Reactivity 1.2 | Energy Cycles

📋 IB Content Statements (R1.2)

This topic covers the following syllabus points from the IB Chemistry 2025 guide:

R1.2.1: Hess's law states that the enthalpy change for a reaction is independent of the pathway taken.
R1.2.2: Enthalpy changes of reactions can be calculated from standard enthalpy changes of formation or combustion using Hess's law.
R1.2.3: Bond enthalpy is the energy needed to break one mole of a particular bond in gaseous molecules. Average bond enthalpies are used because the same bond can have different energies in different molecules.
R1.2.4: $\Delta H$ for a reaction can be estimated using average bond enthalpies: $\Delta H = \Sigma E(\text{bonds broken}) - \Sigma E(\text{bonds formed})$.

HL Extension

R1.2.5 (HL): Born-Haber cycles can be used to calculate lattice enthalpies for ionic compounds.
R1.2.6 (HL): Entropy ($S$) is a measure of the number of possible microstates. $\Delta S$ can be calculated from standard entropy values.
R1.2.7 (HL): Gibbs free energy: $\Delta G = \Delta H - T\Delta S$. A reaction is spontaneous when $\Delta G < 0$.

🔹 Hess's Law

Statement

The total enthalpy change for a chemical reaction is independent of the route taken, provided the initial and final conditions are the same.

This is a direct consequence of the law of conservation of energy.

Using Formation Data

$$\Delta H_{rxn} = \Sigma \Delta H_f^{\circ}(\text{products}) - \Sigma \Delta H_f^{\circ}(\text{reactants})$$

Key rule: The standard enthalpy of formation ($\Delta H_f^{\circ}$) of an element in its standard state is zero by definition. Example: $\Delta H_f^{\circ}[O_2(g)] = 0$, $\Delta H_f^{\circ}[C(graphite)] = 0$.

Using Combustion Data

$$\Delta H_{rxn} = \Sigma \Delta H_c^{\circ}(\text{reactants}) - \Sigma \Delta H_c^{\circ}(\text{products})$$

⚠️ Notice: For combustion data, it's reactants minus products (the reverse of formation). This is because the cycle goes "down then up" through combustion products as intermediates.

Hess's Law energy cycle diagram showing direct and indirect routes

⚡ Bond Enthalpies

Definition

Bond enthalpy is the energy required to break one mole of a particular covalent bond in gaseous molecules under standard conditions. All bond enthalpy values are positive (endothermic).

Calculating ΔH from Bond Enthalpies

$$\Delta H = \Sigma E(\text{bonds broken}) - \Sigma E(\text{bonds formed})$$

Bond	Average BE (kJ/mol)	Bond	Average BE (kJ/mol)
$C-H$	414	$O=O$	498
$C-C$	346	$C=O$ (in $CO_2$)	804
$C=C$	614	$O-H$	463
$C-O$	358	$N≡N$	945

⚠️ Why "average"? The same bond can have different strengths in different molecules. For example, $O-H$ in water (459 kJ/mol for the first, 428 for the second) differs from $O-H$ in ethanol. Therefore, bond enthalpy calculations give estimates, not exact values.

🏗️ Born-Haber Cycles (HL)

What is a Born-Haber Cycle?

A special application of Hess's law used to calculate the lattice enthalpy of an ionic compound — the enthalpy change when one mole of an ionic compound is formed from its gaseous ions under standard conditions.

Steps in a Born-Haber Cycle (for NaCl)

Step	Process	Term	Sign
1	$Na(s) \rightarrow Na(g)$	Atomization of Na	Endothermic (+)
2	$\frac{1}{2}Cl_2(g) \rightarrow Cl(g)$	Atomization of Cl / ½ Bond dissociation	Endothermic (+)
3	$Na(g) \rightarrow Na^+(g) + e^-$	1st Ionization Energy	Endothermic (+)
4	$Cl(g) + e^- \rightarrow Cl^-(g)$	1st Electron Affinity	Exothermic (−)
5	$Na^+(g) + Cl^-(g) \rightarrow NaCl(s)$	Lattice Enthalpy	Exothermic (−)

$$\Delta H_f = \Delta H_{at}(Na) + \Delta H_{at}(Cl) + IE_1(Na) + EA_1(Cl) + \Delta H_{lat}$$

🌀 Entropy & Gibbs Free Energy (HL)

📺 R1.4 — Standard Entropy Change (HL)

📺 R1.4.1 — Predicting Entropy Changes (HL)

📺 R1.4.2 — How Temperature Changes Delta G (HL)

📺 R1.4.3 — Sign of Delta G (HL)

Entropy ($S$)

Entropy is a measure of the disorder (number of possible microstates) in a system. Higher entropy = more ways to arrange the particles.

$$\Delta S_{rxn} = \Sigma S^{\circ}(\text{products}) - \Sigma S^{\circ}(\text{reactants})$$

Change	ΔS	Example
Solid → Liquid	Positive (+)	Ice melting
Liquid → Gas	Large Positive (++)	Water boiling
Fewer → More moles of gas	Positive (+)	$CaCO_3(s) → CaO(s) + CO_2(g)$
More → Fewer moles of gas	Negative (−)	$N_2 + 3H_2 → 2NH_3$

Gibbs Free Energy

$$\Delta G = \Delta H - T\Delta S$$

$\Delta H$	$\Delta S$	$\Delta G$	Spontaneity
Negative (−)	Positive (+)	Always negative	Always spontaneous
Positive (+)	Negative (−)	Always positive	Never spontaneous
Negative (−)	Negative (−)	Depends on T	Spontaneous at low T
Positive (+)	Positive (+)	Depends on T	Spontaneous at high T

🧠 Memory Aids

🔤 Bond Enthalpies — "Break = Buy, Form = Free"

Breaking bonds costs energy (endothermic, positive). Forming bonds releases energy (exothermic, negative). $\Delta H = \text{Break} - \text{Form}$.

🔤 Hess's Law Routes — "FPC" (Formation/Products first, Combustion/Products last)

Formation: $\Delta H = \Sigma \Delta H_f(\text{Products}) - \Sigma \Delta H_f(\text{Reactants})$
Combustion: $\Delta H = \Sigma \Delta H_c(\text{Reactants}) - \Sigma \Delta H_c(\text{Products})$

For Formation: Products first (P minus R). For Combustion: Reactants first (R minus P).

🔤 Gibbs — "GHiTS" ($\Delta G = \Delta H - T \Delta S$)

Gibbs = Heat minus Temperature times Scatter (entropy). Negative G = spontaneous (Go!).

🔤 Born-Haber Steps — "AAIIE + Lattice"

Atomize metal + Atomize non-metal + Ionize metal + Ionize non-metal (electron affinity) + Ends at lattice enthalpy.

🌍 Real-World Applications

🔋 Hand Warmers — Exothermic Crystallization

Context: Reusable hand warmers contain supersaturated sodium acetate. Clicking a metal disc triggers crystallization.

Science: The crystallization is exothermic ($\Delta H < 0$) and has negative $\Delta S$ (solid is more ordered). Using $\Delta G=\Delta H - T\Delta S$: both terms contribute to negative $\Delta G$ at room temperature → spontaneous.

Impact: Understanding Gibbs helps explain why some endothermic reactions can still occur spontaneously (at high T) when $T\Delta S$ overcomes $\Delta H$.

🧊 Instant Cold Packs — Entropy-Driven

Context: Instant cold packs dissolve ammonium nitrate ($NH_4NO_3$) in water. The pack feels cold.

Science: Dissolving is endothermic ($\Delta H > 0$), but the large increase in entropy ($\Delta S >> 0$) as the ordered crystal dissolves makes $T\Delta S > \Delta H$ → $\Delta G < 0$ → spontaneous despite being endothermic.

Impact: This is a classic example of an entropy-driven endothermic process. First-aid kits rely on this thermodynamics principle.

🏭 Haber Process — Gibbs Compromise

Context: $N_2 + 3H_2 \rightleftharpoons 2NH_3$ ($\Delta H = -92$ kJ/mol).

Science: The reaction is exothermic but $\Delta S < 0$ (4 mol gas → 2 mol gas). At low T, $\Delta G < 0$ (spontaneous) but the rate is too slow. At high T, $T\Delta S$ term dominates and $\Delta G$ becomes positive. The compromise: 450°C with an iron catalyst (kinetics) at 200 atm (Le Chatelier).

Impact: Produces 150 million tonnes of ammonia per year for fertilizers — feeding roughly half the world's population.

⚠️ Common Mistakes

❌ Swapping the formula for formation vs combustion → ✅ Formation: Products − Reactants. Combustion: Reactants − Products. Draw the cycle if unsure.
❌ "Breaking bonds releases energy" → ✅ Breaking bonds requires energy (endothermic). Only forming bonds releases energy.
❌ Forgetting to account for stoichiometric coefficients → ✅ Multiply bond/formation/combustion enthalpies by their coefficients in the balanced equation.
❌ "Spontaneous means fast" → ✅ Spontaneous ($\Delta G < 0$) means thermodynamically favorable — it says nothing about rate. Diamond → graphite is spontaneous but infinitely slow.
❌ Using °C instead of K for Gibbs → ✅ Temperature in $\Delta G = \Delta H - T\Delta S$ must be in Kelvin. Also, if $\Delta H$ is in kJ, convert $\Delta S$ from J/K to kJ/K.

📝 Exam-Style Questions

Question 1: State Hess's Law. [1 mark]

Mark Scheme:

[1 mark] The enthalpy change for a reaction is independent of the pathway / route taken (provided initial and final conditions are the same).

Question 2: Using bond enthalpies, estimate $\Delta H$ for the combustion of methane: $CH_4 + 2O_2 → CO_2 + 2H_2O$. [3 marks]

Mark Scheme:

[1 mark] Bonds broken: 4(C-H) + 2(O=O) = 4(414) + 2(498) = 2652 kJ.
[1 mark] Bonds formed: 2(C=O in CO₂) + 4(O-H) = 2(804) + 4(463) = 3460 kJ.
[1 mark] $\Delta H = 2652 - 3460 = -808$ kJ/mol.

Question 3: Explain why bond enthalpy calculations give only estimates. [2 marks]

Mark Scheme:

[1 mark] Average bond enthalpies are used, not exact values for specific molecules.
[1 mark] The same bond type has slightly different energies in different molecular environments.

Question 4 (HL): Predict the sign of $\Delta S$ for: $CaCO_3(s) \rightarrow CaO(s) + CO_2(g)$. [1 mark]

Mark Scheme:

[1 mark] Positive. A gas is produced from a solid → increase in disorder / number of microstates.

Question 5 (HL): Calculate $\Delta G$ at 298 K if $\Delta H = -100$ kJ/mol and $\Delta S = +50$ J/(K·mol). [2 marks]

Mark Scheme:

[1 mark] Convert: $\Delta S = 0.050$ kJ/(K·mol).
[1 mark] $\Delta G = -100 - (298)(0.050) = -100 - 14.9 = -114.9$ kJ/mol. Spontaneous.

Question 6 (HL): Determine the temperature at which a reaction becomes spontaneous if $\Delta H = +50$ kJ/mol and $\Delta S = +100$ J/(K·mol). [2 marks]

Mark Scheme:

[1 mark] At equilibrium: $\Delta G = 0$, so $T = \Delta H / \Delta S = 50000 / 100 = 500$ K (227°C).
[1 mark] Spontaneous above 500 K.

Question 7: Calculate $\Delta H$ for the reaction $C_2H_4 + H_2 → C_2H_6$ using formation data: $\Delta H_f[C_2H_4] = +52$, $\Delta H_f[C_2H_6] = -85$ kJ/mol. [2 marks]

Mark Scheme:

[1 mark] $\Delta H_f[H_2] = 0$ (element in standard state).
[1 mark] $\Delta H = (-85) - (52 + 0) = -137$ kJ/mol.

Question 8 (HL): List the steps in a Born-Haber cycle for $MgO$ and state whether each is endothermic or exothermic. [3 marks]

Mark Scheme:

[1 mark] Atomization of Mg and O (both endothermic).
[1 mark] IE₁ + IE₂ of Mg (endothermic), EA₁ + EA₂ of O (EA₁ exothermic, EA₂ endothermic).
[1 mark] Lattice enthalpy (exothermic — ions come together).