Chemical
Energetics I

H2 Chemistry (9476), JC1

By the end of this topic, you can answer

1

Why does a cold pack feel cold and a hand warmer feel hot?

2

Why do chemists insist on standard conditions when they compare enthalpy changes?

3

How can a polystyrene cup and a thermometer measure the energy released when an acid is neutralised?

4

How can you find the enthalpy change of a reaction you can never carry out in the lab, such as making methane from carbon and hydrogen?

5

Why does magnesium oxide melt at 2852 °C, while sodium chloride melts at 801 °C?

6

Why do some salts warm the water as they dissolve, while others cool it?

Recall: what you know from O levels

1A reaction makes its test tube feel hot. Is it exothermic or endothermic?

Exothermic: heat is released to the surroundings.[1]

2Does breaking a bond absorb or release energy?

Absorb. Bond forming releases energy.[1]

3Give one exothermic and one endothermic process.

Exothermic: combustion, neutralisation. Endothermic: thermal decomposition, dissolving NH₄NO₃.[1]

1

Enthalpy Changes

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

AnalogyA ball rolling down into a well: it ends up lower in energy, releases energy on the way, and needs energy to be lifted out again.

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

1

Enthalpy Changes

	Exothermic	Endothermic
ΔH	negative	positive
Heat	released to surroundings	absorbed from surroundings
Surroundings	temperature ↑	temperature ↓
Products	more stable	less stable

1

Enthalpy Changes

1.1Energy profile diagrams

DefinitionActivation energy, E_a: the minimum amount of energy reacting particles must possess for an effective collision to take place.

1

Enthalpy Changes

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).

Your turnN₂(g) + 3H₂(g) → 2NH₃(g) ΔH = −92 kJ mol⁻¹. Find ΔH for NH₃(g) → ½N₂(g) + 1½H₂(g).

Reverse the equation: the sign changes, +92

Halve it: ΔH = +46 kJ mol⁻¹[1]

1

Enthalpy Changes

Quick checkWhich is true of an exothermic reaction?

AΔH is positive.

BThe temperature of the surroundings rises.

CThe products are less stable than the reactants.

DMore energy is absorbed breaking bonds than is released forming bonds.

NoteB: heat is released to the surroundings. ΔH is negative, the products are more stable, and more energy is released forming bonds than is absorbed breaking them.

2

Standard Enthalpy Changes

Term	Definition	Example equation	Sign
Reaction, ΔH_r^⦵	the enthalpy change when molar quantities of reactants, as specified by the balanced equation, react to form products under standard conditions	N₂(g) + 3H₂(g) → 2NH₃(g)	exo / endo
Formation, ΔH_f^⦵	the enthalpy change when one mole of the substance is formed from its constituent elements in their standard states under standard conditions	2C(s) + 3H₂(g) + ½O₂(g) → C₂H₅OH(l)	exo / endo
Combustion, ΔH_c^⦵	the enthalpy change when one mole of the substance is completely burnt in oxygen under standard conditions	CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)	exo

2

Standard Enthalpy Changes

Term	Definition	Example equation	Sign
Neutralisation, ΔH_neut^⦵	the enthalpy change when one mole of water is formed from the reaction of an acid and an alkali under standard conditions	H⁺(aq) + OH⁻(aq) → H₂O(l)	exo
Bond energy, BE	the average enthalpy change when one mole of covalent bonds between atoms in gaseous molecules is broken	Cl–Cl(g) → 2Cl(g)	endo
Atomisation, ΔH_at^⦵	the enthalpy change when one mole of gaseous atoms is formed from the element in its standard state under standard conditions	½Cl₂(g) → Cl(g) (= ½BE)	endo

2

Standard Enthalpy Changes

Term	Definition	Example equation	Sign
1st / 2nd IE	the enthalpy change when one mole of electrons is removed from one mole of gaseous atoms / singly charged cations	Mg(g) → Mg⁺(g) + e⁻	endo
1st / 2nd EA	the enthalpy change when one mole of gaseous atoms / singly charged anions accepts one mole of electrons	O(g) + e⁻ → O⁻(g)	1st exo, 2nd endo
Lattice energy, LE	the energy released when one mole of the solid ionic compound is formed from its constituent gaseous ions under standard conditions	Na⁺(g) + Cl⁻(g) → NaCl(s)	exo

2

Standard Enthalpy Changes

Term	Definition	Example equation	Sign
Hydration, ΔH_hyd^⦵	the enthalpy change when one mole of gaseous ions is hydrated under standard conditions	Na⁺(g) + aq → Na⁺(aq)	exo
Solution, ΔH_sol^⦵	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 solvent	NaCl(s) + aq → Na⁺(aq) + Cl⁻(aq)	exo / endo

Hydration is always exothermic: ion–dipole interactions form between the ions and water molecules
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

2

Standard Enthalpy Changes

Why is 1st EA negative but 2nd EA positive?

PredictAn electron is added to O, then a second electron to O⁻. Which step needs energy?

1st EA: energy is given off when electrostatic attraction forms between the positive nucleus and the electron added
2nd EA: energy is needed to overcome the repulsion between the electron added and the singly charged anion, both negatively charged

2

Standard Enthalpy Changes

ΔH_f^⦵ measures stability relative to the elements: negative → compound more stable than its elements
Strong acid + strong base: always −57.0 kJ mol⁻¹, as both fully dissociate → the reaction is just H⁺ + OH⁻ → H₂O

NoteWeak acid or base: ΔH_neut is less exothermic than −57.0 (e.g. CH₃CO₂H, −55.2)

Weak acid is only partially dissociated
Energy is absorbed to dissociate the undissociated acid → less energy released overall

2

Standard Enthalpy Changes

Your turnWrite the equation, with state symbols, for each standard enthalpy change.

(a) ΔH_f of methanol, CH₃OH(l)

(a)

C(s) + 2H₂(g) + ½O₂(g) → CH₃OH(l)[1]

(b) ΔH_c of ethane

(b)

C₂H₆(g) + 3½O₂(g) → 2CO₂(g) + 3H₂O(l)[1]

(c) lattice energy of MgO

(c)

Mg²⁺(g) + O²⁻(g) → MgO(s)[1]

(d) ΔH_at of bromine

(d)

½Br₂(l) → Br(g)[1]

2

Standard Enthalpy Changes

Quick checkWhich equation represents the standard enthalpy change of formation of water?

A2H₂(g) + O₂(g) → 2H₂O(l)

BH₂(g) + ½O₂(g) → H₂O(l)

CH₂(g) + ½O₂(g) → H₂O(g)

D2H(g) + O(g) → H₂O(l)

NoteB: one mole of H₂O, from its elements in their standard states (H₂(g), O₂(g)), giving water in its standard state, liquid, at 298 K.

3

Calorimetry

Calorimeter: the insulated container in which the reaction takes place, while its temperature change is measured to find ΔH.

How it works

The reaction takes place in the solution and gives out heat → the heat goes into the solution
→ the solution warms: its temperature ↑ by ΔT, read on the thermometer
The cup and lid insulate → almost no heat escapes, so the heat the solution gains = the heat the reaction gives out
Heat gained by the solution, mcΔT → divide by the moles that reacted to get ΔH

3

Calorimetry

The two equations

q = −mcΔT ΔH = q / n

The minus sign: heat the solution gains is heat the reaction loses. A temperature rise gives a negative ΔH, exothermic.

m = mass of solution (g), c = 4.18 J g⁻¹ K⁻¹, ΔT = T_final − T_initial, n = amount of limiting reagent (or the specified substance)

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

3

Calorimetry

Temperature correction

DescribeAfter mixing, the temperature rises to a maximum and then falls slowly. Why does it fall?

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

3

Calorimetry

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 ΔH_neut.

Initial T = 40.0(22.3) + 50.0(22.5)90.0 = 22.41 °C (weighted average)
q = −(90.0)(4.18)(34.6 − 22.41) = −4586 J
n_HCl = 0.08000 mol (limiting) = n_H₂O formed
ΔH_neut = −4586 ÷ 0.08000 = −57.3 kJ mol⁻¹

3

Calorimetry

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

q = −250 × 4.18 × 10.4 = −10.87 kJ

n = 0.530 ÷ 46.0 = 0.01152 mol

ΔH_c = −10.87 ÷ 0.01152 = −944 kJ mol⁻¹

PredictThe data book value is −1368 kJ mol⁻¹. Why is the experimental value less exothermic?

Heat loss to surroundings, heat absorbed by the calorimeter, or incomplete combustion.

If only x% of the heat reaches the water: q′ = (x / 100) q.

3

Calorimetry

Your turn25.0 cm³ of 1.00 mol dm⁻³ HCl is mixed with 25.0 cm³ of 1.00 mol dm⁻³ NaOH. The temperature rises by 6.7 °C. Find ΔH_neut.

q = −(50.0)(4.18)(6.7) = −1400 J[1]

n_H₂O = 0.0250 × 1.00 = 0.02500 mol[1]

ΔH_neut = −1400 ÷ 0.02500 = −56.0 kJ mol⁻¹[1]

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.

AnalogyTaking the MRT from Ang Mo Kio, which is above ground, to Dhoby Ghaut, deep underground: straight down the North–South Line, or changing to the Circle Line at Bishan and going round. Either way you go down by the same height, because that depends only on the two stations, not on the route.

ΔH₁ = ΔH₂ − ΔH₃ + ΔH₄

4

Hess' Law

From ΔH_c

ΔH_r = ΣΔH_c(reactants) − ΣΔH_c(products)

From ΔH_f

ΔH_r = ΣΔH_f(products) − ΣΔH_f(reactants)

From BE

ΔH_r = ΣBE(bonds broken) − ΣBE(bonds formed)

Exam tipCombustion goes "reactants − products"; formation goes "products − reactants". Draw the cycle if in doubt.

4

Hess' Law

Energy level diagram: ΔH_f(CO)

CO cannot be made cleanly from C and O₂. Use combustion data:

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

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

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

4

Hess' Law

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

3C(s) + 4H₂(g) → C₃H₈(g)

ΔH_f = ΣΔH_c(reactants) − ΣΔH_c(products)

= 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.

Broken: 1 N≡N, 3 H–H. Formed: 6 N–H

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.

4

Hess' Law

Your turnFind ΔH for C₂H₄(g) + H₂(g) → C₂H₆(g), given ΔH_f / kJ mol⁻¹: C₂H₄ +52, C₂H₆ −85.

ΔH = ΣΔH_f(products) − ΣΔH_f(reactants) = −85 − (+52 + 0) = −137 kJ mol⁻¹[1]

ChallengeWhy is ΔH_f(H₂) not in the sum?

H₂(g) is an element in its standard state, so its ΔH_f^⦵ = 0.[1]

4

Hess' Law

Quick checkUsing bond energies, ΔH of a reaction is

AΣBE(bonds broken) − ΣBE(bonds formed)

BΣBE(bonds formed) − ΣBE(bonds broken)

CΣBE(products) + ΣBE(reactants)

DΣBE(bonds broken) only

NoteA: breaking bonds absorbs energy (+), forming bonds releases energy (−).

5

Lattice Energy and the Born–Haber Cycle

5.1Factors affecting lattice energy

LE ∝ q₊ × q₋r₊ + r₋ NaCl −776, NaBr −742, MgCl₂ −2375 kJ mol⁻¹

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

Lattice Energy and the Born–Haber Cycle

Building a Born–Haber cycle: Na₂O

Draw the elements in their standard states at 0
Atomise each element (up): ΔH_at, times the number of atoms
Ionise the metal atoms (up): sum of IEs
Add electrons to the non-metal: EA₁ down, EA₂ up
Gaseous ions → solid: LE (down). ΔH_f joins elements to solid

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

LE = −2567 kJ mol⁻¹

5

Lattice Energy and the Born–Haber Cycle

Your turnFind the lattice energy of NaCl, given / kJ mol⁻¹: ΔH_f(NaCl) −411, ΔH_at(Na) +107, ΔH_at(Cl) +122, IE(Na) +494, EA(Cl) −349.

−411 = 107 + 122 + 494 − 349 + LE[1]

LE = −411 − 374 = −785 kJ mol⁻¹[1]

ChallengeWhy is ΔH_at(Cl) half the Cl–Cl bond energy (244 kJ mol⁻¹)?

ΔH_at forms one mole of Cl atoms: ½Cl₂(g) → Cl(g). Breaking one mole of Cl–Cl bonds gives two moles of atoms.[1]

5

Lattice Energy and the Born–Haber Cycle

Quick checkWhich compound has the most exothermic lattice energy?

ANaCl

BKBr

CMgO

DCaO

NoteC: both ions carry charge 2, and Mg²⁺ is smaller than Ca²⁺, so q₊q₋ / (r₊ + r₋) is largest.

6

Hydration and Solution

6.1Using hydration and lattice energies

ΔH_hyd ∝ q / r ΔH_sol = −LE + ΣΔH_hyd

Dissolving = break up the lattice (−LE, endothermic),

then hydrate the gaseous ions (ΣΔH_hyd, exothermic).

The overall ΔH_sol is the small difference between two large numbers, so it can be either sign.

6

Hydration and Solution

Worked exampleGiven ΔH_sol(NaCl) = +5, ΔH_hyd(Na⁺) = −390 and LE(NaCl) = −776 kJ mol⁻¹, find ΔH_hyd(Cl⁻).

ΔH_sol = −LE + ΔH_hyd(Na⁺) + ΔH_hyd(Cl⁻)

+5 = +776 + (−390) + ΔH_hyd(Cl⁻)

ΔH_hyd(Cl⁻) = −381 kJ mol⁻¹

Why is ΔH_hyd(I⁻) = −308 less exothermic?

I⁻ has the same charge but a larger ionic radius than Cl⁻
→ lower charge density → weaker ion–dipole interactions with water molecules

6

Hydration and Solution

Your turnFind ΔH_sol(KCl), given LE(KCl) = −711, ΔH_hyd(K⁺) = −322 and ΔH_hyd(Cl⁻) = −381 kJ mol⁻¹.

ΔH_sol = +711 + (−322) + (−381) = +8 kJ mol⁻¹[1]

ChallengeWhich ion has the more exothermic ΔH_hyd, Mg²⁺ or Na⁺? Explain.

Mg²⁺. Mg²⁺ has a higher charge and a smaller ionic radius than Na⁺, so it has a higher charge density.[1]

Hence, Mg²⁺ forms stronger ion–dipole interactions with water molecules.[1]

Exit ticket

1Define standard enthalpy change of combustion.

The enthalpy change when one mole of the substance is completely burnt in oxygen under standard conditions.[1]

2State Hess' law.

The enthalpy change of a reaction depends only on the initial and final states, and is independent of the pathway taken.[1]

3Why is the 2nd EA of oxygen positive?

Energy is needed to overcome the repulsion between the electron added and the negatively charged O⁻ ion.[1]

4Give two reasons a calorimetry value of ΔH_c is less exothermic than the data book value.

Heat loss to the surroundings[1]

Incomplete combustion or heat absorbed by the calorimeter[1]

ChemicalEnergetics I

By the end of this topic, you can answer

Recall: what you know from O levels

Enthalpy Changes

Enthalpy Changes

Enthalpy Changes

1.1Energy profile diagrams

Enthalpy Changes

Enthalpy Changes

Standard Enthalpy Changes

Standard Enthalpy Changes

Standard Enthalpy Changes

Standard Enthalpy Changes

Standard Enthalpy Changes

Why is 1st EA negative but 2nd EA positive?

Standard Enthalpy Changes

Standard Enthalpy Changes

Standard Enthalpy Changes

Calorimetry

Calorimetry

The two equations

Assumptions

Calorimetry

Temperature correction

Calorimetry

Calorimetry

Calorimetry

Hess' Law

Hess' Law

From ΔHc

From ΔHf

From BE

Hess' Law

Energy level diagram: ΔHf(CO)

Hess' Law

Hess' Law

Hess' Law

Lattice Energy and the Born–Haber Cycle

5.1Factors affecting lattice energy

Lattice Energy and the Born–Haber Cycle

Building a Born–Haber cycle: Na₂O

Lattice Energy and the Born–Haber Cycle

Lattice Energy and the Born–Haber Cycle

Hydration and Solution

6.1Using hydration and lattice energies

Hydration and Solution

Why is ΔHhyd(I⁻) = −308 less exothermic?

Hydration and Solution

Exit ticket

Chemical
Energetics I

From ΔH_c

From ΔH_f

Energy level diagram: ΔH_f(CO)

Why is ΔH_hyd(I⁻) = −308 less exothermic?