Chemistry Physical How to Solve: Thermodynamics & Thermochemistry (ΔH, ΔS, ΔG, Hess’s Law, Born-Haber Cycle) – IIT JEE Guide

How to Solve: Thermodynamics & Thermochemistry (ΔH, ΔS, ΔG, Hess’s Law, Born-Haber Cycle) – IIT JEE Guide

Introduction

Mastering thermodynamics unlocks 10-15 marks in IIT JEE (Main + Advanced)—enough to push you from a 90 to a 100+ percentile. It also explains why ice melts, why batteries work, and how rockets launch—real-world chemistry you’ll use in engineering.

WHAT YOU NEED TO KNOW FIRST

1. First Law of Thermodynamics – Energy is conserved (ΔU = q + w).
2. Enthalpy (H) – Heat content at constant pressure (H = U + PV).
3. Standard States – 1 atm pressure, 298 K temperature, 1 M concentration.

(If you’re shaky on these, pause and review them first.)

KEY TERMS & FORMULAS

1. Enthalpy Change (ΔH)

• Formula: ΔH = H_products – H_reactants
• Meaning:
• ΔH = heat absorbed/released at constant pressure.
• ΔH < 0 → Exothermic (heat released).
• ΔH > 0 → Endothermic (heat absorbed).
• MEMORISE THIS: ΔH°_reaction = Σ ΔH°_{f (products)} – Σ ΔH°_{f (reactants)}

2. Entropy Change (ΔS)

• Formula: ΔS = S_products – S_reactants
• Meaning:
• ΔS = measure of disorder.
• ΔS > 0 → More disorder (favored at high T).
• ΔS < 0 → Less disorder.
• MEMORISE THIS: ΔS°_reaction = Σ S°_(products) – Σ S°_(reactants)

3. Gibbs Free Energy (ΔG)

• Formula: ΔG = ΔH – TΔS
• Meaning:
• ΔG = spontaneity predictor.
• ΔG < 0 → Spontaneous (favored).
• ΔG = 0 → Equilibrium.
• ΔG > 0 → Non-spontaneous.
• MEMORISE THIS: ΔG° = –RT ln K (for equilibrium constant K).

4. Hess’s Law

• Statement: ΔH for a reaction is independent of path—only depends on initial and final states.
• How to use:
• Flip reactions → Change ΔH sign.
• Multiply reactions → Multiply ΔH by same factor.
• Add reactions → Add ΔH values.

5. Born-Haber Cycle

• Purpose: Calculates lattice energy of ionic compounds.
• Steps (for NaCl):
• Sublimation of Na(s) → Na(g) (ΔH_sub).
• Ionization of Na(g) → Na⁺(g) + e⁻ (IE).
• Dissociation of Cl₂(g) → 2Cl(g) (ΔH_diss).
• Electron affinity of Cl(g) → Cl⁻(g) (EA).
• Formation of NaCl(s) → Na⁺(g) + Cl⁻(g) → NaCl(s) (ΔH_lattice).
• MEMORISE THIS: ΔH_f = ΔH_sub + IE + ½ ΔH_diss + EA + ΔH_lattice

STEP-BY-STEP METHOD

Step 1: Identify What’s Given & What’s Asked

• Given: ΔH, ΔS, ΔG values, reactions, or Born-Haber data.
• Asked: Calculate ΔH, ΔS, ΔG, or lattice energy.

Step 2: Choose the Right Formula

• ΔH? → Use Hess’s Law or ΔH°_f formula.
• ΔS? → Use ΔS° = Σ S°_(products) – Σ S°_(reactants).
• ΔG? → Use ΔG = ΔH – TΔS or ΔG° = –RT ln K.
• Lattice energy? → Use Born-Haber cycle.

Step 3: Manipulate Reactions (Hess’s Law)

• Flip reactions? → Reverse ΔH sign.
• Multiply reactions? → Multiply ΔH by same factor.
• Add reactions? → Add ΔH values.

Step 4: Plug in Values & Solve

• Units matter! (kJ/mol, J/mol·K, kJ).
• Check signs! (Exothermic = negative ΔH).

Step 5: Interpret the Answer

• ΔG < 0? → Spontaneous.
• ΔH < 0? → Exothermic.
• ΔS > 0? → More disorder.

WORKED EXAMPLES

Example 1 – Basic (ΔH Calculation)

Question: Calculate ΔH for: C(s) + O₂(g) → CO₂(g) (ΔH = ?) Given: - C(s) + ½ O₂(g) → CO(g) (ΔH = –110 kJ/mol) - CO(g) + ½ O₂(g) → CO₂(g) (ΔH = –283 kJ/mol)

Solution:
1. Given reactions: - (1) C + ½ O₂ → CO (ΔH = –110 kJ) - (2) CO + ½ O₂ → CO₂ (ΔH = –283 kJ)
2. Add (1) + (2): C + ½ O₂ + CO + ½ O₂ → CO + CO₂ → C + O₂ → CO₂
3. ΔH = (–110) + (–283) = –393 kJ/mol

What we did and why: - Used Hess’s Law to combine reactions. - No flipping/multiplying needed—just added ΔH values.

Example 2 – Medium (ΔG Calculation)

Question: For the reaction: N₂(g) + 3H₂(g) → 2NH₃(g) ΔH = –92 kJ/mol, ΔS = –198 J/mol·K at 298 K. Is the reaction spontaneous?

Solution:
1. Convert ΔS to kJ: –198 J/mol·K = –0.198 kJ/mol·K.
2. Use ΔG = ΔH – TΔS: ΔG = (–92) – (298)(–0.198) ΔG = –92 + 59.004 = –32.996 kJ/mol
3. ΔG < 0 → Spontaneous.

What we did and why: - Converted units (J → kJ) to match ΔH. - Plugged into ΔG formula to check spontaneity.

Example 3 – Exam-Style (Born-Haber Cycle)

Question: Calculate the lattice energy of NaCl(s) given: - ΔH_f(NaCl) = –411 kJ/mol - ΔH_sub(Na) = +108 kJ/mol - IE(Na) = +496 kJ/mol - ΔH_diss(Cl₂) = +242 kJ/mol - EA(Cl) = –349 kJ/mol

Solution:
1. Write Born-Haber equation: ΔH_f = ΔH_sub + IE + ½ ΔH_diss + EA + ΔH_lattice
2. Plug in values: –411 = 108 + 496 + ½(242) + (–349) + ΔH_lattice
3. Simplify: –411 = 108 + 496 + 121 – 349 + ΔH_lattice –411 = 376 + ΔH_lattice
4. Solve for ΔH_lattice: ΔH_lattice = –411 – 376 = –787 kJ/mol

What we did and why: - Used Born-Haber cycle to isolate lattice energy. - Careful with signs (EA is negative).

COMMON MISTAKES

EXAM TRAPS

1-MINUTE RECAP

"Listen up—this is your last-minute thermo cheat sheet. Memorise these three things:
1. ΔG = ΔH – TΔS → Negative ΔG = spontaneous.
2. Hess’s Law → Flip reactions? Flip ΔH. Multiply? Multiply ΔH.
3. Born-Haber → Sublimation → Ionization → Dissociation → EA → Lattice. For ΔH, use standard formation values. For ΔS, subtract reactants from products. For ΔG, plug into the formula. And always—ALWAYS—check units. You’ve got this!