JEE (Main) · Chemistry, Class 11
Thermodynamics: JEE Main PYQ, 2020–2025
Thermodynamics (Chemistry, Class 11) has been asked 119 times in the 111 JEE (Main) shift papers from 2020–2025: about 1.1 questions a paper, 5% of all Chemistry questions. It is the 11th most-asked of 21 Chemistry chapters. The 2025 shifts had 24 from it in all. Last checked 28 October 2026.
Questions per year, all shifts
Year by year.
| Year | Questions |
|---|---|
| 2020 | 11in 14 papers |
| 2021 | 26in 26 papers |
| 2022 | 25in 24 papers |
| 2023 | 13in 12 papers |
| 2024 | 20in 18 papers |
| 2025 | 24in 17 papers |
JEE (Main) is sat in many shifts a year, each with its own paper; the counts add up every shift. 2021–2024 papers had 30 questions a subject, the others 25.
Inside the chapter
Topics asked most.
- First law and state functions43
- Entropy and Gibbs energy31
- Enthalpy25
- Hess law20
Try it
10 Thermodynamics questions to practise.
Our own questions, written in the JEE Main pattern and checked, each with an explanation. Pick an option to see the answer.
Q1 · Chemistry · Thermodynamics
Given C(s) + O₂(g) → CO₂(g), ΔH = −393.5 kJ, and CO(g) + ½O₂(g) → CO₂(g), ΔH = −283.0 kJ, what is ΔH for C(s) + ½O₂(g) → CO(g)?
Answer: (A)
Adding C + O₂ → CO₂ (−393.5) and CO₂ → CO + ½O₂ (+283.0) gives C + ½O₂ → CO, with ΔH = −393.5 + 283.0 = −110.5 kJ.
Q2 · Chemistry · Thermodynamics
The enthalpy of solution of anhydrous CuSO₄(s) in excess water is −66.5 kJ mol⁻¹, while that of CuSO₄·5H₂O(s) is +11.7 kJ mol⁻¹. What is the enthalpy change for CuSO₄(s) + 5H₂O(l) → CuSO₄·5H₂O(s)?
Answer: (D)
Both solids give the same solution of CuSO₄(aq). Take the dissolution of anhydrous CuSO₄ (ΔH = −66.5) and reverse the dissolution of the pentahydrate (ΔH = −11.7). The solution cancels, leaving CuSO₄ + 5H₂O → CuSO₄·5H₂O. ΔH = −66.5 + (−11.7) = −78.2 kJ mol⁻¹.
Q3 · Chemistry · Thermodynamics
Given at 298 K: ΔfH°(H₂O, l) = −286 kJ mol⁻¹; ΔvapH°(H₂O) = +44 kJ mol⁻¹; bond dissociation enthalpy of H–H = 436 kJ mol⁻¹; bond dissociation enthalpy of O=O = 498 kJ mol⁻¹. Taking the O–H bond enthalpy in gaseous water as the average of its two O–H bonds, what is its value in kJ mol⁻¹?
Answer: (A)
Bond enthalpies apply to gaseous species, so first convert to H₂O(g): H₂(g) + ½O₂(g) → H₂O(l) is −286 kJ, and H₂O(l) → H₂O(g) is +44 kJ, giving ΔfH(H₂O, g) = −242 kJ mol⁻¹. For H₂(g) + ½O₂(g) → H₂O(g), ΔH = (bonds broken) − (bonds formed) = 436 + ½(498) − 2x. So −242 = 685 − 2x, giving 2x = 927 and x = 463.5 kJ mol⁻¹. Mistakes arise from skipping the liquid-to-gas correction, using a whole O=O instead of half, or flipping the sign of ΔH.
Q4 · Chemistry · Thermodynamics
Which of the following quantities depends only on the initial and final states of a system, and not on the path taken between them?
Answer: (C)
A state function depends only on the current state of the system, so its change depends only on the initial and final states. Internal energy, enthalpy, pressure, volume and temperature are state functions. Heat and work are not: they describe energy transferred during a process, and that depends on the route. Because ΔU is a state function, the first law ΔU = q + w means that q and w can vary along different paths while their sum stays the same.
Q5 · Chemistry · Thermodynamics
The enthalpy of neutralisation of HCN with NaOH is −12.1 kJ mol⁻¹, whereas that of a strong acid with a strong base (H⁺(aq) + OH⁻(aq) → H₂O(l)) is −57.1 kJ mol⁻¹. What is the enthalpy of ionisation of HCN, HCN(aq) → H⁺(aq) + CN⁻(aq)?
Answer: (C)
Neutralising HCN with NaOH can be split into two steps: (1) HCN → H⁺ + CN⁻ (ΔH_ion) and (2) H⁺ + OH⁻ → H₂O (−57.1). By Hess's law the sum equals the observed −12.1 kJ mol⁻¹. So ΔH_ion = −12.1 − (−57.1) = +45.0 kJ mol⁻¹. Part of the energy released by forming water is spent in ionising the weak acid.
Q6 · Chemistry · Thermodynamics
At 298 K the following enthalpy changes are known: (i) C(graphite) + O₂(g) → CO₂(g), ΔH = −393.5 kJ; (ii) 2CO(g) + O₂(g) → 2CO₂(g), ΔH = −566.0 kJ. What is ΔH, in kJ, for the reaction C(graphite) + CO₂(g) → 2CO(g)?
Answer: (A)
Hess's law: ΔH depends only on the initial and final states, so thermochemical equations can be added, reversed (sign of ΔH changes) or multiplied (ΔH scales). The target needs C and CO₂ on the left and 2CO on the right. Equation (i) already has C on the left and one CO₂ on the right. CO₂ must be on the left, so keep equation (i) as written and reverse equation (ii): 2CO₂ → 2CO + O₂, ΔH = +566.0 kJ. Adding: C + O₂ + 2CO₂ → CO₂ + 2CO + O₂. Cancelling O₂ and one CO₂ gives C + CO₂ → 2CO. ΔH = −393.5 + 566.0 = +172.5 kJ.
Q7 · Chemistry · Thermodynamics
An ideal gas is compressed at a constant external pressure of 2 atm. Its volume decreases from 10 L to 4 L. Taking 1 L·atm = 101.3 J, what is the work done on the gas?
Answer: (A)
At constant pressure, work is w = -PΔV, where w is the work done on the system. Compression means ΔV is negative, so w is positive. Substituting P = 2 atm and ΔV = -6 L gives w = 12 L·atm. Converting with 1 L·atm = 101.3 J gives about +1216 J.
Q8 · Chemistry · Thermodynamics
Given: 2Al(s) + 3/2 O₂(g) → Al₂O₃(s), ΔH = −1676 kJ; and 2Fe(s) + 3/2 O₂(g) → Fe₂O₃(s), ΔH = −824 kJ. What is ΔH, in kJ, for the thermite reaction Fe₂O₃(s) + 2Al(s) → Al₂O₃(s) + 2Fe(s)?
Answer: (C)
Keep the first equation as written, since Al₂O₃ is a product. Reverse the second, since Fe₂O₃ is a reactant, which changes its ΔH to +824 kJ. Adding them cancels 3/2 O₂, giving Fe₂O₃ + 2Al → Al₂O₃ + 2Fe with ΔH = −1676 + 824 = −852 kJ.
Q9 · Chemistry · Thermodynamics
For a reaction carried out at 300 K and constant pressure, ΔH = −60 kJ mol⁻¹ and the entropy change of the system is −150 J K⁻¹ mol⁻¹. What is the total entropy change of the system and surroundings?
Answer: (C)
ΔS_surr = −ΔH/T = −(−60 000 J mol⁻¹)/300 K = +200 J K⁻¹ mol⁻¹. ΔS_total = ΔS_sys + ΔS_surr = −150 + 200 = +50 J K⁻¹ mol⁻¹. The total is positive, so the reaction is spontaneous although the entropy of the system falls.
Q10 · Chemistry · Thermodynamics
At 298 K: (i) C(s) + O₂(g) → CO₂(g), ΔH = −393.5 kJ mol⁻¹; (ii) CO(g) + ½O₂(g) → CO₂(g), ΔH = −283.0 kJ mol⁻¹; (iii) H₂(g) + ½O₂(g) → H₂O(g), ΔH = −242.0 kJ mol⁻¹. What is the enthalpy change for the water-gas reaction C(s) + H₂O(g) → CO(g) + H₂(g)?
Answer: (A)
Build the target from the given equations. Start with (i) − (ii): C(s) + ½O₂ → CO(g), ΔH = −110.5. The target also needs H₂O → H₂ + ½O₂, which is the reverse of (iii), so ΔH = +242.0. Sum: C(s) + H₂O(g) → CO(g) + H₂(g), ΔH = −110.5 + 242.0 = +131.5 kJ mol⁻¹. Each equation is used once, with the sign set by its direction.
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All 119 real JEE Main questions from Thermodynamics.
The questions from the actual papers, with the official answer and an explanation for each — chapter by chapter, or as full timed papers.
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About Thermodynamics in JEE Main.
How many questions come from Thermodynamics in JEE Main?
Thermodynamics (Chemistry, Class 11) has been asked 119 times in the 111 JEE (Main) shift papers from 2020–2025: about 1.1 questions a paper, 5% of all Chemistry questions. It is the 11th most-asked of 21 Chemistry chapters. The 2025 shifts had 24 from it in all.
Which topics of Thermodynamics are asked most in JEE Main?
First law and state functions (43), Entropy and Gibbs energy (31), Enthalpy (25), Hess law (20), counted over the papers from 2020–2025.
Is Thermodynamics important for JEE Main 2027?
It is in the syllabus in force, so it can be asked. By past papers it ranks 11 of 21 in Chemistry; use that to decide how much time it gets, not whether to study it.
Where can I solve Thermodynamics JEE Main previous year questions?
On Nikaalle, free: sign up and practise every past JEE Main question from Thermodynamics with an explanation for each, or sit the full papers timed. The ten questions on this page are our own, written in the same pattern.
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