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IITJEE CHEMISTRY: Chemical Thermodynamics, Born-Haber Cycle & Ellingham Diagram Visualizer

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Ashish Kumar Bahekar

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Apex Class • IITJEE Physical Chemistry Lab

Chemical Thermodynamics, Born-Haber Cycle & Ellingham Diagram

Master Chemistry 1st Law Sign Conventions, Gibbs Spontaneity (ΔG), Bomb Calorimetry (ΔU), and Lattice Enthalpy!

🔬 Heat (q) & Work (W) Transfers. Select a Concept:
Work ON Gas System Heat IN 🚨 IUPAC Chemistry Sign Convention (vs. Physics!) ΔU = q + w (Work formula: w = -Pext ΔV) • Work done ON the system (Compression ΔV < 0): w is POSITIVE (+)! Energy enters the gas! • Heat absorbed BY the system (Endothermic): q is POSITIVE (+)! 1. Bomb Calorimeter (Rigid Steel) Ignition ΔV = 0 Constant Volume (w = 0) Heat Measured: qv qv = ΔU Measures Internal Energy! 2. Open Coffee Cup (Atmospheric Pressure) Solution ΔP = 0 Constant Pressure Heat Measured: qp qp = ΔH ΔH = ΔU + Δng R T Born-Haber Cycle for NaCl(s) Formation Na(s) + ½Cl₂(g) ΔHsub Na(g) ½ΔHdiss Cl(g) IE₁ Na⁺(g) + Cl(g) + e⁻ EA₁ (Exo!) Cl⁻(g) Lattice ΔH NaCl(s) Hess’s Law Application Total ΔHf = Sum of all individual steps! ΔHf = ΔHsub + IE₁ + ½ΔHdiss + EA₁ + ΔHlattice • Electron Gain Enthalpy (EA₁) is usually Negative (Exo). • Lattice Formation is highly Negative (Provides stability). Bond Enthalpy Rule: The Only Time It’s “Reactants Minus Products” ΔHrxn = Σ(Bonds Broken in Reactants) – Σ(Bonds Formed in Products) • Why? Breaking bonds REQUIRES energy (Endothermic, +). • Forming bonds RELEASES energy (Exothermic, -). Example CH₄ + 2O₂ → CO₂ + 2H₂O: ΔH = [4(C-H) + 2(O=O)] – [2(C=O) + 4(O-H)]
Extensive Properties
Mass, Volume, Enthalpy (H)
Depends on quantity of matter
Intensive Properties
Temp (T), Pressure (P), Density
Independent of quantity
State Functions
U, H, S, G, T, P, V
Path Independent (Exact Differentials)
Path Functions
Heat (q) & Work (w)
Depend on the route taken!
🌌 ΔG = ΔH – TΔS. Select a Second-Law Concept:
Entropy (S): Measure of Disordered Microstates Solid (Smin) +ΔS Liquid ++ΔS Gas (Smax) Thermodynamic Def: ΔS = qrev / T Phase Change (T constant): ΔSfus = ΔHfus / Tm ΔSvap = ΔHvap / Tb Exothermic (ΔH < 0) Endothermic (ΔH > 0) More Disordered (ΔS > 0) Less Disordered (ΔS < 0) ΔG < 0 ALWAYS Spontaneous at ALL Temperatures ΔG < 0 at HIGH T Entropy-driven (Needs TΔS > ΔH) ΔG < 0 at LOW T Enthalpy-driven (Needs |ΔH| > |TΔS|) ΔG > 0 ALWAYS Non-Spontaneous at ALL Temperatures Relationship Between ΔG° and Equilibrium Constant K ΔG° = – R T ln K = -2.303 R T log₁₀ K • If K > 1 (Products Favored): ln K is POSITIVE ⇒ ΔG° < 0 (Exergonic) • If K < 1 (Reactants Favored): ln K is NEGATIVE ⇒ ΔG° > 0 (Endergonic) At Equilibrium, the NON-standard free energy change ΔG = 0! ΔG° Temperature (T) → 0 2M + O₂ → 2MO 2C + O₂ → 2CO Tintersect Ellingham Diagram Rules • Plots ΔG° of formation of oxides vs T. • Most metals slope UP (ΔS < 0, consuming O₂). • Carbon to CO slopes DOWN (ΔS > 0). 🏆 Golden Rule of Reduction: An element lower on the graph can reduce the oxide of an element above it!
2nd Law of Thermodynamics
ΔSuniverse > 0
For any spontaneous process
Gibbs Free Energy Def
G = H – T S
Useful non-expansion work capacity
Entropy of Vaporization
Trouton’s Rule ≈ 85 J/mol·K
For many non-polar liquids
Residual Entropy (At 0 K)
S = kB ln(W)
W = number of microstates (e.g. CO)
🔥 High-Yield JEE Thermo Trap:
In an Isothermal Reversible Expansion of an ideal gas, internal energy change ΔU = 0, so Heat q = -w (Work done BY gas = Heat absorbed). But in Free Expansion (Vacuum), external pressure is zero, so w = 0, q = 0, and ΔU = 0 (Isothermal)!
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