.0NH30.400.33366.7total1.201.000200.0
Fig. 5Each partial pressure is the mole fraction times the total pressure; these feed directly into the Kp expression.

Key points

For an equilibrium involving only gases, it is convenient to work with partial pressures rather than concentrations, giving the equilibrium constant KpK_pKp​. The partial pressure of a gas is the pressure it would exert if it alone occupied the container; the sum of the partial pressures equals the total pressure (Dalton's law). The partial pressure of a gas is its mole fraction multiplied by the total pressure: pA=xA×Ptotalp_{\text{A}} = x_{\text{A}} \times P_{\text{total}}pA​=xA​×Ptotal​, where the mole fraction xAx_{\text{A}}xA​ is the moles of A divided by the total moles of gas.
The KpK_pKp​ expression has the same form as KcK_cKc​, with each partial pressure raised to the power of its balancing number: for N2+3H2⇌2NH3\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3N2​+3H2​⇌2NH3​, Kp=pNH32pN2 pH23K_p = \dfrac{p_{\text{NH}_3}^2}{p_{\text{N}_2}\, p_{\text{H}_2}^{3}}Kp​=pN2​​pH2​3​pNH3​2​​. Its units are deduced by substituting the pressure unit (kPa or Pa) and cancelling, exactly as for KcK_cKc​. Only gases appear in a KpK_pKp​ expression; pure solids and liquids are omitted because their 'pressure' is effectively constant.
To calculate KpK_pKp​, first find the equilibrium amount of each gas (often from an ICE table), then the total moles, then each mole fraction, then each partial pressure, and finally substitute into the expression. It is a multi-step calculation where careful layout prevents slips - keep a clear table of moles, mole fractions and partial pressures. A common check is that the mole fractions must add up to 111 and the partial pressures to the total pressure.
KpK_pKp​ behaves like KcK_cKc​: it is constant at a given temperature and changes only with temperature (in the direction predicted by Le Chatelier for the enthalpy change). Changing the total pressure or adding a catalyst does not change KpK_pKp​ - increasing the pressure shifts the position of equilibrium but the system readjusts so that KpK_pKp​ stays the same. This subtlety - that pressure changes the position but not KpK_pKp​ - is a favourite discussion point.
pA=xA×Ptotal,xA=nAntotalp_{\text{A}} = x_{\text{A}} \times P_{\text{total}}, \qquad x_{\text{A}} = \dfrac{n_{\text{A}}}{n_{\text{total}}}pA​=xA​×Ptotal​,xA​=ntotal​nA​​

Partial pressure

The partial pressure of a gas is its mole fraction multiplied by the total pressure.

Kp=pNH32pN2 pH23K_p = \dfrac{p_{\text{NH}_3}^2}{p_{\text{N}_2}\, p_{\text{H}_2}^{3}}Kp​=pN2​​pH2​3​pNH3​2​​

Kp expression

Same form as Kc but using partial pressures; deduce the units by cancelling.

Worked example

Kp for the ammonia equilibrium

At equilibrium a mixture contains 0.20 mol N2, 0.60 mol H2 and 0.40 mol NH3 at a total pressure of 200 kPa. Calculate Kp for N2 + 3H2 = 2NH3.

  1. 01Mole fractions

    Total = 1.20 mol; x(N2)=0.167, x(H2)=0.500, x(NH3)=0.333.

  2. 02Partial pressures

    p(N2)=0.167x200=33.3 kPa; p(H2)=0.500x200=100.0 kPa; p(NH3)=0.333x200=66.7 kPa.

  3. 03Substitute

    Kp=(66.7)233.3×(100.0)3=44493.33×107=1.33×10−4 kPa−2K_p = \dfrac{(66.7)^2}{33.3 \times (100.0)^3} = \dfrac{4449}{3.33\times10^{7}} = 1.33\times10^{-4}\ \text{kPa}^{-2}Kp​=33.3×(100.0)3(66.7)2​=3.33×1074449​=1.33×10−4 kPa−2.

    Kp=(66.7)233.3×(100.0)3=1.33×10−4 kPa−2K_p = \dfrac{(66.7)^2}{33.3 \times (100.0)^3} = 1.33\times10^{-4}\ \text{kPa}^{-2}Kp​=33.3×(100.0)3(66.7)2​=1.33×10−4 kPa−2

Result: Kp = 1.33 x 10^-4 kPa^-2 (units from three pressure terms on the bottom, two on top).

Exam focus

  • Calculate mole fractions and partial pressures, then Kp with its correct units, for a gaseous equilibrium.
  • Explain that changing the total pressure shifts the position of equilibrium but does not change Kp.

Typical mistakes

  • Including solids or liquids in the Kp expression, or forgetting the powers from the equation.
  • Using moles instead of partial pressures, or letting the mole fractions fail to sum to 1.

Active revision

At equilibrium a mixture contains 0.30 mol0.30\ \text{mol}0.30 mol of PCl5\text{PCl}_5PCl5​, 0.10 mol0.10\ \text{mol}0.10 mol of PCl3\text{PCl}_3PCl3​ and 0.10 mol0.10\ \text{mol}0.10 mol of Cl2\text{Cl}_2Cl2​ at a total pressure of 250 kPa250\ \text{kPa}250 kPa. For PCl5⇌PCl3+Cl2\text{PCl}_5 \rightleftharpoons \text{PCl}_3 + \text{Cl}_2PCl5​⇌PCl3​+Cl2​, calculate KpK_pKp​.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Chemistry 7405 specification (AQA)

§ 05

Electrode potentials and electrochemical cells#

●●●AdvancedLPAQA 7405 3.1.11.1LPDfE GCE Chemistry - electrode potentials

A zinc-copper electrochemical cell

Zn | Zn2+ || Cu2+ | CuSchematic diagram with 9 elements, Zn2+(aq), Cu2+(aq), Zn (-), Cu (+), V, salt bridge, electrons flow Zn to CuZn2+(aq)Cu2+(aq)Zn (-)Cu (+)Vsalt bridgeelectrons flowZn to Cu
Fig. 6Electrons flow from the zinc (negative) to the copper (positive) electrode; the salt bridge completes the circuit. EMF = +0.34 - (-0.76) = +1.10 V.

Key points

When a metal is placed in a solution of its ions, an equilibrium is set up between the metal and its ions, Mn++ne−⇌M\text{M}^{n+} + ne^- \rightleftharpoons \text{M}Mn++ne−⇌M, which sets up an electrode potential. Because a single electrode potential cannot be measured in isolation, all values are quoted relative to the standard hydrogen electrode, which is defined as zero. The standard electrode (reduction) potential E⊖E^{\ominus}E⊖ is measured under standard conditions (298 K298\ \text{K}298 K, 100 kPa100\ \text{kPa}100 kPa, 1 mol dm−31\ \text{mol dm}^{-3}1 mol dm−3 solutions) against this reference.
Two half-cells connected by a wire (for electrons) and a salt bridge (to complete the circuit and balance charge) make an electrochemical cell. The more negative electrode is where oxidation occurs (the negative terminal, losing electrons), and the more positive electrode is where reduction occurs (the positive terminal). The cell EMF (electromotive force) is the difference between the two standard electrode potentials: Ecell⊖=Epositive⊖−Enegative⊖E^{\ominus}_{\text{cell}} = E^{\ominus}_{\text{positive}} - E^{\ominus}_{\text{negative}}Ecell⊖​=Epositive⊖​−Enegative⊖​, always giving a positive value for a spontaneous cell.
Electrode potentials predict the feasibility and direction of redox reactions. The half-cell with the more positive E⊖E^{\ominus}E⊖ undergoes reduction (its species is the better oxidising agent); the more negative one undergoes oxidation (its species is the better reducing agent). A reaction is feasible if the cell EMF calculated for it is positive. This lets you predict, for example, that Cu2+\text{Cu}^{2+}Cu2+ (+0.34 V+0.34\ \text{V}+0.34 V) will oxidise zinc (−0.76 V-0.76\ \text{V}−0.76 V) but not silver (+0.80 V+0.80\ \text{V}+0.80 V).
Cells have practical importance and limitations. Non-rechargeable and rechargeable cells (batteries) and hydrogen-oxygen fuel cells all use these principles; a fuel cell continuously oxidises a fuel to generate a voltage with water as the only product, offering high efficiency. A key limitation of E⊖E^{\ominus}E⊖ predictions is that they indicate only thermodynamic feasibility, not rate - a reaction with a positive EMF may still be immeasurably slow because of a high activation energy - and they apply strictly only under standard conditions, so changing concentration shifts the potentials (as Le Chatelier predicts).
Ecell⊖=Epositive⊖−Enegative⊖E^{\ominus}_{\text{cell}} = E^{\ominus}_{\text{positive}} - E^{\ominus}_{\text{negative}}Ecell⊖​=Epositive⊖​−Enegative⊖​

Cell EMF

The more positive electrode potential minus the more negative; a positive EMF means the cell reaction is feasible.

Worked example

EMF of the Daniell cell

Calculate the EMF of a cell made from Zn2+/Zn (E = -0.76 V) and Cu2+/Cu (E = +0.34 V), and state the direction of electron flow.

  1. 01Identify the electrodes

    Cu2+/Cu is more positive, so copper is the positive electrode (reduction); Zn is the negative electrode (oxidation).

  2. 02Calculate the EMF

    Ecell⊖=(+0.34)−(−0.76)=+1.10 VE^{\ominus}_{\text{cell}} = (+0.34) - (-0.76) = +1.10\ \text{V}Ecell⊖​=(+0.34)−(−0.76)=+1.10 V.

    Ecell⊖=(+0.34)−(−0.76)=+1.10 VE^{\ominus}_{\text{cell}} = (+0.34) - (-0.76) = +1.10\ \text{V}Ecell⊖​=(+0.34)−(−0.76)=+1.10 V
  3. 03Direction of electrons

    Electrons flow through the external wire from the negative (Zn) to the positive (Cu) electrode.

Result: The EMF is +1.10 V, with electrons flowing from zinc to copper; the positive value confirms the reaction is feasible.

Exam focus

  • Calculate the cell EMF from two standard electrode potentials and identify the positive and negative electrodes.
  • Use standard electrode potentials to predict whether a redox reaction is feasible, noting that this does not predict the rate.

Typical mistakes

  • Subtracting the electrode potentials the wrong way and obtaining a negative EMF for a feasible cell.
  • Assuming a positive EMF means the reaction is fast - it indicates feasibility only, not rate.

Active revision

Given E⊖E^{\ominus}E⊖: Fe2+/Fe=−0.44 V\text{Fe}^{2+}/\text{Fe} = -0.44\ \text{V}Fe2+/Fe=−0.44 V and Ag+/Ag=+0.80 V\text{Ag}^+/\text{Ag} = +0.80\ \text{V}Ag+/Ag=+0.80 V, calculate the EMF of the cell and state which metal is oxidised.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Chemistry 7405 specification (AQA)

§ 06

pH, Ka and the ionic product of water#

●●●AdvancedLPAQA 7405 3.1.12.1LPDfE GCE Chemistry - acids and bases

The pH scale

The pH scale (298 K)Number line, 0.1 M HCl, neutral, 0.1 M NaOH0246781012140.1 M HClneutral0.1 M NaOH
Fig. 7pH = -log[H+]; each unit is a tenfold change in [H+]. Neutral is pH 7 at 298 K.

Key points

A Bronsted-Lowry acid is a proton (H+\text{H}^+H+) donor and a base is a proton acceptor; every acid-base reaction transfers a proton from the acid to the base, forming a conjugate acid-base pair on each side. The acidity of a solution is measured by pH, defined as pH=−log⁡10[H+]\text{pH} = -\log_{10}[\text{H}^+]pH=−log10​[H+], where [H+][\text{H}^+][H+] is the hydrogen-ion concentration in mol dm−3\text{mol dm}^{-3}mol dm−3. Because it is a logarithmic scale, a change of one pH unit is a tenfold change in [H+][\text{H}^+][H+], and the concentration is recovered by [H+]=10−pH[\text{H}^+] = 10^{-\text{pH}}[H+]=10−pH.
A strong acid is fully dissociated, so for a monoprotic strong acid such as HCl\text{HCl}HCl, [H+][\text{H}^+][H+] equals the acid concentration and the pH follows directly. A weak acid is only partially dissociated, and its extent of dissociation is measured by the acid dissociation constant Ka=[H+][A−][HA]K_a = \dfrac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}Ka​=[HA][H+][A−]​. For a weak acid the approximations [H+]=[A−][\text{H}^+] = [\text{A}^-][H+]=[A−] and [HA]≈[\text{HA}] \approx[HA]≈ its initial concentration give [H+]=Ka×[HA][\text{H}^+] = \sqrt{K_a \times [\text{HA}]}[H+]=Ka​×[HA]​. Because KaK_aKa​ values span many orders of magnitude, they are often quoted as pKa=−log⁡10Ka\text{p}K_a = -\log_{10}K_apKa​=−log10​Ka​; a smaller pKa\text{p}K_apKa​ means a stronger acid.
Water itself slightly ionises, H2O⇌H++OH−\text{H}_2\text{O} \rightleftharpoons \text{H}^+ + \text{OH}^-H2​O⇌H++OH−, characterised by the ionic product of water Kw=[H+][OH−]=1.0×10−14 mol2dm−6K_w = [\text{H}^+][\text{OH}^-] = 1.0 \times 10^{-14}\ \text{mol}^2\text{dm}^{-6}Kw​=[H+][OH−]=1.0×10−14 mol2dm−6 at 298 K298\ \text{K}298 K. This links [H+][\text{H}^+][H+] and [OH−][\text{OH}^-][OH−] in any aqueous solution, and is the key to the pH of alkalis: for a strong base you find [OH−][\text{OH}^-][OH−] from the concentration, then use [H+]=Kw/[OH−][\text{H}^+] = K_w/[\text{OH}^-][H+]=Kw​/[OH−] to get the pH. In neutral water at 298 K298\ \text{K}298 K, [H+]=[OH−]=1.0×10−7[\text{H}^+] = [\text{OH}^-] = 1.0 \times 10^{-7}[H+]=[OH−]=1.0×10−7, giving pH 777.
KwK_wKw​ is temperature-dependent because the ionisation of water is endothermic, so warming water increases KwK_wKw​ and lowers the pH of neutral water below 777 - yet the water is still neutral because [H+][\text{H}^+][H+] still equals [OH−][\text{OH}^-][OH−]. This is a subtle but examinable point: neutrality means equal [H+][\text{H}^+][H+] and [OH−][\text{OH}^-][OH−], not necessarily pH 777. Careful significant figures matter throughout: pH values are conventionally quoted to two decimal places, because only the digits after the decimal point of a logarithm are significant.
pH=−log⁡10[H+],[H+]=10−pH\text{pH} = -\log_{10}[\text{H}^+], \qquad [\text{H}^+] = 10^{-\text{pH}}pH=−log10​[H+],[H+]=10−pH

Definition of pH

A logarithmic measure of hydrogen-ion concentration; one pH unit is a tenfold change.

Ka=[H+][A−][HA],[H+]=Ka [HA]K_a = \dfrac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}, \qquad [\text{H}^+] = \sqrt{K_a\,[\text{HA}]}Ka​=[HA][H+][A−]​,[H+]=Ka​[HA]​

Weak-acid dissociation

For a weak acid, assuming [H+] = [A-] and [HA] approximately its initial value.

Kw=[H+][OH−]=1.0×10−14 mol2dm−6 (298 K)K_w = [\text{H}^+][\text{OH}^-] = 1.0\times10^{-14}\ \text{mol}^2\text{dm}^{-6}\ (298\,\text{K})Kw​=[H+][OH−]=1.0×10−14 mol2dm−6 (298K)

Ionic product of water

Links [H+] and [OH-]; use it to find the pH of a strong base from its [OH-].

Worked example

pH of a weak acid

Calculate the pH of 0.100 mol dm^-3 ethanoic acid, given Ka = 1.75 x 10^-5 mol dm^-3.

  1. 01Use the weak-acid approximation

    [H+]=Ka×[HA]=1.75×10−5×0.100[\text{H}^+] = \sqrt{K_a \times [\text{HA}]} = \sqrt{1.75\times10^{-5} \times 0.100}[H+]=Ka​×[HA]​=1.75×10−5×0.100​.

  2. 02Evaluate [H+]

    =1.75×10−6=1.323×10−3 mol dm−3= \sqrt{1.75\times10^{-6}} = 1.323\times10^{-3}\ \text{mol dm}^{-3}=1.75×10−6​=1.323×10−3 mol dm−3.

  3. 03Take the pH

    pH=−log⁡10(1.323×10−3)=2.88\text{pH} = -\log_{10}(1.323\times10^{-3}) = 2.88pH=−log10​(1.323×10−3)=2.88.

    pH=−log⁡10(1.323×10−3)=2.88\text{pH} = -\log_{10}(1.323\times10^{-3}) = 2.88pH=−log10​(1.323×10−3)=2.88

Result: The pH is 2.88 - higher than the 1.00 a strong acid of the same concentration would give, because the weak acid is only partly dissociated.

Exam focus

  • Calculate the pH of a strong acid, a strong base (via Kw) and a weak acid (via Ka), to two decimal places.
  • Explain that neutrality means [H+] = [OH-], which is pH 7 only at 298 K because Kw is temperature-dependent.

Typical mistakes

  • Treating a weak acid as fully dissociated, or forgetting to use Kw for a base.
  • Quoting a pH to the wrong precision - two decimal places is conventional.

Active revision

Calculate the pH of (a) 0.050 mol dm−30.050\ \text{mol dm}^{-3}0.050 mol dm−3 hydrochloric acid and (b) 0.100 mol dm−30.100\ \text{mol dm}^{-3}0.100 mol dm−3 sodium hydroxide (Kw = 1.0 x 10^-14).

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Chemistry 7405 specification (AQA)

§ 07

Buffers, titration curves and indicators#

●●●AdvancedLPAQA 7405 3.1.12.5LPDfE GCE Chemistry - buffers and titrations

Titration curves

Function graph, strong-strong = 7 + 6*tanh((x-25)/2); weak-strong = 8 + 5*tanh((x-25)/3.2)Graph of strong-strong, minimum at (0.05, 1), maximum at (0.1, 1), maximum at (0.225, 1), maximum at (0.263, 1), minimum at (0.325, 1), maximum at (0.375, 1), maximum at (0.4, 1), minimum at (0.45, 1), minimum at (0.488, 1), minimum at (0.738, 1), maximum at (0.75, 1), minimum at (1.075, 1), maximum at (48.223, 13), maximum at (48.375, 13), maximum at (48.625, 13), maximum at (48.65, 13), maximum at (48.688, 13), maximum at (48.7, 13), minimum at (48.824, 13), minimum at (48.922, 13), minimum at (49.05, 13), minimum at (49.075, 13), minimum at (49.125, 13), minimum at (49.15, 13), maximum at (49.223, 13), minimum at (49.275, 13), minimum at (49.375, 13), minimum at (49.474, 13), minimum at (49.5, 13), minimum at (49.525, 13), minimum at (49.6, 13), maximum at (49.675, 13), maximum at (49.7, 13), minimum at (49.775, 13), minimum at (49.8, 13), minimum at (49.9, 13), maximum at (49.95, 13), y-intercept at y = 1, on the interval x from 0 to 50, Graph of weak-strong, y-intercept at y = 3, increasing, on the interval x from 0 to 5010203040502468101214equivalencevolumestrong-strongweak-strongpHvolume of base added / cm3
Fig. 8Illustrative curves: the strong-strong equivalence is at pH 7; the weak acid-strong base equivalence is above 7. The indicator must change within the steep section.

Key points

A buffer solution resists changes in pH when small amounts of acid or alkali are added. An acidic buffer is made from a weak acid and one of its salts (for example ethanoic acid and sodium ethanoate), providing a reservoir of the weak acid HA\text{HA}HA and its conjugate base A−\text{A}^-A−. If acid is added, the extra H+\text{H}^+H+ reacts with A−\text{A}^-A− to form HA\text{HA}HA; if alkali is added, the OH−\text{OH}^-OH− reacts with HA\text{HA}HA to form A−\text{A}^-A− and water. In each case the equilibrium shifts to remove most of the added ions, so the pH changes only slightly.
The pH of a buffer is found by rearranging the KaK_aKa​ expression to [H+]=Ka×[HA][A−][\text{H}^+] = K_a \times \dfrac{[\text{HA}]}{[\text{A}^-]}[H+]=Ka​×[A−][HA]​. Because the weak acid and its salt are present in comparable amounts, the ratio [HA]/[A−][\text{HA}]/[\text{A}^-][HA]/[A−] is close to one and the buffer's pH is close to the pKa\text{p}K_apKa​ of the acid; equal amounts give [H+]=Ka[\text{H}^+] = K_a[H+]=Ka​ exactly, so pH =pKa= \text{p}K_a=pKa​. This is why a buffer of a chosen pH is made by selecting a weak acid whose pKa\text{p}K_apKa​ is near the target pH. Blood is buffered near pH 7.47.47.4 by the carbonic acid-hydrogencarbonate system.
A titration curve plots the pH of the reaction mixture against the volume of titrant added. Its shape depends on whether the acid and base are strong or weak. A strong acid-strong base titration starts near pH 111, has a long near-vertical section at the equivalence point centred on pH 777, and ends near pH 131313. A weak acid-strong base titration starts higher (around pH 333), rises through a buffer region where the curve is gentle, and its equivalence point lies above pH 777 (around 888-999) because the salt formed is slightly alkaline.
The indicator chosen must change colour within the steep, near-vertical part of the titration curve, so that a single drop of titrant triggers the colour change at the equivalence point. Methyl orange (changing around pH 3.13.13.1-4.44.44.4) suits titrations with a low equivalence pH (strong acid with weak base), while phenolphthalein (changing around pH 8.38.38.3-10.010.010.0) suits those with a high equivalence pH (weak acid with strong base); either works for a strong-strong titration because its steep section spans both ranges. A weak acid-weak base titration has no sharp vertical section, so no indicator gives a clear end point.
[H+]=Ka×[HA][A−][\text{H}^+] = K_a \times \dfrac{[\text{HA}]}{[\text{A}^-]}[H+]=Ka​×[A−][HA]​

Buffer pH

For equal amounts of weak acid and its salt, [H+] = Ka and pH = pKa.

Worked example

pH of a buffer solution

Calculate the pH of a buffer containing 0.100 mol dm^-3 ethanoic acid and 0.100 mol dm^-3 sodium ethanoate (Ka = 1.75 x 10^-5 mol dm^-3).

  1. 01Use the buffer expression

    [H+]=Ka×[HA][A−]=1.75×10−5×0.1000.100[\text{H}^+] = K_a \times \dfrac{[\text{HA}]}{[\text{A}^-]} = 1.75\times10^{-5} \times \dfrac{0.100}{0.100}[H+]=Ka​×[A−][HA]​=1.75×10−5×0.1000.100​.

  2. 02Evaluate

    The ratio is 1, so [H+]=1.75×10−5 mol dm−3[\text{H}^+] = 1.75\times10^{-5}\ \text{mol dm}^{-3}[H+]=1.75×10−5 mol dm−3.

  3. 03Take the pH

    pH=−log⁡10(1.75×10−5)=4.76\text{pH} = -\log_{10}(1.75\times10^{-5}) = 4.76pH=−log10​(1.75×10−5)=4.76.

    pH=−log⁡10(1.75×10−5)=4.76\text{pH} = -\log_{10}(1.75\times10^{-5}) = 4.76pH=−log10​(1.75×10−5)=4.76

Result: The pH is 4.76, equal to the pKa of ethanoic acid because the acid and its salt are in equal amounts.

Exam focus

  • Explain how a buffer resists pH change on adding small amounts of acid or alkali, referring to HA and A-.
  • Calculate the pH of a buffer, and choose an indicator whose colour change falls within the steep part of the titration curve.

Typical mistakes

  • Saying a buffer keeps the pH constant - it only minimises the change.
  • Choosing an indicator whose range does not lie within the vertical section of the curve (e.g. phenolphthalein for a weak acid-weak base titration).

Active revision

A buffer is made from 0.20 mol dm−30.20\ \text{mol dm}^{-3}0.20 mol dm−3 ethanoic acid and 0.10 mol dm−30.10\ \text{mol dm}^{-3}0.10 mol dm−3 sodium ethanoate (Ka = 1.75 x 10^-5). Calculate its pH.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Chemistry 7405 specification (AQA)

Contents

Section -- / 07

    • 01Born-Haber cycles and enthalpies of solution and hydration●
    • 02Entropy and Gibbs free energy●
    • 03Rate equations, orders and the Arrhenius equation●
    • 04Kp for homogeneous gaseous equilibria●
    • 05Electrode potentials and electrochemical cells●
    • 06pH, Ka and the ionic product of water●
    • 07Buffers, titration curves and indicators●

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  • AQA A-level Chemistry 7405 specification

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