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NEET Chemical Thermodynamics MCQs with answers

508 questions on Chemical Thermodynamics for NEET 2027, 1 of them from previous NEET papers. Below are 15 to try; tap a question to see its answer and explanation.

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15 sample questions

hardNEET 2023

1. ΔH<0 and ΔS<0. Reaction is spontaneous:

  1. All temperatures
  2. High T
  3. Low T
  4. Never
Show answer and explanation

Answer: (C) Low T

ΔG = ΔH − TΔS. At low T, TΔS small → ΔG negative → spontaneous.

Quick tip: ΔH− ΔS− → low T. ΔH+ ΔS+ → high T. Use ΔG = ΔH − TΔS.

NCERT Class 11, Chapter 6, page 180

easy

2. A sealed glass ampoule contains liquid water in equilibrium with its vapour at 25 °C. Which type of thermodynamic system does the content of the ampoule represent?

  1. Open system
  2. Adiabatic system
  3. Isolated system
  4. Closed system
Show answer and explanation

Answer: (D) Closed system

A sealed glass ampoule prevents any exchange of matter with the surroundings, but its diathermic glass walls allow the exchange of heat energy. Therefore, it is a closed system.

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3. Match the items in List-I with their corresponding description in List-II: List-I: (A) Rigid diathermic container (B) Perfectly insulated closed vessel (C) Boiling of water in an open beaker (D) Free expansion of an ideal gas List-II: (I) Isolated system (II) Open system (III) Closed system capable of heat exchange (IV) w = 0 Choose the correct answer from the options given below:

  1. A-III, B-II, C-I, D-IV
  2. A-I, B-III, C-II, D-IV
  3. A-III, B-I, C-II, D-IV
  4. A-II, B-I, C-III, D-IV
Show answer and explanation

Answer: (C) A-III, B-I, C-II, D-IV

A rigid diathermic container does not allow matter transfer or volume change, but allows heat exchange (Closed system, III). An insulated closed vessel allows neither mass nor heat transfer (Isolated system, I). Boiling in an open beaker exchanges both matter and energy (Open system, II). In free expansion against vacuum (P_ext = 0), work done w = 0 (IV).

hard

4. Given the following thermochemical equations: (1) FeO(s) + C(graphite) → Fe(s) + CO(g); ΔH° = +156.0 kJ (2) C(graphite) + O₂(g) → CO₂(g); ΔH° = -393.5 kJ (3) 2CO(g) + O₂(g) → 2CO₂(g); ΔH° = -566.0 kJ What is the standard enthalpy of formation (ΔfH°) of FeO(s)?

  1. +266.5 kJ mol⁻¹
  2. -266.5 kJ mol⁻¹
  3. -318.0 kJ mol⁻¹
  4. +318.0 kJ mol⁻¹
Show answer and explanation

Answer: (B) -266.5 kJ mol⁻¹

From (2) and (3), combustion of CO: C(s) + 1/2 O₂(g) → CO(g) has ΔH° = ΔH°(2) - (1/2)ΔH°(3) = -393.5 - (-283.0) = -110.5 kJ. For FeO(s) + C(s) → Fe(s) + CO(g), ΔH° = ΔfH°(CO) - ΔfH°(FeO) = 156.0 kJ. Thus, -110.5 - ΔfH°(FeO) = 156.0, which gives ΔfH°(FeO) = -110.5 - 156.0 = -266.5 kJ mol⁻¹.

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5. Consider the following thermochemical equations: (1) 2Zn(s) + O₂(g) → 2ZnO(s); ΔH° = -696 kJ (2) 2ZnS(s) + 3O₂(g) → 2ZnO(s) + 2SO₂(g); ΔH° = -880 kJ What is the standard enthalpy of the reaction: Zn(s) + SO₂(g) → ZnS(s) + O₂(g)?

  1. -92 kJ
  2. +92 kJ
  3. +184 kJ
  4. -184 kJ
Show answer and explanation

Answer: (B) +92 kJ

Divide equation (1) by 2: Zn(s) + 1/2 O₂(g) → ZnO(s), ΔH = -348 kJ. Divide equation (2) by 2 and reverse it: ZnO(s) + SO₂(g) → ZnS(s) + 3/2 O₂(g), ΔH = +440 kJ. Adding these two gives Zn(s) + SO₂(g) → ZnS(s) + O₂(g) with ΔH = -348 + 440 = +92 kJ.

easy

6. The standard enthalpy of formation of gaseous ammonia, NH₃(g), is -46.1 kJ mol⁻¹. What is the standard enthalpy change for the reaction: 2NH₃(g) → N₂(g) + 3H₂(g)?

  1. -46.1 kJ mol⁻¹
  2. +92.2 kJ mol⁻¹
  3. +46.1 kJ mol⁻¹
  4. -92.2 kJ mol⁻¹
Show answer and explanation

Answer: (B) +92.2 kJ mol⁻¹

The formation reaction is 1/2 N₂(g) + 3/2 H₂(g) → NH₃(g) with ΔfH° = -46.1 kJ mol⁻¹. The reverse reaction for 2 moles of NH₃ is 2NH₃(g) → N₂(g) + 3H₂(g), so ΔrH° = -2 × ΔfH° = -2 × (-46.1) = +92.2 kJ mol⁻¹.

hard

7. The standard Gibbs free energy change (Δ_r G°) for the decomposition reaction 2SO₃(g) ⇌ 2SO₂(g) + O₂(g) is +140 kJ mol⁻¹ at 1000 K. What is the partial pressure of O₂(g) at equilibrium if P(SO₃) = P(SO₂) = 1.0 bar? (Given: 2.303 RT at 1000 K ≈ 19.14 kJ mol⁻¹)

  1. 7.3 × 10⁻⁵ bar
  2. 2.1 × 10⁻⁶ bar
  3. 4.8 × 10⁻⁸ bar
  4. 1.0 × 10⁻⁴ bar
Show answer and explanation

Answer: (C) 4.8 × 10⁻⁸ bar

At equilibrium, Δ_r G° = -2.303 RT log₁₀ K_p. Thus, log₁₀ K_p = -140 / 19.14 = -7.315. Therefore, K_p = 10⁻⁷·³¹⁵ = 10⁰.685 × 10⁻⁸ ≈ 4.8 × 10⁻⁸. Since K_p = [P(SO₂)]² × P(O₂) / [P(SO₃)]² and P(SO₂) = P(SO₃) = 1.0 bar, K_p = P(O₂), giving P(O₂) ≈ 4.8 × 10⁻⁸ bar.

easy

8. Which of the following sets contains ONLY intensive properties of a thermodynamic system?

  1. Gibbs free energy, concentration, surface tension, volume
  2. Volume, enthalpy, entropy, heat capacity
  3. Mass, internal energy, pressure, temperature
  4. Density, refractive index, molar volume, viscosity
Show answer and explanation

Answer: (D) Density, refractive index, molar volume, viscosity

Intensive properties do not depend on the amount of matter in the system. Density, refractive index, molar volume, and viscosity are all intensive properties because their values remain unchanged regardless of the size or sample mass.

easy

9. According to Hess's law of constant heat summation, which of the following expressions is correct for a closed cyclic process 1 → 2 → 3 → 1 consisting of three steps with enthalpy changes ΔH₁, ΔH₂, and ΔH₃?

  1. ΔH₁ + ΔH₂ + ΔH₃ = 0
  2. ΔH₁ + ΔH₂ = ΔH₃
  3. ΔH₁ × ΔH₂ × ΔH₃ = 1
  4. ΔH₁ - ΔH₂ + ΔH₃ = 0
Show answer and explanation

Answer: (A) ΔH₁ + ΔH₂ + ΔH₃ = 0

Since enthalpy (H) is a state function, the net change in enthalpy for any cyclic process returning to its initial state must be identically zero: ∮ dH = 0, which gives ΔH₁ + ΔH₂ + ΔH₃ = 0.

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10. Given below are two statements: Statement I: When an ideal gas undergoes free expansion in an insulated container, its final temperature is equal to its initial temperature. Statement II: In an adiabatic free expansion of an ideal gas, q = 0, w = 0, and therefore ΔU = 0. In light of the above statements, choose the most appropriate 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
Show answer and explanation

Answer: (A) Both Statement I and Statement II are correct

In free expansion against vacuum (P_ext = 0), work done w = 0. In an insulated container, q = 0. By the first law, ΔU = q + w = 0. Since U is a function of temperature alone for an ideal gas, ΔT = 0, so T_final = T_initial. Both statements are correct.

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11. A gas expands against a constant external pressure of 1 atm by 10 L. Taking 1 L atm = 101.3 J, the work done on the gas is:

  1. 1013 J
  2. −1013 J
  3. −10 J
  4. −10130 J
Show answer and explanation

Answer: (B) −1013 J

w = −P_ext ΔV = −1 × 10 L atm = −1013 J (negative: the gas does work).

easy

12. Which of the following statements is INCORRECT regarding enthalpies of physical and chemical changes?

  1. Bond dissociation enthalpy of a diatomic molecule is always negative because bond formation releases energy.
  2. The enthalpy of sublimation of a solid is approximately equal to the sum of its enthalpy of fusion and enthalpy of vaporisation at the same temperature.
  3. Enthalpy of combustion (ΔcH°) of any combustible substance is always an exothermic quantity.
  4. The standard enthalpy of fusion of a substance is always endothermic.
Show answer and explanation

Answer: (A) Bond dissociation enthalpy of a diatomic molecule is always negative because bond formation releases energy.

Bond dissociation enthalpy is defined as the enthalpy change required to break one mole of bonds in gaseous molecules into constituent gaseous atoms or radicals; hence bond breaking is always an endothermic process with a positive enthalpy value (ΔH > 0).

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13. Given below are two statements: Statement I: For an ideal gas undergoing a reversible isothermal process, the enthalpy change ΔH is equal to zero. Statement II: Enthalpy of an ideal gas is a function of both temperature and pressure. In the light of the above statements, choose the correct answer from the options given below:

  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
Show answer and explanation

Answer: (B) Statement I is correct but Statement II is incorrect

For an ideal gas, both internal energy and enthalpy depend solely on temperature (ΔH = n C_p ΔT). Therefore, in an isothermal process (ΔT = 0), ΔH = 0. Statement II is incorrect because for an ideal gas, enthalpy does not depend on pressure.

hard

14. An ideal gas with molar heat capacity Cv = 2.5 R is heated reversibly from 300 K to 500 K. When 2 moles of this gas are heated at constant pressure, how much more heat is absorbed compared to when it is heated at constant volume?

  1. 4988 J
  2. 1663 J
  3. 3326 J
  4. 8314 J
Show answer and explanation

Answer: (C) 3326 J

At constant pressure, qp = n × Cp × ΔT. At constant volume, qv = n × Cv × ΔT. The difference is qp - qv = n × (Cp - Cv) × ΔT = n × R × ΔT. Substituting the values: 2 mol × 8.314 J K⁻¹ mol⁻¹ × (500 - 300) K = 2 × 8.314 × 200 = 3325.6 J ≈ 3326 J.

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15. Consider the following statements: Statement I: The molar heat capacity of any substance cannot be negative for any process. Statement II: During a polytropic expansion in which the work done by an ideal gas exceeds the heat absorbed, the temperature drops, resulting in a negative heat capacity. In the light of the above statements, choose the correct option:

  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
Show answer and explanation

Answer: (C) Statement I is incorrect but Statement II is correct

Statement I is incorrect because molar heat capacity depends on the thermodynamic path and can take any value from -∞ to +∞, including negative values. Statement II is correct: when work done during expansion exceeds the heat supplied (q > 0 and w < -q), ΔU = q + w is negative, so temperature decreases (ΔT < 0), giving C = q/ΔT < 0.

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