Chemical Thermodynamics — vidhyapath 2027

Concept Core

From heat of reaction to spontaneity — the chemical thermodynamics framework.

First Law Applied to Chemistry: Enthalpy

For chemical reactions at constant pressure (most lab conditions), heat exchanged = enthalpy change ΔH:

ΔH = H_products − H_reactants

Exothermic: ΔH < 0 (heat released, products more stable). Endothermic: ΔH > 0 (heat absorbed).

Relation: ΔH = ΔU + ΔnᵍRT where Δnᵍ = moles of gaseous products − gaseous reactants.

Hess's Law

Enthalpy is a state function — ΔH is independent of the path. Therefore:

ΔH_reaction = Σ ΔH_f°(products) − Σ ΔH_f°(reactants)

You can add/subtract thermochemical equations to obtain ΔH for any target reaction. Standard enthalpy of formation ΔH_f° of an element in its standard state = 0.

Bond Enthalpies

ΔH can be estimated from bond energies:

ΔH ≈ Σ(bonds broken) − Σ(bonds formed)

Breaking bonds requires energy (+ve). Forming bonds releases energy (−ve). If bonds formed are stronger than broken: exothermic overall.

Entropy (S)

Entropy = measure of disorder/randomness. Unit: J/mol/K.

ΔS > 0: more disordered (gases, dissolved, more particles, higher T)

ΔS < 0: more ordered (fewer particles, solid, lower T)

For a reaction: ΔS° = Σ S°(products) − Σ S°(reactants)

2nd Law: ΔS_universe = ΔS_system + ΔS_surroundings ≥ 0

Gibbs Free Energy — Spontaneity Criterion

At constant T and P, the criterion for spontaneity:

ΔG = ΔH − TΔS

ΔG < 0: spontaneous. ΔG > 0: non-spontaneous. ΔG = 0: equilibrium.

ΔH	ΔS	Spontaneous?
−	+	Always
+	−	Never
−	−	Low T only
+	+	High T only

Standard Gibbs Energy & Equilibrium

At equilibrium: ΔG = 0. The standard Gibbs energy relates to the equilibrium constant:

ΔG° = −RT ln K = −2.303 RT log K

K > 1 → ΔG° < 0 → forward reaction favoured at standard conditions. K < 1 → ΔG° > 0 → reverse reaction favoured.

Formula Vault

Chemical thermodynamics formulas for EAPCET.

Enthalpy Change

ΔH = H_prod − H_react

<0 exothermic; >0 endothermic

ΔH vs ΔU

ΔH = ΔU + Δnᵍ RT

Δnᵍ = gaseous Δmoles

Hess's Law

ΔH_rxn = ΣΔH_f°(prod) − ΣΔH_f°(react)

State function; path independent

Bond Enthalpy Method

ΔH ≈ Σ(broken) − Σ(formed)

Average bond energies

Gibbs Free Energy

ΔG = ΔH − TΔS

ΔG<0: spontaneous

Standard ΔG°

ΔG° = −RT ln K

R = 8.314 J/mol/K; T in Kelvin

Entropy of Universe

ΔS_univ = ΔS_sys + ΔS_surr ≥ 0

Spontaneous: ΔS_univ > 0

Kirchhoff's Law

ΔH_T2 = ΔH_T1 + ΔCₚ(T₂−T₁)

ΔH temperature dependence

Worked Examples

5 problems — Hess's law, Gibbs energy, bond enthalpy, and spontaneity.

EasyFind ΔG given ΔH = −40 kJ, ΔS = −100 J/K, T = 300 K▾

Calculate ΔG for a reaction with ΔH = −40 kJ/mol, ΔS = −100 J/K/mol at T = 300 K.

Convert ΔS to kJ: ΔS = −0.1 kJ/K/mol

ΔG = ΔH − TΔS = −40 − (300)(−0.1) = −40 + 30 = −10 kJ/mol

ΔG < 0 → spontaneous at 300 K.

✓ ΔG = −10 kJ/mol (spontaneous)

EasyPredict spontaneity: ΔH = −ve, ΔS = +ve▾

A reaction has ΔH = −200 kJ/mol (exothermic) and ΔS = +150 J/K/mol. Predict spontaneity.

ΔG = ΔH − TΔS. With ΔH negative and ΔS positive:

ΔG = (−) − T(+) = negative − positive = always negative

Reaction is spontaneous at ALL temperatures.

✓ Spontaneous at all temperatures

MediumUse Hess's Law: find ΔH for C + ½O₂ → CO▾

Given: (1) C + O₂ → CO₂, ΔH₁ = −393 kJ; (2) CO + ½O₂ → CO₂, ΔH₂ = −283 kJ. Find ΔH for C + ½O₂ → CO.

Target: C + ½O₂ → CO.

Use equation (1) forward: C + O₂ → CO₂ [ΔH₁ = −393]

Reverse equation (2): CO₂ → CO + ½O₂ [ΔH = +283]

Add: C + O₂ + CO₂ → CO₂ + CO + ½O₂ → simplify: C + ½O₂ → CO

ΔH = −393 + 283 = −110 kJ/mol

✓ ΔH(C + ½O₂ → CO) = −110 kJ/mol

EAPCET LevelFind ΔH of combustion of CH₄ from bond energies▾

Find ΔH for CH₄ + 2O₂ → CO₂ + 2H₂O using bond energies: C−H=414, O=O=498, C=O=804, O−H=460 kJ/mol.

Bonds broken: 4(C−H) + 2(O=O) = 4(414) + 2(498) = 1656 + 996 = 2652 kJ

Bonds formed: 2(C=O in CO₂) + 4(O−H in 2H₂O) = 2(804) + 4(460) = 1608 + 1840 = 3448 kJ

ΔH = Bonds broken − Bonds formed = 2652 − 3448 = −796 kJ/mol

✓ ΔH combustion ≈ −796 kJ/mol (exothermic)

Trap QuestionΔH < 0 guarantees spontaneity — True or False?▾

A student claims: 'Since ΔH < 0 (exothermic), the reaction is always spontaneous.' Is this correct?

The trap: Many reactions are exothermic but non-spontaneous at high temperatures.

Spontaneity is determined by ΔG = ΔH − TΔS, not by ΔH alone.

If ΔH < 0 and ΔS < 0: ΔG = (−) − T(−). At high T, the TΔS term dominates → ΔG becomes positive → non-spontaneous.

Example: 2H₂O(g) → 2H₂ + O₂ is endothermic but becomes spontaneous at very high temperatures.

✓ False — ΔG = ΔH − TΔS determines spontaneity. ΔH < 0 alone is not sufficient at high temperatures.

Mistake DNA

4 chemical thermodynamics errors that EAPCET distractors exploit.

🔢

ΔH vs ΔU — Forgetting the Δnᵍ RT Correction

ΔH and ΔU are only equal when Δnᵍ = 0. For reactions changing moles of gas, they differ by Δnᵍ RT.

❌ Wrong

N₂ + 3H₂ → 2NH₃ ΔH = ΔU (ignoring gas mole change) ✗

✓ Correct

Δnᵍ = 2 − (1+3) = −2 ΔH = ΔU + (−2)RT ✓ ΔH < ΔU here

ΔH = ΔU + ΔnᵍRT where Δnᵍ = moles of gaseous products − moles of gaseous reactants. Check sign carefully.

🌡️

Celsius in ΔG = ΔH − TΔS

T must be in Kelvin in all thermodynamics equations. Using °C gives wrong ΔG.

❌ Wrong

At 27°C: ΔG = ΔH − 27ΔS ✗ (must use 300 K)

✓ Correct

T = 27 + 273 = 300 K ΔG = ΔH − 300ΔS ✓ Always Kelvin

Every thermodynamics formula involving T requires Kelvin. ΔG = ΔH − TΔS, ΔG° = −RT ln K — all need T in K.

➕

Hess's Law: Wrong Sign When Reversing an Equation

When reversing a thermochemical equation, the sign of ΔH must be flipped.

❌ Wrong

Reverse: A → B ΔH=−100 Become: B → A ΔH=−100 ✗ (sign NOT flipped)

✓ Correct

Reverse: B → A ΔH=+100 ✓ Forward exothermic → Reverse endothermic

Enthalpy is path-independent but direction-dependent. If forming compound A releases heat, decomposing it absorbs the same amount.

🎯

Confusing ΔS > 0 with Spontaneity

ΔS > 0 for a system doesn't guarantee spontaneity — the universe's entropy must increase, not just the system's.

❌ Wrong

ΔS_system = +50 J/K → spontaneous ✗ (only ΔS_universe determines this)

✓ Correct

ΔS_univ = ΔS_sys + ΔS_surr ✓ ΔS_surr = −ΔH/T ✓ If ΔS_univ > 0: spontaneous

ΔS_surroundings = −ΔH_system/T. For exothermic reactions, surroundings gain entropy even if system loses it. ΔG < 0 is the simplest combined criterion.

Chapter Intelligence

Chemical Thermodynamics is tested in both Physics and Chemistry — master it once, use it twice.

EAPCET Weightage (2019–2024)

Gibbs free energy / spontaneity

Hess's law calculations

ΔH, ΔU relationship

Bond enthalpy method

ΔG° and equilibrium K

High-Yield PYQ Patterns

Predict spontaneity from ΔH, ΔSHess's law — 3 equationsΔH = ΔU + ΔnᵍRT calculationBond enthalpy ΔH estimateTemperature for ΔG = 0 (equilibrium)ΔG° from equilibrium constant

Exam Strategy

Spontaneity table (ΔH, ΔS combinations) is the most tested concept. Memorise the 4 cases — especially the temperature-dependent ones (−,−) and (+,+).
Hess's Law: write the target equation, then manipulate given equations (reverse, multiply) to add up to the target. Flip the sign when reversing.
ΔH = ΔU when Δnᵍ = 0 (reactions with no gaseous species or equal moles of gas on both sides).
Temperature for non-spontaneous to become spontaneous: set ΔG = 0 → T = ΔH/ΔS (both in same units). This is the transition temperature.
ΔH_f° for elements in standard state = 0. This is used to compute ΔH° of reactions using Hess's Law.