CBSE Class 11 Chemistry Thermodynamics Important Questions 2026

For students preparing for CBSE Class 11 Chemistry, mastering Thermodynamics is essential for achieving top marks in the 2026 annual examinations. In the CBSE Chemistry syllabus, Chemical Thermodynamics carries substantial weightage (typically 9 to 10 marks) and establishes core principles that recur in Class 12 Electrochemistry, Chemical Kinetics, and competitive entrance tests. Working through curated CBSE Class 11 Chemistry Thermodynamics Important Questions 2026 allows you to clarify sign conventions, master core derivations like C_p − C_v = R, and solve multi-step thermochemical numericals with complete confidence. This guide provides a syllabus-aligned concept review followed by high-yield CBSE questions with detailed, step-by-step solutions.

Key Concepts

Thermodynamics deals with energy transformations in chemical and physical processes. To tackle both conceptual questions and numerical problems in the board exam 11 assessment, focus on these five core pillars:

1. System, Surroundings, and State Functions

• System: The part of the universe under thermodynamic observation. A system can be:
  • Open System: Exchanges both matter and energy with surroundings (e.g., boiling water in an open beaker).
  • Closed System: Exchanges energy but not matter (e.g., water heated in a sealed copper vessel).
  • Isolated System: Exchanges neither matter nor energy with surroundings (e.g., hot coffee stored in an idealized thermos flask).
• State Functions: Thermodynamic variables that depend solely on the initial and final states of the system, not on the path taken. Examples include internal energy (U), enthalpy (H), entropy (S), Gibbs free energy (G), pressure (P), temperature (T), and volume (V).
• Path Functions: Quantities whose values depend on the specific pathway followed during the change. Heat (q) and work (w) are path functions.
• Extensive vs. Intensive Properties: Extensive properties depend on the quantity of matter present (e.g., mass, volume, internal energy, heat capacity). Intensive properties are independent of the amount of substance (e.g., density, molar heat capacity, refractive index, temperature, pressure).

2. First Law of Thermodynamics and Sign Conventions

The First Law states that energy can neither be created nor destroyed, only transformed from one form to another. Its mathematical formulation is:

ΔU = q + w

According to standard IUPAC conventions adopted by CBSE:

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

For pressure-volume work against a constant external pressure (P_ext):

w = −P_ext ΔV

For an isothermal reversible expansion of n moles of an ideal gas from volume V₁ to V₂:

w_rev = −2.303 nRT log₁₀(V₂ / V₁) = −2.303 nRT log₁₀(P₁ / P₂)

3. Enthalpy (H) and Heat Capacities (C_p and C_v)

Enthalpy is the total heat content of a system at constant pressure, defined as H = U + PV. For any process at constant pressure:

ΔH = ΔU + PΔV

For chemical reactions involving ideal gases, substituting the ideal gas equation (PV = nRT) yields:

ΔH = ΔU + Δn_gRT

Where Δn_g = (Moles of gaseous products) − (Moles of gaseous reactants).

The relationship between molar heat capacity at constant pressure (C_p) and constant volume (C_v) for one mole of an ideal gas is derived as follows:

1. By definition: H = U + PV = U + RT (for 1 mole of an ideal gas).
2. Differentiating with respect to temperature T: dH/dT = dU/dT + R.
3. Since C_p = dH/dT and C_v = dU/dT, substituting these relations gives:

C_p − C_v = R

4. Hess's Law of Constant Heat Summation

Hess's Law states that whether a chemical reaction takes place in one step or in several successive steps, the total enthalpy change (ΔH) is identical. Because enthalpy is a state function, thermochemical equations can be added, subtracted, and multiplied algebraically just like mathematical equations. This law is extensively utilized to compute standard enthalpies of formation, lattice enthalpies using the Born-Haber cycle, and bond dissociation energies.

5. Second Law of Thermodynamics, Entropy (S), and Gibbs Energy (G)

• Second Law: The entropy of the universe always increases in the course of any spontaneous process (ΔS_total > 0).
• Entropy: A quantitative measure of molecular randomness or disorder:

  ΔS = q_rev / T

• Gibbs-Helmholtz Equation: Determines reaction spontaneity at constant temperature and pressure:

  ΔG = ΔH − TΔS

• Spontaneity Criteria:
  • If ΔG < 0: The process is spontaneous in the forward direction.
  • If ΔG = 0: The system is at dynamic equilibrium.
  • If ΔG > 0: The process is non-spontaneous in the forward direction.
• Relation to Equilibrium Constant: ΔG° = −2.303 RT log₁₀ K

Important CBSE Questions with Answers

Very Short Answer Questions (1 and 2 Marks)

Question 1: Classify the following properties into extensive and intensive: Internal energy, Density, Temperature, Molar heat capacity, Enthalpy, and Volume.

Answer:

• Extensive Properties: Internal energy, Enthalpy, Volume (these depend on the amount or mass of the substance).
• Intensive Properties: Density, Temperature, Molar heat capacity (these are independent of the amount or mass of the substance).

Question 2: In a process, 701 J of heat is absorbed by a system and 394 J of work is done by the system. Calculate the change in internal energy (ΔU) for the process.

Answer:

Using the First Law of Thermodynamics: ΔU = q + w

• Heat absorbed by the system (q) = +701 J
• Work done by the system (w) = −394 J

ΔU = (+701 J) + (−394 J) = +307 J

The internal energy of the system increases by 307 J.

Question 3: State the signs of ΔH and ΔS for the reaction: 2 Cl(g) → Cl₂(g). Give reasons.

Answer:

• Sign of ΔH: Negative (ΔH < 0). When chlorine atoms combine to form chlorine molecules, a covalent bond is formed. Bond formation always releases energy (exothermic process).
• Sign of ΔS: Negative (ΔS < 0). Two moles of independent gaseous atoms combine to form one mole of diatomic gaseous molecules, leading to decreased randomness and translational freedom.

Question 4: State the Third Law of Thermodynamics and write its principal practical application.

Answer:

Statement: The entropy of any perfectly crystalline pure substance approaches zero as the absolute temperature approaches absolute zero (0 Kelvin or −273.15 °C).

Application: The Third Law provides an absolute baseline for entropy, enabling chemists to calculate the absolute molar entropies (S°) of pure elements and chemical compounds at any specified temperature.

Short Answer Questions (3 Marks)

Question 5: Derive the relation ΔH = ΔU + Δn_gRT for a gaseous chemical reaction. Under what conditions is ΔH = ΔU?

Answer:

By definition, enthalpy is given by:

H = U + PV

For an initial state (reactants): H₁ = U₁ + P₁V₁

For a final state (products): H₂ = U₂ + P₂V₂

At constant pressure (P₁ = P₂ = P):

ΔH = H₂ − H₁ = (U₂ − U₁) + P(V₂ − V₁) = ΔU + PΔV

Assuming gaseous reactants and products behave ideally, from the ideal gas law: PV₁ = n₁RT and PV₂ = n₂RT

P(V₂ − V₁) = (n₂ − n₁)RT = Δn_gRT

Substituting this into the enthalpy expression yields:

ΔH = ΔU + Δn_gRT

Conditions under which ΔH = ΔU:

1. When the reaction is conducted in a closed vessel at constant volume (ΔV = 0).
2. When the reaction involves only liquids or solids, where the volume change is negligible.
3. When the number of moles of gaseous products equals the number of moles of gaseous reactants (Δn_g = 0), such as in H₂(g) + I₂(g) → 2 HI(g).

Question 6: Predict the effect of temperature on the spontaneity of a chemical reaction under the following conditions using the Gibbs-Helmholtz equation:

Question 7: Calculate the standard enthalpy of formation (Δ_fH°) of liquid methanol, CH₃OH(l), given the following thermochemical data:

1. CH₃OH(l) + ³/₂ O₂(g) → CO₂(g) + 2 H₂O(l); Δ_cH° = −726.0 kJ mol⁻¹
2. C(graphite) + O₂(g) → CO₂(g); Δ_cH° = −393.5 kJ mol⁻¹
3. H₂(g) + ¹/₂ O₂(g) → H₂O(l); Δ_fH° = −285.8 kJ mol⁻¹

Answer:

The target thermochemical equation for the formation of one mole of CH₃OH(l) from its constituent elements in standard states is:

C(graphite) + 2 H₂(g) + ¹/₂ O₂(g) → CH₃OH(l)     (Δ_fH° = ?)

Applying Hess's Law, we combine the given equations as follows:

Target = [Equation (2)] + 2 × [Equation (3)] − [Equation (1)]

Substituting the corresponding enthalpy values:

• ΔH = (−393.5 kJ) + 2(−285.8 kJ) − (−726.0 kJ)
• ΔH = −393.5 − 571.6 + 726.0
• ΔH = −965.1 + 726.0 = −239.1 kJ mol⁻¹

The standard enthalpy of formation of liquid methanol is −239.1 kJ mol⁻¹.

Long Answer Question (5 Marks)

Question 8:

(a) Derive an expression for the work done in the reversible isothermal expansion of an ideal gas.

(b) The equilibrium constant for a reversible reaction is 10 at 300 K. Calculate the standard Gibbs free energy change (ΔG°) of the reaction. (Given: R = 8.314 J K⁻¹ mol⁻¹, log₁₀ 10 = 1).

(c) Dissolution of ammonium chloride (NH₄Cl) in water is an endothermic process (ΔH > 0), yet it dissolves spontaneously at room temperature. Explain using thermodynamic principles.

Answer:

(a) Derivation of Reversible Isothermal Work:

For an infinitesimal expansion against an opposing external pressure P_ext, the small work done is: dw = −P_ext dV.

In a reversible expansion, the external pressure differs infinitesimally from the internal pressure of the gas: P_ext = P_int − dP.

dw = −(P − dP) dV = −P dV + dP dV. Neglecting the second-order differential term dP dV: dw = −P dV.

For n moles of an ideal gas, P = nRT / V. Integrating from initial volume V₁ to final volume V₂ at constant temperature T:

w_rev = − ∫_V1^V₂ (nRT / V) dV = − nRT ln(V₂ / V₁)

Converting natural log to base-10 logarithm:

w_rev = −2.303 nRT log₁₀(V₂ / V₁)

(b) Numerical Calculation of ΔG°:

The relation between standard Gibbs free energy change and the equilibrium constant is:

ΔG° = −2.303 RT log₁₀ K

Substitute the given numerical parameters:

• T = 300 K
• R = 8.314 J K⁻¹ mol⁻¹
• K = 10 ⇒ log₁₀(10) = 1

ΔG° = −2.303 × 8.314 × 300 × 1 = −5744.14 J mol⁻¹ = −5.74 kJ mol⁻¹

The standard Gibbs energy change is −5.74 kJ mol⁻¹.

(c) Thermodynamic Explanation for NH₄Cl Dissolution:

According to the Gibbs-Helmholtz equation: ΔG = ΔH − TΔS.

When solid NH₄Cl dissolves in water, the crystalline lattice breaks apart into freely moving aqueous ions: NH₄Cl(s) → NH₄⁺(aq) + Cl⁻(aq). This transition results in a marked increase in molecular disorder, so the entropy change is strongly positive (ΔS > 0).

At room temperature, the magnitude of the positive term TΔS exceeds the positive enthalpy of solution (ΔH). Consequently, TΔS > ΔH, which makes ΔG < 0 (negative). The negative value of ΔG makes the dissolution process spontaneous despite being endothermic.

How to Prepare for This Topic

To score full marks in Thermodynamics on your CBSE Class 11 Chemistry annual exam, incorporate these exam-tested preparation habits:

1. Strictly Apply IUPAC Sign Conventions: Always specify signs before inserting numbers into ΔU = q + w. A sign error on work (w) or heat (q) invalidates the entire numerical answer.
2. Master Key Derivations: Practice writing out the step-by-step derivations for C_p − C_v = R and reversible isothermal expansion work (w_rev) on unlined paper until you can reproduce them without referencing your textbook.
3. Double-Check Δn_g in Reactions: When evaluating ΔH = ΔU + Δn_gRT, count only gaseous phase species. Always ensure the chemical equation is correctly balanced first.
4. Master Thermochemical Algebraic Manipulations: Write target equations explicitly at the top of your answer sheet when solving Hess's Law problems. Reverse the sign of ΔH when inverting an equation, and multiply ΔH proportionally when scaling stoichiometric coefficients.
5. Analyze Spontaneity with Gibbs-Helmholtz: Practice explaining real-world phenomena (such as evaporation of water at room temperature or freezing of water below 0 °C) using the signs of ΔH, ΔS, and TΔS.

Where to Practice More

Consistently practicing high-yield numericals and conceptual questions is the most reliable way to secure top grades in Class 11 chemistry. Explore targeted practice material and blueprint-aligned papers using Theorify QPTool. You can practice chapter-wise question sets from the comprehensive CBSE Class 11 Chemistry question bank or generate customized test papers using the QPTool custom paper generator to test your timing and accuracy before your 2026 exams.