Can an endothermic process be spontaneous?

Yes. Spontaneity is not determined by enthalpy change alone. Even if a process is endothermic (ΔH > 0), it can proceed spontaneously if it is accompanied by a sufficiently large increase in entropy (ΔS > 0) such that the total entropy change of the universe (ΔS_total = ΔS_sys + ΔS_surr) remains positive (or ΔG = ΔH - TΔS < 0 at that temperature), as seen in the melting of ice above 273 K or evaporation of water at room temperature.

Why is the reversible heat (q_rev) used in the definition of entropy (ΔS = q_rev / T)?

Entropy (S) is a state function whose change depends only on initial and final states. Irreversible processes generate additional entropy internally through friction, turbulence, and unrestrained expansion (q_irrev < q_rev). Only a reversible path provides the precise measure of thermal energy transferred without internal entropy generation.

What is Trouton's Rule?

Trouton's rule states that the standard entropy of vaporization (Δ_vap S°) for most non-associated liquids at their normal boiling point is approximately constant, having a value of around 85 to 88 J K^-1 mol^-1. Liquids that form extensive hydrogen bonds (like water or ethanol) deviate significantly with higher entropy values.

§ 5.4NCERT Class 11 · Chemistry · Chapter 5

Spontaneity, Entropy & Second Law of Thermodynamics

Q: Why is the reversible heat (q_rev) used in the definition of entropy (ΔS = q_rev / T)?

Entropy (S) is a state function whose change depends only on initial and final states. Irreversible processes generate additional entropy internally through friction, turbulence, and unrestrained expansion (q_irrev < q_rev). Only a reversible path provides the precise measure of thermal energy transferred without internal entropy generation.

Q: What is Trouton's Rule?

Trouton's rule states that the standard entropy of vaporization (Δ_vap S°) for most non-associated liquids at their normal boiling point is approximately constant, having a value of around 85 to 88 J K^-1 mol^-1. Liquids that form extensive hydrogen bonds (like water or ethanol) deviate significantly with higher entropy values.

QCC Notes · Class 11 Chemistry NCERT-aligned Updated 26 Sep 2026 8 min read

The First Law of Thermodynamics tells us that energy is conserved in every physical and chemical change, but it provides no clue regarding the direction of natural processes. Why does heat flow spontaneously only from a hot body to a cold body? Why does a gas expand to fill an evacuated flask but never re-compress on its own? The answers lie in the thermodynamic concepts of Spontaneity and Entropy.

1. The Concept of Spontaneity & Driving Forces

A Spontaneous Process (or feasible process) is an irreversible process that occurs on its own under a specified set of conditions, without requiring continuous external work. Once initiated, or proceeding by itself, it moves towards a state of dynamic equilibrium.

Examples of spontaneous processes: Water flowing down a hill, gas diffusing into vacuum, dissolving sugar in hot water, rusting of iron in moist air.
Driving forces behind spontaneity:
1. Tendency towards minimum energy: Many exothermic reactions (ΔH < 0) are spontaneous because lower potential energy states are more stable.
2. Tendency towards maximum disorder/randomness: Even endothermic processes (ΔH > 0, such as melting of ice at 25 °C or evaporation of alcohol) occur spontaneously because the system gains molecular randomness.

2. Entropy (S): Definition & Physical Meaning

Entropy (S) is a thermodynamic state function that serves as a quantitative measure of molecular disorder, randomness, or the number of microstates (Ω) accessible to a macroscopic system: S = k_B ln(Ω).

Thermodynamic Definition of Entropy Change:
For an infinitesimal reversible transfer of heat dq_rev at absolute temperature T:
dS = dq_rev / T ⇒ ΔS = ∫ (dq_rev / T)
Units of Entropy: J K⁻¹ mol⁻¹ (Joules per Kelvin per mole).

Physical State / Transformation	Relative Entropy	Reasoning
Solid Phase	Lowest entropy (S_solid)	Particles are locked in rigid crystal lattices with only vibrational motion.
Liquid Phase	Intermediate entropy (S_liquid)	Molecules possess translational and rotational freedom of movement.
Gaseous Phase	Highest entropy (S_gas)	Molecules move completely independently in vast empty space.

Order of Entropy: S(solid) < S(liquid) << S(gas)

Entropy Transformation Across Phase Transitions

Visual molecular randomness during heating from absolute zero to gas phase

Entropy (S)
  ^                                                      / Vapor State
  |                                                     /  (High Entropy)
  |                                           +--------*
  |                                           | Δ_vap S = Δ_vap H / T_b
  |                               Liquid State|
  |                             +-------------*
  |                             | Δ_fus S = Δ_fus H / T_m
  |                 Solid State |
  |               /
  |              /
  +-------------+---------------+-------------+----------> Temperature (T)
  0 K           T_melting       T_boiling

Sharp Entropy Jumps: At the melting point (T_m) and boiling point (T_b), latent heat is absorbed reversibly at constant temperature. This produces vertical jumps in entropy: Δ_fusS = Δ_fusH / T_m and Δ_vapS = Δ_vapH / T_b. Because expansion into the gas phase creates a vastly larger volume of accessible microstates, Δ_vapS >> Δ_fusS.

3. Mathematical Formulae for Entropy Changes (ΔS)

A. Phase Transitions at Equilibrium (Constant T and P)

Entropy of Fusion: Δ_fusS = Δ_fusH / T_f
Entropy of Vaporization: Δ_vapS = Δ_vapH / T_b
Entropy of Sublimation: Δ_subS = Δ_subH / T_sub = Δ_fusS + Δ_vapS

B. Entropy Change for an Ideal Gas (n moles)

From First Law (dq_rev = dU + PdV = nC_vdT + (nRT/V)dV):

General Formula:
ΔS = n C_v ln(T₂ / T₁) + n R ln(V₂ / V₁)
In terms of pressure:
ΔS = n C_p ln(T₂ / T₁) − n R ln(P₂ / P₁)

Isothermal Process (T₁ = T₂): ΔS = 2.303 n R log₁₀(V₂ / V₁) = 2.303 n R log₁₀(P₁ / P₂)
Isochoric Process (V₁ = V₂): ΔS = 2.303 n C_v log₁₀(T₂ / T₁)
Isobaric Process (P₁ = P₂): ΔS = 2.303 n C_p log₁₀(T₂ / T₁)
Reversible Adiabatic Process (q_rev = 0): ΔS = 0 (Isentropic Process)

4. The Second Law of Thermodynamics

The Second Law of Thermodynamics establishes the universal criterion for spontaneous change:

      Second Law Statement: The entropy of an isolated system (such as the entire universe) always increases during any spontaneous, natural process.

      ΔStotal = ΔSsystem + ΔSsurroundings > 0   (Spontaneous / Irreversible process)

      ΔStotal = ΔSsystem + ΔSsurroundings = 0   (Reversible process at dynamic equilibrium)

      ΔStotal < 0   (Non-spontaneous / Impossible process)

Entropy Change of the Surroundings

Because the surroundings act as an infinitely large thermal reservoir at constant temperature T:

ΔS_surr = q_surr / T = −q_sys / T = −ΔH_sys / T

Therefore:

ΔS_total = ΔS_sys − (ΔH_sys / T)

Multiplying both sides by −T gives: −TΔS_total = ΔH_sys − TΔS_sys, which directly leads to Gibbs Free Energy (ΔG)!

5. Solved High-Yield Numerical Examples & Exam MCQs

Example 1 (NEET): The enthalpy of vaporization (Δ_vapH) of water at its normal boiling point (373 K) is 40.66 kJ mol⁻¹. Calculate the entropy change (Δ_vapS) when 1 mole of liquid water turns to steam at 373 K.

Step-by-step Solution:
1. Identify the given values:
Δ_vapH = 40.66 kJ mol⁻¹ = 40,660 J mol⁻¹
T_b = 373 K
2. Apply the entropy of vaporization formula at equilibrium:
Δ_vapS = Δ_vapH / T_b
Δ_vapS = 40,660 J mol⁻¹ / 373 K = +109.0 J K⁻¹ mol⁻¹.
Notice that Δ_vapS is positive, reflecting the tremendous increase in randomness when molecules leave the liquid phase and expand into vapour.

Practice MCQ 1 (JEE Main): For which of the following reactions is ΔS° positive?

(A) N₂(g) + 3H₂(g) → 2NH₃(g)

(B) CaCO₃(s) → CaO(s) + CO₂(g)

(D) H₂O(g) → H₂O(l)

Show answer

Correct Answer: (B) CaCO₃(s) → CaO(s) + CO₂(g)

Explanation:
In reaction (B), a solid reactant decomposes to produce 1 mole of gas: Δn_g = 1 − 0 = +1. Gaseous molecules have vastly greater entropy than solids; hence ΔS° > 0. In (A) and (C), Δn_g is negative, causing an entropy decrease. In (D), condensation of gas to liquid also decreases entropy (ΔS < 0).

Practice MCQ 2 (JEE Advanced): One mole of an ideal gas expands isothermally and reversibly from 10 L to 100 L at 300 K. What is the entropy change of the gas? (R = 8.314 J K⁻¹ mol⁻¹)

(A) +19.14 J K⁻¹

(B) −19.14 J K⁻¹

(D) zero

Show answer

Correct Answer: (A) +19.14 J K⁻¹

Explanation:
For an isothermal process of an ideal gas:
ΔS = 2.303 n R log₁₀(V₂ / V₁)
ΔS = 2.303 × 1 × 8.314 × log₁₀(100 / 10) = 2.303 × 8.314 × 1 = +19.15 J K⁻¹.

6. Frequently Asked Questions (FAQs)

1. What is an isentropic process?

An isentropic process is a thermodynamic process in which entropy remains constant throughout (ΔS = 0). A reversible adiabatic process is strictly isentropic because q_rev = 0 and no entropy is generated internally.

2. What happens to entropy when an egg is boiled?

Upon boiling, the soluble globular proteins in egg white undergo denaturation and coagulation. The folded compact tertiary structures unfold into random, tangled peptide coils, leading to an increase in molecular randomness and entropy (ΔS > 0). This is a classic conceptual question in NEET and CBSE!

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