IB Chemistry · Unit 5 · Reactivity

Energy.
Entropy.
Direction.

Energy never disappears. Reactions only go where the entropy wins. Two laws that decide what's possible in chemistry.

Lessons HL extension Key terms
18Lessons
5HL extensions
42Key terms
SL+HLLevel
A B DYNAMIC EQUILIBRIUM
Unit 5 · Standard Level

13 lessons to work through.

The required syllabus content for Unit 5, in order. Each card is one lesson-sized checkpoint.

Lesson 1

Dynamic Equilibrium

Brainstorm everything you already know about equilibrium (reversible reactions)

Lesson 2

Position of Equilibrium Kc

Lesson 2 of Unit 5.

Lesson 3

Le Châtelliers principle

Lesson 3 of Unit 5.

Lesson 4

Equilibrium Case Studies

Lesson 4 of Unit 5.

Lesson 7

Energy changes in reactions

Lesson 7 of Unit 5.

Lesson 8

Measuring Enthalpy Changes

In chemical transformations, energy can neither be created nor destroyed (the first law of thermodynamics).

Lesson 9

Measuring Enthalpy Changes- Practical

Lesson 9 of Unit 5.

Lesson 10

Bond Enthalpies

Average bond energies can be found in the data booklet!

Lesson 11

Hess’s Law

This law is a statement of the conservation of energy.

Lesson 14

Combustion

The products have less enthalpy than the reactants and are therefore energetically more stable.

Lesson 15

Comparing Fuels

The shorter the chain of a hydrocarbon or fuel, the higher the amount of specific energy (amount of energy released per unit mass), and the more likely that the fuel combusts completely.

Lesson 16

Entropy (HL)

Lesson 16 of Unit 5.

Lesson 17

Spontaneity (HL)

Lesson 17 of Unit 5.

Lessons in detail

The unit, lesson by lesson.

Each lesson card below mirrors the original teacher deck — syllabus refs, content, worked examples and practice questions in order.

Lesson 1

Dynamic equilibrium

Reactivity 2.3.1

Lesson outcomes

A reversible reaction can proceed in either direction. In a closed system it reaches a point where the forward and reverse rates are equal — dynamic equilibrium.

Four features of equilibrium

Evidence for the dynamic nature

Isotopic labelling: introduce a radioactive isotope of one element into the reactant. Over time, the label shows up in both reactant and product. Both reactions are happening — concentrations are stable because forward and reverse cancel.

Try these

  1. List four features of a system at dynamic equilibrium.
    Show answer
    (1) Forward and reverse reactions both occur (dynamic). (2) Forward rate = reverse rate. (3) Concentrations of all species are constant (but not zero). (4) Only achievable in a closed system.
  2. What evidence do we have that equilibrium is dynamic rather than static?
    Show answer
    Isotope-labelling experiments: introduce a radioactive isotope into the reactant; over time the label appears in both reactant and product, confirming continuous interconversion in both directions.
  3. Why can equilibrium not be reached in an open beaker for the reaction H₂CO₃ ⇌ H₂O + CO₂?
    Show answer
    CO₂ escapes from the open container. The reverse reaction (CO₂ + H₂O → H₂CO₃) requires both species present — without trapped CO₂, the reverse reaction effectively stops. The forward reaction continues until all H₂CO₃ is consumed.
  4. Give an example of a chemical equilibrium and a physical equilibrium.
    Show answer
    Chemical: N₂(g) + 3H₂(g) ⇌ 2NH₃(g) — reversible reaction in a closed vessel. Physical: a saturated NaCl(aq) in contact with solid NaCl(s) — dissolution and crystallisation occur at the same rate, [NaCl] constant.
Lesson 2

The equilibrium constant K_c

Reactivity 2.3.2Reactivity 2.3.3

Lesson outcomes

For the general reaction aA + bB ⇌ cC + dD at equilibrium:

Kc = ([C]c[D]d) / ([A]a[B]b)

Kc depends only on temperature.

Interpreting Kc

What to include / exclude

Pure solids and pure liquids are excluded from the K expression. Their "concentration" is essentially constant (= their density / M), absorbed into K.

For CaCO₃(s) ⇌ CaO(s) + CO₂(g): Kc = [CO₂] only.

Worked example

Calculating K_c from equilibrium concentrations

Problem. For N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g), equilibrium concentrations are [N₂] = 0.50, [H₂] = 1.50, [NH₃] = 0.40 mol dm⁻³. Calculate Kc.
Solution. Kc = [NH₃]² / ([N₂][H₂]³) = (0.40)² / (0.50 × 1.50³) = 0.16 / (0.50 × 3.375) = 0.16 / 1.6875 = 0.0948 mol⁻² dm⁶.

Try these

  1. Write the Kc expression for 2 SO₂(g) + O₂(g) ⇌ 2 SO₃(g).
    Show answer
    Kc = [SO₃]² / ([SO₂]² · [O₂]).
  2. Write the Kc expression for CaCO₃(s) ⇌ CaO(s) + CO₂(g). Why are CaCO₃ and CaO excluded?
    Show answer
    Kc = [CO₂]. The two solids are pure substances — their 'concentration' is effectively constant (equal to density / molar mass) and is absorbed into K.
  3. Write the Kc expression for N₂O₄(g) ⇌ 2 NO₂(g) and explain the units.
    Show answer
    Kc = [NO₂]² / [N₂O₄]. Units: (mol dm⁻³)² / (mol dm⁻³) = mol dm⁻³.
  4. What does Kc = 1.5 × 10⁻⁵ tell you about a reaction?
    Show answer
    Kc << 1: equilibrium lies far to the left; reactants strongly favoured. Very little product forms at this T.
  5. What does Kc = 4.0 × 10⁸ tell you about a reaction?
    Show answer
    Kc >> 1: equilibrium lies far to the right; products strongly favoured. The reaction goes essentially to completion.
Lesson 3

Le Chatelier's principle

Reactivity 2.3.4

Lesson outcomes

Le Chatelier's principle: when a system at equilibrium is disturbed, it shifts in a direction that partially counteracts the disturbance.

DisturbanceDirection of shiftKc
Add reactantForward (consume the added reactant)Unchanged
Remove productForward (replace the removed product)Unchanged
Increase p (gases)Toward fewer moles of gasUnchanged
Increase TEndothermic directionChanges
Decrease TExothermic directionChanges
Add catalystNo shift (forward and reverse rates ↑ equally)Unchanged
Worked example

Predicting shifts in the Haber process

Problem. N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g), ΔH = −92 kJ mol⁻¹. Predict the effect on the equilibrium position and yield of NH₃ when: (a) p is increased, (b) T is increased, (c) NH₃ is removed.
Solution. (a) p ↑: 4 mol of gas on the left, 2 mol on the right. Shift to the side with fewer moles = right → more NH₃.
(b) T ↑: forward reaction is exothermic, so shift in the endothermic direction = left → less NH₃. Kc also decreases.
(c) Remove NH₃: shift right to replace product → more NH₃ continually forms. Kc unchanged.

Try these

  1. State Le Chatelier's principle.
    Show answer
    When a system at dynamic equilibrium is disturbed, it shifts in a direction that partially counteracts the disturbance, re-establishing equilibrium.
  2. For N₂ + 3H₂ ⇌ 2NH₃ (ΔH = −92 kJ mol⁻¹), predict the effect of: (a) adding more H₂, (b) removing NH₃, (c) increasing pressure, (d) increasing temperature, (e) adding a catalyst.
    Show answer
    (a) Shift right (consume H₂) — more NH₃. K unchanged. (b) Shift right (replace NH₃) — more NH₃. K unchanged. (c) Shift to fewer moles of gas — right (4 → 2) — more NH₃. K unchanged. (d) Shift in endothermic direction — left — less NH₃. K decreases. (e) No shift. Equilibrium reached faster but at same position.
  3. Why is the industrial Haber process run at ~450 °C rather than at low T, even though low T favours product formation?
    Show answer
    Compromise. Low T does favour NH₃ formation thermodynamically (K higher), but kinetically the reaction is too slow at low T. ~450 °C with Fe catalyst gives an acceptable yield at an acceptable rate.
  4. For an endothermic reaction, would you increase or decrease T to get more product? Justify with Le Chatelier.
    Show answer
    Increase T. Endothermic = absorbs heat in the forward direction. Treating heat as a reactant, adding it shifts equilibrium forward (consume the added heat) → more product. K also increases with T for endothermic reactions.
  5. Why does adding an inert gas (e.g. He) at constant volume not shift an equilibrium of gases?
    Show answer
    Adding inert gas at constant V increases the total pressure but does not change the partial pressures (or concentrations) of the reacting species. So Q is unchanged, no shift. (Only if you increase total p by compressing — reducing V — do partial pressures and concentrations rise, triggering a shift.)
Lesson 4

Equilibrium case studies

Reactivity 2.3.4

Lesson outcomes

The Contact process · making H₂SO₄

2 SO₂(g) + O₂(g) ⇌ 2 SO₃(g), ΔH = −197 kJ mol⁻¹. Compromise conditions: ~450 °C and atmospheric pressure with V₂O₅ catalyst. Low p would shift right but the small mole change (3 → 2) means high p isn't worth the engineering cost.

Carbonated drinks

CO₂(g) ⇌ CO₂(aq) (with subsequent H₂CO₃ formation). Bottled under high CO₂ pressure → high [CO₂(aq)]. Opening the bottle releases CO₂(g), equilibrium shifts left, CO₂ escapes as bubbles.

Blood pH and CO₂

CO₂ + H₂O ⇌ H₂CO₃ ⇌ HCO₃⁻ + H⁺. Hyperventilation removes CO₂ → equilibrium shifts left → [H⁺] drops → blood pH rises (respiratory alkalosis). Breathing into a bag re-inhales CO₂ to restore balance.

Try these

  1. For the Contact process (2 SO₂ + O₂ ⇌ 2 SO₃, ΔH = −197 kJ mol⁻¹), low T and high p both favour SO₃. Why is it operated at ~450 °C and only modest pressure (~1 atm)?
    Show answer
    T compromise: too low → too slow even with V₂O₅ catalyst. ~450 °C balances rate and yield. p compromise: only 3 → 2 mol of gas (modest reduction), so the gain from compression isn't worth the engineering cost. High p only helps if there's a big mol change.
  2. Why are carbonated drinks bottled under pressure? What happens when you open the bottle?
    Show answer
    CO₂(g) ⇌ CO₂(aq). High p forces equilibrium right → high [CO₂(aq)]. Opening releases pressure → equilibrium shifts left → CO₂(g) bubbles out. The drink eventually goes 'flat'.
  3. Why does climbing to high altitude initially cause hyperventilation, which then leads to dizziness?
    Show answer
    Low atmospheric pO₂ at altitude triggers increased breathing rate (to absorb more O₂). But this also exhales more CO₂, shifting the CO₂/HCO₃⁻ equilibrium left → blood [H⁺] drops → blood pH rises. Alkalosis causes dizziness, tingling. The body slowly compensates by adjusting renal HCO₃⁻ excretion.
  4. Why does breathing into a paper bag help someone hyperventilating?
    Show answer
    Re-breathing exhaled CO₂ raises blood CO₂, shifting the CO₂/HCO₃⁻ equilibrium right → restores [H⁺] → blood pH returns to ~7.4. A simple Le Chatelier remedy.
Lesson 7

Energy changes in reactions

Reactivity 1.1.1Reactivity 1.1.2

Lesson outcomes

Enthalpy H is the internal energy + pV term. Almost all chemistry happens at constant pressure, so ΔH = q (heat exchanged with surroundings).

TypeSign of ΔHDirection of heatSurroundings
Exothermic−Released by reactionGet warmer
Endothermic+Absorbed by reactionGet cooler

Bond breaking vs bond making

Bond breaking requires energy (endothermic). Bond making releases energy (exothermic). For a complete reaction, ΔH = (energy needed to break reactant bonds) − (energy released forming product bonds).

Try these

  1. Why does dissolving ammonium nitrate in water feel cold?
    Show answer
    Dissolution of NH₄NO₃ is endothermic (positive ΔH). The reaction absorbs heat from the water, which loses thermal energy and feels cold. Used commercially in cold packs.
  2. For a reaction with ΔH = −250 kJ mol⁻¹, sketch an enthalpy profile diagram and label reactants, products, ΔH and Ea.
    Show answer
    Curve starts at reactant level, rises through a peak (transition state, Ea above reactants), falls to products at level 250 kJ mol⁻¹ below reactants. ΔH = −250 marked as the vertical drop from reactants to products.
  3. List three everyday exothermic and three everyday endothermic reactions.
    Show answer
    Exo: combustion (burning fuel), neutralisation (acid + base), respiration. Endo: photosynthesis, dissolving NH₄NO₃ in water (cold pack), thermal decomposition of CaCO₃ in a kiln.
  4. ΔHrxn can be calculated from bond enthalpies as Σ(broken) − Σ(formed). Why does the sign convention work?
    Show answer
    Breaking bonds requires energy in (+). Making bonds releases energy out (−). For an exothermic reaction, energy released forming new bonds exceeds energy needed to break old → net negative ΔH.
Lesson 8

Measuring enthalpy changes · calorimetry

Reactivity 1.2.1

Lesson outcomes

Calorimetry measures the heat absorbed or released by a reaction:

q = m × c × ΔT

where m is the mass of the solution (g), c is the specific heat capacity (4.18 J g⁻¹ K⁻¹ for water), ΔT is the temperature change (K or °C — same numerical magnitude).

Then ΔH = ±q / n, where n is the moles of the limiting reactant. The sign convention:

Worked example

Neutralisation enthalpy

Problem. 50.0 cm³ of 1.00 mol dm⁻³ HCl is mixed with 50.0 cm³ of 1.00 mol dm⁻³ NaOH. The temperature rises from 19.5 °C to 26.3 °C. Calculate ΔHneut in kJ mol⁻¹.
Solution. Assume solution density 1.00 g cm⁻³, so m = 100 g. c = 4.18.
q = 100 × 4.18 × (26.3 − 19.5) = 100 × 4.18 × 6.8 = 2842 J released.
n(HCl) = 0.0500 × 1.00 = 0.0500 mol. (Equimolar, so this is the limiting reactant.)
ΔH = −q / n = −2842 / 0.0500 = −56 850 J mol⁻¹ = −56.8 kJ mol⁻¹. Compares well to the literature value (−57.3 kJ mol⁻¹).

Try these

  1. Why is the actual measured ΔHneut always slightly less exothermic (less negative) than the literature value?
    Show answer
    Heat loss to the surroundings (the calorimeter, the air). Some of q escapes before being measured, so ΔT is underestimated. Improvements: insulated calorimeter, lid, temperature extrapolation back to t=0 of mixing.
  2. 100 cm³ of 1.0 M HCl is added to 100 cm³ of 1.0 M NaOH; ΔT = +6.8 °C. Calculate ΔHneut.
    Show answer
    m = 200 g (combined). q = 200 × 4.18 × 6.8 = 5685 J. n(HCl) = n(NaOH) = 0.10 mol. ΔH = −q/n = −5685/0.10 = −56.9 kJ mol⁻¹.
  3. Why must we assume that the solution density and heat capacity are equal to water in calorimetry calculations?
    Show answer
    Dilute aqueous solutions are dominated by water. Density ≈ 1.00 g cm⁻³ and c ≈ 4.18 J g⁻¹ K⁻¹ within ~1% — close enough for the precision of school-level calorimetry. For very concentrated solutions or non-aqueous solvents, you'd need actual values.
  4. In a calorimetry experiment for the heat of solution, you measure m = 2.50 g of NH₄NO₃ in 50.0 g of water. ΔT = −5.5 °C. Calculate ΔHsol per mole.
    Show answer
    q = m(solution) × c × ΔT = 52.5 × 4.18 × (−5.5) = −1207 J. Note: q is negative because the solution lost heat (T went down). For the dissolution reaction (which absorbed heat from the water): ΔHsol = +1207 J for the dissolved sample. n(NH₄NO₃) = 2.50/80.06 = 0.0312 mol. ΔH = +1207/0.0312 = +38.7 kJ mol⁻¹.
Lesson 9

Measuring enthalpy changes · practical

Reactivity 1.2.1Tool 1

Lesson outcomes

A simple coffee-cup calorimeter (polystyrene cup with lid) is good enough for ΔH measurements at the IB level.

Temperature extrapolation

Some heat is always lost during the reaction. To minimise the error, plot T vs t every 30 s for ~5 minutes before mixing (baseline) and ~5 minutes after. Extrapolate the post-reaction line back to the moment of mixing — this is your "true" Tmax. Subtract initial T to get a more accurate ΔT.

Common experiments

Try these

  1. Why is temperature extrapolation needed in ΔHneutralisation experiments?
    Show answer
    Mixing isn't instantaneous, and heat is lost to the surroundings during the time it takes for T to peak. Extrapolating the cooling curve back to t = 0 of mixing estimates what the T would have been if the reaction were instantaneous and adiabatic — a more accurate ΔT.
  2. Why is polystyrene a good calorimeter material? What are its limitations?
    Show answer
    Excellent thermal insulator → minimises heat loss to surroundings. Cheap and disposable. Limitations: Low maximum T (~70 °C); some heat absorbed by the cup itself; can react with strong acids/bases over time.
  3. In a displacement reaction Zn + CuSO₄, why do you use excess Zn?
    Show answer
    To ensure CuSO₄ is the limiting reactant. This way, the moles of reaction equals the moles of CuSO₄ used (a known quantity), so ΔH per mol is calculated from accurate stoichiometry rather than estimating the unknown excess of Zn.
Lesson 10

Bond enthalpies

Reactivity 1.2.2

Lesson outcomes

Bond enthalpy: the energy required to break one mole of a specific bond in the gaseous state. Values are averages across many molecules.

ΔHrxn ≈ Σ(bonds broken in reactants) − Σ(bonds formed in products)

Limitations: only approximate; only really applies to gas-phase reactions (not liquids or solids where intermolecular forces also matter).

Worked example

Combustion of methane

Problem. CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(g). Bond enthalpies (kJ mol⁻¹): C–H = 414, O=O = 498, C=O = 804, O–H = 463. Calculate ΔH.
Solution. Bonds broken (reactants): 4 C–H + 2 O=O = 4(414) + 2(498) = 1656 + 996 = 2652 kJ.
Bonds formed (products): 2 C=O + 4 O–H = 2(804) + 4(463) = 1608 + 1852 = 3460 kJ.
ΔH = 2652 − 3460 = −808 kJ mol⁻¹. Strongly exothermic, as expected for combustion.

Try these

  1. Why are bond enthalpy calculations only approximate?
    Show answer
    Tabulated values are averages over many molecules. The actual bond strength varies with molecular environment (e.g. C–H is different in CH₄, in CH₃Cl, in benzene). Real ΔH from bond enthalpies can be ±20 kJ mol⁻¹ from the true value.
  2. Why do bond enthalpies only really apply to gas-phase reactions?
    Show answer
    In condensed phases (liquid, solid), intermolecular forces also contribute to ΔH. Bond enthalpies only account for the energy of breaking/forming covalent bonds — they ignore the IMFs that would also need to be overcome to vaporise reactants or condense products.
  3. Calculate ΔH for the hydrogenation of ethene: C₂H₄(g) + H₂(g) → C₂H₆(g). Bond enthalpies (kJ mol⁻¹): C=C 614, C–H 414, H–H 436, C–C 346.
    Show answer
    Bonds broken: 1 C=C + 4 C–H + 1 H–H = 614 + 4(414) + 436 = 614 + 1656 + 436 = 2706 kJ. Bonds formed: 1 C–C + 6 C–H = 346 + 6(414) = 346 + 2484 = 2830 kJ. ΔH = 2706 − 2830 = −124 kJ mol⁻¹. Exothermic (as expected for an addition forming new bonds).
Lesson 11

Hess's law

Reactivity 1.2.3

Lesson outcomes

Hess's law: the enthalpy change of a reaction depends only on the initial and final states, not on the route taken. Enthalpy is a state function.

So if reaction A→D has ΔH unknown, but you know A→B→C→D with known ΔH for each step, simply sum.

From enthalpies of formation

ΔHrxn = Σ ΔHf(products) − Σ ΔHf(reactants)

ΔHf = standard enthalpy of formation, per mole of compound from its elements in their standard states. ΔHf of an element in its standard state = 0 by definition.

From enthalpies of combustion

ΔHrxn = Σ ΔHc(reactants) − Σ ΔHc(products)

Useful when formation data isn't available (combustion is easier to measure for many organic compounds).

Worked example

ΔH from ΔH_f values

Problem. Calculate ΔH for CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(l). Given: ΔHf(CH₄) = −74.8, ΔHf(CO₂) = −393.5, ΔHf(H₂O) = −285.8 kJ mol⁻¹.
Solution. ΔHf(O₂) = 0 (element in standard state). ΔHrxn = [(−393.5) + 2(−285.8)] − [(−74.8) + 2(0)] = (−393.5 − 571.6) − (−74.8) = −965.1 + 74.8 = −890.3 kJ mol⁻¹.

Try these

  1. State Hess's law and explain why it works.
    Show answer
    The enthalpy change of a reaction depends only on the initial and final states, not on the route taken. It works because enthalpy is a state function — its value depends only on the current state, not how the state was reached.
  2. Why is ΔHf of an element in its standard state defined as zero?
    Show answer
    Because the 'formation' of an element from itself is no reaction at all. Setting elements to zero gives a consistent reference baseline — the ΔHf of any compound is the energy released or absorbed compared to its constituent elements.
  3. Calculate ΔH for: 2 H₂S(g) + 3 O₂(g) → 2 SO₂(g) + 2 H₂O(l). Given ΔHf(H₂S) = −20.6, ΔHf(SO₂) = −296.8, ΔHf(H₂O,l) = −285.8 kJ mol⁻¹.
    Show answer
    Products: 2(−296.8) + 2(−285.8) = −593.6 − 571.6 = −1165.2. Reactants: 2(−20.6) + 3(0) = −41.2. ΔH = −1165.2 − (−41.2) = −1124.0 kJ mol⁻¹.
  4. Why might you choose ΔHc data over ΔHf data when calculating ΔH for an organic reaction?
    Show answer
    Many organic ΔHf values are not directly measurable (you can't synthesise toluene from its elements at standard conditions). But you can burn almost anything in a bomb calorimeter. So ΔHc data is more readily available for organic compounds, and Hess's law lets us convert to ΔHrxn.
Lesson 14

Combustion

Reactivity 1.1

Lesson outcomes

Complete combustion

Hydrocarbon + sufficient O₂ → CO₂ + H₂O. Highly exothermic.

Example: C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O.

Incomplete combustion

Insufficient O₂ → CO (carbon monoxide, toxic) and/or C(s) (soot).

Examples: 2 C₃H₈ + 7 O₂ → 6 CO + 8 H₂O (incomplete, all the way to CO); 2 C₃H₈ + 3 O₂ → 6 C + 8 H₂O (extreme — only water and soot).

Why CO is dangerous

CO binds to haemoglobin ~250× more strongly than O₂. A small amount of CO permanently displaces O₂ from the blood, suffocating from the inside.

Try these

  1. Why might a Bunsen burner with the air hole closed give a soft yellow flame, while opening the air hole gives a sharper blue flame?
    Show answer
    Closed hole = restricted O₂ → incomplete combustion → glowing soot particles emit yellow light. Open hole = excess O₂ → complete combustion → no soot, all energy goes to heating the gas → blue flame from heated combustion gases.
Lesson 15

Comparing fuels

Reactivity 1.3

Lesson outcomes

Two metrics for comparing fuels:

FuelSpecific energy (MJ/kg)Energy density (MJ/dm³)
H₂ (liquid)14210
Methane (LNG)5523
Petrol4734
Diesel4538
Ethanol3024
Coal2435
Wood (dry)1610

Try these

  1. Hydrogen has the highest specific energy of any chemical fuel. Why isn't it the dominant fuel for cars?
    Show answer
    Low energy density. At room T and pressure H₂ is a gas — needing huge tanks or extreme pressure (700 bar) to carry useful amounts. Storage, transport and refuelling infrastructure are not yet practical for mass use. Also: pure H₂ production today is largely from CH₄ (CO₂ emissions), undermining its 'green' credentials.
HL extension

Free energy
as a
compass.

HL: Gibbs free energy, equilibrium constants from ΔG°, and the relationship between Kc and ΔG.

HL only

Kc and reaction quotients (HL Only)

Lesson 5 of Unit 5.

HL only

Equilibrium Problems (HL Only)

Lesson 6 of Unit 5.

HL only

Enthalpy of Combustion and formation (HL Only)

Lesson 12 of Unit 5.

HL only

(HL ONLY) – Born Haber cycles

Lesson 13 of Unit 5.

HL only

(HL ONLY)

Definition: The reaction quotient (Q) is a measure of the relative amounts of products and reactants present in a chemical reaction at any given time.

Lesson 5 HL only

K_c and reaction quotient Q

Reactivity 2.3.6 (HL)

Lesson outcomes

  • Calculate the reaction quotient Q from current (non-equilibrium) concentrations.
  • Compare Q to Kc to predict the direction of net reaction.
  • Recognise that ΔG° = −RT ln Kc.

The reaction quotient Q has the same form as Kc but uses current concentrations (not necessarily at equilibrium):

Q = [products]... / [reactants]... (same powers as Kc)

ComparisonMeaningDirection
Q < KToo few productsForward (→)
Q = KAt equilibriumNo net change
Q > KToo many productsReverse (←)

Link to ΔG

ΔG° = −RT ln K: a positive K (> 1) → negative ΔG° (spontaneous). For non-equilibrium states: ΔG = ΔG° + RT ln Q.

Try these

  1. For 2 NO₂ ⇌ N₂O₄, Kc = 200 at 298 K. A mixture currently has [NO₂] = 0.10, [N₂O₄] = 0.50. Calculate Q and predict the direction.
    Show answer
    Q = [N₂O₄]/[NO₂]² = 0.50/0.01 = 50. Q (50) < K (200), so the reaction proceeds forward to make more N₂O₄.
  2. If Q = K, what does this tell you about the system?
    Show answer
    The system is at equilibrium — no net change occurs. Both forward and reverse rates are equal.
  3. If K = 1.0, can you have any combination of reactant and product concentrations and still be at equilibrium?
    Show answer
    Yes — any combination where [products]/[reactants] (with appropriate powers) = 1. Many equilibrium positions are possible for the same K.
Lesson 6 HL only

Equilibrium problems · ICE tables

Reactivity 2.3.5 (HL)

Lesson outcomes

  • Use an ICE (Initial, Change, Equilibrium) table to solve equilibrium concentration problems.
  • Calculate Kc from initial concentrations and one equilibrium concentration.
  • Calculate equilibrium concentrations from Kc and initial concentrations.

The ICE method

  1. Initial: write the initial concentration of each species.
  2. Change: in terms of x (let x be the change in moles per litre of one species). Use stoichiometry for the others.
  3. Equilibrium: I + C for each species.
  4. Plug equilibrium values into the Kc expression and solve for x.
Worked example

Find equilibrium concentrations

Problem. For H₂ + I₂ ⇌ 2 HI, Kc = 50 at 700 K. Initially [H₂] = [I₂] = 0.100 M, [HI] = 0. Calculate the equilibrium concentrations.
Solution. ICE table
    H₂   +   I₂   ⇌   2 HI
I: 0.100   0.100   0
C: −x   −x   +2x
E: 0.100−x   0.100−x   2x
Kc = (2x)² / [(0.100−x)(0.100−x)] = 4x² / (0.100−x)² = 50
Take √: 2x / (0.100 − x) = √50 = 7.07.
2x = 0.707 − 7.07x → 9.07x = 0.707 → x = 0.0780.
[H₂] = [I₂] = 0.022 M; [HI] = 0.156 M.

Try these

  1. Why is the 'x ≪ initial concentration' approximation often used in ICE problems? When does it fail?
    Show answer
    Simplifies the algebra by approximating (initial − x) ≈ initial. Valid when K is small (typically < 10⁻³ × initial concentration). Fails when K is large enough that significant amounts react — then you must solve the full quadratic.
  2. For 2 NOCl ⇌ 2 NO + Cl₂, Kc = 1.6 × 10⁻⁵ at 35 °C. Start with 1.0 mol NOCl in 1.0 L. Find equilibrium [NOCl] using the small-x approximation.
    Show answer
    ICE: [NOCl] = 1.0 − 2x, [NO] = 2x, [Cl₂] = x. K = (2x)²(x)/(1.0−2x)² = 4x³/(1.0−2x)². Approximate (1.0−2x) ≈ 1.0: 4x³ = 1.6 × 10⁻⁵, x³ = 4 × 10⁻⁶, x = 0.0159. [NOCl] ≈ 1.0 − 0.032 = 0.97 M. Check: 0.032 ≪ 1.0 ✓.
Lesson 12 HL only

Enthalpy of combustion and formation

Reactivity 1.2.4 (HL)

Lesson outcomes

  • Define standard enthalpy of formation (ΔH°f) and combustion (ΔH°c).
  • Calculate ΔHf of a compound from its combustion data.
  • Convert between formation, combustion and overall reaction enthalpies.

Standard enthalpy of formation ΔH°f: enthalpy change when 1 mol of a compound forms from its elements in their standard states under standard conditions (298 K, 100 kPa).

Standard enthalpy of combustion ΔH°c: enthalpy change when 1 mol of a substance is completely burned in O₂ under standard conditions.

Both are tabulated in the IB data booklet for hundreds of compounds.

Conversion between them

For organic compounds, ΔHf is often calculated indirectly via Hess from ΔHc values (because direct synthesis from elements is rarely practical):

ΔHc(compound) = ΣΔHf(products) − ΔHf(compound)

Worked example

ΔH_f of ethanol from ΔH_c

Problem. C₂H₅OH(l) + 3 O₂(g) → 2 CO₂(g) + 3 H₂O(l). ΔHc(C₂H₅OH) = −1367 kJ mol⁻¹. Given ΔHf(CO₂) = −393.5, ΔHf(H₂O) = −285.8 kJ mol⁻¹, calculate ΔHf(C₂H₅OH).
Solution. From Hess: −1367 = [2(−393.5) + 3(−285.8)] − [ΔHf(C₂H₅OH) + 3(0)]
−1367 = (−787 − 857.4) − ΔHf(C₂H₅OH)
−1367 = −1644.4 − ΔHf(C₂H₅OH)
ΔHf(C₂H₅OH) = −1644.4 + 1367 = −277.4 kJ mol⁻¹. (Matches data booklet −277.7.) ✓

Try these

  1. Why is ΔHf of an element in its standard state defined as zero?
    Show answer
    By definition: 'formation' of an element from itself isn't a reaction. Setting elements to zero provides a consistent zero point for measuring all ΔHf values.
  2. Define standard conditions in the IB context.
    Show answer
    T = 298.15 K (25 °C); pressure = 100 kPa (1 bar). All solutions at 1 mol dm⁻³. The standard state of an element/compound is its most stable form at these conditions (C = graphite; H₂O = liquid at 25 °C).
  3. Why is it usually easier to measure ΔHc than ΔHf directly for an organic compound?
    Show answer
    Combustion goes essentially to completion in a bomb calorimeter — easy to measure. Direct synthesis from elements often requires multiple steps, side reactions, slow kinetics. ΔHf is typically derived from ΔHc via Hess's law.
Lesson 13 HL only

Born-Haber cycles

Reactivity 1.2.5 (HL)

Lesson outcomes

  • Construct a Born-Haber cycle for an ionic compound.
  • Identify the steps: atomisation, ionisation energy, electron affinity, lattice enthalpy.
  • Use Hess's law to calculate the lattice enthalpy from the cycle.
  • Relate lattice enthalpy to ionic charge and ionic radius (Coulomb's law).

Lattice enthalpy can't be measured directly. A Born-Haber cycle decomposes the formation of an ionic solid into measurable steps, then uses Hess to extract the lattice enthalpy.

The five steps to form 1 mol of solid MX from elements

  1. Atomise metal: M(s) → M(g). +ΔHat.
  2. Ionise metal: M(g) → M⁺(g) + e⁻. +IE.
  3. Atomise non-metal: ½X₂(g) → X(g). +½ bond enthalpy.
  4. Add electron to non-metal: X(g) + e⁻ → X⁻(g). +EA (usually exothermic).
  5. Form lattice: M⁺(g) + X⁻(g) → MX(s). ΔHlattice (always exothermic).

Sum of all 5 steps = ΔHf(MX). Rearrange to find ΔHlattice.

Coulomb's law: lattice enthalpy ∝ charge × charge / sum of radii

So MgO (2+ and 2−, small radii) has a much larger lattice enthalpy than NaCl (1+ and 1−).

Worked example

Lattice enthalpy of NaCl

Problem. Calculate the lattice enthalpy of NaCl given: ΔHat(Na) = +108, IE(Na) = +496, ½ bond(Cl₂) = +122, EA(Cl) = −349, ΔHf(NaCl) = −411 kJ mol⁻¹.
Solution. Sum of all five steps = ΔHf. 108 + 496 + 122 + (−349) + ΔHlattice = −411. 377 + ΔHlattice = −411. ΔHlattice = −788 kJ mol⁻¹. (Matches literature.)

Try these

  1. Predict whether MgCl₂ or NaCl has the larger (more exothermic) lattice enthalpy. Explain.
    Show answer
    MgCl₂. Mg²⁺ has a higher charge than Na⁺ and a smaller radius. By Coulomb's law, this gives stronger ionic attractions → more exothermic lattice enthalpy. Typical values: NaCl ≈ −788; MgCl₂ ≈ −2526 kJ mol⁻¹.
  2. Predict the order of lattice enthalpy for: NaF, NaCl, NaBr, NaI. Explain.
    Show answer
    Decreasing magnitude (less exothermic): NaF > NaCl > NaBr > NaI. Anion radius increases down the halogen group, so the cation-anion distance increases — weaker electrostatic attraction → less exothermic lattice enthalpy.
  3. Why can't lattice enthalpy be measured directly?
    Show answer
    It requires bringing isolated gaseous ions together to form a crystal — impossible to set up directly in the lab. Instead, we use a Born-Haber cycle to derive it indirectly from measurable steps.
  4. List the five steps in a Born-Haber cycle for forming MX from the elements.
    Show answer
    (1) Atomisation of metal: M(s) → M(g). (2) Ionisation of metal: M(g) → M⁺(g) + e⁻. (3) Atomisation of non-metal: ½X₂(g) → X(g). (4) Electron affinity: X(g) + e⁻ → X⁻(g). (5) Lattice formation: M⁺(g) + X⁻(g) → MX(s).
Lesson 16 HL only

Entropy

Reactivity 1.4.1 (HL)

Lesson outcomes

  • Define entropy S as a measure of dispersal of energy / disorder.
  • Predict the sign of ΔS for common reactions.
  • Calculate ΔS° from standard entropies of products and reactants.

Entropy S measures the dispersal of energy in a system — often described loosely as "disorder". Higher entropy = more ways for energy to be distributed.

Trends

  • State: gas > liquid > solid (gases have far more accessible microstates).
  • Mixing: solution > pure components (always ΔSmix > 0).
  • Complexity: more atoms or more vibrational modes → higher S. C₂H₆ > CH₄.
  • Temperature: S of any substance ↑ with T (more thermal motion → more accessible states).

Calculate ΔS° = ΣS°(products) − ΣS°(reactants). Tabulated S° values are absolute (positive) — unlike ΔHf values which are relative changes.

Worked example

Sign of ΔS

Problem. Predict the sign of ΔS for: (a) N₂(g) + 3 H₂(g) → 2 NH₃(g), (b) CaCO₃(s) → CaO(s) + CO₂(g), (c) H₂O(l) → H₂O(s).
Solution. (a) 4 mol gas → 2 mol gas: ΔS < 0. (b) 0 mol gas → 1 mol gas: ΔS > 0. (c) Liquid → solid: ΔS < 0 (water freezing is ordered).

Try these

  1. Why is entropy described as 'disorder' a useful but imperfect analogy?
    Show answer
    Disorder is intuitive (a tidy room → messy room ↑ entropy). But strictly entropy is about energy dispersal — the number of microstates available. The 'disorder' label can mislead (e.g. a 'tidy' ice crystal isn't lower entropy than 'messy' water because the relevant question is the number of microstates, not visual arrangement).
  2. Predict the sign of ΔS for: (a) dissolving NaCl in water, (b) condensing water vapour, (c) precipitation of AgCl from Ag⁺(aq) + Cl⁻(aq), (d) photosynthesis (6 CO₂ + 6 H₂O → C₆H₁₂O₆ + 6 O₂).
    Show answer
    (a) + (mixing entropy). (b) − (g → l, fewer microstates). (c) − (aq → s; ions becoming ordered crystal). (d) − (12 mol gas + reactants → 6 mol gas + 1 mol complex molecule; complexity falls).
  3. Calculate ΔS° for: N₂(g) + 3 H₂(g) → 2 NH₃(g). Given S° (J K⁻¹ mol⁻¹): N₂ = 192, H₂ = 131, NH₃ = 193.
    Show answer
    ΔS° = 2(193) − [192 + 3(131)] = 386 − (192 + 393) = 386 − 585 = −199 J K⁻¹ mol⁻¹. Negative — consistent with 4 mol gas → 2 mol gas.
Lesson 17 HL only

Spontaneity · Gibbs free energy

Reactivity 1.4.2 (HL)

Lesson outcomes

  • State the Gibbs free energy equation ΔG = ΔH − TΔS.
  • Predict spontaneity (ΔG < 0 = spontaneous).
  • Determine the temperature range over which a reaction is spontaneous.

ΔG = ΔH − TΔS

A reaction is spontaneous at temperature T if ΔG < 0. Note: spontaneous means thermodynamically favoured — says nothing about rate.

ΔHΔSSpontaneous when
− (exo)+Always (at all T)
+ (endo)−Never
− (exo)−Only at low T
+ (endo)+Only at high T
Worked example

Crossover temperature

Problem. For CaCO₃(s) → CaO(s) + CO₂(g), ΔH = +178 kJ mol⁻¹, ΔS = +161 J K⁻¹ mol⁻¹. Above what temperature does the decomposition become spontaneous?
Solution. Spontaneity needs ΔG < 0, i.e. ΔH < TΔS, i.e. T > ΔH/ΔS.
T > 178 000 / 161 = 1106 K ≈ 833 °C.
This is why limestone needs to be roasted hot in industry to make lime — the reaction is only spontaneous above ~830 °C.

Try these

  1. Does 'spontaneous' mean 'fast'? Give an example to illustrate.
    Show answer
    No. Spontaneity is about thermodynamics (ΔG < 0), not kinetics. Diamond → graphite has ΔG < 0 at 298 K — it's spontaneous — but the rate is so slow that diamond is effectively permanent. Spontaneity tells us about the destination, not the journey time.
  2. For a reaction with ΔH = −100 kJ mol⁻¹ and ΔS = −150 J K⁻¹ mol⁻¹, calculate ΔG at 298 K and predict spontaneity.
    Show answer
    ΔG = ΔH − TΔS = −100 − (298)(−0.150) = −100 − (−44.7) = −100 + 44.7 = −55.3 kJ mol⁻¹. Negative → spontaneous at 298 K. (Note: exo and ΔS negative — sweet spot at low T.)
  3. Above what T does the same reaction become non-spontaneous?
    Show answer
    ΔG = 0 when ΔH = TΔS. T = ΔH/ΔS = −100 000 / (−150) = 667 K. Above 667 K, TΔS becomes more negative than ΔH and ΔG turns positive. Below 667 K, the reaction is spontaneous.
  4. Why is the melting of ice spontaneous at room T but not at −20 °C?
    Show answer
    ΔH(melting) > 0 (endothermic — input heat to break ice lattice). ΔS > 0 (solid → liquid, more microstates). ΔG = ΔH − TΔS. Crossover T where ΔG = 0 is 273 K (0 °C). Above 273 K, TΔS exceeds ΔH and ΔG < 0 (spontaneous). Below 273 K, ΔG > 0 (non-spontaneous; ice stays solid).
Lesson 18 HL only

ΔG° and equilibrium constants

Reactivity 1.4.3 (HL)

Lesson outcomes

  • Link ΔG° to K via ΔG° = −RT ln K.
  • Calculate K from ΔG° and vice versa.
  • Use ΔG = ΔG° + RT ln Q to predict direction away from equilibrium.

ΔG° = −RT ln K

This single equation links thermodynamics (ΔG°) to equilibrium (K):

  • ΔG° < 0 (spontaneous at standard conditions) → ln K > 0 → K > 1 (products favoured).
  • ΔG° = 0 → K = 1.
  • ΔG° > 0 → K < 1 (reactants favoured).

Away from equilibrium

ΔG = ΔG° + RT ln Q. When Q < K, ΔG < 0 and the forward reaction is favoured; when Q > K, ΔG > 0 and the reverse is favoured.

Worked example

K from ΔG°

Problem. For a reaction at 298 K, ΔG° = −34.2 kJ mol⁻¹. Calculate K.
Solution. ln K = −ΔG°/(RT) = −(−34 200) / (8.31 × 298) = 34 200 / 2476 = 13.81.
K = e^13.81 = 1.0 × 10⁶. Strongly products-favoured.

Try these

  1. If ΔG° = 0 for a reaction at 298 K, what is the value of K?
    Show answer
    ΔG° = −RT ln K. So 0 = −RT ln K → ln K = 0 → K = 1. Equal amounts of products and reactants at equilibrium (or such that the activity ratio = 1).
  2. For NH₃(g) + HCl(g) ⇌ NH₄Cl(s), ΔH° = −176 kJ mol⁻¹ and ΔS° = −285 J K⁻¹ mol⁻¹. Calculate K at 298 K.
    Show answer
    ΔG° = ΔH° − TΔS° = −176 − (298)(−0.285) = −176 + 84.9 = −91.1 kJ mol⁻¹. ln K = −ΔG°/(RT) = 91 100/(8.31 × 298) = 36.8. K = e^36.8 = ~10¹⁶. Strongly products-favoured.
  3. If Q > K for a reaction, which way does it shift and what is the sign of ΔG?
    Show answer
    Q > K means there are too many products (or too few reactants) relative to equilibrium. Reaction shifts reverse (←). ΔG = ΔG° + RT ln Q. When Q > K, RT ln Q is more positive than RT ln K, so ΔG > 0 → reverse reaction is spontaneous.
  4. A reaction has ΔG° = +25 kJ mol⁻¹ at 298 K. Comment on whether it can take place spontaneously at standard conditions, and whether changing conditions might help.
    Show answer
    ΔG° > 0 means not spontaneous at standard conditions (where all species at 1 mol dm⁻³). But ΔG = ΔG° + RT ln Q — at non-standard conditions with low Q (i.e. high reactant, low product concentrations), ΔG can become negative and the reaction proceeds. So yes — adjust concentrations or temperature to shift the balance.
Vocabulary

42 terms to own.

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elementcompoundmixturehomogeneousheterogeneousatomelectronmolestoichiometrylimiting reactantstpconcentrationsolutesolventsolutionstoichiometric coefficientbalanced equationaccuracygraphgradientrate of reactionactivation energycatalystpressuretemperatureintermediateequilibriumdynamic equilibriumreversible reactionle chatelier's principleequilibrium constantkckpkwhomogeneous equilibriumheterogeneous equilibriumendothermicexothermicenthalpy changestandard enthalpy of formationbond enthalpycalorimetry