Topic 7 · Chemical Energetics I · spread

Chemical Energetics I

1

Enthalpy Changes

DefinitionEnthalpy, H: the measure of the energy content of a substance. A substance with lower enthalpy has greater energetic stability.

Bond breaking absorbs energy, bond forming releases energy. ΔH is the heat change at constant pressure (kJ mol⁻¹).

ExothermicEndothermic
ΔHnegativepositive
Heatreleased to surroundingsabsorbed from surroundings
Surroundingstemperature ↑temperature ↓
Productsmore stableless stable

1.1Energy profile diagrams

DefinitionActivation energy, Ea: the minimum amount of energy reacting particles must possess for an effective collision to take place.
  • Standard conditions: 298 K, 1 bar (10⁵ Pa), 1 mol dm⁻³ for solutions. ΔH⦵(element in standard state) = 0
  • Thermochemical equations need state symbols, ΔH scales with the amounts: doubling the equation doubles ΔH
NoteStandard conditions (298 K, 1 bar) are not s.t.p. (273 K, 1 bar).
exothermic, ΔH < 0reaction coordinateHreactantsproductsEaΔH endothermic, ΔH > 0reaction coordinateHreactantsproductsEaΔH
2

Standard Enthalpy Changes

TermDefinitionExample equationSign
Reaction, ΔHr⦵the enthalpy change when molar quantities of reactants, as specified by the balanced equation, react to form products under standard conditionsN₂(g) + 3H₂(g) → 2NH₃(g)exo / endo
Formation, ΔHf⦵the enthalpy change when one mole of the substance is formed from its constituent elements in their standard states under standard conditions2C(s) + 3H₂(g) + ½O₂(g) → C₂H₅OH(l)exo / endo
ΔHf⦵ measures stability relative to the elements: negative → compound more stable than its elements
Combustion, ΔHc⦵the enthalpy change when one mole of the substance is completely burnt in oxygen under standard conditionsCH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)exo
Neutralisation, ΔHneut⦵the enthalpy change when one mole of water is formed from the reaction of an acid and an alkali under standard conditionsH⁺(aq) + OH⁻(aq) → H₂O(l)exo
Strong acid + strong base: always −57.0 kJ mol⁻¹, as both fully dissociate → the reaction is just H⁺ + OH⁻ → H₂O.
Weak acid or base: ΔHneut is less exothermic than −57.0 (e.g. CH₃CO₂H, −55.2): (1) weak acid is only partially dissociated, (2) energy is absorbed to dissociate the undissociated acid → less energy released overall
28Topic 07 · Chemical Energetics I
2

Standard Enthalpy Changes (continued)

TermDefinitionExample equationSign
Bond energy, BEthe average enthalpy change when one mole of covalent bonds between atoms in gaseous molecules is brokenCl–Cl(g) → 2Cl(g)endo
Atomisation, ΔHat⦵the enthalpy change when one mole of gaseous atoms is formed from the element in its standard state under standard conditions½Cl₂(g) → Cl(g)  (= ½BE)endo
1st / 2nd IEthe enthalpy change when one mole of electrons is removed from one mole of gaseous atoms / singly charged cationsMg(g) → Mg⁺(g) + e⁻endo
1st / 2nd EAthe enthalpy change when one mole of gaseous atoms / singly charged anions accepts one mole of electronsO(g) + e⁻ → O⁻(g)1st exo,
2nd endo
Why is 1st EA negative but 2nd EA positive? (1) 1st EA: energy is given off when electrostatic attraction forms between the positive nucleus and the electron added. (2) 2nd EA: energy is needed to overcome the repulsion between the electron added and the singly charged anion, both negatively charged
Lattice energy, LEthe energy released when one mole of the solid ionic compound is formed from its constituent gaseous ions under standard conditionsNa⁺(g) + Cl⁻(g) → NaCl(s)exo
Hydration, ΔHhyd⦵the enthalpy change when one mole of gaseous ions is hydrated under standard conditionsNa⁺(g) + aq → Na⁺(aq)exo
Hydration is always exothermic: ion–dipole interactions form between the ions and water molecules
Solution, ΔHsol⦵the enthalpy change when one mole of the substance is completely dissolved to give an infinitely dilute solution under standard conditions, so that no further enthalpy change takes place on adding more solventNaCl(s) + aq → Na⁺(aq) + Cl⁻(aq)exo / endo

IE, EA and LE values are per mole: e.g. IE(Mg) +736, EA₁(O) −141, EA₂(O) +844, LE(NaCl) −776 kJ mol⁻¹.

Exam tipRead the definition off the term itself. The term names what there is one mole of: formation → per mole of compound formed, combustion → per mole of substance burnt, neutralisation → per mole of water formed, atomisation → per mole of gaseous atoms formed, hydration → per mole of gaseous ions hydrated.
Answer structureThe enthalpy change when one mole of (what the term names) is (formed / burnt / hydrated …) under standard conditions.Equation: give that species a coefficient of 1, then balance the rest, using fractions if needed, e.g. formation: 2C(s) + 3H₂(g) + ½O₂(g) → 1 C₂H₅OH(l).
Topic 07 · Chemical Energetics I29
3

Calorimetry

q = −mcΔT
mmass of solution / g
cspecific heat capacity, 4.18 J g⁻¹ K⁻¹
ΔTTfinal − Tinitial
ΔH = q / n
namount of limiting reagent (or the specified substance) / mol
Assumptions
  • Negligible heat loss to the surroundings
  • Negligible heat absorbed by the calorimeter
  • Density of solution = 1.00 g cm⁻³, c of solution = c of water
Temperature correction

For slow reactions, plot T against time and extrapolate back to the time of mixing to account for heat loss.

timeTmixingΔT

Worked example: NeutralisationA student mixes 40.0 cm³ of 2.0 mol dm⁻³ HCl at 22.3 °C with 50.0 cm³ of 2.0 mol dm⁻³ NaOH at 22.5 °C. The highest temperature reached is 34.6 °C. Find ΔHneut.

  1. Initial T = 40.0(22.3) + 50.0(22.5)90.0 = 22.41 °C (weighted average)
  2. q = −(90.0)(4.18)(34.6 − 22.41) = −4586 J
  3. nHCl = 0.08000 mol (limiting) = nH₂O formed
  4. ΔHneut = −4586 ÷ 0.08000 = −57.3 kJ mol⁻¹

Worked example: CombustionBurning 0.530 g of ethanol in a spirit burner heats 250 g of water by 10.4 °C. Find ΔHc of ethanol.

  1. q = −250 × 4.18 × 10.4 = −10 870 J = −10.87 kJ
  2. nethanol = 0.530 ÷ 46.0 = 0.01152 mol
  3. ΔHc = −10.87 ÷ 0.01152 = −944 kJ mol⁻¹

Less exothermic than the data value (−1368) because of heat loss to surroundings, heat absorbed by the calorimeter, or incomplete combustion. Efficiency: q′ = (x / 100) q.

30Topic 07 · Chemical Energetics I
4

Hess' Law

DefinitionHess' Law: the enthalpy change of a chemical reaction depends only on the initial and final states of the system and is independent of the pathway taken.
From ΔHc
ΔHr = ΣΔHc(reactants) − ΣΔHc(products)
From ΔHf
ΔHr = ΣΔHf(products) − ΣΔHf(reactants)
From BE
ΔHr = ΣBE(bonds broken) − ΣBE(bonds formed)
ΔH₁ (direct)ΔH₂ΔH₃ΔH₄ADBC

ΔH₁ = ΔH₂ − ΔH₃ + ΔH₄: going against an arrow changes its sign

4.1Energy level diagram: ΔHf(CO)

C(s) + O₂(g) → CO₂(g)ΔHc(C) = −394
CO(g) + ½O₂(g) → CO₂(g)ΔHc(CO) = −283

ΔHf(CO) = −394 − (−283) = −111 kJ mol⁻¹

Check: arrows down for exothermic, elements at 0, state symbols.

energy / kJ mol⁻¹CO₂(g)−394−394CO(g) + ½O₂(g)−283ΔH(CO)C(s) + O₂(g)0

4.2Using combustion and bond energies

Worked example: From ΔHcPropane cannot be made directly from its elements. Find ΔHf(C₃H₈) given ΔHc / kJ mol⁻¹: C −394, H₂ −286, C₃H₈ −2220.

3C(s) + 4H₂(g) → C₃H₈(g)
ΔHf = 3(−394) + 4(−286) − (−2220)
= −106 kJ mol⁻¹

Worked example: From BEEstimate ΔH for the Haber process, N₂ + 3H₂ → 2NH₃, given BE / kJ mol⁻¹: N≡N 944, H–H 436, N–H 390.

ΔH = [944 + 3(436)] − [6(390)]
= −88 kJ mol⁻¹

BE values are averages over many molecules, so ΔH from bond energies differs from values found by other methods.

Topic 07 · Chemical Energetics I31
5

Lattice Energy and the Born–Haber Cycle

5.1Factors affecting lattice energy

LE ∝q+ × q−r+ + r−
CompoundLE / kJ mol⁻¹
NaCl−776
NaBr−742
MgCl₂−2375
  • Br⁻ has the same charge but a larger ionic radius than Cl⁻ → LE(NaBr) less exothermic than LE(NaCl)
  • Mg²⁺: higher charge, smaller radius than Na⁺ → LE(MgCl₂) more exothermic
Exam tipCompare lattice energies as "more / less exothermic", never "larger / smaller".
NoteExperimental LE more exothermic than theoretical (e.g. AgI, MgCl₂) → covalent character from polarisation of the anion.

5.2Born–Haber cycle for Na₂O

How to build a Born–Haber cycle (not to scale)
  1. Draw the elements in their standard states at 0
  2. Atomise each element (up): ΔHat, times the number of atoms
  3. Ionise the metal atoms (up): sum of IEs
  4. Add electrons to the non-metal: EA₁ down, EA₂ up
  5. Gaseous ions → solid: LE (down), ΔHf joins elements to solid
Calculating LE: ΔHf = sum of the other steps

−414 = 2(107) + ½(496) + 2(494) − 141 + 844 + LE

LE = −2567 kJ mol⁻¹

energy2Na(s) + ½O₂(g)02Na(g) + ½O₂(g)2Na(g) + O(g)2ΔHat(Na)2(+107)ΔHat(O)½(+496)2Na⁺(g) + 2e⁻ + O(g)2 × IE(Na)2(+494)2Na⁺(g) + e⁻ + O⁻(g)2Na⁺(g) + O²⁻(g)EA₁(O)−141EA₂(O)+844Na₂O(s)LE?ΔHf(Na₂O)−414
32Topic 07 · Chemical Energetics I
6

Hydration and Solution

6.1Using hydration and lattice energies

ΔHhyd ∝ q / r Small, highly charged ions attract water molecules more strongly → more exothermic hydration.
ΔHsol = −LE + ΣΔHhyd Dissolving happens in two steps: (1) break the lattice into gaseous ions, which takes in energy (−LE), then (2) surround the gaseous ions with water, which gives out energy (ΣΔHhyd).
Na⁺(g) + Cl⁻(g)NaCl(s)Na⁺(aq) + Cl⁻(aq)−LE+776ΣΔHhydΔHsolnot to scale

Worked exampleThe enthalpy change of solution of NaCl is +5 kJ mol⁻¹. The lattice energy of NaCl is −776 kJ mol⁻¹ and the enthalpy change of hydration of Na⁺ is −390 kJ mol⁻¹.
(a) Calculate the enthalpy change of hydration of Cl⁻.
(b) ΔHhyd(I⁻) is −308 kJ mol⁻¹. Explain why it is less exothermic than ΔHhyd(Cl⁻).

(a)

  1. ΔHsol = −LE + ΔHhyd(Na⁺) + ΔHhyd(Cl⁻)
  2. +5 = −(−776) + (−390) + ΔHhyd(Cl⁻)
  3. ΔHhyd(Cl⁻) = 5 − 776 + 390 = −381 kJ mol⁻¹

(b)

  1. I⁻ has the same charge but a larger ionic radius than Cl⁻
  2. → lower charge density
  3. → weaker ion–dipole interactions with water molecules → less energy released
Topic 07 · Chemical Energetics I33