Question 1: Enthalpy and Spontaneity
For the melting of ice at 273 K, ΔH_fus = 6010 J/mol. The entropy of fusion is about:
- A 22 J/mol K
- B 2.2 J/mol K
- C 1.6 × 10⁶ J/mol K
- D 6010 J/mol K
Answer: A. 22 J/mol K
Why: ΔS = ΔH/T = 6010/273.
Thermodynamics is a Class 11 Chemistry chapter that carries about 3.2% of the Chemistry questions in NEET. Practise it with 44 questions covering 2 topics, each with an explanation.
Sets of 15 questions that adapt to your level, with an explanation for every answer.
Across the 13 previous year papers, Thermodynamics had 24 questions, about 1.8 a year. The most in one paper was 3, in NEET 2020.
| Paper | Questions from this chapter |
|---|---|
| NEET 2026 | 2 |
| NEET 2025 | 2 |
| NEET 2024 | 3 |
| NEET 2023 | 2 |
| NEET 2022 | 1 |
| NEET 2021 | 2 |
| NEET 2020 | 3 |
| NEET 2019 | 2 |
| NEET 2018 | 1 |
| NEET 2017 | 2 |
| NEET 2016 | 1 |
| Re-AIPMT 2015 | 1 |
| AIPMT 2014 | 2 |
8 questions from the practice bank, spread across the chapter's topics. The full set is in the app.
For the melting of ice at 273 K, ΔH_fus = 6010 J/mol. The entropy of fusion is about:
Answer: A. 22 J/mol K
Why: ΔS = ΔH/T = 6010/273.
At constant pressure, the enthalpy change is related to the internal energy change by:
Answer: A. ΔH = ΔU + PΔV
Why: H = U + PV.
For an exothermic reaction, the enthalpy change ΔH is:
Answer: B. negative
Why: Exothermic reactions release heat to the surroundings, so the enthalpy of the products is lower than that of the reactants, giving a negative ΔH.
Which of the following is an intensive property?
Answer: B. density
Why: Intensive properties do not depend on the quantity of matter.
The enthalpy of formation of CO₂ is equal to the enthalpy of combustion of:
Answer: C. graphite
Why: C(graphite) + O₂ → CO₂ is both the formation and the combustion reaction.
The work done by a gas expanding into a vacuum (free expansion) is:
Answer: B. zero
Why: The external pressure is zero, so w = −P_ext ΔV = 0.
The entropy of a system is a measure of its:
Answer: C. disorder (randomness)
Why: S = k ln W, where W is the number of microstates.
The heat absorbed by a system at constant volume equals:
Answer: C. the change in its internal energy
Why: With ΔV = 0 there is no pressure–volume work, so q_v = ΔU.
Thermodynamics carries about 3.2% of the Chemistry questions in NEET. Across the 13 previous year papers on NeetVictory it appeared 24 times, about 1.8 questions a year.
Thermodynamics is a Class 11 Chemistry chapter in the NCERT syllabus.
NeetVictory has 44 practice questions on Thermodynamics, each with an explanation, in sets of 15 that adapt to your level. This page shows a free sample.