NCERT Solutions for Class 11 Chemistry Chapter 6: Thermodynamics
Chapter 6 of Class 11 Chemistry, Thermodynamics, deals with one of the most profound subjects in science — the study of energy and how it transforms. Every reaction in chemistry, every process in biology, every engine in engineering is governed by the laws of thermodynamics. Yet for many Class 11 students, this chapter feels abstract and difficult. The NCERT solutions on Myclass24 are specifically designed to change that. Each concept — from internal energy and enthalpy to entropy and Gibbs free energy — is explained clearly with worked examples and connections to real chemistry.
You will not just learn that ΔG = ΔH − TΔS; you will understand what it means for a reaction to be spontaneous, and under what conditions endothermic reactions can still occur naturally. The chapter also covers the laws of thermodynamics, standard enthalpy of various processes, Hess's Law of constant heat summation, and Kirchhoff's equation. Students preparing for CBSE boards, JEE, or NEET will find thermodynamics questions to be among the most formula-intensive — but with the right preparation, they are also among the most scoring. The NCERT solutions at Myclass24 are updated to match the current NCERT syllabus and are accessible 24/7 from anywhere in India, whether you are in Varanasi, Coimbatore, Amritsar, or Ranchi.
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NCERT Solutions — Class 11 Chemistry Chapter 6: Thermodynamics
Enthalpy calculations, Hess's Law, Gibbs energy, spontaneity criteria | Free PDF on Myclass24
Chapter 6 Detailed Notes: Thermodynamics Laws, Enthalpy & Spontaneity
Basic Terminology in Thermodynamics
Before diving into laws, it is important to be clear about the vocabulary. The system is the part of the universe being studied; everything else is the surroundings. Systems can be open (exchange of energy and matter), closed (only energy exchange), or isolated (no exchange). A process is isothermal if temperature is constant, isobaric if pressure is constant, isochoric if volume is constant, and adiabatic if there is no heat exchange with the surroundings. State functions (like internal energy U, enthalpy H, entropy S) depend only on the current state of the system, not the path taken. One can check out all chapters of NCERT Solutions for Class 11 Chemistry and all subjects of NCERT Solutions for Class 11 from the Myclass24 page.
First Law of Thermodynamics
The First Law states that energy can neither be created nor destroyed — only converted from one form to another. Mathematically: ΔU = q + w, where ΔU is the change in internal energy, q is heat absorbed by the system, and w is work done on the system. Work done by expansion against constant external pressure: w = −PₑₓₜΔV. At constant volume (isochoric process), no work is done, so ΔU = qᵥ.
| Process | Condition | Key Relationship |
|---|---|---|
| Isothermal | ΔT = 0 | ΔU = 0 (ideal gas); q = −w |
| Adiabatic | q = 0 | ΔU = w |
| Isobaric | ΔP = 0 | q_p = ΔH |
| Isochoric | ΔV = 0 | w = 0; ΔU = q_v |
Enthalpy (H) and Thermochemistry
Enthalpy is defined as H = U + PV. At constant pressure: ΔH = ΔU + PΔV. Since most reactions in chemistry occur at constant (atmospheric) pressure, ΔH is the heat exchanged. If ΔH is negative, the reaction is exothermic (releases heat); if positive, it is endothermic (absorbs heat). The relationship between ΔH and ΔU is: ΔH = ΔU + Δnᵍ × RT, where Δnᵍ is the change in moles of gaseous species.
| Type of Enthalpy | Definition | Example |
|---|---|---|
| Enthalpy of Formation (ΔH°f) | Formation of 1 mol compound from elements | H₂ + ½O₂ → H₂O; ΔH°f = −286 kJ/mol |
| Enthalpy of Combustion (ΔH°c) | Complete combustion of 1 mol substance | CH₄ + 2O₂ → CO₂ + 2H₂O |
| Enthalpy of Neutralisation | Neutralisation of strong acid + strong base | ≈ −57.1 kJ/mol always |
| Enthalpy of Atomisation | Breaking 1 mol substance into gaseous atoms | Na(s) → Na(g); ΔH = 108.4 kJ/mol |
| Bond Enthalpy | Energy to break 1 mol of bond in gaseous phase | C−H bond ≈ 413 kJ/mol |
Hess's Law of Constant Heat Summation
Hess's Law states that the total enthalpy change of a reaction is the same, regardless of whether it occurs in one step or multiple steps. This allows calculation of ΔH for reactions that are difficult to measure directly. It is essentially a consequence of enthalpy being a state function. Using Hess's Law, we can calculate lattice enthalpies (Born-Haber cycle), bond enthalpies, and enthalpy of formation for complex compounds.
Second Law and Entropy
The Second Law states that the total entropy of an isolated system always increases during a spontaneous process. Entropy (S) is a measure of disorder or randomness in a system. ΔS = q_rev / T. For a spontaneous process: ΔS_universe = ΔS_system + ΔS_surroundings > 0. Melting ice, dissolving salt, and expanding gases are all examples of entropy increasing.
Gibbs Free Energy and Spontaneity
Gibbs free energy (G) combines enthalpy and entropy to predict whether a process is spontaneous at a given temperature: ΔG = ΔH − TΔS. A reaction is spontaneous if ΔG < 0, non-spontaneous if ΔG > 0, and at equilibrium if ΔG = 0.
| ΔH | ΔS | Spontaneity |
|---|---|---|
| Negative (−) | Positive (+) | Always spontaneous |
| Negative (−) | Negative (−) | Spontaneous at low T |
| Positive (+) | Positive (+) | Spontaneous at high T |
| Positive (+) | Negative (−) | Never spontaneous |
Third Law of Thermodynamics: The entropy of a perfect crystal at absolute zero (0 K) is zero. This establishes an absolute scale for entropy.
- R (gas constant) = 8.314 J mol⁻¹ K⁻¹
- Standard state: 298 K (25°C) and 1 bar pressure
- For a reaction: ΔH°rxn = ΣΔH°f (products) − ΣΔH°f (reactants)
- ΔG° = −RT ln K (connects thermodynamics with equilibrium constant)
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