NCERT Solutions for Class 11 Chemistry Chapter 5 Chemical Thermodynamics
NCERT Solutions for Class 11 Chemistry Chapter 5 Chemical Thermodynamics covers thermodynamic systems and surroundings, state functions, internal energy, work, heat, First Law of Thermodynamics, enthalpy (ΔH), heat capacity (Cp and Cv), Hess's Law of Constant Heat Summation, spontaneity, entropy (S), Second and Third Laws of Thermodynamics, Gibbs free energy change (ΔG), and its relation with equilibrium constants.
Overview of Topics & Subtopics (Click to Jump)
1. System, Surroundings, State Functions & First Law of Thermodynamics
Short Note:
• Types of Systems:
- Open System: Can exchange both matter and energy with surroundings (e.g., hot water in an open beaker).
- Closed System: Can exchange energy but not matter with surroundings (e.g., hot water in a closed steel vessel).
- Isolated System: Can exchange neither energy nor matter with surroundings (e.g., liquid in a thermos flask).
• State Functions vs Path Functions: State functions depend only on the initial and final states of the system (e.g., U, H, S, G, P, V, T). Path functions depend on the path taken (e.g., q, w).
• First Law of Thermodynamics: Energy can neither be created nor destroyed; it can only be transformed from one form to another.
- Mathematical relation: ΔU = q + w, where ΔU = change in internal energy, q = heat absorbed, w = work done on the system.
- Expansion work: w = -Pext ΔV. For isothermal reversible expansion: wrev = -2.303 nRT log(V2 / V1).
2. Enthalpy Changes (ΔH), Heat Capacities & Hess's Law
Short Note:
• Enthalpy (H): Total heat content of a system at constant pressure: H = U + PV.
- Relation between ΔH and ΔU for gaseous reactions: ΔH = ΔU + Δng RT (where Δng = moles of gaseous products - moles of gaseous reactants).
• Heat Capacities (Cp and Cv): Heat required to raise the temperature of a system by 1 K.
- At constant volume: qv = Cv ΔT = ΔU.
- At constant pressure: qp = Cp ΔT = ΔH.
- Relationship for 1 mole of an ideal gas: Cp - Cv = R.
• Hess's Law of Constant Heat Summation: The total enthalpy change of a reaction is the same whether the reaction takes place in one step or in several steps.
3. Spontaneity, Entropy (S) & Second Law of Thermodynamics
Short Note:
• Spontaneous Process: A process that takes place on its own without requiring continuous external assistance.
• Entropy (S): Measure of randomness or disorder of a system. State function and extensive property.
- Units: J·K-1·mol-1. For phase changes: ΔS = qrev / T.
- Order of entropy: Solid < Liquid < Gas.
• Second Law of Thermodynamics: The total entropy of the universe (system + surroundings) increases in any spontaneous process (ΔStotal = ΔSsystem + ΔSsurroundings > 0).
• Third Law of Thermodynamics: The entropy of a perfectly crystalline substance approaches zero as the absolute temperature approaches zero (T = 0 K).
4. Gibbs Free Energy (ΔG) & Relationship with Equilibrium Constant
Short Note:
• Gibbs Free Energy (G): Amount of energy available from a system that can be converted into useful work: G = H - TS.
• Gibbs-Helmholtz Equation: ΔG = ΔH - T ΔS (at constant T and P).
• Criteria for Spontaneity based on ΔG:
- If ΔG < 0 (negative): Process is spontaneous.
- If ΔG > 0 (positive): Process is non-spontaneous.
- If ΔG = 0: Process is at equilibrium.
• Relation with Equilibrium Constant (K): ΔG° = -2.303 RT log K.
Intext & Conceptual Questions
Proof:
At constant volume, heat absorbed qv = Cv ΔT = ΔU.
At constant pressure, heat absorbed qp = Cp ΔT = ΔH.
For 1 mole of an ideal gas, enthalpy is defined as:
H = U + PV
Since PV = RT for 1 mole of an ideal gas:
H = U + RT
Differentiating with respect to temperature T:
ΔH = ΔU + R ΔT
Substituting ΔH = Cp ΔT and ΔU = Cv ΔT:
Cp ΔT = Cv ΔT + R ΔT
Dividing both sides by ΔT gives:
Cp - Cv = R.
Textbook Chapter End Exercise Questions
Given:
ΔU° = -10.5 kJ = -10500 J
ΔS° = -44.1 J·K-1
T = 298 K
R = 8.314 J·K-1·mol-1
Step 1: Calculate Δng:
Δng = moles of gaseous products - moles of gaseous reactants
Δng = 2 - (2 + 1) = -1
Step 2: Calculate ΔH°:
ΔH° = ΔU° + Δng RT
ΔH° = -10500 + (-1 × 8.314 × 298)
ΔH° = -10500 - 2477.57 = -12977.57 J (or -12.98 kJ)
Step 3: Calculate ΔG° using Gibbs-Helmholtz equation:
ΔG° = ΔH° - T ΔS°
ΔG° = -12977.57 - [298 × (-44.1)]
ΔG° = -12977.57 + 13141.8 = +164.23 J (or +0.164 kJ)
Since ΔG° is positive (+0.164 kJ), the reaction is non-spontaneous at 298 K.
Given:
K = 10
T = 300 K
R = 8.314 J K-1 mol-1
Formula:
ΔG° = -2.303 R T log K
ΔG° = -2.303 × 8.314 J K-1 mol-1 × 300 K × log(10)
Since log(10) = 1:
ΔG° = -2.303 × 8.314 × 300 × 1
ΔG° = -5744.14 J/mol = -5.744 kJ/mol.
The standard Gibbs free energy change ΔG° is -5.744 kJ/mol.
Read More & Chapter Resources
These short notes are designed for quick revision. Read the main points once, then use the related material below to practise the same chapter. Revising a concept and checking it with questions can make your preparation more focused.