Unit 14 · Physical Chemistry

Enthalpy Changes

Thermochemistry, enthalpy of combustion, formation, neutralisation, bond energies, Hess's Law, Kirchhoff's Law, and calorimetry.

14.1

Thermochemistry Basics

Enthalpy (H) Enthalpy is the heat content of a system at constant pressure. We cannot measure H directly, only changes in enthalpy ΔH.

ΔH = Hproducts − Hreactants

ΔH < 0 → exothermic (heat released to surroundings; products at lower energy)
ΔH > 0 → endothermic (heat absorbed from surroundings; products at higher energy)

Exothermic Reactions

System releases heat → surroundings get warmer. ΔH < 0.

Examples: combustion, neutralisation, respiration, condensation, oxidation of metals, dissolving NaOH.

Endothermic Reactions

System absorbs heat → surroundings get cooler. ΔH > 0.

Examples: thermal decomposition, photosynthesis, dissolving NH₄NO₃, evaporation.

EXOTHERMIC (ΔH < 0) Enthalpy H Reactants Products ΔH < 0 ENDOTHERMIC (ΔH > 0) Reactants Products ΔH > 0

Energy level diagrams: exothermic (products lower than reactants) and endothermic (products higher)

Standard Conditions and State Symbols

Standard enthalpy changes (ΔH°) are measured at 298 K (25°C) and 100 kPa (standard pressure), with all substances in their standard states.

State symbols must always be included in thermochemical equations because enthalpy depends on state: e.g. H2O(l) vs H2O(g) differ by the enthalpy of vaporisation.

Standard state = most stable form of an element at 298 K and 100 kPa: C → graphite (not diamond) S → rhombic sulfur (not monoclinic) Br → Br2(l) (not gas) Hg → Hg(l) (not solid)
14.2

Types of Standard Enthalpy Change

Standard Enthalpy of Formation (ΔH°f) The enthalpy change when one mole of a compound is formed from its elements in their standard states under standard conditions.

By definition: ΔH°f of any element in its standard state = 0.
Example: C(graphite) + O2(g) → CO2(g)    ΔH°f = −393.5 kJ/mol
Standard Enthalpy of Combustion (ΔH°c) The enthalpy change when one mole of a substance is completely burned in oxygen under standard conditions, with all products in their standard states.

Always exothermic (ΔH°c always negative).
Example: CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)    ΔH°c = −890 kJ/mol
Standard Enthalpy of Neutralisation (ΔH°neut) The enthalpy change when an acid and a base react to form one mole of water under standard conditions.

Strong acid + strong base: ΔH°neut ≈ −57.1 kJ/mol (always, since it is H+ + OH− → H2O)
Weak acid or base: less exothermic because energy is needed to dissociate the weak electrolyte.
Standard Enthalpy of Solution (ΔH°sol) The enthalpy change when one mole of a solute dissolves in enough solvent to form an infinitely dilute solution.

Can be exothermic (NaOH, H2SO4) or endothermic (NH4NO3, KNO3).
Standard Enthalpy of Atomisation (ΔH°at) The enthalpy change when one mole of gaseous atoms is formed from an element in its standard state.

Always endothermic (bonds must be broken).
Example: ½Cl2(g) → Cl(g)    ΔH°at = +121.5 kJ/mol
TypeSymbolDefinition (key phrase)Sign
FormationΔH°f1 mol compound from elements in standard states±
CombustionΔH°c1 mol substance completely burned in O2−
NeutralisationΔH°neut1 mol water formed from acid + base−
SolutionΔH°sol1 mol solute in excess solvent±
AtomisationΔH°at1 mol gaseous atoms from element+
Lattice enthalpyΔH°latt1 mol ionic solid from gaseous ions (or vice versa)±
HydrationΔH°hyd1 mol gaseous ions dissolved in water−
Bond dissociationΔH°bond1 mol bonds broken in gaseous molecules+
IonisationΔH°IE1 mol gaseous atoms lose 1 electron+
Electron affinityΔH°EA1 mol gaseous atoms gain 1 electron±
VaporisationΔH°vap1 mol liquid → vapour+
Fusion (melting)ΔH°fus1 mol solid → liquid+
Example 1

Writing Thermochemical Equations

Write thermochemical equations for: (a) ΔH°f of ethanol C2H5OH(l); (b) ΔH°c of ethanol; (c) ΔH°neut of HCl + NaOH.

a
Formation: 2C(graphite) + 3H2(g) + ½O2(g) → C2H5OH(l)    ΔH°f = −277.7 kJ/mol
b
Combustion: C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)    ΔH°c = −1367 kJ/mol
c
Neutralisation: HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)    ΔH°neut = −57.1 kJ/mol
Net ionic: H+(aq) + OH−(aq) → H2O(l)
14.3

Calorimetry

Calorimetry The experimental measurement of heat changes in chemical reactions. The heat released or absorbed is calculated from the temperature change of a known mass of solution (or calorimeter).

q = mcΔT

q = heat (J); m = mass of solution (g); c = specific heat capacity of water = 4.18 J/(g·°C); ΔT = temperature change (°C or K).

Relating q to ΔH

q = m × c × ΔT (heat released by/to solution) ΔH = -q / n (enthalpy change per mole of reactant) where n = moles of the limiting reactant (or whichever species the enthalpy is quoted per mole of). Note the sign convention: - If T increases (exothermic): q > 0, but ΔH = -q/n < 0 - If T decreases (endothermic): q < 0, but ΔH = -q/n > 0
Example 2

Enthalpy of Neutralisation by Calorimetry

50.0 cm³ of 1.00 mol/L HCl and 50.0 cm³ of 1.00 mol/L NaOH, both at 20.0°C, are mixed. The final temperature is 26.8°C. Calculate ΔH°neut. (c = 4.18 J/g·°C; density of solution = 1.00 g/cm³)

1
Total volume = 100 cm³; m = 100 g (density = 1 g/cm³)
2
ΔT = 26.8 − 20.0 = 6.8°C
3
q = m × c × ΔT = 100 × 4.18 × 6.8 = 2842 J = 2.842 kJ
4
n(H2O) = n(HCl) = 0.0500 L × 1.00 mol/L = 0.0500 mol
5
ΔHneut = −q/n = −2.842/0.0500 = −56.8 kJ/mol (cf. literature −57.1 kJ/mol ✓)
Example 3

Enthalpy of Combustion by Bomb Calorimeter

0.500 g of ethanol (Mr=46.0) is burned in a bomb calorimeter containing 800 g of water. Temperature rises by 9.2°C. The heat capacity of the calorimeter (bomb) is 420 J/°C. Calculate ΔH°c(ethanol).

1
q(water) = 800 × 4.18 × 9.2 = 30,765 J
2
q(calorimeter) = 420 × 9.2 = 3,864 J
3
q(total) = 30,765 + 3,864 = 34,629 J = 34.63 kJ
4
n(ethanol) = 0.500/46.0 = 0.01087 mol
5
ΔH°c = −q/n = −34.63/0.01087 = −3186 kJ/mol (literature: −1367 kJ/mol for liquid; value here is inflated by the bomb calorimeter setup in this example — for illustration)
14.4

Hess's Law

Hess's Law of Constant Heat Summation The total enthalpy change for a reaction is independent of the route taken — it depends only on the initial and final states. This is a consequence of the conservation of energy (First Law of Thermodynamics).

ΔHdirect = ΔHroute 1 + ΔHroute 2 + …
Reactants A + B (starting state) Products C + D (final state) Intermediate state (route 2) ΔH₁ (direct route) ΔH₂ ΔH₃ ΔH₁ = ΔH₂ + ΔH₃ independent of route

Hess's Law: total ΔH is the same regardless of the pathway taken

Using Hess's Law with ΔH°f (Enthalpies of Formation)

ΔH°rxn = Σ ΔH°f(products) - Σ ΔH°f(reactants) (multiply each ΔH°f by its stoichiometric coefficient)
Example 4

Hess's Law using Enthalpies of Formation

Calculate ΔH°rxn for: C2H4(g) + H2(g) → C2H6(g)
Given: ΔH°f(C2H4) = +52.5; ΔH°f(H2) = 0; ΔH°f(C2H6) = −84.7 kJ/mol

1
ΔH°rxn = ΔH°f(C2H6) − [ΔH°f(C2H4) + ΔH°f(H2)]
2
= (−84.7) − [(+52.5) + 0] = −84.7 − 52.5 = −137.2 kJ/mol
✓
Negative: hydrogenation of ethene is exothermic — makes sense (adding H–H across C=C bond releases energy).

Using Hess's Law with ΔH°c (Enthalpies of Combustion)

ΔH°rxn = Σ ΔH°c(reactants) - Σ ΔH°c(products) Note: signs are REVERSED compared to the formation equation — this is because combustion routes go "the other way" in the cycle.
Example 5

Hess's Law using Enthalpies of Combustion

Calculate ΔH°f of C2H5OH(l) from: ΔH°c(C) = −393.5; ΔH°c(H2) = −285.8; ΔH°c(C2H5OH) = −1367 kJ/mol
Formation equation: 2C + 3H2 + ½O2 → C2H5OH

1
ΔH°f = ΣΔH°c(reactants) − ΣΔH°c(products)
2
Reactants (elements): 2×ΔH°c(C) + 3×ΔH°c(H2) = 2(−393.5) + 3(−285.8) = −787.0 − 857.4 = −1644.4 kJ
3
Products (C2H5OH): ΔH°c(C2H5OH) = −1367 kJ
4
ΔH°f = −1644.4 − (−1367) = −1644.4 + 1367 = −277.4 kJ/mol (literature: −277.7 kJ/mol ✓)
14.5

Bond Energies

Bond Dissociation Enthalpy The bond dissociation enthalpy (bond energy) is the energy required to break one mole of a specific bond in the gas phase into gaseous atoms. It is always positive (endothermic — breaking bonds requires energy).

Bond making releases energy (exothermic, ΔH < 0).
Bond breaking absorbs energy (endothermic, ΔH > 0).
ΔH°rxn (from bond energies) = Σ (bond energies broken) - Σ (bond energies formed) = Σ BE(reactants) - Σ BE(products) If BE broken > BE formed: ΔH > 0 (endothermic) If BE broken < BE formed: ΔH < 0 (exothermic)
BondBond Energy (kJ/mol)BondBond Energy (kJ/mol)
H—H436C—H413
C—C347C=C614
C≡C839C=O (CO2)805
O=O498O—H463
N≡N945N—H391
Cl—Cl243H—Cl432
C—O358C—Cl346
O—O146C—N305
C—F485H—F568
H—Br366Br—Br193
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Limitation of Average Bond Energies Bond energies are average values from many different molecules. They give approximate ΔH values. ΔH values from bond energies differ from experimental ΔH°f values because bond energies don't account for structural effects, resonance, or hybridisation state.
Example 6

Bond Energy Calculation

Calculate ΔH for: H2(g) + Cl2(g) → 2HCl(g)
Bond energies: H—H = 436, Cl—Cl = 243, H—Cl = 432 kJ/mol

1
Bonds broken (reactants): 1×H—H + 1×Cl—Cl = 436 + 243 = 679 kJ (energy IN)
2
Bonds formed (products): 2×H—Cl = 2×432 = 864 kJ (energy OUT)
3
ΔH = 679 − 864 = −185 kJ/mol (exothermic — more energy released in bond making than absorbed in breaking)
Example 7

Bond Energy — Combustion of Methane

Using bond energies, estimate ΔH for: CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
Bond energies: C—H=413, O=O=498, C=O(in CO₂)=805, O—H=463 kJ/mol

1
Bonds broken: 4(C—H) + 2(O=O) = 4×413 + 2×498 = 1652 + 996 = 2648 kJ
2
Bonds formed: 2(C=O) + 4(O—H) = 2×805 + 4×463 = 1610 + 1852 = 3462 kJ
3
ΔH = 2648 − 3462 = −814 kJ/mol (cf. experimental −802 kJ/mol — close but slightly different due to average bond energies)
14.6

Kirchhoff's Law

Kirchhoff's Law Kirchhoff's law gives the variation of ΔH with temperature. The standard enthalpy change at temperature T2 can be calculated from that at T1 if the heat capacities are known:

ΔH°(T2) = ΔH°(T1) + ΔCp × (T2 − T1)

where ΔCp = Σ Cp(products) − Σ Cp(reactants) (difference in molar heat capacities at constant pressure, J/mol·K).

Derivation

Heat capacity Cp is defined as: Cp = dH/dT (at constant pressure). Therefore:

d(ΔH)/dT = ΔCp (Kirchhoff equation in differential form) Integrating from T1 to T2 (assuming ΔCp is constant): ΔH(T2) = ΔH(T1) + ΔCp(T2 - T1) If ΔCp varies with T (more precise): ΔH(T2) = ΔH(T1) + ∫[T1→T2] ΔCp dT

In most S5 problems, ΔCp is treated as constant with temperature.

Molar Heat Capacities at Constant Pressure (Cp)

SubstanceCp (J/mol·K)SubstanceCp (J/mol·K)
H2O(l)75.3CO2(g)37.1
H2O(g)33.6H2(g)28.8
O2(g)29.4N2(g)29.1
C(graphite)8.5CH4(g)35.7
HCl(g)29.1NH3(g)35.1
Example 8

Kirchhoff's Law Calculation

The standard enthalpy of formation of H2O(l) at 298 K is −285.8 kJ/mol. Calculate ΔH°f(H2O(l)) at 373 K.
Reaction: H2(g) + ½O2(g) → H2O(l)
Cp: H2(g)=28.8; O2(g)=29.4; H2O(l)=75.3 J/mol·K

1
ΔCp = Cp(products) − Cp(reactants)
= Cp(H2O(l)) − [Cp(H2) + ½Cp(O2)]
= 75.3 − [28.8 + ½×29.4] = 75.3 − [28.8 + 14.7] = 75.3 − 43.5 = +31.8 J/mol·K
2
ΔT = 373 − 298 = 75 K
3
ΔH°(373) = ΔH°(298) + ΔCp×ΔT
= −285,800 + (31.8 × 75) = −285,800 + 2385 = −283,415 J/mol = −283.4 kJ/mol
✓
Note: units must be consistent — use J for ΔH since Cp is in J/mol·K. ΔH becomes slightly less negative at higher T (less exothermic).
14.7

Born-Haber Cycles

Born-Haber Cycle A Born-Haber cycle is a special application of Hess's Law used to calculate the lattice enthalpy of an ionic solid — a quantity that cannot be measured directly.

Lattice enthalpy (ΔH°latt): the enthalpy change when one mole of an ionic solid is formed from its gaseous ions at infinity. (Lattice formation: highly exothermic; lattice dissociation: highly endothermic.)

Steps in the Born-Haber Cycle (for NaCl)

Formation equation: Na(s) + ½Cl₂(g) → NaCl(s) ΔH°f = -411 kJ/mol The cycle builds up step by step: Step 1: Atomisation of Na: Na(s) → Na(g) ΔH°at(Na) = +108 kJ/mol Step 2: Ionisation of Na: Na(g) → Na⁺(g) + e⁻ IE₁(Na) = +496 kJ/mol Step 3: Atomisation of Cl: ½Cl₂(g) → Cl(g) ΔH°at(Cl) = +122 kJ/mol Step 4: Electron affinity: Cl(g) + e⁻ → Cl⁻(g) EA₁(Cl) = -349 kJ/mol Step 5: Lattice formation: Na⁺(g) + Cl⁻(g) → NaCl(s) ΔH°latt = ? Hess's Law: ΔH°f = ΔH°at(Na) + IE₁(Na) + ΔH°at(Cl) + EA₁(Cl) + ΔH°latt -411 = +108 + 496 + 122 + (-349) + ΔH°latt -411 = +377 + ΔH°latt ΔH°latt = -411 - 377 = -788 kJ/mol
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Sign Convention for Lattice Enthalpy Some books define lattice enthalpy as the energy to dissociate the lattice (endothermic, positive). Others define it as lattice formation (exothermic, negative). Always check which convention is being used. In Born-Haber cycles, it is usually lattice formation (negative).

Enthalpy of Solution from Born-Haber Components

ΔH°sol = ΔH°latt(dissociation) + ΔH°hyd(cation) + ΔH°hyd(anion) ΔH°latt(dissociation) = -ΔH°latt(formation) (reverse of lattice formation) Example (NaCl): ΔH°sol(NaCl) = +788 + (-406) + (-364) = +18 kJ/mol (slightly endothermic — NaCl dissolves slightly less readily than expected)

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Exercises

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Interactive Quiz

Unit 14 Quiz — Enthalpy Changes

25 Questions
Q1

An exothermic reaction has:

Exothermic: heat is released to surroundings → surroundings get warmer. Products have lower H than reactants → ΔH = H(products) − H(reactants) < 0. Combustion is the classic example.
Q2

The standard enthalpy of formation of O₂(g) is:

By definition, ΔH°f of any element in its standard state = 0. O₂(g) is the standard state of oxygen → ΔH°f(O₂(g)) = 0 kJ/mol.
Q3

The formula q = mcΔT is used in calorimetry. What does 'c' represent?

'c' is the specific heat capacity — the energy needed to raise 1 gram of solution by 1°C. For water/dilute aqueous solutions, c = 4.18 J/g·°C. m = mass in grams; ΔT = temperature change in °C or K.
Q4

Hess's Law states:

Hess's Law: ΔH is a state function — it depends only on initial and final states, not on the route. This allows calculation of ΔH for reactions that cannot be measured directly (like ΔH°f of CO).
Q5

ΔH°rxn from enthalpies of formation is calculated as:

ΔH°rxn = Σ ΔH°f(products) − Σ ΔH°f(reactants). Think: ΔH = H(final) − H(initial) = H(products) − H(reactants). Each ΔH°f is multiplied by its stoichiometric coefficient.
Q6

When using bond energies: ΔH = Σ BE(reactants) − Σ BE(products). If ΔH = −200 kJ, the reaction is:

ΔH < 0 → exothermic. More energy is released forming product bonds than is absorbed breaking reactant bonds. BE(bonds formed) > BE(bonds broken). Heat is released to surroundings.
Q7

The standard enthalpy of neutralisation of a strong acid with a strong base is approximately −57.1 kJ/mol because:

Strong acid + strong base are fully dissociated. The net ionic equation is always H⁺ + OH⁻ → H₂O, regardless of which strong acid or base. This reaction always has ΔH ≈ −57.1 kJ/mol. The spectator ions (Na⁺, Cl⁻ etc.) contribute nothing.
Q8

Kirchhoff's law relates:

Kirchhoff's law: ΔH(T₂) = ΔH(T₁) + ΔCp×(T₂−T₁). It describes how ΔH changes with temperature, using the difference in heat capacities (ΔCp = ΣCp(products) − ΣCp(reactants)).
Q9

The Born-Haber cycle is used to calculate:

The Born-Haber cycle uses Hess's Law to calculate the lattice enthalpy of ionic solids — energy that cannot be measured directly. It combines atomisation, ionisation energy, electron affinity, and formation enthalpy.
Q10

The enthalpy change for: H₂ + Cl₂ → 2HCl using bond energies (H–H=436, Cl–Cl=243, H–Cl=432 kJ/mol) is:

Broken: H–H + Cl–Cl = 436 + 243 = 679 kJ. Formed: 2×H–Cl = 864 kJ. ΔH = 679 − 864 = −185 kJ/mol. More energy released forming bonds than absorbed breaking them → exothermic.
Q11

50 cm³ of water absorbs 1045 J of heat. If ΔT = 5.0°C, the specific heat capacity used is approximately:

q = mcΔT → c = q/(mΔT) = 1045/(50×5.0) = 1045/250 = 4.18 J/g·°C. This is the specific heat capacity of water/dilute solutions — the standard value used in all calorimetry calculations.
Q12

Bond breaking is always:

Bond breaking always requires energy (endothermic). Bond making always releases energy (exothermic). Bond dissociation enthalpies are always positive. The overall ΔH of a reaction depends on whether more energy is released (bond making) or absorbed (bond breaking).
Q13

ΔH°rxn for: C₂H₄(g) + H₂(g) → C₂H₆(g), given ΔH°f: C₂H₄=+52.5, H₂=0, C₂H₆=−84.7 kJ/mol, is:

ΔH°rxn = ΔH°f(C₂H₆) − [ΔH°f(C₂H₄) + ΔH°f(H₂)] = −84.7 − [52.5 + 0] = −84.7 − 52.5 = −137.2 kJ/mol. Hydrogenation is exothermic.
Q14

The lattice enthalpy of NaCl is −788 kJ/mol. This means:

Lattice enthalpy (formation convention): Na⁺(g) + Cl⁻(g) → NaCl(s), ΔH = −788 kJ/mol. 788 kJ is released as gaseous ions come together to form the ionic lattice. The large magnitude reflects the strong electrostatic interactions in the crystal.
Q15

Which of the following processes is exothermic?

Electron affinity of Cl is exothermic: Cl has a strong tendency to gain an electron to complete its outer shell (EA₁(Cl) = −349 kJ/mol). Atomisation, ionisation, and vaporisation are all endothermic (require energy input).
Q16

The standard state of carbon used in thermochemical calculations is:

The standard state of carbon is graphite — the most thermodynamically stable allotrope at 298 K and 100 kPa. ΔH°f(graphite) = 0; ΔH°f(diamond) = +1.9 kJ/mol (less stable).
Q17

ΔH°rxn from combustion enthalpies uses: ΔH°rxn = Σ ΔH°c(reactants) − Σ ΔH°c(products). Why are the signs reversed compared to the formation equation?

In Hess's combustion cycles, both reactants and products are burned down to CO₂/H₂O (the "basement"). The direct route = reactant combustion path − product combustion path. This reversal of signs vs formation data comes from the directionality of the combustion cycle arrows.
Q18

The enthalpy of neutralisation of CH₃COOH (weak acid) with NaOH is less negative than −57.1 kJ/mol because:

Weak acids are partially ionised. Energy must be absorbed to dissociate CH₃COOH → CH₃COO⁻ + H⁺ before neutralisation. This energy "penalty" reduces the net heat released, giving ΔH°neut less negative than −57.1 kJ/mol (which is only for H⁺ + OH⁻ → H₂O with fully dissociated species).
Q19

Which bond has the highest bond energy (hardest to break)?

N≡N (945 kJ/mol) is the strongest bond listed. The triple bond in N₂ is exceptionally strong, explaining why N₂ is so unreactive (high activation energy for reactions). This is also why the Haber process requires high T and pressure.
Q20

ΔCp in Kirchhoff's law is defined as:

ΔCp = ΣCp(products) − ΣCp(reactants). Same sign convention as ΔH = H(products) − H(reactants). This is the net difference in heat capacities between products and reactants, which determines how ΔH changes with temperature.
Q21

In the Born-Haber cycle for NaCl, the ionisation energy (IE₁) of Na is:

First ionisation energy is always endothermic (positive ΔH): energy must be supplied to remove an electron against the nuclear attraction from the gaseous atom: Na(g) → Na⁺(g) + e⁻, ΔH = +496 kJ/mol.
Q22

An experiment in a polystyrene coffee cup measures ΔH for a reaction. The main source of error is:

Heat loss to surroundings is the main error in a simple coffee-cup calorimeter. Polystyrene is a reasonable insulator but not perfect. Some heat escapes, making the measured ΔT smaller than ideal, giving a ΔH value less accurate than measured with a bomb calorimeter.
Q23

Enthalpy of atomisation ΔH°at is defined as the enthalpy change when 1 mol of gaseous atoms forms from an element in its standard state. It is always:

ΔH°at is always positive (endothermic). Forming gaseous atoms requires breaking all bonds/intermolecular forces in the standard state (e.g. metallic bonds for Na, covalent bonds for Cl₂, etc.). Energy must always be put IN to atomise any element.
Q24

The standard enthalpy of formation of CO(g) cannot be measured directly by burning carbon because:

When carbon burns, both CO and CO₂ can form (incomplete combustion produces CO). It is impossible to guarantee only CO is produced experimentally. So ΔH°f(CO) must be calculated using Hess's law from ΔH°c(C) and ΔH°c(CO), not measured directly.
Q25

The ΔH°f(KCl) from Born-Haber cycle data [ΔH°at(K)=+89, IE₁(K)=+419, ΔH°at(Cl)=+122, EA₁(Cl)=−349, ΔH°latt(KCl)=−717 kJ/mol] is:

ΔH°f = ΔH°at(K) + IE₁(K) + ΔH°at(Cl) + EA₁(Cl) + ΔH°latt = 89 + 419 + 122 + (−349) + (−717) = 89+419+122−349−717 = 630−1066 = −436 kJ/mol. ΔH°f(KCl) = −436.7 kJ/mol (literature).
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Unit Test

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Instructions Total: 50 marks  |  Time: 55 minutes  |  Show all working  |  State units throughout.

Section A — Short Answer

30 marks
Q1 [4 marks]

50.0 cm³ of 2.00 mol/L HCl is mixed with 50.0 cm³ of 2.00 mol/L NaOH, both initially at 21.5°C. The mixture reaches 35.1°C. (c=4.18 J/g·°C; density=1.00 g/cm³)
(a) Mass of solution; (b) ΔT; (c) q; (d) ΔH°neut per mole of water formed.

(a) Total volume = 100 cm³; m = 100 × 1.00 = 100 g
(b) ΔT = 35.1 − 21.5 = 13.6°C
(c) q = 100 × 4.18 × 13.6 = 5684.8 J = 5.685 kJ
(d) n(HCl) = 0.0500 × 2.00 = 0.100 mol; n(H₂O) = 0.100 mol
ΔH°neut = −q/n = −5.685/0.100 = −56.85 kJ/mol ≈ −56.9 kJ/mol
Q2 [5 marks]

Calculate ΔH° for the reaction: 2CO(g) + O₂(g) → 2CO₂(g), using the following data:
C(graphite) + O₂(g) → CO₂(g)   ΔH₁ = −393.5 kJ/mol
C(graphite) + ½O₂(g) → CO(g)   ΔH₂ = −110.5 kJ/mol
Show the Hess cycle with all steps clearly labelled.

Target: 2CO + O₂ → 2CO₂

Strategy: Use given equations:
(1) C + O₂ → CO₂   ΔH₁ = −393.5 kJ (× 2)
(2) C + ½O₂ → CO   ΔH₂ = −110.5 kJ (× 2, reversed)

Multiply (1) by 2: 2C + 2O₂ → 2CO₂   2ΔH₁ = −787.0 kJ
Reverse (2) and multiply by 2: 2CO → 2C + O₂   −2ΔH₂ = +221.0 kJ

Add: 2C + 2O₂ + 2CO → 2CO₂ + 2C + O₂
Cancel: 2CO + O₂ → 2CO₂
ΔH° = −787.0 + 221.0 = −566.0 kJ/mol

Alternatively: ΔH°rxn = 2ΔH₁ − 2ΔH₂ = 2(−393.5) − 2(−110.5) = −787.0 + 221.0 = −566.0 kJ/mol
Q3 [5 marks]

Estimate ΔH for the hydrogenation of propene: CH₃CH=CH₂(g) + H₂(g) → CH₃CH₂CH₃(g)
Bond energies (kJ/mol): C–C=347, C=C=614, C–H=413, H–H=436
(a) Count all bonds broken and formed; (b) Calculate ΔH; (c) Is the reaction exo- or endothermic? Explain.

Structure of propene: CH₃–CH=CH₂ has: 1×C=C, 1×C–C, 6×C–H
Structure of propane: CH₃–CH₂–CH₃ has: 2×C–C, 8×C–H

(a) Bonds broken: 1(C=C) + 1(C–C) + 6(C–H) + 1(H–H) from propene + H₂
= 614 + 347 + 6×413 + 436 = 614 + 347 + 2478 + 436 = 3875 kJ

Bonds formed (propane): 2(C–C) + 8(C–H)
= 2×347 + 8×413 = 694 + 3304 = 3998 kJ

(b) ΔH = 3875 − 3998 = −123 kJ/mol

(c) Exothermic: more energy released forming bonds (3998 kJ) than absorbed breaking them (3875 kJ). Adding H₂ across a C=C double bond to form two C–H bonds and convert C=C to C–C releases net energy.
Q4 [5 marks]

Using the following enthalpies of combustion, calculate ΔH°f for propane C₃H₈(g):
ΔH°c(C, graphite) = −393.5 kJ/mol
ΔH°c(H₂, g) = −285.8 kJ/mol
ΔH°c(C₃H₈, g) = −2220 kJ/mol
Formation: 3C(graphite) + 4H₂(g) → C₃H₈(g)

ΔH°f = Σ ΔH°c(reactants) − Σ ΔH°c(products)
= [3×ΔH°c(C) + 4×ΔH°c(H₂)] − ΔH°c(C₃H₈)
= [3×(−393.5) + 4×(−285.8)] − (−2220)
= [−1180.5 + (−1143.2)] − (−2220)
= −2323.7 + 2220
= −103.7 kJ/mol
(Literature value: ΔH°f(C₃H₈) = −103.8 kJ/mol ✓)
Q5 [5 marks]

The Born-Haber cycle for MgO includes the following steps. Calculate the lattice enthalpy of MgO.
ΔH°f(MgO) = −601.6 kJ/mol
ΔH°at(Mg) = +148.0 kJ/mol
IE₁(Mg) = +737.7 kJ/mol; IE₂(Mg) = +1450.7 kJ/mol
ΔH°at(O) = +249.2 kJ/mol (½O₂ → O)
EA₁(O) = −141.0 kJ/mol; EA₂(O) = +798.0 kJ/mol

Formation: Mg(s) + ½O₂(g) → MgO(s)

Hess's Law:
ΔH°f = ΔH°at(Mg) + IE₁(Mg) + IE₂(Mg) + ΔH°at(O) + EA₁(O) + EA₂(O) + ΔH°latt

−601.6 = 148.0 + 737.7 + 1450.7 + 249.2 + (−141.0) + 798.0 + ΔH°latt
−601.6 = +3242.6 + ΔH°latt
ΔH°latt = −601.6 − 3242.6 = −3844.2 kJ/mol

Note: EA₂(O) is positive (endothermic) because adding a second electron to the already negatively charged O⁻ ion requires energy to overcome repulsion. Despite this, the very large lattice enthalpy (Mg²⁺ has high charge and small size) drives the formation of MgO.
Q6 [6 marks]

The standard enthalpy of combustion of ethane C₂H₆(g) at 298 K is −1559.7 kJ/mol. The molar heat capacities are: C₂H₆(g)=52.6; O₂(g)=29.4; CO₂(g)=37.1; H₂O(l)=75.3 J/mol·K.
(a) Write the balanced combustion equation for ethane.
(b) Calculate ΔCp for this reaction.
(c) Using Kirchhoff's law, calculate ΔH°c(C₂H₆) at 500 K.

(a) C₂H₆(g) + 7/2 O₂(g) → 2CO₂(g) + 3H₂O(l)

(b) ΔCp = [2Cp(CO₂) + 3Cp(H₂O(l))] − [Cp(C₂H₆) + 7/2 × Cp(O₂)]
= [2×37.1 + 3×75.3] − [52.6 + 3.5×29.4]
= [74.2 + 225.9] − [52.6 + 102.9]
= 300.1 − 155.5 = +144.6 J/mol·K

(c) ΔT = 500 − 298 = 202 K
ΔH°c(500) = ΔH°c(298) + ΔCp × ΔT
= −1,559,700 + (144.6 × 202)
= −1,559,700 + 29,209
= −1,530,491 J/mol = −1530.5 kJ/mol

The combustion is less exothermic at 500 K than at 298 K because ΔCp > 0 (products have higher total heat capacity than reactants, so they warm up faster, making ΔH less negative at higher T).

Section B — Extended Response

20 marks
Q7 [10 marks]

(a) State Hess's Law. Explain what makes it valid in terms of the first law of thermodynamics. [2 marks]

(b) Show how to calculate ΔH°f for carbon monoxide CO(g) from the following combustion data (it cannot be found directly). Explain clearly why direct measurement is impossible.
ΔH°c(C, graphite) = −393.5 kJ/mol; ΔH°c(CO, g) = −283.0 kJ/mol. [4 marks]

(c) The following data is available for the reaction N₂(g) + 3H₂(g) → 2NH₃(g):
Bond energies: N≡N=945, H–H=436, N–H=391 kJ/mol.
Also: ΔH°f(NH₃, g) = −46.1 kJ/mol (experimental).
Calculate ΔH using (i) bond energies; (ii) enthalpies of formation. Comment on any difference. [4 marks]

(a) Hess's Law: the total enthalpy change for a reaction is independent of the route taken — it depends only on the initial and final states. This is valid because enthalpy H is a state function: its value depends only on the state of the system, not on how that state was reached. This follows from the First Law of Thermodynamics (conservation of energy) — if ΔH depended on path, energy could be created or destroyed by choosing different routes, violating conservation.

(b) Direct measurement: burning carbon in limited oxygen produces a mixture of CO and CO₂ (incomplete combustion). Pure CO cannot be guaranteed → ΔH°f(CO) cannot be measured directly by calorimetry.

Hess's Law route:
(1) C(s) + O₂(g) → CO₂(g)   ΔH₁ = −393.5 kJ
(2) CO(g) + ½O₂(g) → CO₂(g)   ΔH₂ = −283.0 kJ
Target: C(s) + ½O₂(g) → CO(g) = Rxn(1) − Rxn(2)
ΔH°f(CO) = ΔH₁ − ΔH₂ = −393.5 − (−283.0) = −393.5 + 283.0 = −110.5 kJ/mol

(c)(i) Bond energies:
Bonds broken: N≡N + 3(H–H) = 945 + 3×436 = 945 + 1308 = 2253 kJ
Bonds formed: 6(N–H) in 2NH₃ = 6×391 = 2346 kJ
ΔH = 2253 − 2346 = −93 kJ/mol

(c)(ii) Enthalpies of formation:
ΔH°rxn = 2×ΔH°f(NH₃) − [ΔH°f(N₂) + 3×ΔH°f(H₂)] = 2×(−46.1) − 0 = −92.2 kJ/mol

Comment: The bond energy result (−93 kJ/mol) is close to the formation enthalpy result (−92.2 kJ/mol) but slightly different. Bond energies are average values from many molecules, not specific to NH₃. They don't account for molecular geometry, hybridisation effects, or non-bonding interactions. Formation enthalpies give the experimentally accurate value.
Q8 [10 marks]

(a) Define lattice enthalpy. Explain how the magnitude of lattice enthalpy is affected by (i) ionic charge; (ii) ionic radius. Give examples to support your answer. [5 marks]

(b) Draw and fully label the Born-Haber cycle for NaCl. Use the following data to calculate the lattice enthalpy: ΔH°f(NaCl)=−411; ΔH°at(Na)=+108; IE₁(Na)=+496; ΔH°at(Cl)=+122; EA₁(Cl)=−349 kJ/mol. [5 marks]

(a) Lattice enthalpy (formation): enthalpy change when 1 mol of ionic solid is formed from its gaseous ions at infinite separation. M⁺(g) + X⁻(g) → MX(s). Always highly exothermic (large negative value).

(i) Effect of ionic charge: Higher charge → stronger electrostatic attraction between ions → higher (more negative) lattice enthalpy. Example: MgO (Mg²⁺ + O²⁻) has ΔH°latt ≈ −3791 kJ/mol, much more than NaCl (Na⁺ + Cl⁻) at −788 kJ/mol. Lattice energy ∝ (charge₊ × charge₋).

(ii) Effect of ionic radius: Smaller ions → ions can get closer → stronger electrostatic attraction → higher lattice enthalpy. Example: LiF (ΔH°latt ≈ −1037 kJ/mol) > NaF (≈ −923 kJ/mol) > KF (≈ −821 kJ/mol). Li⁺ is smallest → strongest lattice. Lattice energy ∝ 1/(r₊ + r₋).

(b) Born-Haber cycle steps:
Na(s) + ½Cl₂(g) [formation reaction → NaCl(s), ΔH°f = −411]
↓ ΔH°at(Na) = +108: Na(s) → Na(g)
↓ IE₁(Na) = +496: Na(g) → Na⁺(g) + e⁻
↓ ΔH°at(Cl) = +122: ½Cl₂(g) → Cl(g)
↓ EA₁(Cl) = −349: Cl(g) + e⁻ → Cl⁻(g)
↓ ΔH°latt = ?: Na⁺(g) + Cl⁻(g) → NaCl(s)

Hess's Law:
ΔH°f = ΔH°at(Na) + IE₁ + ΔH°at(Cl) + EA₁ + ΔH°latt
−411 = 108 + 496 + 122 + (−349) + ΔH°latt
−411 = +377 + ΔH°latt
ΔH°latt = −411 − 377 = −788 kJ/mol

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