Physical chemistry II: A-level Chemistry AQA revision notes

3.1.8

Thermodynamics

Born–Haber cycles 3.1.8.1

Definitions

Lattice enthalpy of formation: the enthalpy change when one mole of a solid ionic compound is formed from its gaseous ions.
Enthalpy of atomisation: the enthalpy change when one mole of gaseous atoms is formed from the element in its standard state.
First electron affinity: the enthalpy change when one mole of gaseous atoms each gain one electron to form one mole of gaseous $1 -$ ions.

Key results

For NaCl: $Δ_{f} H^{\circ} = Δ_{a t} H (Na) + Δ_{a t} H (Cl) + I E_{1} (Na) + E A_{1} (Cl) + Δ_{L} H^{\circ}$ .
$- 411.3 = 107.3 + 121.7 + 495.8 - 348.8 + Δ_{L} H^{\circ}$ , so $Δ_{L} H^{\circ} (formation) = - 787.3 kJ mol^{- 1}$ .

Notes

Lattice enthalpy cannot be measured directly, so a Born–Haber cycle (Hess's law) calculates it from measurable steps.
Chlorine's atomisation forms one mole of Cl atoms from $\frac{1}{2} Cl_{2}$ , so it equals half the Cl–Cl bond enthalpy. Do not double it.
Upward arrows are endothermic (atomisation, ionisation); downward arrows are exothermic (first electron affinity, lattice formation).
Worked example with a 2+ ion: for MgO the cycle needs both ionisation energies of Mg (738 + 1451) and both electron affinities of O (−141 then +798). With $Δ_{f} H^{\circ} = - 602$ , $Δ_{a t} H$ (Mg) = +148 and $Δ_{a t} H$ (O) = +249, the lattice enthalpy of formation is about $- 3845 kJ mol^{- 1}$ , roughly five times that of NaCl because both ions are doubly charged and small.
The second electron affinity of oxygen is endothermic: the incoming electron is repelled by the negative $O^{-}$ ion, so energy must be supplied to force it on.
When drawing a cycle, write the species and state symbols on every level, keep electrons in the equations until the electron affinity step, and label each arrow with the name of the enthalpy change. Many marks are for correct labelling rather than arithmetic.

Lattice dissociation, hydration and solution 3.1.8.1

Definitions

Lattice enthalpy of dissociation: the enthalpy change to separate one mole of a solid ionic compound into gaseous ions (endothermic; equal and opposite to the lattice enthalpy of formation).
Enthalpy of hydration: the enthalpy change when one mole of gaseous ions dissolves to form aqueous ions (always exothermic).
Enthalpy of solution: the enthalpy change when one mole of an ionic solid dissolves in enough water that further dilution has no effect.

Key results

$Δ_{so l} H = Δ_{L} H (dissociation) + Δ_{h y d} H (M^{+}) + Δ_{h y d} H (X^{-})$ .
Illustrative: $+ 800 - 400 - 380 = + 20 kJ mol^{- 1}$ , a slightly endothermic dissolving.

Notes

Smaller ions or higher charges give stronger electrostatic attraction, so both lattice formation and hydration become more exothermic.
The perfect ionic model assumes spherical ions and purely ionic bonding. When the Born–Haber value is more exothermic than the model predicts, the bonding has some covalent character, because a small, highly charged cation distorts (polarises) the anion's electron cloud.
Whether a salt dissolves depends on the balance between lattice and hydration enthalpies, and also on entropy.
Worked example: for KCl, with typical data values $Δ_{L} H$ (dissociation) = +711, $Δ_{h y d} H (K^{+}) = - 322$ and $Δ_{h y d} H (Cl^{-}) = - 372$ , $Δ_{so l} H = + 17 kJ mol^{- 1}$ ; KCl dissolves endothermically, and the solution cools.
Hydration is exothermic because water's $δ -$ oxygen atoms are attracted to cations and its $δ +$ hydrogen atoms to anions. Draw water molecules the right way round around each ion.
Comparing Born–Haber and ionic-model values: for NaCl they agree within about 3%, so the bonding is close to purely ionic; for silver chloride the Born–Haber value is markedly more exothermic, showing significant covalent character.
Polarising power increases with charge density (charge ÷ size), so $Mg^{2 +}$ and $Al^{3 +}$ polarise anions far more than $Na^{+}$ ; large anions such as $I^{-}$ are the most easily polarised.

Entropy 3.1.8.2

Definitions

Entropy, $S$ : a measure of the number of ways particles and their energy can be arranged; greater disorder means greater entropy. Units: $J K^{- 1} mol^{- 1}$ .

Key results

$S (solid) < S (liquid) < S (gas)$ for a substance.
$Δ S^{\circ} = Σ S^{\circ} (products) - Σ S^{\circ} (reactants)$ , each multiplied by its coefficient.
$N_{2} + 3 H_{2} \to 2 NH_{3}$ : $Δ S^{\circ} = 2 (192.8) - [191.6 + 3 (130.7)] = - 198.1 J K^{- 1} mol^{- 1}$ .

Notes

Unlike $Δ_{f} H^{\circ}$ , the standard entropy of an element is not zero: every substance above 0 K has some entropy.
A reaction that reduces the number of moles of gas (here 4 → 2) almost always has a negative $Δ S$ , because gases dominate entropy.
Entropy increases when a solid melts or a liquid boils, when a solid dissolves, when the number of moles of gas increases, and when a substance is heated. Molecules with more atoms, and so more ways of storing energy, have higher standard entropies.
Worked example: $CaCO_{3} (s) \to CaO(s) + CO_{2} (g)$ , using $S^{\circ}$ values of 92.9, 39.7 and 213.6 J K⁻¹ mol⁻¹: $Δ S^{\circ} = 39.7 + 213.6 - 92.9 = + 160.4 J K^{- 1} mol^{- 1}$ , positive because a gas is produced from a solid.
At 0 K a perfect crystal has zero entropy: there is only one way of arranging its particles and energy. Entropy rises sharply at each change of state, the largest step being on boiling.

Gibbs free energy and feasibility 3.1.8.2

Definitions

Gibbs free-energy change: $Δ G = Δ H - T Δ S$ . A process is feasible when $Δ G \leq 0$ .

Key results

$Δ H < 0$ , $Δ S > 0$ : feasible at all temperatures. $Δ H > 0$ , $Δ S < 0$ : never feasible.
$Δ H > 0$ , $Δ S > 0$ : feasible above $T = Δ H /Δ S$ . $Δ H < 0$ , $Δ S < 0$ : feasible below that temperature.

Notes

Use consistent units: $Δ H$ in kJ mol⁻¹ needs $Δ S$ in kJ K⁻¹ mol⁻¹ (divide J values by 1000).
The analysis assumes $Δ H$ and $Δ S$ stay roughly constant with temperature.
Endothermic reactions can be feasible when the $T Δ S$ term outweighs $Δ H$ ; that is why some salts dissolve with cooling.
Worked example: decomposition of calcium carbonate has $Δ H^{\circ} = + 178 kJ mol^{- 1}$ and $Δ S^{\circ} = + 0.1604 kJ K^{- 1} mol^{- 1}$ . At 298 K, $Δ G = 178 - 298 (0.1604) = + 130 kJ mol^{- 1}$ , not feasible. It becomes feasible above $T = 178/0.1604 = 1110$ K, which is why lime kilns run at over 1100 K.
Endothermic dissolving, such as ammonium nitrate in water, is feasible at room temperature because the large increase in entropy as the lattice breaks up makes $T Δ S$ larger than $Δ H$ .
Feasibility means a reaction can occur without a continuous input of energy from outside; whether it happens at a measurable rate is a question for kinetics.

Reading a ΔG–temperature graph 3.1.8.2

Key results

A plot of $Δ G$ against $T$ is a straight line with intercept $Δ H$ and gradient $- Δ S$ .
Ammonia formation: $Δ H = - 91.9 kJ mol^{- 1}$ , $Δ S = - 0.1981 kJ K^{- 1} mol^{- 1}$ ; $Δ G = 0$ at $T = Δ H /Δ S \approx 464 K$ . Feasible below about 464 K.

Notes

The line crosses zero where the reaction changes from feasible to not feasible.
Thermodynamically feasible does not mean fast: a large activation energy can make a feasible reaction immeasurably slow. That is why the Haber process uses a catalyst and a compromise temperature above 464 K.
A change of state shows as a kink in the line, because $Δ S$ changes there.
To read $Δ S$ from the graph, take the gradient over a large interval: e.g. if $Δ G$ rises from −60 to +40 kJ mol⁻¹ between 160 K and 664 K, the gradient is $100/504 = 0.198$ , so $Δ S = - 0.198 kJ K^{- 1} mol^{- 1}$ .
For a reaction with positive $Δ S$ the line slopes downwards. If that reaction is also endothermic, the line starts above zero at low $T$ and the reaction becomes feasible above the temperature where it crosses the axis.
Extrapolating the line back to $T = 0$ gives $Δ H$ , because at 0 K the $T Δ S$ term vanishes.

Worked examples

Worked example

The thermal decomposition:

CaCO_{3} (s) \to CaO (s) + CO_{2} (g)

has:

Δ H = + 178 kJ mol^{- 1}

and:

Δ S = + 161 J K^{- 1} mol^{- 1}

Calculate the minimum temperature at which this reaction becomes thermodynamically feasible.

Show worked solution

Feasibility begins where $Δ G = 0$ , i.e.

$Δ H = T Δ S$ , so:

T = \frac{Δ H}{Δ S}

Converting $Δ H$ to J mol⁻¹:

T = \frac{178 000}{161} = 1106 K

(to 4 s.f.), approximately 833 °C.

Below this temperature $Δ G$ is positive and the reaction is not thermodynamically feasible; above it, $Δ G$ is negative. (In practice, an industrial lime kiln can decompose calcium carbonate at somewhat lower observed temperatures than this calculated minimum, partly because continuously removing the CO₂ product shifts the position of equilibrium and partly because of other practical factors — the value calculated here is the thermodynamic threshold, not a precise prediction of observed industrial conditions.)

Worked example

Use the following data to construct a Born–Haber cycle for sodium chloride and calculate its lattice enthalpy of formation:

Δ H_{at} (Na) = + 107

Δ H_{at} (Cl) = + 121

(half the Cl–Cl bond enthalpy),

I E_{1} (Na) = + 496

E A_{1} (Cl) = - 349

Δ H_{f} (NaCl) = - 411

(all $kJ mol^{- 1}$ ).

Show worked solution

By Hess's law, the direct route (ΔH_f) equals the sum of the indirect route's steps around the cycle:

Δ H_{f} = Δ H_{at} (Na) + Δ H_{at} (Cl) + I E_{1} (Na) + E A_{1} (Cl) + Δ H_{lat}

Rearranging for the one unknown:

Δ H_{lat} = Δ H_{f} - [Δ H_{at} (Na) + Δ H_{at} (Cl) + I E_{1} (Na) + E A_{1} (Cl)] = - 411 - (107 + 121 + 496 - 349) = - 411 - 375 = - 786 kJ mol^{- 1}

3.1.9

Rate equations

The rate equation and orders 3.1.9.1

Definitions

Rate equation: $rate = k [A]^{m} [B]^{n}$ , where $m$ and $n$ are the orders with respect to A and B and $k$ is the rate constant.
Overall order: $m + n$ .

Key results

Zero order: doubling [A] leaves the rate unchanged. First order: doubling [A] doubles the rate. Second order: doubling [A] quadruples the rate.
On a rate–concentration graph: zero order is a horizontal line, first order a straight line through the origin, second order an upward curve.

Notes

Orders are found by experiment. They cannot be read from the coefficients of the overall equation.
The units of $k$ depend on the overall order: $s^{- 1}$ (first), $mol^{- 1} dm^{3} s^{- 1}$ (second), $mol^{- 2} dm^{6} s^{- 1}$ (third).
Worked example (units): for rate $= k [A]^{2} [B]$ , $k = \frac{rate}{[ A ] ^{2} [ B ]}$ , so the units are $\frac{mol dm ^{- 3} s ^{- 1}}{( mol dm ^{- 3} ) ^{3}} = mol^{- 2} dm^{6} s^{- 1}$ .
The rate constant is constant only at a fixed temperature. It is the rate when every concentration in the rate equation is 1 mol dm⁻³.
A reactant that is zero order still has to be present for the reaction to happen; it simply takes part after the rate-determining step, so its concentration does not affect the rate.
Orders are usually 0, 1 or 2 at A-level, but fractional orders exist for complex reactions; they always come from experiment.

Orders from initial rates 3.1.9.2

Key results

Doubling [A] multiplies the rate by 4, so $2^{m} = 4$ and $m = 2$ ; doubling [B] doubles the rate, so $n = 1$ .
$rate = k [A]^{2} [B]$ ; $k = \frac{2.00 \times 1 0 ^{- 4}}{( 0.100 ) ^{2} ( 0.100 )} = 0.200 dm^{6} mol^{- 2} s^{- 1}$ .

Method

Find two experiments in which only one concentration changes.
Compare the factor change in rate with the factor change in concentration: rate factor = (concentration factor) $^{order}$ .
Repeat for each reactant, write the rate equation, then substitute one experiment's data to find $k$ .

In practiceExperiment 1: [A] = 0.10, [B] = 0.10, rate $2.0 \times 1 0^{- 4}$ . Experiment 2: [A] = 0.20, [B] = 0.10, rate $8.0 \times 1 0^{- 4}$ . Experiment 3: [A] = 0.10, [B] = 0.20, rate $4.0 \times 1 0^{- 4}$ (mol dm $^{- 3}$ , mol dm $^{- 3}$ s $^{- 1}$ ).

$Exps 1, 2: [A] \times 2, rate \times 4 = 2^{2}$ second order in A
$Exps 1, 3: [B] \times 2, rate \times 2 = 2^{1}$ first order in B
$rate = k [A]^{2} [B]$
$k = \frac{2.0 \times 1 0 ^{- 4}}{( 0.10 ) ^{2} ( 0.10 )} = 0.20 mol^{- 2} dm^{6} s^{- 1}$ units: rate ÷ (concentration)³

Notes

All experiments must be at the same temperature, since $k$ depends on temperature.
If both concentrations change between two experiments, divide out the effect of the one whose order you already know.
A clock reaction gives initial rates directly. In the iodine clock, $H_{2} O_{2} + 2 I^{-} + 2 H^{+} \to I_{2} + 2 H_{2} O$ , a small fixed amount of thiosulfate removes the first iodine formed; once it is used up, iodine turns starch blue-black. The time $t$ is for a fixed small extent of reaction, so initial rate $\propto 1/ t$ .
This is Required Practical 7: vary one concentration at a time by dilution (keeping total volume constant), keep temperature constant, and plot $lo g (1/ t)$ against $lo g [reactant]$ ; the gradient is the order.
Worked example with both concentrations changing: if [A] doubles and [B] triples and the rate rises 12-fold, and the reaction is already known to be second order in A (factor 4), the remaining factor 3 shows first order in B.

Rates from concentration–time graphs 3.1.9.2

Key results

Rate at any time = −(gradient of the tangent) on a [reactant]–time graph; the initial rate is the tangent at $t = 0$ . Example: $0.100/200 = 5.00 \times 1 0^{- 4} mol dm^{- 3} s^{- 1}$ .
Zero order: a straight line. First order: an exponential fall with constant half-life. Second order: half-lives that get longer.

Notes

A constant half-life is the quickest test for first order: the time to halve from $[A]_{0}$ to $\frac{1}{2} [A]_{0}$ equals the time from $\frac{1}{2}$ to $\frac{1}{4}$ .
A tangent gives an instantaneous rate; a chord between two points gives only an average rate.
For first order, $k = \frac{ln 2}{t _{1/2}}$ .
Worked example: a first-order reaction with a half-life of 200 s has $k = 0.693/200 = 3.47 \times 1 0^{- 3} s^{- 1}$ . After 600 s (three half-lives), $\frac{1}{8}$ of the reactant remains.
To find an order from one continuous-monitoring run, draw tangents at several concentrations, calculate the rates, and plot rate against concentration: a horizontal line means zero order, a straight line through the origin first order, an upward curve second order.
Radioactive decay is the classic first-order process: every nucleus has the same chance of decaying per unit time, so the half-life is independent of the amount present.

Rate equations and mechanisms 3.1.9.1

Definitions

Rate-determining step: the slowest step in a mechanism, which limits the overall rate.

Key results

Species in the rate equation appear in (or before) the rate-determining step, in the numbers shown by their orders.
Mechanism $X + Y \to I$ (slow), $I + Y \to P$ (fast) predicts $rate = k [X] [Y]$ .

Notes

An intermediate is formed in one step and used up in a later one, so it does not appear in the overall equation.
A rate equation can rule a mechanism out but cannot prove one: different mechanisms can give the same rate equation.
Example: for $(CH_{3})_{3} CBr$ with $OH^{-}$ , rate $= k [(CH_{3})_{3} CBr]$ , consistent with a slow ionisation step ( $S_{N} 1$ ).
Worked example: $2 NO_{2} + F_{2} \to 2 NO_{2} F$ has rate $= k [NO_{2}] [F_{2}]$ . A consistent mechanism is $NO_{2} + F_{2} \to NO_{2} F + F$ (slow), then $NO_{2} + F \to NO_{2} F$ (fast). The slow step contains one of each reactant.
The acid-catalysed iodination of propanone, $CH_{3} COCH_{3} + I_{2} \to CH_{3} COCH_{2} I + HI$ , has rate $= k [CH_{3} COCH_{3}] [H^{+}]$ . Iodine is zero order, so it reacts after the rate-determining step; $H^{+}$ appears in the rate equation even though it is a catalyst.
Primary halogenoalkanes such as bromoethane follow rate $= k [C_{2} H_{5} Br] [OH^{-}]$ , consistent with a single step in which hydroxide attacks as bromide leaves ( $S_{N} 2$ ). Tertiary ones are first order overall ( $S_{N} 1$ ).

The Arrhenius equation 3.1.9.1

Definitions

Arrhenius equation: $k = A e^{- E_{a} / RT}$ , where $A$ is the pre-exponential factor, $E_{a}$ the activation energy (J mol⁻¹) and $T$ the temperature in K.

Key results

Logarithmic form: $ln k = - \frac{E _{a}}{R} \cdot \frac{1}{T} + ln A$ , a straight line of $ln k$ against $1/ T$ with gradient $- E_{a} / R$ and intercept $ln A$ .
Example: $A = 1.00 \times 1 0^{8} s^{- 1}$ , $E_{a} = 50.0 kJ mol^{- 1}$ gives $k = 0.197 s^{- 1}$ at 300 K and $2.08 s^{- 1}$ at 340 K.

Notes

Raising $T$ makes $- E_{a} / RT$ less negative, so $k$ (and the rate) increases. A larger $E_{a}$ makes $k$ more sensitive to temperature.
From the graph: $E_{a} = - gradient \times R$ . Check the axis scale, since $1/ T$ is often plotted ×10³.
Concentration changes the rate but not $k$ ; temperature and catalysts change $k$ .
Worked example from two temperatures: $ln \frac{k _{2}}{k _{1}} = - \frac{E _{a}}{R} (\frac{1}{T _{2}} - \frac{1}{T _{1}})$ . With $k$ rising from 0.197 to 2.08 s⁻¹ between 300 K and 340 K: $ln 10.56 = 2.357$ , and $1/300 - 1/340 = 3.92 \times 1 0^{- 4}$ , so $E_{a} = 2.357 \times 8.31/3.92 \times 1 0^{- 4} = 49.9 kJ mol^{- 1}$ .
The factor $e^{- E_{a} / RT}$ is the fraction of collisions with at least $E_{a}$ , linking the equation directly to the area under the Maxwell–Boltzmann curve. $A$ reflects the collision frequency and the fraction of collisions with the right orientation.
A catalyst lowers $E_{a}$ , so on an Arrhenius plot the catalysed line is less steep, and $k$ is larger at every temperature.
Keep $E_{a}$ in J mol⁻¹ when using $R = 8.31$ , and convert the final answer to kJ mol⁻¹ if asked.

Worked example

A rate constant is $2.0 \times 1 0^{- 3} s^{- 1}$ at 300 K and $8.0 \times 1 0^{- 3} s^{- 1}$ at 310 K.

Calculate the activation energy (:

R = 8.31 J K^{- 1} mol^{- 1}

Show worked solution

ln (\frac{k _{2}}{k _{1}}) = ln (4.0) = 1.386

\frac{1}{T _{1}} - \frac{1}{T _{2}} = \frac{1}{300} - \frac{1}{310} = 1.075 \times 1 0^{- 4} K^{- 1}

E_{a} = \frac{R \times 1.386}{1.075 \times 1 0 ^{- 4}} = \frac{8.31 \times 1.386}{1.075 \times 1 0 ^{- 4}} \approx 1.07 \times 1 0^{5} J mol^{- 1} \approx 107 kJ mol^{- 1}

3.1.10

Equilibrium constant Kp

Mole fractions and partial pressures 3.1.10

Definitions

Mole fraction: $x_{i} = \frac{n _{i}}{n _{total}}$ .
Partial pressure: $p_{i} = x_{i} P_{total}$ , the pressure a gas would exert if it alone occupied the container.

Key results

0.750 mol $N_{2} O_{4}$ and 0.500 mol $NO_{2}$ at 250 kPa: $x = 0.600$ and $0.400$ ; $p = 150$ and $100 kPa$ .
Partial pressures add up to the total pressure.

Notes

Include every gas in $n_{total}$ , including any inert gas present.
Mole fractions have no units and always sum to 1, a useful check on arithmetic before multiplying by the total pressure.
Dalton's law of partial pressures underlies the method: in an ideal mixture each gas behaves as if the others were absent, so its pressure is proportional to its amount.
Air is about 78% nitrogen and 21% oxygen by moles, so at 100 kPa the partial pressure of oxygen is about 21 kPa.

Writing Kp and its units 3.1.10

Definitions

For $a A(g) + b B(g) ⇌ c C(g) + d D(g)$ : $K_{p} = \frac{p ( C ) ^{c} p ( D ) ^{d}}{p ( A ) ^{a} p ( B ) ^{b}}$ , using equilibrium partial pressures.

Key results

$N_{2} O_{4} ⇌ 2 NO_{2}$ : $K_{p} = \frac{p ( NO _{2} ) ^{2}}{p ( N _{2} O _{4} )}$ , units kPa.
$N_{2} + 3 H_{2} ⇌ 2 NH_{3}$ : $K_{p} = \frac{p ( NH _{3} ) ^{2}}{p ( N _{2} ) p ( H _{2} ) ^{3}}$ , units kPa⁻².
Units of $K_{p}$ are $(pressure unit)^{Δ n}$ , where $Δ n$ = gas moles of products − gas moles of reactants.

Notes

For heterogeneous equilibria, only gases appear in $K_{p}$ .
When $Δ n = 0$ , $K_{p}$ has no units.
For $CaCO_{3} (s) ⇌ CaO(s) + CO_{2} (g)$ , $K_{p} = p (CO_{2})$ : the equilibrium pressure of carbon dioxide over the solids depends only on temperature, not on how much solid is present.
Square brackets mean concentration and belong only in $K_{c}$ ; use $p (X)$ or $p_{X}$ in $K_{p}$ . Mixing the two notations loses marks.
Pressure may be in Pa, kPa or atm; quote units consistent with the data, e.g. kPa⁻² for the ammonia equilibrium.

Calculating Kp 3.1.10

Key results

1.000 mol $N_{2} O_{4}$ , 0.250 mol dissociates: 0.750 mol $N_{2} O_{4}$ and 0.500 mol $NO_{2}$ ; at 250 kPa, $K_{p} = \frac{10 0 ^{2}}{150} = 66.7 kPa$ .

Method

Use the stoichiometry to find equilibrium amounts (initial, change, equilibrium).
Find the total amount and each mole fraction.
Multiply by the total pressure for each partial pressure, then substitute into $K_{p}$ .

In practice1.00 mol of $N_{2} O_{4}$ forms 0.40 mol of $NO_{2}$ at equilibrium, $N_{2} O_{4} ⇌ 2 NO_{2}$ , at a total pressure of 200 kPa. Find $K_{p}$ .

$N_{2} O_{4} 0.80, NO_{2} 0.40, total 1.20 mol$ 0.20 mol of N₂O₄ reacted
$p (N_{2} O_{4}) = \frac{0.80}{1.20} \times 200 = 133.3, p (NO_{2}) = \frac{0.40}{1.20} \times 200 = 66.7 kPa$ mole fraction × total pressure
$K_{p} = \frac{p ( NO _{2} ) ^{2}}{p ( N _{2} O _{4} )} = \frac{66. 7 ^{2}}{133.3} = 33.3 kPa$

Notes

Each mole of $N_{2} O_{4}$ that dissociates gives two of $NO_{2}$ , so the total amount rises.
Worked example: for $2 SO_{2} + O_{2} ⇌ 2 SO_{3}$ , equilibrium amounts 0.40 mol $SO_{2}$ , 0.20 mol $O_{2}$ and 1.40 mol $SO_{3}$ at 200 kPa give mole fractions 0.20, 0.10 and 0.70 and partial pressures 40, 20 and 140 kPa. $K_{p} = \frac{14 0 ^{2}}{4 0 ^{2} \times 20} = 0.613 kPa^{- 1}$ .
If the question gives the percentage of a reactant that has reacted, convert it to moles in the change row of the table first.
Questions may run in reverse: given $K_{p}$ and some partial pressures, rearrange to find the unknown pressure, then its mole fraction.

Effect of pressure 3.1.10

Notes

Increasing the total pressure at constant temperature shifts the equilibrium towards fewer moles of gas: for $N_{2} O_{4} ⇌ 2 NO_{2}$ , towards $N_{2} O_{4}$ .
$K_{p}$ is unchanged: the partial pressures readjust until the expression returns to the same value.
If both sides have the same number of gas moles, compression has no effect on the equilibrium composition.
The shift can be shown quantitatively. For $N_{2} O_{4} ⇌ 2 NO_{2}$ , $K_{p} = \frac{x ( NO _{2} ) ^{2}}{x ( N _{2} O _{4} )} P$ . If $P$ doubles, the ratio of mole fractions must halve to keep $K_{p}$ constant, so the mixture shifts towards $N_{2} O_{4}$ .
Adding an inert gas at constant volume leaves every partial pressure unchanged, so the equilibrium does not move. Adding it at constant total pressure lowers the partial pressures, like an expansion.
Industrial processes that reduce the number of gas moles (ammonia, methanol, sulfur trioxide) favour high pressure, balanced against the cost of compressors and thick-walled plant.

Effect of temperature and catalysts 3.1.10

Key results

Endothermic forward reaction: heating increases $K_{p}$ . Exothermic forward reaction: heating decreases $K_{p}$ .

Notes

Temperature is the only factor that changes $K_{p}$ .
A catalyst speeds up the approach to equilibrium but changes neither the composition nor $K_{p}$ .
Worked example: for $N_{2} O_{4} ⇌ 2 NO_{2}$ ( $Δ H = + 57 kJ mol^{- 1}$ ), heating increases $K_{p}$ , so the mixture darkens as more brown $NO_{2}$ forms.
When $K_{p}$ changes, the composition must change to match it; when $K_{p}$ stays constant (pressure changes), the composition changes only if the expression includes the total pressure.
Explain temperature effects in terms of rates: heating speeds up both directions, but the endothermic direction, with its higher activation energy, speeds up more.

3.1.11

Electrode potentials and electrochemical cells

Electrochemical cells 3.1.11.1

Definitions

Electrochemical cell: two half-cells connected by a wire and a salt bridge, in which a redox reaction drives electrons through an external circuit.

Key results

Zinc–copper cell: $Zn(s) \to Zn^{2 +} (a q) + 2 e^{-}$ (negative electrode, oxidation) and $Cu^{2 +} (a q) + 2 e^{-} \to Cu(s)$ (positive electrode, reduction).

Notes

Electrons flow through the wire from the negative to the positive electrode; ions move through the salt bridge (here $KNO_{3}$ ) to keep each solution electrically neutral.
The salt bridge's ions must not react with the solutions, which is why potassium nitrate is often used rather than a chloride with silver ions.
Over time the zinc electrode loses mass and copper is deposited on the copper electrode.
Half-cells can be metal/metal-ion, two ions of the same element in solution with an inert platinum electrode (e.g. $Fe^{3 +} / Fe^{2 +}$ ), or a gas with its ions over platinum (e.g. $Cl_{2} / Cl^{-}$ ).
The conventional representation puts phase boundaries as single lines and the salt bridge as a double line, with the reduced form furthest from the bridge on the left: $Pt ∣ H_{2} (g) ∣ H^{+} (a q) ∥ Fe^{3 +} (a q), Fe^{2 +} (a q) ∣ Pt$ .
A salt bridge is commonly filter paper soaked in saturated potassium nitrate. A wire cannot replace it, because ions, not electrons, must move between the solutions.

The standard hydrogen electrode 3.1.11.1

Definitions

Standard electrode potential, $E^{\circ}$ : the EMF of a half-cell connected to a standard hydrogen electrode under standard conditions (298 K, 100 kPa, 1.00 mol dm⁻³ ions).
Standard hydrogen electrode: $H_{2}$ at 100 kPa bubbled over platinised platinum in 1.00 mol dm⁻³ $H^{+}$ ; assigned $E^{\circ} = 0.00 V$ .

Notes

A single electrode potential cannot be measured on its own, so every value is relative to the hydrogen electrode.
A high-resistance voltmeter draws negligible current, so it measures the maximum potential difference (the EMF).
Platinum is inert and provides a surface for the $H^{+} / H_{2}$ equilibrium; platinising it increases the surface area.
Because the hydrogen electrode is awkward to use (gas supply, platinum that is easily poisoned), secondary standards such as the silver/silver chloride or calomel electrode, whose potentials are known accurately, are used in practice.
All three standard conditions matter: changing temperature, gas pressure or ion concentration changes the measured potential, so values are only standard under 298 K, 100 kPa and 1.00 mol dm⁻³.
The sign of $E^{\circ}$ shows whether the half-cell is more (positive) or less (negative) readily reduced than $H^{+}$ : magnesium, at −2.37 V, is a far better reducing agent than hydrogen; fluorine, at +2.87 V, is the strongest oxidising agent.

Using electrode potentials 3.1.11.1

Key results

$E_{cell} = E^{\circ} (right) - E^{\circ} (left) = E^{\circ} (positive) - E^{\circ} (negative)$ . For Zn–Cu: $+ 0.34 - (- 0.76) = + 1.10 V$ .
Cell notation: $Zn(s) ∣ Zn^{2 +} (a q) ∥ Cu^{2 +} (a q) ∣ Cu(s)$ , with the oxidised half-cell on the left.

Notes

Tabulated $E^{\circ}$ values refer to reduction. The more positive half-cell is reduced; the more negative one is oxidised.
When combining half-equations, balance the electrons but do not multiply $E^{\circ}$ values: they are per electron, not per mole of equation.
A positive $E_{cell}$ means the reaction is feasible under standard conditions; it says nothing about rate, and non-standard concentrations or temperatures change the potential.
Predicting reactions: the species on the left of a half-equation with the more positive $E^{\circ}$ oxidises the species on the right of one with a less positive $E^{\circ}$ . $Fe^{3 +} / Fe^{2 +}$ (+0.77 V) oxidises iodide (+0.54 V) to iodine, but cannot oxidise bromide (+1.09 V).
Changing concentration shifts the equilibrium in a half-cell: increasing $[Cu^{2 +}]$ in $Cu^{2 +} + 2 e^{-} ⇌ Cu$ shifts it right, making the electrode potential more positive and the cell EMF larger.
A predicted reaction may not be observed if its activation energy is too high, or if the conditions are far from standard.
Useful values: $MnO_{4}^{-} / Mn^{2 +}$ +1.51, $Cl_{2} / Cl^{-}$ +1.36, $Cr_{2} O_{7}^{2 -} / Cr^{3 +}$ +1.33, $Ag^{+} / Ag$ +0.80, $Cu^{2 +} / Cu$ +0.34, $Zn^{2 +} / Zn$ −0.76 V.

Commercial cells 3.1.11.2

Key results

Lithium cell (simplified) during discharge: negative electrode $Li \to Li^{+} + e^{-}$ ; positive electrode $Li^{+} + CoO_{2} + e^{-} \to Li^{+} [CoO_{2}]^{-}$ .

Notes

Non-rechargeable cells stop when the reactants are used up. Rechargeable cells can be reversed by an external supply because their reactions are reversible and the products stay on the electrodes.
Lithium cells use a non-aqueous electrolyte because lithium reacts with water.
Fuel cells are not recharged; the fuel and oxidant are supplied continuously.
The lithium cell's EMF can be calculated from its half-cells: $Li^{+} / Li$ at −3.04 V and the cobalt oxide electrode at about +0.56 V give $E_{cell} \approx 3.6$ V, far more than any aqueous cell could sustain.
On recharging, the electrode reactions are reversed: lithium ions move back to the graphite-based negative electrode, where they are reduced and stored.
Benefits of cells include portable power and, for rechargeables, less waste; drawbacks include the energy and mining involved in making them, the toxicity or flammability of some components, and the need for careful recycling.

The alkaline hydrogen–oxygen fuel cell 3.1.11.2

Key results

Negative electrode: $2 H_{2} + 4 OH^{-} \to 4 H_{2} O + 4 e^{-}$ . Positive electrode: $O_{2} + 2 H_{2} O + 4 e^{-} \to 4 OH^{-}$ .
Overall: $2 H_{2} + O_{2} \to 2 H_{2} O$ , with a standard EMF of about 1.23 V.

Notes

Water is the only product at the cell, but hydrogen is mostly made from fossil fuels or with electricity, so the overall environmental benefit depends on how it was produced.
Disadvantages include storing and transporting a flammable gas, and the alkaline electrolyte reacting with carbon dioxide from the air.
Porous electrodes with a catalyst give a large surface for the gas reactions.
The EMF can be checked from the half-cells: $O_{2} + 2 H_{2} O + 4 e^{-} \to 4 OH^{-}$ at +0.40 V and $2 H_{2} O + 2 e^{-} \to H_{2} + 2 OH^{-}$ at −0.83 V give $0.40 - (- 0.83) = 1.23$ V.
A fuel cell converts chemical energy straight into electrical energy, so it can be more efficient than burning hydrogen in an engine, where much of the energy is lost as heat.
An ion-exchange membrane separates the two gases while letting ions through, so hydrogen and oxygen never meet directly.
The EMF is fixed by the half-cell reactions, so it does not fall as the reactants are used, provided the gases keep being supplied.

3.1.12

Acids and bases

Brønsted–Lowry acids and bases 3.1.12.1

Definitions

Brønsted–Lowry acid: a proton donor. Brønsted–Lowry base: a proton acceptor.
Conjugate pair: two species differing by one proton, e.g. HA / $A^{-}$ and $H_{3} O^{+}$ / $H_{2} O$ .
Strong acid: essentially fully dissociated in water. Weak acid: only partly dissociated.

Key results

$HA + H_{2} O ⇌ H_{3} O^{+} + A^{-}$ : HA and $H_{3} O^{+}$ are acids; $H_{2} O$ and $A^{-}$ are bases.

Notes

Strength describes the extent of dissociation; concentration describes how much acid is dissolved per dm³. A dilute strong acid and a concentrated weak acid are both possible.
Acid–base reactions are proton transfers, so every acid–base equilibrium involves two conjugate pairs.
Water is amphoteric: it acts as a base with HCl ( $HCl + H_{2} O \to H_{3} O^{+} + Cl^{-}$ ) and as an acid with ammonia ( $NH_{3} + H_{2} O ⇌ NH_{4}^{+} + OH^{-}$ ).
The stronger an acid, the weaker its conjugate base: $Cl^{-}$ has no tendency to accept a proton back, whereas the ethanoate ion is a moderately good base.
Diprotic acids such as $H_{2} SO_{4}$ can donate two protons per molecule, so 1 mol of sulfuric acid neutralises 2 mol of NaOH.
$H^{+} (a q)$ is shorthand for $H_{3} O^{+}$ : a bare proton is too small and highly charged to exist alone in water.

pH and the ionic product of water 3.1.12.2–3.1.12.3

Definitions

$pH = - lo g_{10} [H^{+}]$ ; $[H^{+}] = 1 0^{- pH}$ .
Ionic product of water: $K_{w} = [H^{+}] [OH^{-}] = 1.00 \times 1 0^{- 14} mol^{2} dm^{- 6}$ at 298 K.

Key results

0.0100 mol dm⁻³ HCl: pH = 2.00. 0.0100 mol dm⁻³ NaOH: $[H^{+}] = K_{w} / [OH^{-}] = 1.00 \times 1 0^{- 12}$ , pH = 12.00.
A fall of 1 pH unit is a tenfold rise in $[H^{+}]$ .

Notes

Neutral means $[H^{+}] = [OH^{-}]$ , not pH 7. Water's dissociation is endothermic, so $K_{w}$ rises with temperature and the neutral pH falls below 7 when hot.
For a strong base, find $[OH^{-}]$ first, then use $K_{w}$ to get $[H^{+}]$ .
Worked example: 0.0500 mol dm⁻³ $H_{2} SO_{4}$ , treated as fully dissociated, gives $[H^{+}] = 0.100$ and pH = 1.00. 0.0500 mol dm⁻³ $Ba(OH)_{2}$ gives $[OH^{-}] = 0.100$ , so pH = 13.00.
Diluting a strong acid tenfold raises its pH by exactly 1 unit (until the pH approaches 7, where water's own ions matter).
$K_{w}$ at 323 K is about $5.5 \times 1 0^{- 14}$ , so pure water there has pH 6.63 and is still neutral.
Mixing problems: calculate the moles of $H^{+}$ and $OH^{-}$ , subtract to find which is in excess, divide by the total volume, then find the pH.

Weak acids and Ka 3.1.12.4

Definitions

Acid dissociation constant: $K_{a} = \frac{[ H ^{+} ] [ A ^{-} ]}{[ HA ]}$ (mol dm⁻³); $p K_{a} = - lo g_{10} K_{a}$ .

Key results

For a weak acid alone: $[H^{+}] \approx K_{a} c$ .
$c = 0.100$ , $K_{a} = 1.60 \times 1 0^{- 5}$ : $[H^{+}] = 1.26 \times 1 0^{- 3}$ , pH = 2.90; only about 1.3% is ionised.

Notes

Assumptions: $[H^{+}] = [A^{-}]$ (water's contribution is negligible) and $[HA]$ at equilibrium ≈ the initial concentration (dissociation is slight).
A larger $K_{a}$ (smaller p $K_{a}$ ) means a stronger acid.
Worked example (reverse): 0.100 mol dm⁻³ ethanoic acid has pH 2.88, so $[H^{+}] = 1.32 \times 1 0^{- 3}$ and $K_{a} = (1.32 \times 1 0^{- 3})^{2} /0.100 = 1.74 \times 1 0^{- 5} mol dm^{- 3}$ (p $K_{a}$ 4.76).
Diluting a weak acid tenfold raises its pH by only about 0.5, because $[H^{+}] \propto c$ and the acid also dissociates further.
The approximations fail for relatively strong weak acids or very dilute solutions; at A-level they are always stated as assumptions when used.
Required Practical 9 measures pH during a weak acid–strong base titration; $K_{a}$ is found from the pH at half-neutralisation.

Buffers 3.1.12.6

Definitions

Buffer: a solution that resists changes in pH when small amounts of acid or base are added, or on dilution.

Key results

Acidic buffer: $[H^{+}] = K_{a} \frac{[ HA ]}{[ A ^{-} ]}$ .
Partial neutralisation: 0.0100 mol HA + 0.00400 mol $OH^{-}$ leaves 0.00600 mol HA and 0.00400 mol $A^{-}$ ; $[H^{+}] = 1.60 \times 1 0^{- 5} \times 1.5 = 2.40 \times 1 0^{- 5}$ , pH = 4.62.

Notes

An acidic buffer (weak acid plus its salt) removes added $H^{+}$ by $A^{-} + H^{+} \to HA$ and added $OH^{-}$ by $HA + OH^{-} \to A^{-} + H_{2} O$ . Both members of the pair must be present in large amounts.
A basic buffer works the same way with $NH_{3}$ and $NH_{4}^{+}$ .
Because the acid and salt share one volume, moles can be used in the ratio. Blood is buffered mainly by carbonic acid and hydrogencarbonate ions.
Worked example showing buffering: 1.00 dm³ containing 0.100 mol HA and 0.100 mol $A^{-}$ ( $K_{a} = 1.60 \times 1 0^{- 5}$ ) has pH 4.80. Adding 0.0100 mol HCl changes the amounts to 0.110 and 0.090, giving pH 4.71, a change of 0.09. The same acid added to 1.00 dm³ of water would give pH 2.00.
A buffer works best when [HA] and [ $A^{-}$ ] are similar (pH within about 1 unit of p $K_{a}$ ) and both are concentrated; its capacity is exhausted once either component is used up.
For a basic buffer use $K_{a}$ of the ammonium ion, $5.6 \times 1 0^{- 10}$ , with $[H^{+}] = K_{a} [NH_{4}^{+}] / [NH_{3}]$ .
Blood is kept at pH 7.35–7.45 by $H_{2} CO_{3} ⇌ H^{+} + HCO_{3}^{-}$ ; buffers are also used in shampoos, food and biological experiments where enzymes need a steady pH.

Titration curves and indicators 3.1.12.5

Key results

Strong acid + strong base: equivalence at pH 7, with a long vertical section (about pH 3–11).
Weak acid + strong base: equivalence above pH 7; at half-equivalence, $pH = p K_{a}$ .
Strong acid + weak base: equivalence below pH 7. Weak acid + weak base: no sharp vertical section.

Notes

Choose an indicator whose colour-change range lies within the vertical section: phenolphthalein (8.2–10.0) for a weak acid with a strong base, methyl orange (3.1–4.4) for a strong acid with a weak base.
No visual indicator suits a weak acid–weak base titration; use a pH meter.
The half-equivalence point gives $K_{a}$ directly, because there $[HA] = [A^{-}]$ .
Indicators are themselves weak acids, $HIn ⇌ H^{+} + In^{-}$ , with different colours for HIn and $In^{-}$ . The colour changes over roughly p $K_{a}$ (HIn) ± 1, which is why each indicator has its own range.
Only a few drops of indicator are used, because it is an acid and would otherwise react with the titrant and affect the end point.
The equivalence point is the midpoint of the vertical section; read the volume there for the titration calculation.
The starting pH identifies the acid: a strong acid at 0.100 mol dm⁻³ starts at pH 1, a weak one of the same concentration nearer pH 3. The curve's final pH tends towards that of the excess base.

Worked example

Calculate the pH of a buffer containing 0.100 mol dm⁻³ ethanoic acid and 0.150 mol dm⁻³ sodium ethanoate (:

K_{a} = 1.74 \times 1 0^{- 5} mol dm^{- 3}

Show worked solution

p K_{a} = - lo g_{10} (1.74 \times 1 0^{- 5}) = 4.76

lo g_{10} (\frac{0.150}{0.100}) = lo g_{10} (1.5) = 0.18

pH = p K_{a} + lo g_{10} (\frac{[ salt ]}{[ acid ]}) = 4.76 + 0.18 = 4.94