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Second law, entropy and spontaneity: NEET MCQs with answers

76 questions on this topic of Chemical Thermodynamics (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. In which of the following physical transformations is the entropy change of the system negative (ΔS < 0)?

  1. Sublimation of solid iodine to iodine vapour
  2. Dissolution of sucrose crystals in liquid water
  3. Condensation of steam into liquid water
  4. Stretching of a rubber band at room temperature
Show answer and explanation

Answer: (C) Condensation of steam into liquid water

Condensation of steam into liquid water involves a transition from a highly disordered gaseous state to a more ordered liquid state, leading to a decrease in entropy (ΔS < 0). Sublimation and dissolution both lead to an increase in molecular disorder and thus have ΔS > 0. Stretching a rubber band aligns polymer chains causing ΔS < 0, but condensation is a standard physical phase change showing ΔS < 0 directly from gas to liquid.

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2. For a reaction, ΔH = +20 kJ mol⁻¹ and ΔS = +100 J K⁻¹ mol⁻¹. Assuming both are constant, the reaction becomes spontaneous above:

  1. 5000 K
  2. 0.2 K
  3. 200 K
  4. 400 K
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Answer: (C) 200 K

At the threshold ΔG = 0, so T = ΔH/ΔS = 20000/100 = 200 K.

hard

3. A heat engine absorbs 600 J of heat from a reservoir at 400 K and rejects 450 J of heat to a sink at 300 K. What is the total entropy change of the universe (ΔS_total) for this cyclic process?

  1. +0.5 J K⁻¹
  2. 0 J K⁻¹
  3. -0.5 J K⁻¹
  4. +1.5 J K⁻¹
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Answer: (B) 0 J K⁻¹

For the cyclic engine itself, ΔS_engine = 0 because entropy is a state function. For the surroundings: ΔS_surr = -Q_in/T_hot + Q_out/T_cold = -600/400 + 450/300 = -1.5 + 1.5 = 0 J K⁻¹. Thus, ΔS_total = ΔS_engine + ΔS_surr = 0 J K⁻¹.

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4. For a spontaneous process occurring in an isolated system, which of the following statements is always correct?

  1. ΔS_sys > 0 and ΔS_surr > 0
  2. ΔS_sys = 0 and ΔS_surr > 0
  3. ΔS_sys < 0 and ΔS_total = 0
  4. ΔS_total > 0 and ΔS_surr = 0
Show answer and explanation

Answer: (D) ΔS_total > 0 and ΔS_surr = 0

For an isolated system, no heat or matter is exchanged with the surroundings, so q_surr = 0, which gives ΔS_surr = 0. According to the Second Law of Thermodynamics, any spontaneous process must have ΔS_total > 0, which means ΔS_sys = ΔS_total > 0.

easy

5. A reaction has a standard reaction enthalpy of Δ_r H° and standard entropy change of Δ_r S°. If the equilibrium constant K_p equals 1.0 at 400 K, which relationship must hold true at this temperature?

  1. Δ_r H° = -400 Δ_r S°
  2. Δ_r H° = 400 Δ_r S°
  3. Δ_r H° = Δ_r S° / 400
  4. Δ_r H° = 0
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Answer: (B) Δ_r H° = 400 Δ_r S°

At equilibrium, Δ_r G° = -R T ln K_p. Since K_p = 1.0, ln K_p = 0, which means Δ_r G° = 0. Because Δ_r G° = Δ_r H° - T Δ_r S°, setting this to zero gives Δ_r H° = T Δ_r S° = 400 Δ_r S°.

hard

6. Two blocks of the same metal with identical heat capacity C = 50 J K⁻¹ are initially at temperatures T₁ = 400 K and T₂ = 200 K. They are brought into thermal contact in an isolated enclosure until they reach thermal equilibrium at 300 K. What is the net entropy change of the system during this process? (Given: ln(1.5) = 0.405, ln(0.75) = -0.288)

  1. -5.85 J K⁻¹
  2. 0 J K⁻¹
  3. +5.85 J K⁻¹
  4. +11.70 J K⁻¹
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Answer: (C) +5.85 J K⁻¹

For the hot block, ΔS₁ = C ln(T_f / T₁) = 50 ln(300 / 400) = 50 ln(0.75) = 50 × (-0.288) = -14.4 J K⁻¹. For the cold block, ΔS₂ = C ln(T_f / T₂) = 50 ln(300 / 200) = 50 ln(1.5) = 50 × 0.405 = +20.25 J K⁻¹. The net change is ΔS_total = -14.4 + 20.25 = +5.85 J K⁻¹.

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7. For a reaction, ΔH = +50 kJ mol⁻¹ and ΔS = +100 J K⁻¹ mol⁻¹. Assuming both are constant, the reaction becomes spontaneous above:

  1. 0.5 K
  2. 500 K
  3. 227 K
  4. 2000 K
Show answer and explanation

Answer: (B) 500 K

At the threshold ΔG = 0, so T = ΔH/ΔS = 50000/100 = 500 K.

easy

8. If the enthalpy of fusion of ice at 273 K is 6.0 kJ mol⁻¹, what is the molar entropy change for melting ice into water at 273 K?

  1. +22.0 J K⁻¹ mol⁻¹
  2. +2.20 J K⁻¹ mol⁻¹
  3. -22.0 J K⁻¹ mol⁻¹
  4. +16.4 J K⁻¹ mol⁻¹
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Answer: (A) +22.0 J K⁻¹ mol⁻¹

The melting of ice at 273 K is an equilibrium phase transition, so Δ_fus S = Δ_fus H / T = 6000 J mol⁻¹ / 273 K = +21.98 J K⁻¹ mol⁻¹ ≈ +22.0 J K⁻¹ mol⁻¹.

easy

9. Given the following standard entropy values at 298 K: S°[H₂(g)] = 130.6 J K⁻¹ mol⁻¹ S°[Cl₂(g)] = 223.0 J K⁻¹ mol⁻¹ S°[HCl(g)] = 186.8 J K⁻¹ mol⁻¹ What is the standard entropy change (Δ_r S°) for the reaction: H₂(g) + Cl₂(g) → 2HCl(g)?

  1. -40.0 J K⁻¹ mol⁻¹
  2. +40.0 J K⁻¹ mol⁻¹
  3. -20.0 J K⁻¹ mol⁻¹
  4. +20.0 J K⁻¹ mol⁻¹
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Answer: (D) +20.0 J K⁻¹ mol⁻¹

Δ_r S° = Σ S°(products) - Σ S°(reactants) = 2 × S°[HCl(g)] - [S°[H₂(g)] + S°[Cl₂(g)]] = 2(186.8) - [130.6 + 223.0] = 373.6 - 353.6 = +20.0 J K⁻¹ mol⁻¹.

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10. For an endothermic reaction A → B, the value of ΔH is +120 kJ mol⁻¹. If the reaction is non-spontaneous at 300 K but becomes spontaneous at 500 K, what is the most plausible value of ΔS for this reaction (assuming ΔH and ΔS are temperature independent)?

  1. +150 J K⁻¹ mol⁻¹
  2. +280 J K⁻¹ mol⁻¹
  3. +450 J K⁻¹ mol⁻¹
  4. -280 J K⁻¹ mol⁻¹
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Answer: (B) +280 J K⁻¹ mol⁻¹

For the reaction to be non-spontaneous at 300 K (ΔG > 0) and spontaneous at 500 K (ΔG < 0), the transition temperature T = ΔH / ΔS must lie between 300 K and 500 K. Thus, 300 < 120000 / ΔS < 500, which gives 240 J K⁻¹ mol⁻¹ < ΔS < 400 J K⁻¹ mol⁻¹. Among the choices, +280 J K⁻¹ mol⁻¹ falls within this range.

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11. Consider the following statements regarding entropy: Statement I: The entropy of a perfectly crystalline solid approaches zero as the absolute temperature approaches zero. Statement II: Dissolving table sugar (sucrose) in liquid water leads to a decrease in entropy. In light of the above statements, choose the correct answer:

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
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Answer: (B) Statement I is correct but Statement II is incorrect

Statement I is the Third Law of Thermodynamics and is correct. Statement II is incorrect because dissolution of a solid solute into a solvent disrupts the crystal lattice and creates more disordered molecular arrangements, so ΔS > 0.

hard

12. For a hypothetical reaction A(s) + B(g) → C(g) + D(g), the value of Δ_r U° is -10.5 kJ mol⁻¹ at 300 K. If Δ_r S° is +100 J K⁻¹ mol⁻¹, what is the value of Δ_r G° for the reaction at 300 K? (Take R = 8.314 J K⁻¹ mol⁻¹)

  1. -30.0 kJ mol⁻¹
  2. -22.0 kJ mol⁻¹
  3. -38.0 kJ mol⁻¹
  4. -42.5 kJ mol⁻¹
Show answer and explanation

Answer: (C) -38.0 kJ mol⁻¹

First, calculate Δn_g = moles of gaseous products - gaseous reactants = 2 - 1 = 1. Then Δ_r H° = Δ_r U° + Δn_g R T = -10.5 kJ + (1 × 8.314 × 300 × 10⁻³) kJ = -10.5 + 2.494 = -8.0 kJ mol⁻¹. Finally, Δ_r G° = Δ_r H° - T Δ_r S° = -8.0 kJ - (300 × 0.100) kJ = -8.0 - 30.0 = -38.0 kJ mol⁻¹.

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