CBSE Class 11 Chemistry Chapter 6 Notes: Thermodynamics

Chapter Introduction

Thermodynamics is the branch of physical chemistry that deals with energy transformations in chemical and physical processes. In CBSE Class 11 Chemistry, Chapter 6 introduces you to the laws governing heat, work, and energy — concepts that form the backbone of physical chemistry and are essential for NEET and JEE preparation.

This chapter connects the macroscopic properties of systems (temperature, pressure, volume) to energy changes. You will learn to predict whether a reaction will occur spontaneously using enthalpy, entropy, and Gibbs energy.

Key Definitions

System Types — Comparison

State Functions vs Path Functions

Key point: In a cyclic process, ΔU = 0, ΔH = 0, ΔS = 0, but q and w are generally non-zero.

Internal Energy and Work

Internal Energy (U)

• Sum of kinetic energy (translational, rotational, vibrational) and potential energy (intermolecular forces) of all particles.
• An extensive property (depends on amount of substance).
• For an ideal gas, U depends only on temperature.

PV Work (Expansion Work)

Work done by the system during expansion against external pressure:

w = −Pex × ΔV (constant external pressure)

For reversible isothermal expansion of an ideal gas:

w = −nRT ln(V₂/V₁) = −nRT ln(P₁/P₂)

Sign convention (IUPAC):

• Work done by the system is negative (w < 0)
• Work done on the system is positive (w > 0)
• Heat absorbed by the system is positive (q > 0)
• Heat released by the system is negative (q < 0)

The First Law of Thermodynamics

ΔU = q + w

Energy can neither be created nor destroyed; it can only be converted from one form to another.

Enthalpy (H)

H = U + PV

At constant pressure: ΔH = ΔU + PΔV = ΔU + Δn(g)RT

Where Δn(g) = moles of gaseous products − moles of gaseous reactants.

Heat Capacity

For n moles: qp = nCpΔT and qv = nCvΔT

Standard Enthalpy of Reactions

Hess's Law of Constant Heat Summation

The total enthalpy change for a reaction is the same whether it takes place in one step or in multiple steps.

Applications: 1. Calculate ΔH for reactions that cannot be measured directly. 2. Calculate ΔfH using known enthalpies of combustion. 3. Calculate lattice enthalpy using Born-Haber cycle.

Example: To find ΔH for C(s) + ½O₂(g) → CO(g):

• C(s) + O₂(g) → CO₂(g); ΔH = −393.5 kJ/mol
• CO(g) + ½O₂(g) → CO₂(g); ΔH = −283.0 kJ/mol
• Subtracting: C(s) + ½O₂(g) → CO(g); ΔH = −110.5 kJ/mol

Entropy (S)

Second Law of Thermodynamics

In any spontaneous process, the total entropy of the system and surroundings always increases.

For a reversible process: ΔS = qrev / T

For an irreversible process: ΔStotal > 0

Standard entropy change for a reaction:

ΔrS° = ΣS°(products) − ΣS°(reactants)

Gibbs Energy and Spontaneity

ΔG = ΔH − TΔS (at constant temperature and pressure)

At equilibrium: ΔG = 0, so ΔH = TΔS → T = ΔH / ΔS

Third Law of Thermodynamics

The entropy of a perfectly crystalline substance is zero at absolute zero (0 K).

This law provides a reference point for calculating absolute entropies: S° at 298 K for any substance can be determined.

Important Numerical Patterns

Pattern 1: Calculating ΔU from q and w

If 500 J of heat is supplied to a system and the system does 200 J of work on the surroundings:

• q = +500 J (heat absorbed), w = −200 J (work done by system)
• ΔU = q + w = 500 + (−200) = +300 J

Pattern 2: Calculating ΔH from ΔU

For the reaction N₂(g) + 3H₂(g) → 2NH₃(g) at 298 K, if ΔU = −87.42 kJ/mol:

• Δn(g) = 2 − (1 + 3) = 2 − 4 = −2
• ΔH = ΔU + Δn(g)RT = −87.42 + (−2)(8.314 × 10⁻³)(298) = −87.42 − 4.96 = −92.38 kJ/mol

Pattern 3: Hess's Law — Born-Haber Cycle for NaCl

• ΔfH°(NaCl) = ΔsubH°(Na) + ΔionH°(Na) + ½ΔbondH°(Cl₂) + ΔegH°(Cl) + ΔlatticeH°(NaCl)
• Use this to calculate lattice enthalpy when other values are known.

Pattern 4: Spontaneity from ΔG

For a reaction with ΔH = +80 kJ/mol and ΔS = +150 J/(mol·K) at 400 K:

• ΔG = ΔH − TΔS = 80000 − 400 × 150 = 80000 − 60000 = +20000 J/mol = +20 kJ/mol
• ΔG > 0, so the reaction is non-spontaneous at 400 K.
• It becomes spontaneous when T > ΔH/ΔS = 80000/150 = 533.3 K.

Pattern 5: Entropy change

For melting of ice at 273 K (ΔfusH = 6.0 kJ/mol):

• ΔS = ΔH / T = 6000 / 273 = 21.98 J/(mol·K)

Common Mistakes

1. Confusing sign conventions: Remember IUPAC — work done BY the system is negative, heat absorbed BY the system is positive. Many students reverse these.

2. Using ΔH = ΔU + ΔnRT instead of Δn(g)RT: Only count gaseous moles, not all moles. Solids and liquids contribute negligibly to volume change.

3. Forgetting to convert units: ΔS is usually in J/(mol·K), but ΔH is in kJ/mol. Always convert to the same unit before using ΔG = ΔH − TΔS.

4. Assuming ΔH = ΔU always: They are equal only when Δn(g) = 0 (equal moles of gas on both sides).

5. Confusing Cp and Cv: Cp is at constant pressure, Cv at constant volume. For an ideal gas, Cp − Cv = R. Using Cv when constant pressure conditions apply gives wrong answers.

6. Hess's law errors: When manipulating equations, reverse the sign of ΔH when you reverse the equation; multiply ΔH by the same factor when you multiply the equation.

7. Spontaneity confusion: ΔS(system) > 0 does NOT guarantee spontaneity. You must consider the total entropy change (system + surroundings) or use ΔG.

Exam Tips

1. One-mark answers: Memorize definitions of state function, path function, isolated system, entropy, and Gibbs energy. These appear frequently.

2. Numericals: Practice ΔU, ΔH, and ΔG calculations. At least one numerical is almost guaranteed. Always show: formula → substitution → answer with unit.

3. Hess's law: A 3-mark or 5-mark question is very likely. Practice Born-Haber cycle construction for NaCl, MgCl₂.

4. Spontaneity table: The ΔH/ΔS/T table is a frequent 2-mark question. Write the four combinations clearly.

5. Third law: Be ready to explain why entropy is zero at 0 K for a perfect crystal — a common short-answer question.

6. Relation questions: "Derive ΔH = ΔU + Δn(g)RT" and "Show that Cp − Cv = R" are frequent derivations.

7. Graph-based questions: Isothermal and adiabatic PV diagrams can appear. Know the shapes: isothermal is flatter than adiabatic.

Quick Revision Notes

• First law: ΔU = q + w. Energy is conserved.
• Enthalpy: H = U + PV; ΔH = qp (heat at constant pressure).
• ΔH and ΔU: ΔH = ΔU + Δn(g)RT. Equal when Δn(g) = 0.
• Cp − Cv = R for 1 mole of ideal gas.
• Standard formation: ΔfH° of an element in its standard state = 0.
• Hess's law: ΔH is path-independent; add ΔH values algebraically.
• Second law: Total entropy increases for spontaneous processes.
• ΔS = qrev / T (reversible process).
• ΔG = ΔH − TΔS. Spontaneous if ΔG < 0.
• At equilibrium: ΔG = 0, T = ΔH/ΔS.
• Third law: S → 0 as T → 0 K for a perfect crystal.
• Work: w = −PexΔV (irreversible); w = −nRT ln(V₂/V₁) (reversible isothermal).
• Bond enthalpy is average value; differs from bond dissociation enthalpy for polyatomic molecules.
• Spontaneity shortcut: −ve ΔH and +ve ΔS → always spontaneous; +ve ΔH and −ve ΔS → never spontaneous.

NCERT Exercise Focus Areas

The following NCERT exercise question types are most likely to appear in exams. Prioritize these during revision:

Derivations to Master

1. Relationship between ΔH and ΔU

From the definition of enthalpy: H = U + PV

For a process: ΔH = ΔU + Δ(PV)

At constant pressure: ΔH = ΔU + PΔV

For ideal gas: PV = nRT, so PΔV = Δn(g)RT

2. Cp − Cv = R (Mayer's relation)

At constant pressure: qp = nCpΔT = ΔH

At constant volume: qv = nCvΔT = ΔU

From first law: qp − qv = ΔH − ΔU = PΔV = P × nRΔT/P = nRΔT

Dividing by nΔT: Cp − Cv = R

3. Work done in isothermal reversible expansion

w = −∫PexdV = −∫(nRT/V)dV (since PV = nRT for ideal gas)

w = −nRT ln(V₂/V₁) = −nRT × 2.303 × log(V₂/V₁) = −2.303 nRT log(V₂/V₁)

Since P₁V₁ = P₂V₂ for isothermal process: w = −nRT ln(P₁/P₂)