QCC Notes
CLASS 11 · CHEMISTRYJEE MAIN × NEETहिंदी
§ 5.5NCERT Class 11 · Chemistry · Chapter 5

Gibbs Energy, Equilibrium & Third Law of Thermodynamics

QCC Notes · Class 11 Chemistry NCERT-aligned Updated 26 Sep 2026 6 min read

In chemical practice, reactions occur at constant temperature and constant atmospheric pressure. To determine reaction feasibility without needing to compute entropy changes of the universe, J. Willard Gibbs formulated the single most powerful thermodynamic potential: Gibbs Free Energy (G).

1. Gibbs Free Energy (G) & The Gibbs-Helmholtz Equation

Gibbs Free Energy is defined mathematically as:

G = H − TS

Where H is enthalpy, T is absolute temperature in Kelvin, and S is entropy. Because H, T, and S are state functions, Gibbs energy is an extensive state function.

For a change occurring at constant temperature (T = constant) and constant pressure:

ΔG = ΔH − TΔS   (Gibbs-Helmholtz Equation)

Physical Significance of Gibbs Energy Change

The decrease in Gibbs free energy (−ΔG) represents the maximum useful non-expansion work (such as electrical work in an electrochemical cell) that can be extracted from a system at constant temperature and pressure:

−ΔG = wnon-PV, max = nFEcell

2. Criteria for Spontaneity & Equilibrium at Constant T and P

Sign of ΔG Thermodynamic Meaning Physical Reality
ΔG < 0 (Negative) Spontaneous / Feasible Process Reaction proceeds spontaneously in the forward direction.
ΔG = 0 (Zero) Equilibrium State System is in a state of dynamic equilibrium; no net change occurs.
ΔG > 0 (Positive) Non-Spontaneous Process Forward reaction cannot occur spontaneously; reverse reaction is spontaneous.

Gibbs Free Energy Curve vs Reaction Coordinate

Free energy minimum defining the position of dynamic chemical equilibrium
Gibbs Free Energy (G)
  ^
  |  Reactants (G_reactants)
  |        |       \   Spontaneous Forward (ΔG < 0)
  |          |         \                    Products (G_products)
  |          \                  /
  |           \                /  Spontaneous Reverse (ΔG_rev < 0)
  |            \              /
  |             \            /
  |              +----------+
  |                   *
  |              EQUILIBRIUM: (dG/dξ) = 0, ΔG = 0
  +--------------------------------------------------------> Extent of Reaction (ξ)
Equilibrium Principle: A closed thermodynamic system naturally evolves in the direction that lowers its free energy (ΔG < 0). Equilibrium corresponds to the absolute minimum in the Gibbs free energy profile, where ΔG = 0.

3. Effect of Temperature on Spontaneity (Sign Analysis Matrix)

ΔH ΔS ΔG = ΔH − TΔS Reaction Spontaneity Behaviour Example
− (Exothermic) + (Disorder increases) Always − at all temperatures Spontaneous at all temperatures 2H2O2(l) → 2H2O(l) + O2(g)
+ (Endothermic) − (Order increases) Always + at all temperatures Non-spontaneous at all temperatures 3O2(g) → 2O3(g)
− (Exothermic) − (Order increases) − at low T; + at high T Spontaneous at low temperatures (Enthalpy-driven) N2(g) + 3H2(g) → 2NH3(g)
+ (Endothermic) + (Disorder increases) + at low T; − at high T Spontaneous at high temperatures (Entropy-driven) CaCO3(s) → CaO(s) + CO2(g)
NEET/JEE Calculation Trick: The temperature at which a reaction transitions from non-spontaneous to spontaneous is the Equilibrium Temperature (Teq) where ΔG = 0:
Teq = ΔH / ΔS   (Ensure ΔH and ΔS are in matching units: Joules or kiloJoules!)

4. Standard Gibbs Energy & Equilibrium Constant (K)

For any general reaction aA + bB ⇌ cC + dD, the Van 't Hoff reaction isotherm relates ΔG to the reaction quotient Q:

ΔG = ΔG° + RT ln Q

At dynamic equilibrium, ΔG = 0 and Q = K (Equilibrium Constant). Therefore:

ΔrG° = −RT ln K = −2.303 RT log10 K
  • If ΔrG° < 0 ⇒ ln K > 0 ⇒ K > 1 (Products predominate at equilibrium).
  • If ΔrG° > 0 ⇒ ln K < 0 ⇒ K < 1 (Reactants predominate at equilibrium).
  • If ΔrG° = 0 ⇒ ln K = 0 ⇒ K = 1.

5. Third Law of Thermodynamics & Absolute Entropies

Formulated by Walther Nernst and Max Planck: The entropy of any pure, perfectly crystalline substance approaches zero as the absolute temperature approaches absolute zero (0 Kelvin).

limT → 0 K S = 0   (for a perfect crystal)

The Third Law enables the calculation of the absolute entropy of any pure substance at temperature T by measuring its heat capacities from 0 K up to T:

ST = ∫0T (Cp / T) dT

Residual Entropy

Certain substances do not have zero entropy at 0 K due to imperfect crystal orientations (molecular disorder). This remaining entropy at 0 K is called Residual Entropy.

Examples of substances with residual entropy: CO (dipole flip orientation), NO, N2O, and Ice (H-bonding positional configurations).

6. Solved High-Yield Numerical Examples & Exam MCQs

Example 1 (JEE Main): For a chemical reaction, ΔH = +35.5 kJ mol−1 and ΔS = +83.6 J K−1 mol−1. At what temperature does the reaction become spontaneous?
Step-by-step Solution:
1. Condition for spontaneity: ΔG = ΔH − TΔS < 0
2. Set ΔG = 0 to find the equilibrium transition temperature Teq:
Teq = ΔH / ΔS
3. Match the units carefully (ΔH in Joules):
ΔH = 35.5 × 1000 J mol−1 = 35,500 J mol−1
ΔS = 83.6 J K−1 mol−1
Teq = 35,500 / 83.6 = 424.6 K.
Since both ΔH and ΔS are positive, the reaction is spontaneous at temperatures greater than 424.6 K (or above 151.6 °C).
Practice MCQ 1 (NEET): The standard Gibbs free energy change (ΔrG°) for a reversible reaction is −5.71 kJ mol−1 at 300 K. What is the value of the equilibrium constant (K) at this temperature? (Take 2.303 RT = 5.71 kJ mol−1)
(A) 1
(B) 10
(C) 100
(D) 0.1
Show answer
Correct Answer: (B) 10
Explanation:
Formula: ΔrG° = −2.303 RT log10 K
−5.71 kJ mol−1 = −(5.71 kJ mol−1) × log10 K
log10 K = 1 ⇒ K = 101 = 10.
Practice MCQ 2 (JEE Advanced): Which of the following substances possesses a non-zero residual entropy at 0 K?
(A) Perfectly crystalline Diamond
(B) Solid Argon
(C) Carbon monoxide (CO) solid
(D) Pure crystalline Iron
Show answer
Correct Answer: (C) Carbon monoxide (CO) solid
Explanation:
In solid CO, the dipole moment is very small. During rapid freezing, CO molecules freeze into the crystal lattice with random orientations (:C≡O: vs :O≡C:), giving residual entropy S0 = R ln(2) ≈ 5.76 J K−1 mol−1 at 0 K.

7. Frequently Asked Questions (FAQs)

1. Can an endothermic reaction with negative entropy change ever be spontaneous?
No. If ΔH > 0 and ΔS < 0, both the enthalpy term (ΔH > 0) and entropy term (−TΔS > 0) are positive. Consequently, ΔG is always positive at all temperatures, making the reaction non-spontaneous under all conditions.
2. What is the value of ΔG when ice melts at 0 °C (273.15 K) under 1 atm?
At 0 °C and 1 atm, solid ice and liquid water are in dynamic equilibrium. Therefore, ΔG = 0.
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