IB Chemistry How to Solve: IB Chemistry – Energetics/Thermochemistry (Hess, Born-Haber, Entropy, Gibbs)

How to Solve: IB Chemistry – Energetics/Thermochemistry (Hess, Born-Haber, Entropy, Gibbs)

Complete Guide

Introduction

"Mastering Hess’s Law, Born-Haber cycles, entropy, and Gibbs free energy doesn’t just get you 10+ marks on your IB Chemistry exam—it lets you predict whether a reaction will happen in real life, from rust forming on a car to the energy in your phone battery. Miss this, and you’re leaving easy marks on the table."

WHAT YOU NEED TO KNOW FIRST

Before diving in, you must already understand:
1. Enthalpy (ΔH) – Heat energy change at constant pressure. Exothermic (–ΔH) vs. endothermic (+ΔH).
2. Standard conditions (⦵) – 298 K, 100 kPa, 1 mol dm⁻³ solutions.
3. Bond enthalpies – Energy required to break 1 mole of bonds in the gas phase.

If any of these are unclear, pause and review them first.

KEY TERMS & FORMULAS

1. Hess’s Law

Definition: The enthalpy change for a reaction is independent of the pathway taken. Formula: ΔH₁ = ΔH₂ + ΔH₃ (for any pathway) MEMORISE THIS: Hess’s Law lets you rearrange equations like algebra to find unknown ΔH.

2. Born-Haber Cycle

Definition: A special Hess’s Law cycle for lattice enthalpy (ΔHₗₐₜₜᵢcₑ) of ionic compounds. Key terms: - ΔHₐₜₒₘᵢzₐₜᵢₒₙ – Energy to form 1 mole of gaseous atoms from elements. - ΔHᵢₒₙᵢzₐₜᵢₒₙ – Energy to remove 1 mole of electrons (1st IE, 2nd IE, etc.). - ΔHₑₐ – Electron affinity (energy change when 1 mole of electrons is added). - ΔHₗₐₜₜᵢcₑ – Energy released when 1 mole of solid ionic lattice forms from gaseous ions. - ΔH_f – Standard enthalpy of formation (given).

Formula (for NaCl): ΔH_f = ΔHₐₜₒₘ + ΔHᵢₒₙ + ½ΔH_ᵇₒₙd (Cl₂) + ΔHₑₐ + ΔHₗₐₜₜᵢcₑ MEMORISE THIS STRUCTURE – The exact steps vary by compound, but the cycle is always the same.

3. Entropy (S)

Definition: A measure of disorder in a system. Key points: - ΔS > 0 → More disorder (e.g., solid → liquid → gas). - ΔS < 0 → Less disorder (e.g., gas → liquid). - Standard entropy (S⦵) – Given in data booklets (J K⁻¹ mol⁻¹).

Formula: ΔS_system = ΣS⦵(products) – ΣS⦵(reactants) GIVEN ON EXAM SHEET – But you must know how to apply it.

4. Gibbs Free Energy (ΔG)

Definition: Determines if a reaction is spontaneous (ΔG < 0) or non-spontaneous (ΔG > 0). Formula: ΔG = ΔH – TΔS MEMORISE THIS – Units must match (ΔH in kJ, ΔS in kJ K⁻¹).

Key relationships: - ΔG < 0 → Spontaneous (reaction happens on its own). - ΔG = 0 → Equilibrium. - ΔG > 0 → Non-spontaneous (needs energy input).

STEP-BY-STEP METHOD

A. Hess’s Law Problems

Step 1: Write the target equation (the one you need ΔH for). Step 2: List all given equations (with their ΔH values). Step 3: Manipulate given equations to match the target: - Reverse an equation? Flip the sign of ΔH. - Multiply/divide an equation? Do the same to ΔH. Step 4: Add the manipulated equations and their ΔH values. Step 5: Cancel out species that appear on both sides. Step 6: Check that the final equation matches the target. If not, recheck steps.

B. Born-Haber Cycle Problems

Step 1: Write the formation equation (e.g., Na(s) + ½Cl₂(g) → NaCl(s)). Step 2: Draw the cycle with these steps (in order): 1. Atomisation (Na(s) → Na(g), ΔHₐₜₒₘ). 2. Ionisation (Na(g) → Na⁺(g) + e⁻, ΔHᵢₒₙ). 3. Bond breaking (½Cl₂(g) → Cl(g), ½ΔH_ᵇₒₙd). 4. Electron affinity (Cl(g) + e⁻ → Cl⁻(g), ΔHₑₐ). 5. Lattice enthalpy (Na⁺(g) + Cl⁻(g) → NaCl(s), ΔHₗₐₜₜᵢcₑ). Step 3: Apply Hess’s Law: ΔH_f = ΔHₐₜₒₘ + ΔHᵢₒₙ + ½ΔH_ᵇₒₙd + ΔHₑₐ + ΔHₗₐₜₜᵢcₑ Step 4: Rearrange to solve for the unknown (usually ΔHₗₐₜₜᵢcₑ).

C. Entropy & Gibbs Free Energy Problems

Step 1: Calculate ΔS_system using standard entropies: ΔS_system = ΣS⦵(products) – ΣS⦵(reactants) Step 2: Convert ΔS to kJ K⁻¹ (divide by 1000 if in J K⁻¹ mol⁻¹). Step 3: Use the Gibbs equation: ΔG = ΔH – TΔS - ΔH must be in kJ (convert if in J). - T must be in Kelvin (add 273 if given in °C). Step 4: Interpret the sign of ΔG: - ΔG < 0 → Spontaneous. - ΔG > 0 → Non-spontaneous.

WORKED EXAMPLES

Example 1 – Hess’s Law (Basic)

Question: Given:
1. C(s) + O₂(g) → CO₂(g) ΔH = –394 kJ
2. CO(g) + ½O₂(g) → CO₂(g) ΔH = –283 kJ Find ΔH for: C(s) + ½O₂(g) → CO(g)

Solution: Step 1: Target equation: C(s) + ½O₂(g) → CO(g) Step 2: Given equations: - Eq1: C + O₂ → CO₂ ΔH = –394 kJ - Eq2: CO + ½O₂ → CO₂ ΔH = –283 kJ Step 3: Reverse Eq2 to get CO₂ → CO + ½O₂ (ΔH = +283 kJ). Step 4: Add Eq1 and reversed Eq2: C + O₂ → CO₂ (–394 kJ) CO₂ → CO + ½O₂ (+283 kJ)

C + ½O₂ → CO (–394 + 283 = –111 kJ) Step 5: Final ΔH = –111 kJ mol⁻¹.

What we did and why: We reversed Eq2 to cancel CO₂ and isolate CO. Hess’s Law lets us add ΔH values like algebra.

Example 2 – Born-Haber Cycle (Medium)

Question: Calculate the lattice enthalpy of MgO(s) given: - ΔH_f[MgO(s)] = –602 kJ mol⁻¹ - ΔHₐₜₒₘ[Mg(s)] = +148 kJ mol⁻¹ - ΔHᵢₒₙ[Mg(g)] = +738 kJ mol⁻¹ (1st IE) + +1451 kJ mol⁻¹ (2nd IE) - ΔHₐₜₒₘ[½O₂(g)] = +249 kJ mol⁻¹ - ΔHₑₐ[O(g)] = –141 kJ mol⁻¹ (1st EA) + +798 kJ mol⁻¹ (2nd EA)

Solution: Step 1: Formation equation: Mg(s) + ½O₂(g) → MgO(s) ΔH_f = –602 kJ Step 2: Born-Haber steps: 1. Mg(s) → Mg(g) ΔHₐₜₒₘ = +148 kJ 2. Mg(g) → Mg⁺(g) + e⁻ 1st IE = +738 kJ 3. Mg⁺(g) → Mg²⁺(g) + e⁻ 2nd IE = +1451 kJ 4. ½O₂(g) → O(g) ½ΔH_ᵇₒₙd = +249 kJ 5. O(g) + e⁻ → O⁻(g) 1st EA = –141 kJ 6. O⁻(g) + e⁻ → O²⁻(g) 2nd EA = +798 kJ 7. Mg²⁺(g) + O²⁻(g) → MgO(s) ΔHₗₐₜₜᵢcₑ = ? Step 3: Apply Hess’s Law: ΔH_f = ΔHₐₜₒₘ + 1st IE + 2nd IE + ½ΔH_ᵇₒₙd + 1st EA + 2nd EA + ΔHₗₐₜₜᵢcₑ –602 = 148 + 738 + 1451 + 249 + (–141) + 798 + ΔHₗₐₜₜᵢcₑ Step 4: Solve for ΔHₗₐₜₜᵢcₑ: –602 = 3243 + ΔHₗₐₜₜᵢcₑ ΔHₗₐₜₜᵢcₑ = –602 – 3243 = –3845 kJ mol⁻¹

What we did and why: We accounted for two ionisation energies (Mg → Mg²⁺) and two electron affinities (O → O²⁻). Always check the charge of the ions!

Example 3 – Gibbs Free Energy (Exam-Style)

Question: For the reaction: N₂(g) + 3H₂(g) → 2NH₃(g) ΔH = –92 kJ mol⁻¹ S⦵[N₂] = 192 J K⁻¹ mol⁻¹ S⦵[H₂] = 131 J K⁻¹ mol⁻¹ S⦵[NH₃] = 193 J K⁻¹ mol⁻¹ At what temperature (in °C) does the reaction become non-spontaneous?

Solution: Step 1: Calculate ΔS_system: ΔS = ΣS⦵(products) – ΣS⦵(reactants) ΔS = 2(193) – [192 + 3(131)] = 386 – 585 = –199 J K⁻¹ mol⁻¹ Step 2: Convert to kJ K⁻¹: ΔS = –199 / 1000 = –0.199 kJ K⁻¹ mol⁻¹ Step 3: Use ΔG = ΔH – TΔS. At equilibrium (ΔG = 0): 0 = –92 – T(–0.199) T = 92 / 0.199 = 462 K Step 4: Convert to °C: 462 – 273 = 189°C Step 5: Interpretation: - Below 189°C, ΔG < 0 (spontaneous). - Above 189°C, ΔG > 0 (non-spontaneous).

What we did and why: We found the temperature at which ΔG = 0 (equilibrium). Above this, the reaction is non-spontaneous because the –TΔS term dominates.

COMMON MISTAKES

1. MISTAKE: Forgetting to reverse the sign of ΔH when reversing an equation. WHY IT HAPPENS: Students treat ΔH like a normal number, not a state function. CORRECT APPROACH: Always flip the sign when reversing an equation.

2. MISTAKE: Mixing up units (J vs. kJ, °C vs. K). WHY IT HAPPENS: Rushing calculations without checking units. CORRECT APPROACH: Convert all units to kJ and Kelvin before plugging into ΔG = ΔH – TΔS.

3. MISTAKE: Ignoring coefficients in entropy calculations. WHY IT HAPPENS: Students forget to multiply S⦵ by the number of moles. CORRECT APPROACH: Always write ΣS⦵(products) – ΣS⦵(reactants) with coefficients.

4. MISTAKE: Using bond enthalpies for lattice enthalpy in Born-Haber. WHY IT HAPPENS: Confusing bond breaking (covalent) with lattice formation (ionic). CORRECT APPROACH: Lattice enthalpy is only for ionic compounds (e.g., NaCl, MgO).

5. MISTAKE: Assuming ΔG < 0 means "fast reaction." WHY IT HAPPENS: Confusing thermodynamics (spontaneity) with kinetics (rate). CORRECT APPROACH: ΔG tells you if a reaction happens, not how fast.

EXAM TRAPS

1. TRAP: Giving ΔH in kJ but ΔS in J K⁻¹. HOW TO SPOT IT: Units in the question don’t match the formula. HOW TO AVOID IT: Convert ΔS to kJ K⁻¹ before using ΔG = ΔH – TΔS.

2. TRAP: Born-Haber cycle with polyatomic ions (e.g., CO₃²⁻). HOW TO SPOT IT: The question mentions ions like NO₃⁻ or SO₄²⁻. HOW TO AVOID IT: Treat the polyatomic ion as a single species (no atomisation steps for it).

3. TRAP: Gibbs free energy question where ΔH and ΔS have the same sign. HOW TO SPOT IT: Both ΔH and ΔS are positive or both are negative. HOW TO AVOID IT: Calculate the temperature at which ΔG = 0 to find the spontaneity range.

1-MINUTE RECAP

"Listen up—this is your last-minute checklist for Energetics/Thermochemistry:
1. Hess’s Law: Rearrange equations like algebra. Reverse? Flip the sign. Multiply? Multiply ΔH.
2. Born-Haber: Draw the cycle in order—atomisation, ionisation, bond breaking, electron affinity, lattice. Missing a step? You’re wrong.
3. Entropy: Products minus reactants. Coefficients matter! Convert to kJ K⁻¹ for Gibbs.
4. Gibbs: ΔG = ΔH – TΔS. Units must match—kJ for ΔH, kJ K⁻¹ for ΔS. Negative ΔG? Spontaneous. Positive? Non-spontaneous.
5. Exam traps: Watch units, polyatomic ions, and same-sign ΔH/ΔS. Double-check every number. You’ve got this. Now go ace that exam."