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Thermodynamics — NCERT Class 11 Chemistry

Complete NCERT Solutions for CBSE Class 11 Chemistry Chapter 6 — all intext and chapter-end exercise questions on system, surroundings, internal energy, enthalpy, Hess's law, entropy, and Gibbs energy answered clearly.

Updated 11 June 202613 min read

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This page provides complete, step-by-step NCERT solutions for CBSE Class 11 Chemistry Chapter 6 — Thermodynamics. Every intext question (grouped by page number) and every chapter-end exercise question is answered with clear explanations and worked-out numerical solutions. Use the section headings to jump to the question you need help with.

Key Concepts (Quick Recap)

  • System and Surroundings: System = part under study; surroundings = everything else. Types: open (mass + energy exchange), closed (only energy exchange), isolated (no exchange).
  • State Functions: Depend only on initial and final states — U, H, S, G, P, V, T. Path functions: q, w.
  • Internal Energy (U): Total energy of the system. Change: ΔU = q + w (first law of thermodynamics).
  • Work: w = −PₑₓΔV for irreversible expansion; w = −nRT ln(V₂/V₁) for reversible isothermal expansion of ideal gas.
  • First Law: ΔU = q + w. Energy is conserved — cannot be created or destroyed.
  • Enthalpy (H): H = U + PV. At constant pressure, ΔH = qₚ.
  • Heat Capacity: Cₚ − Cᵥ = R (for ideal gas). Cₚ = Cᵥ + R.
  • Standard Enthalpy of Formation (ΔfH°): Enthalpy change when 1 mole of a compound is formed from its elements in their standard states.
  • Hess's Law: ΔH is the same whether the reaction occurs in one step or many — follows from H being a state function.
  • Bond Enthalpy: Average enthalpy change when 1 mole of covalent bonds is broken in gaseous molecules.
  • Spontaneous Processes: Occur on their own; driven by tendency toward minimum energy and maximum disorder.
  • Entropy (S): Measure of disorder. ΔS > 0 for spontaneous processes in isolated systems. Second law: total entropy of the universe increases.
  • Gibbs Energy: ΔG = ΔH − TΔS. ΔG < 0 → spontaneous; ΔG > 0 → non-spontaneous; ΔG = 0 → equilibrium.
  • Third Law: Entropy of a perfectly crystalline solid at 0 K is zero.
  • Absolute Entropy: S° = integral from 0 to T of (Cₚ/T)dT.

Intext Questions

Page 141

Q1. A thermodynamic state function is a quantity which: (a) is used in thermochemistry (b) is a state function dependent only on the initial and final states of the system (c) gives the heat absorbed or released (d) depends on the path

Answer(b) A thermodynamic state function depends only on the initial and final states of the system and not on the path taken. Examples: internal energy (U), enthalpy (H), entropy (S), Gibbs energy (G).

Q2. For the process to occur under adiabatic conditions, the correct condition is: (a) ΔT = 0 (b) Δp = 0 (c) q = 0 (d) w = 0

Answer(c) q = 0. In an adiabatic process, no heat is exchanged between the system and its surroundings.

Q3. The enthalpies of all elements in their standard states are: (a) zero (b) unity (c) negative (d) different for each element

Answer(a) Zero. By convention, the standard enthalpy of formation of all elements in their standard states is taken as zero.

Q4. ΔU° of combustion of methane is −X kJ mol⁻¹. The value of ΔH° is: (a) = ΔU° (b) > ΔU° (c) < ΔU° (d) = 0

Answer(b) > ΔU°. For combustion of methane: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l). Δn(g) = 1 − 3 = −2. ΔH = ΔU + Δn(g)RT. Since Δn is negative, ΔH = ΔU + (negative term). Since ΔU is itself a large negative number, ΔH is less negative than ΔU (i.e., ΔH > ΔU, or mathematically ΔH > ΔU°).

Page 146–147

Q5. A gas expands from 1 L to 5 L against a constant external pressure of 1 atm. Calculate the work done by the gas.
Answer

Work done against constant external pressure:

w = −PₑₓΔV = −Pₑₓ(V₂ − V₁)

w = −1 atm × (5 − 1) L = −4 L·atm

Converting to joules (1 L·atm = 101.3 J):

w = −4 × 101.3 = −405.2 J

The negative sign indicates work is done by the system.

Q6. If the internal energy of the system in Q5 decreases by 200 J, is the process endothermic or exothermic?
Answer

ΔU = q + w

−200 = q + (−405.2)

q = −200 + 405.2 = 205.2 J

Since q is positive, heat is absorbed by the system — the process is endothermic.

Q7. In a process, 701 J of heat is absorbed by a system and 394 J of work is done on the system. What is the change in internal energy?
Answer

ΔU = q + w

q = +701 J (heat absorbed), w = +394 J (work done on the system)

ΔU = 701 + 394 = 1095 J

The internal energy increases by 1095 J.

Q8. For the reaction: 2Cl(g) → Cl₂(g), what are the signs of ΔH and ΔS?
Answer

Two chlorine atoms combine to form a Cl₂ molecule — bonds are formed, so energy is released. ΔH < 0 (negative, exothermic).

The system goes from 2 moles of gaseous atoms to 1 mole of gaseous molecules — disorder decreases. ΔS < 0 (negative).

Page 156–157

Q9. For isolation of the element, what is the condition?
AnswerThe process is spontaneous when ΔG < 0. For the reaction to be feasible, ΔG must be negative. Since ΔG = ΔH − TΔS, spontaneity depends on the values of ΔH, ΔS, and T.

Q10. Predict the sign of ΔS for: (a) freezing of water (b) boiling of water (c) decomposition of CaCO₃

Answer

(a) Freezing: Liquid → Solid, disorder decreases. ΔS < 0 (negative).

(b) Boiling: Liquid → Gas, disorder increases. ΔS > 0 (positive).

(c) Decomposition of CaCO₃: CaCO₃(s) → CaO(s) + CO₂(g) — solid produces a solid plus a gas. ΔS > 0 (positive).

Q11. For the reaction 2A(g) + B(g) → 2D(g), ΔU° = −10.5 kJ and ΔS° = −44.1 J K⁻¹. Calculate ΔG° at 298 K.
Answer

Δn(g) = 2 − 3 = −1

ΔH° = ΔU° + Δn(g)RT = −10.5 + (−1)(8.314 × 10⁻³)(298) = −10.5 − 2.478 = −12.978 kJ

ΔG° = ΔH° − TΔS° = −12.978 − (298)(−44.1 × 10⁻³) = −12.978 + 13.142 = +0.164 kJ

Since ΔG° > 0, the reaction is non-spontaneous at 298 K.

Q12. At 298 K, ΔH° = +5.2 kJ and ΔS° = +34.6 J K⁻¹ for a reaction. At what temperature does the reaction become spontaneous?
Answer

At the crossover point, ΔG = 0:

0 = ΔH − TΔS

T = ΔH/ΔS = 5200/34.6 = 150.3 K

Below 150.3 K, ΔG > 0 (non-spontaneous). Above 150.3 K, ΔG < 0 (spontaneous). The reaction becomes spontaneous above approximately 150.3 K.

Chapter-End Exercise

Q1. Choose the correct answer: A thermodynamic state function is: (a) a quantity which depends on the path (b) a quantity which is always conserved (c) a quantity which depends only on the initial and final states of the system (d) a quantity which is always zero

Answer(c) A state function depends only on the initial and final states of the system and is independent of the path taken.

Q2. For the process to occur under isothermal conditions, the correct condition is: (a) ΔT = 0 (b) Δp = 0 (c) q = 0 (d) w = 0

Answer(a) ΔT = 0. In an isothermal process, the temperature remains constant.

Q3. For an adiabatic process, which is correct? (a) ΔT = 0 (b) Δp = 0 (c) q = 0 (d) ΔV = 0

Answer(c) q = 0. No heat exchange occurs in an adiabatic process.

Q4. Work done by the system is: (a) always positive (b) always negative (c) zero (d) may be positive or negative

Answer(d) may be positive or negative. Work done by the system (expansion) is negative by IUPAC convention; work done on the system (compression) is positive.
Q5. A system does 200 J of work and at the same time, 120 J of heat is absorbed. What is the change in internal energy?
Answer

ΔU = q + w = 120 + (−200) = −80 J

The internal energy decreases by 80 J.

Q6. 1 mole of an ideal gas expands isothermally and reversibly from 10 atm to 1 atm at 300 K. Calculate the work done.
Answer

w = −nRT ln(P₁/P₂) = −1 × 8.314 × 300 × ln(10/1)

ln(10) = 2.303

w = −8.314 × 300 × 2.303 = −5744 J = −5.744 kJ

Q7. Calculate the work done when 2 moles of an ideal gas expand isothermally and irreversibly from volume 10 L to 20 L against a constant external pressure of 1 atm.
Answer

w = −Pₑₓ(V₂ − V₁) = −1 × (20 − 10) = −10 L·atm

= −10 × 101.3 = −1013 J

Q8. The standard enthalpy of formation of CH₄(g) is −74.8 kJ mol⁻¹. Calculate the enthalpy change for the reaction:
Answer

C(s) + 2H₂(g) → CH₄(g)

AnswerBy definition, the standard enthalpy of formation is the enthalpy change when 1 mole of the compound is formed from its elements in their standard states. Therefore, ΔH° = −74.8 kJ mol⁻¹.

Q9. Calculate the enthalpy change for the combustion of 1 mole of methane given: ΔfH°(CH₄) = −74.8 kJ mol⁻¹, ΔfH°(CO₂) = −393.5 kJ mol⁻¹, ΔfH°(H₂O) = −285.8 kJ mol⁻¹

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)

ΔH° = [ΔfH°(CO₂) + 2ΔfH°(H₂O)] − [ΔfH°(CH₄) + 2ΔfH°(O₂)]

= [(−393.5) + 2(−285.8)] − [(−74.8) + 2(0)]

= −964.8 − (−74.8) = −890.0 kJ mol⁻¹

Q10. Use Hess's law to calculate ΔH for: C(s) + ½O₂(g) → CO(g) given: C(s) + O₂(g) → CO₂(g); ΔH = −393.5 kJ CO(g) + ½O₂(g) → CO₂(g); ΔH = −283.0 kJ

Subtracting equation 2 from equation 1:

C(s) + O₂(g) − CO(g) − ½O₂(g) → CO₂(g) − CO₂(g)

C(s) + ½O₂(g) → CO(g)

ΔH = (−393.5) − (−283.0) = −110.5 kJ mol⁻¹

Q11. The bond enthalpy of C–H bond is 413 kJ mol⁻¹. Calculate the enthalpy of formation of CH₄(g).
Answer

CH₄ has 4 C–H bonds. Energy required to break bonds = 4 × 413 = 1652 kJ.

Formation: C(g) + 4H(g) → CH₄(g), ΔH = −1652 kJ

But we need from C(s, graphite) and H₂(g):

C(s) → C(g); ΔH = +715 kJ (sublimation) 2H₂(g) → 4H(g); ΔH = 4 × 218 = +872 kJ (atomization) C(g) + 4H(g) → CH₄(g); ΔH = −1652 kJ

ΔfH° = 715 + 872 − 1652 = −65 kJ mol⁻¹

Q12. Define standard enthalpy of formation. What is the standard enthalpy of formation of an element in its reference form?
AnswerThe standard enthalpy of formation (ΔfH°) is the enthalpy change when 1 mole of a compound is formed from its constituent elements in their standard states (1 atm, 298 K). The standard enthalpy of formation of any element in its reference (standard) form is zero by convention.
Q13. State Hess's law. How does it follow from the first law of thermodynamics?
AnswerHess's law states that the total enthalpy change for a reaction is the same whether it occurs in one step or several steps. This follows from the first law because enthalpy (H) is a state function — its change depends only on initial and final states, not on the path. Since the first law establishes energy conservation (a state function property), Hess's law is a direct consequence.
Q14. What is meant by spontaneous process? Give two examples.
AnswerA spontaneous process is one that can occur on its own without any external intervention. Examples: (1) Water flowing downhill, (2) A gas expanding into a vacuum, (3) Rusting of iron, (4) Ice melting at room temperature.
Q15. Define entropy. What is the sign of ΔS for a spontaneous process in an isolated system?
AnswerEntropy (S) is a measure of the degree of disorder or randomness in a system. For a spontaneous process in an isolated system, ΔS > 0 (positive). The total entropy of the universe always increases for a spontaneous process (second law of thermodynamics).

Q16. Predict the sign of entropy change for: (a) Boiling of an egg (b) 2NaHCO₃(s) → Na₂CO₃(s) + CO₂(g) + H₂O(g)

Answer

(a) Boiling an egg: Proteins denature and become more disordered. ΔS > 0 (positive).

(b) 2NaHCO₃(s) → Na₂CO₃(s) + CO₂(g) + H₂O(g): A solid decomposes to produce gases — disorder increases. ΔS > 0 (positive).

Q17. For the reaction at 298 K: 2A(g) + B(g) → C(g), ΔH° = −400 kJ and ΔS° = −200 J K⁻¹. Is the reaction spontaneous?
Answer

ΔG° = ΔH° − TΔS° = −400 − (298)(−0.200) = −400 + 59.6 = −340.4 kJ

Since ΔG° < 0, the reaction is spontaneous at 298 K.

Q18. At what temperature does the reaction in Q17 become non-spontaneous?
Answer

At the boundary, ΔG = 0:

T = ΔH°/ΔS° = −400/(−0.200) = 2000 K

Above 2000 K, ΔG > 0 and the reaction becomes non-spontaneous.

Q19. State the second law of thermodynamics.
AnswerThe second law states that for any spontaneous process in an isolated system, the total entropy of the system increases. In terms of the universe: ΔS(universe) = ΔS(system) + ΔS(surroundings) > 0 for a spontaneous process. A process will be spontaneous only if the total entropy of the universe increases.
Q20. State the third law of thermodynamics.
AnswerThe third law states that the entropy of a perfectly crystalline solid approaches zero as the temperature approaches absolute zero (0 K). This provides an absolute reference point for entropy calculations.
Q21. Explain: (a) ΔG is useful for predicting spontaneity (b) Absolute value of entropy can be determined but not of internal energy.
Answer

(a) ΔG combines both enthalpy and entropy effects: ΔG = ΔH − TΔS. At constant temperature and pressure (common lab conditions), ΔG alone determines spontaneity — ΔG < 0 means spontaneous, ΔG > 0 means non-spontaneous. It is more practical than ΔS(universe) because it refers only to system properties.

(b) Absolute entropy can be determined because the third law gives a reference point (S = 0 at 0 K for a perfect crystal). We can measure heat capacity from 0 K to T and calculate S(T). However, internal energy includes all forms of energy (kinetic + potential of all particles), which cannot be measured absolutely — only changes (ΔU) can be measured.

Q22. Calculate the standard Gibbs energy change for the reaction: Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g) Given: ΔfG°(Fe₂O₃) = −742.2 kJ mol⁻¹, ΔfG°(CO) = −137.2 kJ mol⁻¹, ΔfG°(CO₂) = −394.4 kJ mol⁻¹

ΔG° = [2ΔfG°(Fe) + 3ΔfG°(CO₂)] − [ΔfG°(Fe₂O₃) + 3ΔfG°(CO)]

= [2(0) + 3(−394.4)] − [(−742.2) + 3(−137.2)]

= −1183.2 − (−742.2 − 411.6)

= −1183.2 + 1153.8 = −29.4 kJ

The reaction is spontaneous (ΔG° < 0).

Q23. The equilibrium constant for a reaction is 10. Calculate ΔG° at 300 K. (R = 8.314 J K⁻¹ mol⁻¹)
Answer

ΔG° = −RT ln K = −8.314 × 300 × ln(10) = −8.314 × 300 × 2.303

= −5743.6 J = −5.74 kJ mol⁻¹

Q24. For the reaction A + B → C + D, ΔH = −20 kJ and ΔS = −30 J K⁻¹. Calculate ΔG at 300 K and predict spontaneity.
Answer

ΔG = ΔH − TΔS = −20 − 300(−0.030) = −20 + 9 = −11 kJ

Since ΔG < 0, the reaction is spontaneous at 300 K.

Internal Links

  • [Chapter 6 Guide — Thermodynamics](/school/cbse/class/11/chemistry/cbse-thermodynamics)
  • [All Class 11 Chemistry Solutions](/school/cbse/class/11/chemistry/solutions)
  • [Class 11 Chemistry Subject Guide](/school/cbse/class/11/chemistry)

Frequently Asked Questions

A state function depends only on the initial and final states of the system, not on the path taken — examples include internal energy (U), enthalpy (H), entropy (S), and Gibbs energy (G). A path function depends on the path taken between the initial and final states — examples are heat (q) and work (w).