Thermodynamics — Class 11 MCQs with Answers

Thermodynamics — important questions & MCQs with answers (Class 11 Chemistry)

14 solved questions from this chapter's difficulty levels, each with its answer and explanation. The other 76 are timed and scored when you sign in.

1. Q1Easy

   Enthalpy H =

   A.U + PV✓ Correct

   B.U − PV

   C.PV

   D.Q

   Answer: A. U + PV

   Explanation: Enthalpy is defined as H = U + PV, internal energy plus the pressure-volume product, so it tracks heat exchanged at constant P. Writing U − PV instead flips the sign of the PV term and gives a different, undefined quantity.

2. Q2Easy

   SI unit of entropy:

   A.J K⁻¹ mol⁻¹✓ Correct

   B.J mol⁻¹

   C.J

   D.K mol⁻¹

   Answer: A. J K⁻¹ mol⁻¹

   Explanation: Entropy behaves like a heat capacity, measured per mole per kelvin, so its SI unit is J K⁻¹ mol⁻¹, e.g. S°(H₂O,l) ≈ 70 J K⁻¹ mol⁻¹. Dropping the K⁻¹ (just J mol⁻¹) confuses entropy's unit with enthalpy's.

3. Q3Easy

   An exothermic reaction has ΔH:

   A.Positive

   B.Negative✓ Correct

   C.Zero

   D.Infinite

   Answer: B. Negative

   Explanation: Exothermic means the system releases heat to the surroundings, so its enthalpy drops and ΔH is negative, e.g. combustion. Picking positive confuses exothermic with endothermic, which absorbs heat instead.

4. Q4Easy

   Third law: entropy of a perfect crystal at 0 K is:

   A.R

   B.Zero✓ Correct

   C.Infinity

   D.1

   Answer: B. Zero

   Explanation: The third law states a perfectly ordered crystal at 0 K has only one possible microstate, so its entropy is exactly zero, the reference point for absolute entropy. Picking infinity confuses order with disorder.

5. Q5Easy

   First law of thermodynamics: ΔU =

   A.q/w

   B.q − w

   C.q × w

   D.q + w✓ Correct

   Answer: D. q + w

   Explanation: The first law states ΔU = q + w using the convention that w is work done ON the system, so compressing a gas (w>0) raises its internal energy. Writing q − w instead flips the sign of the work term.

6. Q6Easy

   Gibbs energy ΔG =

   A.ΔU − PV

   B.ΔH + TΔS

   C.ΔH − TΔS✓ Correct

   D.q − w

   Answer: C. ΔH − TΔS

   Explanation: Gibbs free energy change is defined as ΔG = ΔH − TΔS, combining enthalpy with the temperature-weighted entropy term to predict spontaneity. Writing ΔH + TΔS flips the sign on the entropy contribution.

7. Q7Medium

   Relation ΔH = ΔU + Δn_g RT applies to:

   A.Solids only

   B.Gas-phase reactions✓ Correct

   C.Liquids only

   D.Solutions

   Answer: B. Gas-phase reactions

   Explanation: The added term Δn_g·RT in ΔH = ΔU + Δn_g·RT accounts for PV work from a change in moles of GAS, so it matters only when gases are reactants or products. For solids or liquids, Δn_g ≈ 0 and ΔH ≈ ΔU already.

8. Q8Medium

   Entropy is a measure of:

   A.Mass

   B.Energy released

   C.Disorder/randomness✓ Correct

   D.Volume only

   Answer: C. Disorder/randomness

   Explanation: Entropy quantifies the number of accessible microstates, i.e. the disorder or randomness of a system, e.g. gas molecules spreading through a larger volume. Picking mass or volume only confuses entropy with an extensive property.

9. Q9Medium

   For N₂(g) + 3H₂(g) → 2NH₃(g), Δn_g =

   A.−2✓ Correct

   B.+2

   C.0

   D.−4

   Answer: A. −2

   Explanation: Δn_g counts moles of gas: products (2 mol NH₃) minus reactants (1 mol N₂ + 3 mol H₂) = 2 − 4 = −2. Forgetting to include H₂'s 3 moles in the reactant total gives the wrong answer of −1.

10. Q10Medium

    Spontaneous reaction has ΔG:

    A.Infinity

    B.Positive

    C.Zero only

    D.Negative✓ Correct

    Answer: D. Negative

    Explanation: A reaction proceeds spontaneously in the forward direction only when the Gibbs free energy change is negative, ΔG = ΔH − TΔS < 0. Picking zero describes equilibrium, not an ongoing spontaneous change.

11. Q11Medium

    Work done when 1 mol ideal gas expands isothermally and reversibly from V₁ to V₂:

    A.P(V₂ − V₁)

    B.nRT (V₂ − V₁)

    C.−nRT ln(V₂/V₁)✓ Correct

    D.0

    Answer: C. −nRT ln(V₂/V₁)

    Explanation: For reversible isothermal expansion, the maximum work extracted is w = −nRT ln(V₂/V₁), negative since the gas does work on the surroundings as V₂ > V₁. Using P(V₂−V₁) wrongly assumes constant pressure instead of the changing P along the path.

12. Q12Hard

    Hess's law: ΔH of overall reaction depends only on:

    A.Pressure

    B.Path taken

    C.Temperature alone

    D.Initial and final states✓ Correct

    Answer: D. Initial and final states

    Explanation: Hess's law holds because enthalpy is a state function: the overall ΔH depends only on the initial and final states of the reaction, never on the path or number of steps taken. Picking 'path taken' directly contradicts this state-function property.

13. Q13Hard

    Second law: in an isolated system, entropy:

    A.Tends to increase✓ Correct

    B.Decreases

    C.Stays zero

    D.Becomes negative

    Answer: A. Tends to increase

    Explanation: The second law says the entropy of an isolated system spontaneously tends to increase over time, ΔS_universe ≥ 0, driving processes toward greater disorder. Picking 'decreases' would violate the second law entirely.

14. Q14Hard

    Calculate enthalpy of formation of CH₄ from C(s) + 2H₂(g) → CH₄(g). Given: ΔH_sub(C) = 715 kJ, B.E.(H–H) = 436 kJ, B.E.(C–H) = 414 kJ.

    A.+69 kJ

    B.−69 kJ✓ Correct

    C.−139 kJ

    D.+139 kJ

    Answer: B. −69 kJ

    Explanation: Formation of CH₄ from C(s)+2H₂(g): sublime carbon (+715), break 2 H–H bonds (2×436=872) to get atoms, then form 4 C–H bonds (4×414=1656): ΔH = 715+872−1656 = −69 kJ. Using only 2 C–H bonds instead of 4 for CH₄'s tetrahedral structure understates the energy released.

Thermodynamics — FAQs

More Class 11 Chemistry chapters

• 01Some Basic Concepts of Chemistry
• 02Structure of Atom
• 03Classification of Elements and Periodicity
• 04Chemical Bonding and Molecular Structure
• 06Equilibrium
• 07Redox Reactions

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