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Thermodynamics – Class 11 Chemistry Chapter 5 Notes & NCERT PDF

Class 11ChemistryChapter 5NCERT book: Chemistry Part I - Class 11

Notes and a simple summary of Chapter 5 of the NCERT Class 11 Chemistry book, with the official chapter PDF, flashcards, an MCQ quiz and an AI tutor for your doubts.

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Chapter 5: Thermodynamics

Introduction

Unit 5 of NCERT Chemistry Part I studies energy changes in physical and chemical processes. Thermodynamics does not care how fast or by which path a change happens. It only compares the initial and final states, and uses them to predict the energy exchanged and whether a process is spontaneous. The unit builds from system/surroundings and internal energy to enthalpy, Hess's law, entropy and finally Gibbs energy, which links spontaneity to the equilibrium constant. That last link is the bridge to Unit 6.

Key Concepts

5.1 Thermodynamic Terms

  • System: the part of the universe we observe. Surroundings: everything else. System + surroundings = universe.
  • Types of system (by what crosses the boundary):

    • - Open: both energy and matter can be exchanged (reactants in an open beaker).

    - Closed: energy can be exchanged but not matter (a closed vessel of conducting material).

    - Isolated: neither energy nor matter (reactants in a thermos flask).

  • State of the system is described by state variables such as p, V, T and amount n. A state function depends only on the state, not on the path taken. U, H, S and G are state functions; q and w are not.
  • Internal energy (U) is the total energy of the system (chemical, electrical, mechanical). It changes when heat passes in or out, when work is done on or by the system, or when matter enters or leaves.
  • Adiabatic process: no heat transfer (q = 0), for example a system enclosed by an adiabatic wall.

    • First Law of Thermodynamics

    ΔU = q + w. The energy of an isolated system is constant: energy can be neither created nor destroyed.

    NCERT sign convention: q and w are positive when added to the system.

  • Work done on the system: w positive.
  • Heat absorbed by the system: q positive.

    • 5.2 Applications

    Work (pressure–volume work)

  • Against a constant external pressure: w = −pₑₓ ΔV. Compression makes ΔV negative, so w is positive.
  • Reversible isothermal expansion of an ideal gas: w_rev = −2.303 nRT log(V_f/V_i).
  • Free expansion (expansion into a vacuum, pₑₓ = 0): no work is done, whether the process is reversible or irreversible.
  • For an isothermal ideal-gas process ΔU = 0, so q = −w. At constant volume, ΔU = q_V.

    • Enthalpy, H = U + pV

  • At constant pressure, ΔH = q_p, the heat absorbed at constant pressure.
  • For reactions involving gases: ΔH = ΔU + Δn_g RT, where Δn_g = moles of gaseous products − moles of gaseous reactants.
  • ΔH is negative for exothermic and positive for endothermic reactions.
  • Extensive properties depend on the amount of matter (mass, volume, U, H, heat capacity). Intensive properties do not (temperature, pressure, density, molar heat capacity).
  • Heat capacity: q = C ΔT. Molar heat capacity C_m = C/n. For an ideal gas, C_p − C_V = R.

    • 5.3 Measurement of ΔU and ΔH: Calorimetry

  • A bomb calorimeter works at constant volume, so it measures ΔU.
  • A coffee-cup calorimeter works at constant pressure, so it measures ΔH.

    • 5.4 Enthalpy Change of a Reaction

  • ΔᵣH = Σ aᵢ H(products) − Σ bᵢ H(reactants).
  • Standard enthalpy (ΔᵣH⁰) refers to substances in their standard states: the pure form at 1 bar, usually quoted at 298 K.
  • Thermochemical equations give phases and ΔH. ΔH scales with the coefficients and changes sign when the reaction is reversed.
  • Hess's law of constant heat summation: if a reaction takes place in several steps, its ΔᵣH equals the sum of the enthalpy changes of the steps. This works because H is a state function.

    • 5.5 Enthalpies for Different Types of Reactions

  • Standard enthalpy of formation (Δ_fH⁰): the enthalpy change when 1 mol of a compound forms from its elements in their most stable states. Δ_fH⁰ of an element in its reference state is zero.
  • Combustion (Δ_cH⁰), atomisation (Δ_aH⁰), bond enthalpy, enthalpy of solution, and phase changes (Δ_fusH, Δ_vapH, Δ_subH).
  • Using formation enthalpies: ΔᵣH⁰ = Σ aᵢ Δ_fH⁰(products) − Σ bᵢ Δ_fH⁰(reactants).

    • 5.6 Spontaneity and Entropy

  • A spontaneous process occurs on its own without continuous outside help. A negative ΔH alone does not guarantee spontaneity.
  • Entropy (S) measures the degree of randomness or disorder. ΔS = q_rev/T.
  • Entropy increases when a solid melts, a liquid vaporises, a gas is heated, or the number of gas particles increases.
  • For an isolated system, a spontaneous change has ΔS_total > 0. For a reaction, ΔS_surr = −ΔᵣH/T.
  • Third law: the entropy of a perfectly crystalline substance at 0 K is zero.

    • Gibbs Energy

  • G = H − TS, so at constant T and p: ΔG = ΔH − TΔS.
  • ΔG < 0: spontaneous. ΔG > 0: non-spontaneous. ΔG = 0: at equilibrium.
  • The four sign combinations:

    • - ΔH −, ΔS +: spontaneous at all temperatures.

    - ΔH +, ΔS −: never spontaneous.

    - ΔH −, ΔS −: spontaneous only at low T.

    - ΔH +, ΔS +: spontaneous only at high T, when T > ΔH/ΔS.

    5.7 Gibbs Energy Change and Equilibrium

  • At equilibrium ΔᵣG = 0, and ΔᵣG⁰ = −RT ln K = −2.303 RT log K.
  • A large negative ΔᵣG⁰ means K ≫ 1. A strongly positive ΔᵣG⁰ means K ≪ 1.

    • Worked Examples (from the book)

  • Problem 5.3/5.4. 2 L of an ideal gas at 10 atm and 25 °C expands isothermally to 10 L.

    • - Against a constant 1 atm: q = −w = pₑₓΔV = 1 × 8 = 8 L atm.

    - Reversibly for 1 mol: q = −w = 2.303 × 1 × 0.08206 × 298 × log(10/2) ≈ 39.36 L atm.

    - Into a vacuum (Problem 5.2): no work and no heat.

  • Problem 5.5. ΔvapH of water at 100 °C is 41.00 kJ mol⁻¹, and Δn_g = 1.

    • ΔU = 41.00 − (8.3 × 373)/1000 = 37.904 kJ mol⁻¹.

  • Problem 5.6 (bomb calorimeter). 1 g of graphite raises the temperature by 1 K, and C_V = 20.7 kJ K⁻¹. So q = −20.7 kJ per gram.

    • For 12.0 g: ΔU = −2.48 × 10² kJ mol⁻¹, and since Δn_g = 0, ΔH = ΔU.

  • Problem 5.9 (Hess's law). C₆H₆(l) has Δ_cH⁰ = −3267.0 kJ mol⁻¹. Using Δ_fH⁰(CO₂) = −393.5 and Δ_fH⁰(H₂O, l) = −285.83:

    • Δ_fH⁰(benzene) = 6(−393.5) + 3(−285.83) + 3267.0 = +48.51 kJ mol⁻¹.

  • Problem 5.11. For 4Fe + 3O₂ → 2Fe₂O₃, ΔS_sys = −549.4 J K⁻¹ mol⁻¹ and ΔH = −1648 × 10³ J mol⁻¹.

    • ΔS_surr = 1648 × 10³/298 ≈ 5530 J K⁻¹ mol⁻¹, so ΔS_total ≈ +4980.6 J K⁻¹ mol⁻¹. The reaction is spontaneous.

  • Problems 5.12/5.13.

    • - For 3/2 O₂ → O₃ with Kp = 2.47 × 10⁻²⁹: ΔᵣG⁰ = −2.303RT log Kp ≈ +163 kJ mol⁻¹.

    - If ΔᵣG⁰ = −13.6 kJ mol⁻¹: log K = 2.38, so K ≈ 2.4 × 10².

    Note on Problem 5.14: N₂O₄ is 50% dissociated at 60 °C, so Kp ≈ 1.33. ΔᵣG⁰ = −RT ln Kp then works out to about −0.8 kJ mol⁻¹, roughly −790 J mol⁻¹. The book prints the result as "−763.8 kJ mol⁻¹", which is a slip in both value and unit.

    Formula / Key-Facts Sheet

  • ΔU = q + w; w = −pₑₓΔV; w_rev = −2.303 nRT log(V_f/V_i)
  • ΔH = ΔU + Δn_g RT; q_V = ΔU; q_p = ΔH
  • q = C ΔT = n C_m ΔT; C_p − C_V = R
  • ΔᵣH⁰ = Σ Δ_fH⁰(products) − Σ Δ_fH⁰(reactants)
  • ΔS = q_rev/T; ΔS_surr = −ΔᵣH/T; ΔS_total = ΔS_sys + ΔS_surr
  • ΔG = ΔH − TΔS; ΔᵣG⁰ = −2.303 RT log K
  • R = 8.314 J K⁻¹ mol⁻¹ = 0.08206 L atm K⁻¹ mol⁻¹

    • Exam Focus (from the Exercises)

  • Ex 5.5 (MCQ): the combustion enthalpies of CH₄, graphite and H₂ are −890.3, −393.5 and −285.8. Δ_fH(CH₄) = −393.5 + 2(−285.8) + 890.3 = −74.8 kJ mol⁻¹.
  • Ex 5.7: 701 J absorbed and 394 J of work done by the system. ΔU = 701 − 394 = +307 J.
  • Ex 5.10: freeze 1 mol of water from 10 °C to ice at −10 °C in three steps:

    • - Cool the water: 75.3 × (−10) = −753 J.

    - Freeze it: −6030 J.

    - Cool the ice: 36.8 × (−10) = −368 J.

    - Total ΔH = −7.151 kJ.

  • Ex 5.17: ΔH = 400 kJ and ΔS = 0.2 kJ K⁻¹. The reaction becomes spontaneous above T = ΔH/ΔS = 2000 K.
  • Ex 5.19: ΔU = −10.5 kJ and Δn_g = −1, so ΔH ≈ −12.98 kJ.

    • - With ΔS = −44.1 J K⁻¹: ΔG = ΔH − TΔS ≈ +0.16 kJ, so the reaction is not spontaneous.

    - Outline for this type of question: (1) find Δn_g; (2) convert ΔU to ΔH; (3) keep units consistent, kJ versus J; (4) apply ΔG = ΔH − TΔS and state the conclusion.

    Key Terms

  • System (निकाय)
  • Surroundings (परिवेश)
  • State function (अवस्था फलन)
  • Internal energy (आंतरिक ऊर्जा)
  • Adiabatic (रुद्धोष्म)
  • Isothermal (समतापी)
  • Enthalpy (एन्थैल्पी)
  • Heat capacity (ऊष्मा धारिता)
  • Extensive property (विस्तीर्ण गुण)
  • Intensive property (गहन गुण)
  • Standard state (मानक अवस्था)
  • Hess's law (हेस का नियम)
  • Entropy (एन्ट्रॉपी)
  • Spontaneous (स्वतः प्रवर्तित)
  • Gibbs energy (गिब्ज़ ऊर्जा)

    • Common Mistakes

  • Using the physics sign convention. In NCERT chemistry, work done by the system is negative.
  • Counting liquids or solids in Δn_g. Only gaseous species count.
  • Mixing units. Mixing J and kJ, especially inside ΔG = ΔH − TΔS, is the most common source of wrong answers.
  • Treating q or w as a state function. Only U, H, S and G are state functions.
  • Assuming every exothermic reaction is spontaneous. Always check the TΔS term.
  • Giving elements a non-zero enthalpy of formation. Δ_fH⁰ of an element in its standard state is zero.

    • 💡 Key Learning Points

    • ✓Classify systems as open, closed or isolated, and distinguish state functions (U, H, S, G) from path functions (q, w)
    • ✓Apply the first law ΔU = q + w with the NCERT sign convention, and calculate pressure–volume work for irreversible, reversible and free expansion
    • ✓Relate ΔH and ΔU through ΔH = ΔU + Δn_g RT, and interpret bomb-calorimeter (ΔU) versus constant-pressure (ΔH) measurements
    • ✓Use Hess's law and standard enthalpies of formation, combustion and atomisation to calculate reaction enthalpies
    • ✓Predict spontaneity with entropy and ΔG = ΔH − TΔS, and connect ΔᵣG⁰ = −2.303RT log K to the equilibrium constant

    👨‍🏫 Teaching Tips

    • →Start with a thermos flask, a closed steel bottle and an open cup of tea to fix the idea of isolated, closed and open systems
    • →Draw the NCERT sign-convention diagram (arrows into the system are positive) and make students label q and w before every numerical
    • →Use a mountain-climbing analogy for state functions: the altitude gained (ΔU) is fixed, but the path length (q or w) depends on the route
    • →Build Hess's-law problems as 'equation algebra' on the board: multiply, reverse and add until the target equation appears
    • →Have students fill in the four-quadrant ΔH/ΔS table themselves and predict the temperature at which a reaction switches to spontaneous
    • →Set a unit-checking drill in which every answer must carry J or kJ and a sign, since unit slips cause most marks lost
    • →Link the ΔG⁰ = −RT ln K results to the start of Unit 6 so students see why K ≫ 1 means a 'complete' reaction

    📋 Assessment Questions

    1. Can the student state the first law and apply the correct sign convention to heat and work?
    2. Can the student convert between ΔU and ΔH using Δn_g, counting gases only?
    3. Can the student use Hess's law or formation enthalpies to find an unknown ΔH?
    4. Can the student predict spontaneity from the signs of ΔH and ΔS and find the crossover temperature?
    5. Can the student calculate ΔG⁰ from K, or K from ΔG⁰, with correct units?

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    Keep learning

    All chapters in Chemistry Part I - Class 11

    1. 1. Some Basic Concepts of Chemistry
    2. 2. Structure of Atom
    3. 3. Classification of Elements and Periodicity in Properties
    4. 4. Chemical Bonding and Molecular Structure
    5. 5. Thermodynamics
    6. 6. Equilibrium