NCERT Plus One (Class 11) Chemistry • Chapter 5

Chemical Thermodynamics

Q: Classify the following properties into extensive and intensive: Enthalpy, Temperature, Heat capacity, Density.

Extensive Properties (depend on quantity of matter): Enthalpy, Heat capacity. Intensive Properties (independent of quantity of matter): Temperature, Density.

Q: What is the standard enthalpy of formation of an element in its reference state? Mention the reference states of Carbon and Sulphur.

By convention, Δ f H° of an element in its reference state is zero (0) . Reference state of Carbon: C (graphite) ; Reference state of Sulphur: S (rhombic) .

Q: Write the mathematical statement of the First Law of Thermodynamics. What is ΔU for an isolated system?

Mathematical statement: ΔU = q + w . For an isolated system (q = 0, w = 0), ΔU = 0 (energy remains constant).

Q: Derive the relationship between heat capacities C p and C V for one mole of an ideal gas.

At constant volume: q V = ΔU = C V ΔT. At constant pressure: q p = ΔH = C p ΔT. For 1 mole of an ideal gas (pV = RT): ΔH = ΔU + Δ(pV) = ΔU + RΔT. Substituting ΔH and ΔU: C p ΔT = C V ΔT + RΔT ⇒ C p − C V = R .

Q: State Hess's Law of Constant Heat Summation. Illustrate how it helps find Δ f H° of CO(g).

Statement: If a reaction takes place in several steps, its standard reaction enthalpy is the sum of the standard enthalpies of the intermediate reactions at the same temperature (Δ r H = Δ r H 1 + Δ r H 2 + ...). Illustration: From C(s) + O 2 (g) → CO 2 (g) (ΔH 1 = −393.5 kJ) and reversing CO(g) + ½O 2 (g) → CO 2 (g) (ΔH 2 = −283.0 kJ), we get Δ f H°(CO) = −393.5 + 283.0 = −110.5 kJ mol -1 .

Q: Write the Gibbs equation. Predict the signs of ΔH and ΔS for a reaction that is: (i) spontaneous at all temperatures, (ii) spontaneous only at high temperatures.

Gibbs equation: ΔG = ΔH − TΔS . (i) Spontaneous at ALL temperatures: ΔH 0 (+) . (ii) Spontaneous only at HIGH temperatures: ΔH > 0 (+) and ΔS > 0 (+) (so TΔS > ΔH at high T).

Q: State the Second and Third Laws of Thermodynamics. Predict the entropy change when a crystalline solid is heated from 0 K to 115 K.

Second Law: For any spontaneous process in an isolated system, entropy increases (ΔS total = ΔS sys + ΔS surr > 0). Third Law: The entropy of any pure crystalline substance approaches zero as temperature approaches absolute zero (0 K). When heated from 0 K to 115 K, particles oscillate about lattice sites and disorder increases, so ΔS increases (> 0) .

Systems, First Law, PV Work, Enthalpy, Calorimetry, Hess's Law, Born-Haber Cycle, Entropy & Gibbs Energy

📄 Official NCERT Textbook PDF

Topic 5.1 & 5.2.2(b): Thermodynamic Systems, State vs Path Functions, Extensive vs Intensive & Internal Energy (U)

Picture three cups of hot Sulaimani tea on your study table. Pour one into an open cup: steam escapes into the air and it cools down—both matter and heat leave (an open system). Seal the second inside a steel water bottle: no steam escapes, yet the steel warms your hand—only energy crosses the wall (a closed system). Pour the third into a vacuum thermos flask: neither steam nor heat escapes—that is an isolated system!

Thermodynamics studies energy transformations in macroscopic systems at equilibrium. We divide the universe into two parts: Universe = System + Surroundings, separated by a real or imaginary boundary.

1. Three Types of Thermodynamic Systems

System Type	Matter Exchange	Energy Exchange	NCERT Example
Open System	Yes ✓	Yes ✓	Reactants in an open beaker
Closed System	No ✗	Yes ✓	Reactants in a closed conducting steel/copper vessel
Isolated System	No ✗	No ✗	Reactants in a thermos flask or closed insulated vessel

2. State Functions vs Path Functions

State Variables / State Functions: Depend ONLY on initial and final states, NOT on the path taken (e.g., Pressure p, Volume V, Temperature T, Internal Energy U, Enthalpy H, Entropy S, Gibbs Energy G). Like the height of water in a pond—it doesn't matter whether rain or a tube-well filled it!
Path Functions: Depend on how the change is carried out—Heat (q) and Work (w). Yet remarkably, their sum q + w = ΔU IS a state function!

3. Extensive vs Intensive Properties

Extensive Properties: Depend on the quantity/size of matter (Mass, Volume V, Internal Energy U, Enthalpy H, Heat Capacity C, Entropy S, Gibbs Energy G). Partition a box in half and each half has V/2!
Intensive Properties: Independent of sample size (Temperature T, Pressure p, Density, Molar volume V_m, Molar heat capacity C_m, Concentration).

4. Internal Energy (U) as a State Function

The sum of all chemical, electrical, kinetic, and mechanical energies inside a system is its Internal Energy (U). J. P. Joule (1840–50) proved using an insulated (adiabatic) vessel that doing 1 kJ of mechanical paddle work or 1 kJ of electrical work produces the exact same temperature rise: ΔU = U₂ − U₁ = w_ad.

IUPAC Sign Conventions:
• q > 0: Heat absorbed BY system (endothermic) | q < 0: Heat evolved BY system
• w > 0: Work done ON system (compression) | w < 0: Work done BY system (expansion)

Dividing two extensive properties always yields an intensive property! For example, Mass / Volume = Density (intensive), and Heat Capacity / Moles = Molar Heat Capacity C_m (intensive). Also remember IUPAC chemistry gives +w when work is done ON the system (ΔU = q + w), opposite to older physics textbooks!

Topic 5.1.4 – 5.2.2(a): First Law (ΔU = q + w), Reversible/Irreversible PV Work & Enthalpy (H = U + pV)

Imagine lifting a heavy weight on a gas piston. If you yank away one big brick suddenly, the gas bursts outward in one rapid jump (irreversible)—doing modest work. Instead, if you remove tiny grains of sand one at a time, the gas pushes gently in near-equilibrium at every step (reversible)—and extracts the maximum possible work!

1. First Law of Thermodynamics

ΔU = q + w (Isolated system: q = 0, w = 0 ⇒ ΔU = 0: Energy of an isolated system is constant)

We cannot measure the absolute value of U, only its change ΔU. For adiabatic walls (q = 0), ΔU = w_ad. At constant volume (ΔV = 0), w = 0 so ΔU = q_V.

2. Pressure-Volume (PV) Work: Irreversible vs Reversible

Irreversible Constant External Pressure: w = −p_ex ΔV = −p_ex(V_f − V_i).
Free Expansion (p_ex = 0, vacuum): w = 0 always! For isothermal free expansion of an ideal gas, Joule showed q = 0 and ΔU = 0.
Reversible Process: Proceeds infinitely slowly through equilibrium states (p_ex = p_in ± dp) so it can be reversed at any instant by an infinitesimal change.

Isothermal Reversible Ideal Gas Work (ΔU = 0, q = −w):
w_rev = −nRT ln(V_f / V_i) = −2.303 nRT log(V_f / V_i) = −2.303 nRT log(p_i / p_f)

NCERT Problems 5.2–5.4: 2 L ideal gas expands isothermally at 25 °C (298 K) to 10 L.
• Into vacuum (p_ex = 0): w = 0, q = 0.
• Against constant 1 atm: q = −w = p_ex(10 − 2) = 8 L·atm (w = −8 L·atm).
• 1 mol reversibly: q = −w = 2.303 × 1 × 0.08206 × 298 × log(10/2) = 39.37 L·atm (Maximum work!).

3. Enthalpy (H) & Relation Between ΔH and ΔU

Most lab reactions occur in open flasks at constant atmospheric pressure, where expansion work occurs. Defining Enthalpy: H = U + pV (state function, extensive) gives ΔH = ΔU + pΔV = q_p (heat at constant pressure). ΔH < 0 for exothermic reactions; ΔH > 0 for endothermic reactions.

Gaseous Reaction Bridge: pΔV = Δn_gRT ⇒ ΔH = ΔU + Δn_gRT
where Δn_g = (moles of gaseous products) − (moles of gaseous reactants)

NCERT Problem 5.5: Vaporisation of 1 mol H₂O(l) → H₂O(g) at 1 bar, 100 °C (373 K) has ΔH = 41.00 kJ mol^-1.
Here Δn_g = 1 − 0 = 1. Thus ΔU = ΔH − Δn_gRT = 41.00 − (1)(8.3 × 10^-3)(373) = 41.00 − 3.096 = 37.90 kJ mol^-1.

In ΔH = ΔU + Δn_gRT, count ONLY gaseous species (g)—NEVER include solids (s) or liquids (l)! Also convert R = 8.314 J K^-1 mol^-1 = 8.314 × 10^-3 kJ K^-1 mol^-1 so your units match kJ!

Topic 5.2.2(c)–(d) & 5.3: Heat Capacity (C_p − C_V = R) & Calorimetry (Bomb & Constant-Pressure)

Why does coastal Kerala stay pleasantly breezy while desert sand turns scorching hot by noon? Water has an unusually huge heat capacity—it absorbs vast heat with only a tiny temperature rise! Meanwhile, food scientists burn cashews inside a sealed steel Bomb Calorimeter submerged in water to measure exact food energy (ΔU) without losing a single joule.

1. Heat Capacity, Specific Heat & Molar Heat Capacity

Heat Capacity (C): Heat needed to raise a system's temperature by 1 K (or 1 °C): q = C ΔT (extensive).
Molar Heat Capacity (C_m = C / n): Heat needed for 1 mole per kelvin rise (intensive, J mol^-1 K^-1).
Specific Heat Capacity (c): Heat needed for 1 unit mass (1 g or 1 kg) per kelvin rise: q = m × c × ΔT.

2. Relationship Between C_p and C_V for an Ideal Gas

At constant volume, q_V = C_VΔT = ΔU. At constant pressure, q_p = C_pΔT = ΔH. For 1 mole of an ideal gas:

ΔH = ΔU + Δ(pV) = ΔU + Δ(RT) = ΔU + RΔT
C_pΔT = C_VΔT + RΔT ⇒ C_p − C_V = R (for n moles: C_p − C_V = nR)

Why is C_p > C_V? At constant pressure, part of the supplied heat does expansion work pushing back the atmosphere, so extra heat (R per mole per kelvin) is needed for the same 1 K temperature rise!

3. Calorimetry: Measuring ΔU and ΔH Experimentally

Bomb Calorimeter (Constant Volume → ΔU): Sample burns in pure O₂ inside a sealed steel bomb immersed in a known water bath. Since ΔV = 0, w = 0 and q_V = ΔU!
Constant-Pressure Calorimeter (Atmospheric → ΔH): Reaction occurs in an open insulated vessel; heat measured equals q_p = ΔH.

NCERT Problem 5.6: 1 g of graphite is burnt in a bomb calorimeter (C_V = 20.7 kJ K^-1), raising temperature from 298 K to 299 K.
• Heat released by 1 g graphite: q = −C_V × ΔT = −20.7 × (299 − 298) = −20.7 kJ.
• For 1 mole (12 g) graphite: ΔU = −20.7 × 12 = −248.4 kJ mol^-1 ≈ −2.48 × 10² kJ mol^-1.
• Reaction: C(graphite) + O₂(g) → CO₂(g) has Δn_g = 1 − 1 = 0 ⇒ ΔH = ΔU = −2.48 × 10² kJ mol^-1.

In Bomb Calorimeter problems, water and the bomb absorb heat (+C_VΔT), so the reaction's energy change gets the NEGATIVE sign (−C_VΔT) for exothermic combustion! And always scale from grams burnt (e.g. 1 g) to 1 mole (12 g for C) before reporting kJ mol^-1!

Topic 5.4 & 5.5(a)–(c): Reaction Enthalpies: Phase Changes, Formation (Δ_fH°), Combustion, Atomization & Bond Enthalpy

Step out of a swimming pool with a thin 18 g film of water on your skin and you instantly shiver—because vaporising 1 mole of water steals 44.01 kJ of heat directly from your body! Every physical transition and chemical reaction carries a characteristic enthalpy price tag measured against standard conditions (1 bar, 298 K).

Reaction Enthalpy: Δ_rH = Σ a_i H_m(Products) − Σ b_i H_m(Reactants). Under standard state (pure substance at 1 bar and specified T, usually 298 K), it is written Δ_rH°.

1. Phase Transformation Enthalpies (Always Endothermic, +)

Standard Enthalpy of Fusion (Δ_fusH°): 1 mol solid → liquid at melting point (H₂O: +6.00 kJ mol^-1 at 273 K).
Standard Enthalpy of Vaporisation (Δ_vapH°): 1 mol liquid → vapour at boiling point (H₂O: +40.79 kJ mol^-1 at 373 K; +44.01 kJ mol^-1 at 298 K).
Standard Enthalpy of Sublimation (Δ_subH°): 1 mol solid → vapour directly (Dry ice CO₂: 25.2 kJ mol^-1; Naphthalene: 73.0 kJ mol^-1). Note: Δ_subH° = Δ_fusH° + Δ_vapH°.

2. Standard Enthalpy of Formation (Δ_fH°)

Enthalpy change when 1 mole of a compound is formed from its elements in their most stable reference states at 1 bar & T (e.g., H₂(g), O₂(g), C(graphite), S(rhombic), Br₂(l)). By convention, Δ_fH° = 0 for every element in its reference state!

Δ_rH° = Σ a_i Δ_fH°(Products) − Σ b_i Δ_fH°(Reactants)

Thermochemical Equation Rules: (1) Coefficients represent moles, never molecules, and physical states must be written. (2) Multiplying an equation multiplies ΔH (extensive). (3) Reversing an equation flips the sign of ΔH.
NCERT Example: CaCO₃(s) → CaO(s) + CO₂(g) ⇒ Δ_rH° = (−635.1) + (−393.5) − (−1206.9) = +178.3 kJ mol^-1.

3. Combustion, Atomization & Bond Enthalpies

Enthalpy of Combustion (Δ_cH°): Complete combustion of 1 mole of substance in excess O₂ (always exothermic: butane C₄H₁₀ = −2658 kJ mol^-1; glucose = −2802 kJ mol^-1).
Enthalpy of Atomization (Δ_aH°): Breaking 1 mole of bonds completely to yield gaseous atoms (H₂(g) → 2H(g): +435.0 kJ mol^-1; Na(s) → Na(g): +108.4 kJ mol^-1).
Bond Enthalpy (Δ_bondH°): In polyatomic molecules like CH₄, successive C–H bonds differ (427, 439, 452, 347 kJ mol^-1), so we use Mean Bond Enthalpy = 1665 / 4 = 416 kJ mol^-1.

Gas-Phase Reaction Enthalpy from Bond Enthalpies:
Δ_rH° ≈ Σ Bond Enthalpies(Reactants) − Σ Bond Enthalpies(Products)

Watch the OPPOSITE subtraction order!
• Using Enthalpies of Formation: Products MINUS Reactants.
• Using Bond Enthalpies: Reactants MINUS Products!
Also check stoichiometry: H₂ + Br₂ → 2 HBr gives 2 × Δ_fH°(HBr), and CaO + CO₂ → CaCO₃ is NOT Δ_fH° because CaO and CO₂ are compounds, not elements!

Topic 5.4(e): Hess's Law of Constant Heat Summation & Thermochemical Algebra

Imagine climbing from Kochi (sea level) to Munnar (1,600 m). Whether you helicopter straight to the top or stop for tea at Neriamangalam and Adimali along the winding ghat road, your net gain in altitude is identical! Because Enthalpy H is a state function, Germain Hess realized chemical reactions obey the exact same altitude rule.

Hess's Law of Constant Heat Summation

Statement: If a reaction takes place in several steps, then its standard reaction enthalpy is the algebraic sum of the standard enthalpies of the intermediate reactions into which the overall reaction may be divided at the same temperature.

Direct Route (A → B): Δ_rH = Δ_rH₁ + Δ_rH₂ + Δ_rH₃ + ...

Why Do Chemists Need Hess's Law?

Many reactions cannot be measured directly in a calorimeter—for example, burning graphite in limited O₂ always forms some CO₂ alongside CO, making direct measurement of Δ_fH°(CO) impossible!

NCERT Classic (Formation of CO): Target: C(graphite) + ½O₂(g) → CO(g)
Given: (i) C(graphite) + O₂(g) → CO₂(g); Δ_rH° = −393.5 kJ mol^-1
(ii) CO(g) + ½O₂(g) → CO₂(g); Δ_rH° = −283.0 kJ mol^-1
Reverse (ii) to place CO(g) on the right: CO₂(g) → CO(g) + ½O₂(g); Δ_rH° = +283.0 kJ mol^-1.
Add (i) + reversed (ii) so CO₂ cancels: Δ_fH°(CO) = −393.5 + 283.0 = −110.5 kJ mol^-1!

NCERT Problem 5.9 (Formation of Benzene): Target: 6 C(graphite) + 3 H₂(g) → C₆H₆(l)
• 6 × [C + O₂ → CO₂] = 6 × (−393.5) = −2361.0 kJ
• 3 × [H₂ + ½O₂ → H₂O(l)] = 3 × (−285.83) = −857.49 kJ
• Reverse C₆H₆ combustion: 6CO₂ + 3H₂O → C₆H₆ + ¹⁵/₂O₂ = +3267.0 kJ
• Add all three: Δ_fH°(C₆H₆, l) = −3218.49 + 3267.0 = +48.51 kJ mol^-1.

When manipulating intermediate equations, ALWAYS apply both actions to ΔH too: if you multiply an equation by 3, multiply ΔH by 3; if you reverse (flip) an equation, flip the sign of ΔH! Also check that physical states like H₂O(l) vs H₂O(g) cancel identically.

Topic 5.5(d)–(f): Lattice Enthalpy, Born-Haber Cycle (NaCl), Enthalpy of Solution & Dilution

How much energy holds table salt crystals together? Nobody can pull one mole of solid NaCl cleanly apart into isolated gaseous Na⁺ and Cl⁻ ions in a calorimeter! Instead, Max Born and Fritz Haber applied Hess's Law as a 5-step vertical energy ladder—unlocking Lattice Enthalpy and explaining why table salt barely changes temperature when stirred into water.

1. Lattice Enthalpy (Δ_latticeH°)

The enthalpy change when 1 mole of an ionic compound dissociates into its ions in the gaseous state:

Na⁺Cl⁻(s) → Na⁺(g) + Cl⁻(g); Δ_latticeH° = +788 kJ mol^-1

2. Born-Haber Cycle for NaCl(s)

Equating the direct formation of NaCl(s) (Δ_fH° = −411.2 kJ mol^-1) to a 5-step Hess pathway:

Step	Process	Enthalpy Term	Value (kJ mol^-1)
1	Sublimation: Na(s) → Na(g)	Δ_subH°	+108.4
2	Ionization: Na(g) → Na⁺(g) + e⁻	Δ_iH°	+496.0
3	Dissociation: ½Cl₂(g) → Cl(g)	½ Δ_bondH°	+121.0
4	Electron Gain: Cl(g) + e⁻ → Cl⁻(g)	Δ_egH°	−348.6
5	Lattice Formation: Na⁺(g) + Cl⁻(g) → NaCl(s)	−Δ_latticeH°	−U

Δ_fH° = Δ_subH° + Δ_iH° + ½Δ_bondH° + Δ_egH° − Δ_latticeH°
⇒ Δ_latticeH° = 108.4 + 496 + 121 − 348.6 − (−411.2) = +788 kJ mol^-1

3. Enthalpy of Solution (Δ_solH°) & Hydration (Δ_hydH°)

Dissolving an ionic salt happens in two conceptual steps: breaking the crystal lattice (+Δ_latticeH°) and hydrating the gaseous ions in water (−exothermic Δ_hydH°):

Δ_solH° = Δ_latticeH° + Δ_hydH° | For NaCl: (+788) + (−784) = +4 kJ mol^-1 (nearly zero heat change!)

Enthalpy of Dilution: Heat change when extra solvent is added to a solution (e.g., HCl·25 aq + 15 aq → HCl·40 aq, ΔH = −0.76 kJ mol^-1). As solvent → ∞, Δ_solH approaches the limiting value at infinite dilution.

Watch the sign of Lattice Enthalpy! NCERT defines Δ_latticeH° for dissociation (solid → gaseous ions) as +788 kJ mol^-1 (endothermic), whereas forming the solid crystal from gaseous ions releases −788 kJ mol^-1! Also note why many fluorides are less soluble than chlorides: very high Δ_latticeH° makes Δ_solH° strongly positive.

Topic 5.6(a)–(e): Spontaneity, Entropy (S), Second Law (ΔS_total > 0) & Third Law of Thermodynamics

Drop a spoon of ink into still water—it swirls outward until the whole glass turns tinted, never gathering back into one neat droplet on its own! Why do gases mix across a removed partition even when ΔH = 0, and why does ice melt spontaneously on a warm table despite being endothermic? Because nature is driven by Entropy (S)—the unstoppable spread of disorder!

1. What is a Spontaneous Process?

A process having the potential to proceed on its own without external assistance (irreversible; reversed only by external agency). Caution: Spontaneity tells us direction, NOT speed—H₂ and O₂ react spontaneously yet imperceptibly slowly at room temperature!

Why ΔH < 0 alone fails: Some endothermic processes (ΔH > 0, like ½N₂ + O₂ → NO₂ or ice melting) and zero-enthalpy processes (ΔH = 0, gas diffusion) are spontaneous!

2. Entropy (S) & Second Law of Thermodynamics

Entropy (S): A thermodynamic state function and extensive property measuring the degree of randomness or disorder. Order of entropy: S_gas >> S_liquid > S_solid.
Quantitative Definition: ΔS = q_rev / T (Unit: J K^-1 mol^-1). Heat added at lower T creates greater disorder than the same heat at higher T.
Second Law: For any spontaneous process, total entropy of the universe increases: ΔS_total = ΔS_system + ΔS_surroundings > 0. At equilibrium, entropy is maximum and ΔS_total = 0.

NCERT Problem 5.10 (Predict ΔS Sign):
(i) Liquid crystallises → solid: more ordered → ΔS < 0.
(ii) Crystalline solid heated from 0 K to 115 K: particles vibrate → ΔS > 0.
(iii) 2 NaHCO₃(s) → Na₂CO₃(s) + CO₂(g) + H₂O(g): gases formed → ΔS > 0.
(iv) H₂(g) → 2 H(g): moles of gas double → ΔS > 0.

NCERT Problem 5.11 (Why Rusting is Spontaneous despite ΔS_sys < 0):
4 Fe(s) + 3 O₂(g) → 2 Fe₂O₃(s) has ΔS_sys = −549.4 J K^-1 mol^-1, yet Δ_rH° = −1648 × 10³ J mol^-1 at 298 K.
ΔS_surr = −Δ_rH° / T = +1648000 / 298 = +5530 J K^-1 mol^-1.
⇒ ΔS_total = 5530 + (−549.4) = +4980.6 J K^-1 mol^-1 > 0 → Spontaneous!

3. Third Law of Thermodynamics

Statement: The entropy of any pure crystalline substance approaches zero as temperature approaches absolute zero (T → 0 K, S → 0). Note: Does NOT hold for solutions or supercooled liquids/glasses. Enables measurement of absolute entropies!

A negative ΔS_system does NOT mean a process is non-spontaneous! Always check ΔS_total = ΔS_sys + ΔS_surr > 0—exothermic reactions like iron rusting or water freezing below 0 °C release enough heat to boost ΔS_surr so ΔS_total stays positive!

Topic 5.6(c) & 5.7: Gibbs Energy (ΔG = ΔH − TΔS), Four Temperature Cases & Equilibrium (ΔG° = −RT ln K)

Checking surroundings entropy for every reaction in the universe is exhausting! Josiah Willard Gibbs invented one brilliant system-only referee—Gibbs Free Energy (G)—that pits energy release (ΔH) against disorder gain (TΔS) in a tug-of-war controlled by temperature T, and even tells us the exact equilibrium yield K!

1. Gibbs Energy & Spontaneity at Constant T and P

Define G = H − TS (state function, extensive). Since ΔS_total = ΔS_sys − ΔH_sys/T, multiplying by −T yields the legendary Gibbs Equation:

ΔG = ΔH − TΔS | • ΔG < 0: Spontaneous | • ΔG > 0: Non-spontaneous | • ΔG = 0: Equilibrium!

Physical Meaning: −ΔG represents the maximum useful (non-PV) work available from a reaction—hence called Free Energy.

2. Effect of Temperature on Spontaneity (NCERT Table 5.4)

Δ_rH°	Δ_rS°	Δ_rG° = ΔH° − TΔS°	Spontaneity Verdict
− (Exo)	+ (More disorder)	Always −	Spontaneous at ALL temperatures
− (Exo)	− (Less disorder)	− at low T; + at high T	Spontaneous at LOW temperature (\|ΔH\| > \|TΔS\|)
+ (Endo)	+ (More disorder)	+ at low T; − at high T	Spontaneous at HIGH temperature (TΔS > ΔH)
+ (Endo)	− (Less disorder)	Always +	Non-spontaneous at ALL temperatures

3. Standard Gibbs Energy & Equilibrium Constant (K)

At dynamic equilibrium, free energy reaches its minimum so Δ_rG = 0. Relating standard Gibbs energy change (Δ_rG°) to equilibrium constant K:

Δ_rG° = −RT ln K = −2.303 RT log K | Δ_rG° < 0 ⇒ K > 1 (Products favoured) | Δ_rG° > 0 ⇒ K < 1

NCERT Problems 5.12 & 5.13:
• 5.12: 3/2 O₂(g) → O₃(g) at 298 K has K_p = 2.47 × 10^-29 ⇒ Δ_rG° = −2.303(8.314)(298) log(2.47 × 10^-29) = +163 kJ mol^-1.
• 5.13: If Δ_rG° = −13.6 kJ mol^-1 at 298 K ⇒ log K = 13600 / (2.303 × 8.314 × 298) = 2.38 ⇒ K = 2.4 × 10².

Watch the UNIT MISMATCH TRAP in ΔG = ΔH − TΔS! ΔH is usually given in kJ mol^-1 while ΔS is given in J K^-1 mol^-1! Always divide ΔS by 1000 (or multiply ΔH by 1000) before subtracting! Similarly, in ΔG° = −2.303 RT log K, if R = 8.314 J K^-1 mol^-1, your ΔG° comes out in Joules—divide by 1000 for kJ!

Solved Previous Year Questions (Kerala DHSE & NCERT)

Q1 (Kerala HSE March (Repeated Favorite) • 1 Marks): Classify the following properties into extensive and intensive: Enthalpy, Temperature, Heat capacity, Density.

Extensive Properties (depend on quantity of matter): Enthalpy, Heat capacity.
Intensive Properties (independent of quantity of matter): Temperature, Density.

Q2 (Kerala HSE SAY • 1 Marks): What is the standard enthalpy of formation of an element in its reference state? Mention the reference states of Carbon and Sulphur.

By convention, Δ_fH° of an element in its reference state is zero (0).
Reference state of Carbon: C (graphite); Reference state of Sulphur: S (rhombic).

Q3 (Kerala HSE Model • 1 Marks): Write the mathematical statement of the First Law of Thermodynamics. What is ΔU for an isolated system?

Mathematical statement: ΔU = q + w.
For an isolated system (q = 0, w = 0), ΔU = 0 (energy remains constant).

Q4 (Kerala HSE March (Most Asked) • 2 Marks): Derive the relationship between heat capacities C_p and C_V for one mole of an ideal gas.

At constant volume: q_V = ΔU = C_VΔT.
At constant pressure: q_p = ΔH = C_pΔT.
For 1 mole of an ideal gas (pV = RT): ΔH = ΔU + Δ(pV) = ΔU + RΔT.
Substituting ΔH and ΔU: C_pΔT = C_VΔT + RΔT ⇒ C_p − C_V = R.

Q5 (Kerala HSE SAY (High Yield) • 2 Marks): State Hess's Law of Constant Heat Summation. Illustrate how it helps find Δ_fH° of CO(g).

Statement: If a reaction takes place in several steps, its standard reaction enthalpy is the sum of the standard enthalpies of the intermediate reactions at the same temperature (Δ_rH = Δ_rH₁ + Δ_rH₂ + ...).

Illustration: From C(s) + O₂(g) → CO₂(g) (ΔH₁ = −393.5 kJ) and reversing CO(g) + ½O₂(g) → CO₂(g) (ΔH₂ = −283.0 kJ), we get Δ_fH°(CO) = −393.5 + 283.0 = −110.5 kJ mol^-1.

Q6 (Kerala HSE March • 2 Marks): Write the Gibbs equation. Predict the signs of ΔH and ΔS for a reaction that is: (i) spontaneous at all temperatures, (ii) spontaneous only at high temperatures.

Gibbs equation: ΔG = ΔH − TΔS.
(i) Spontaneous at ALL temperatures: ΔH < 0 (−) and ΔS > 0 (+).
(ii) Spontaneous only at HIGH temperatures: ΔH > 0 (+) and ΔS > 0 (+) (so TΔS > ΔH at high T).

Q7 (Kerala HSE Improvement • 2 Marks): State the Second and Third Laws of Thermodynamics. Predict the entropy change when a crystalline solid is heated from 0 K to 115 K.

Second Law: For any spontaneous process in an isolated system, entropy increases (ΔS_total = ΔS_sys + ΔS_surr > 0).
Third Law: The entropy of any pure crystalline substance approaches zero as temperature approaches absolute zero (0 K).
When heated from 0 K to 115 K, particles oscillate about lattice sites and disorder increases, so ΔS increases (> 0).

Q8 (Kerala HSE March (Standard Numerical) • 2 Marks): For the reaction 2 A + B → C, ΔH = 400 kJ mol^-1 and ΔS = 0.2 kJ K^-1 mol^-1. At what temperature will the reaction become spontaneous?

At equilibrium, ΔG = ΔH − TΔS = 0 ⇒ T = ΔH / ΔS = 400 / 0.2 = 2000 K.
Since both ΔH > 0 and ΔS > 0, ΔG becomes negative above 2000 K. Hence the reaction is spontaneous at T > 2000 K (NCERT Exercise 5.17).

Q9 (Kerala HSE March (Repeated Numerical) • 3 Marks): The enthalpies of combustion of methane, graphite, and dihydrogen at 298 K are −890.3 kJ mol^-1, −393.5 kJ mol^-1, and −285.8 kJ mol^-1 respectively. Calculate the standard enthalpy of formation of CH₄(g).

Target: C(graphite) + 2H₂(g) → CH₄(g).
(i) C(s) + O₂(g) → CO₂(g); ΔH₁ = −393.5 kJ
(ii) 2H₂(g) + O₂(g) → 2H₂O(l); ΔH₂ = 2(−285.8) = −571.6 kJ
(iii) Reverse combustion of CH₄: CO₂(g) + 2H₂O(l) → CH₄(g) + 2O₂(g); ΔH₃ = +890.3 kJ
Adding (i) + (ii) + (iii): Δ_fH°(CH₄) = −393.5 − 571.6 + 890.3 = −74.8 kJ mol^-1.

Q10 (Kerala HSE Model / NCERT 5.6 • 3 Marks): 1 g of graphite is burnt in a bomb calorimeter in excess O₂ at 298 K and 1 atm: C(graphite) + O₂(g) → CO₂(g). Temperature rises from 298 K to 299 K. If the heat capacity of the bomb calorimeter is 20.7 kJ K^-1, find ΔU and ΔH per mole of graphite.

Heat from 1 g graphite: q = −C_V ΔT = −20.7 × (299 − 298) = −20.7 kJ.
For 1 mole (12 g) graphite: ΔU = −20.7 × 12 = −248.4 kJ mol^-1 (−2.48 × 10² kJ mol^-1).
For C(s) + O₂(g) → CO₂(g), Δn_g = 1 − 1 = 0, so ΔH = ΔU = −2.48 × 10² kJ mol^-1.

Q11 (Kerala HSE March / NCERT 5.19 • 3 Marks): For the reaction 2 A(g) + B(g) → 2 D(g), ΔU° = −10.5 kJ and ΔS° = −44.1 J K^-1 at 298 K. Calculate ΔG° and predict whether the reaction may occur spontaneously.

Δn_g = 2 − (2 + 1) = −1.
ΔH° = ΔU° + Δn_gRT = −10.5 + (−1)(8.314 × 10^-3)(298) = −10.5 − 2.478 = −12.98 kJ.
ΔG° = ΔH° − TΔS° = −12.98 − 298(−44.1 × 10^-3) = −12.98 + 13.14 = +0.16 kJ.
Since ΔG° > 0, the reaction is non-spontaneous at 298 K.

Q12 (Kerala HSE March (Essay / Cycle) • 4 Marks): What is Lattice Enthalpy? Explain the Born-Haber Cycle to calculate the lattice enthalpy of NaCl(s) given: Δ_fH° = −411.2, Δ_subH° = +108.4, Δ_iH° = +496, ½Δ_bondH° = +121, Δ_egH° = −348.6 kJ mol^-1.

Lattice Enthalpy (Δ_latticeH°): The enthalpy change when 1 mole of an ionic crystal dissociates completely into gaseous ions: NaCl(s) → Na⁺(g) + Cl⁻(g).

5 Steps of the Born-Haber Cycle:
1. Na(s) → Na(g) [Δ_subH° = +108.4 kJ]
2. Na(g) → Na⁺(g) + e⁻ [Δ_iH° = +496 kJ]
3. ½Cl₂(g) → Cl(g) [½Δ_bondH° = +121 kJ]
4. Cl(g) + e⁻ → Cl⁻(g) [Δ_egH° = −348.6 kJ]
5. Na⁺(g) + Cl⁻(g) → NaCl(s) [−Δ_latticeH°]

By Hess's Law: Δ_fH° = Δ_subH° + Δ_iH° + ½Δ_bondH° + Δ_egH° − Δ_latticeH°
⇒ Δ_latticeH° = 108.4 + 496 + 121 − 348.6 − (−411.2) = +788 kJ mol^-1.