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Chemistry

Chemical Thermodynamics

Heat and work calculations anchor this chapter: the first law, internal energy, enthalpy, Hess's law, entropy, and Gibbs energy for predicting spontaneity. Expect numericals in NEET: first law calculations, converting internal energy change to enthalpy, and computing Gibbs energy to judge spontaneity.

Exam rules change between cycles. Confirm dates, syllabus and pattern against NTA's official site. We are not NTA.

Weightage: what the paper actually did

Chemical Thermodynamics carried 2 questions in the NEET 2026 paper. One paper is a data point, not a destiny: NTA moves weight between chapters every cycle, which is why the weightage tables deserve care. The stable facts are the paper's shape: 180 questions, 720 marks, +4 right, −1 wrong.

Sample PYQs from this chapter

Real questions NTA asked from Chemical Thermodynamics, with the answer and the working. Try before you peek.

Question 1 · NEET 2026

At a certain temperature, T (K), during a process, 500 J is absorbed by the system and work of 200 J is done by the system. Then change in internal energy of the system is:
  1. A.400 J
  2. B.300 J
  3. C.700 J
  4. D.500 J
Answer and explanation

Answer: B

By the first law of thermodynamics, ΔU=q+w\Delta U = q + w. Heat absorbed by the system gives q = +500 J, and work done by the system gives w = -200 J. Therefore ΔU=500−200=300\Delta U = 500 - 200 = 300 J.

Question 2 · NEET 2026

Consider the following reaction: 2A(g)+B(g)→2D(g)2A(g) + B(g) \rightarrow 2D(g). ΔU⊖=−10 kJ mol−1\Delta U^{\ominus} = -10\ \text{kJ mol}^{-1} and ΔS⊖=−44 JK−1\Delta S^{\ominus} = -44\ \text{JK}^{-1} at 298 K. Identify the correct option with ΔG⊖\Delta G^{\ominus} for the reaction and the spontaneity of the reaction at 298 K. (Given: R=8.31 J mol−1 K−1R = 8.31\ \text{J mol}^{-1}\ \text{K}^{-1})
  1. A.−1.635 kJ mol−1-1.635\ \text{kJ mol}^{-1}, spontaneous
  2. B.−0.63568 kJ mol−1-0.63568\ \text{kJ mol}^{-1}, spontaneous
  3. C.+0.63568 kJ mol−1+0.63568\ \text{kJ mol}^{-1}, non-spontaneous
  4. D.+1.635 kJ mol−1+1.635\ \text{kJ mol}^{-1}, non-spontaneous
Answer and explanation

Answer: C

With Δng=2−3=−1\Delta n_g = 2 - 3 = -1, ΔH⊖=ΔU⊖+ΔngRT=−10−298×8.311000=−12.48 kJ\Delta H^{\ominus} = \Delta U^{\ominus} + \Delta n_g RT = -10 - \frac{298\times8.31}{1000} = -12.48\ \text{kJ}. Then ΔG⊖=ΔH⊖−TΔS⊖=−12.48−298×(−44)1000=−12.48+13.112≈+0.636 kJ mol−1\Delta G^{\ominus} = \Delta H^{\ominus} - T\Delta S^{\ominus} = -12.48 - \frac{298\times(-44)}{1000} = -12.48 + 13.112 \approx +0.636\ \text{kJ mol}^{-1}. A positive ΔG⊖\Delta G^{\ominus} means the reaction is non-spontaneous. Why the other options are wrong: - (A) Wrong sign and magnitude; ignores the Δ\DeltangRT correction. - (B) Correct magnitude but wrong sign, so spontaneity is misjudged. - (D) Wrong magnitude (omits the Δ\DeltangRT term when converting Δ\DeltaU to Δ\DeltaH).

How to practise Chemical Thermodynamics

Chapter by chapter, against real past-paper questions, on a screen: NEET is computer based from 2027. Read the chapter once, then attempt its PYQs until your accuracy stops climbing, and let the wrong answers tell you which lines to reread. If the chapter keeps refusing to stick, that has its own guide. There is also a free 12-question mock on the exam screen.

Nearby Chemistry chapters

Questions

How many questions did Chemical Thermodynamics have in NEET 2026?

Chemical Thermodynamics carried 2 questions in the NEET 2026 paper. Weightage shifts between cycles, so treat it as a probability, not a promise.

Where can I practise Chemical Thermodynamics PYQs for NEET?

In the BrainIt app, free until the 2027 exam: its bank holds 50,000 past NTA questions tagged to all 83 chapters, including this one, with accuracy tracked per chapter. Sample questions from this chapter are on this page.

Is Chemical Thermodynamics important for NEET 2027?

It is part of the Chemistry syllabus the paper draws from, and NEET 2027 is computer based, so practise it on a screen. Your own accuracy in the chapter matters more than any weightage table: a heavy chapter you already score well on needs less time than a light one you keep missing.

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