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Chemical Thermodynamics JEE Main PYQs, 2020 to 2025

Chemical Thermodynamics has 104 questions in JEE Main papers from 2020 to 2025, about 17 per year across all shifts. The most asked topic is Enthalpy and Heat Capacity (34 questions), followed by First Law and Internal Energy (23) and Gibbs Free Energy (17).

Updated

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104Total PYQs
6Years covered
7Topics
2025 Apr S2Latest paper

Chemical Thermodynamics questions per year

YearQuestions
202515
202418
202322
202219
202119
202011

Topic-wise weightage

TopicQuestionsShare of chapterLast asked
Enthalpy and Heat Capacity3433%2025
First Law and Internal Energy2322%2025
Gibbs Free Energy1716%2025
Hess's Law1312%2025
Entropy and Spontaneity77%2025
Bond Enthalpy66%2024
Lattice Energy and Born-Haber Cycle44%2023

Sample questions from recent papers

JEE Main 2025 · 4 Apr · Shift 2 · Enthalpy and Heat Capacity

Consider the given data :

(a) HCl(g)+10H2O(l)→HCl.10H2OΔH=−69.01 kJ mol−1\mathrm{HCl}(\mathrm{g})+10 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{HCl} .10 \mathrm{H}_2 \mathrm{O} \Delta \mathrm{H}=-69.01 \mathrm{~kJ} \mathrm{~mol}^{-1}

(b) HCl(g)+40H2O(l)→HCl.40H2OΔH=−72.79 kJ mol−1\mathrm{HCl}(\mathrm{g})+40 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{HCl} .40 \mathrm{H}_2 \mathrm{O} \Delta \mathrm{H}=-72.79 \mathrm{~kJ} \mathrm{~mol}^{-1}

Choose the correct statement :

  1. (A)The heat of dilution for the HCl(HCl.10H2O\mathrm{HCl}\left(\mathrm{HCl} .10 \mathrm{H}_2 \mathrm{O}\right. to HCl.40H2O)\left.\mathrm{HCl} .40 \mathrm{H}_2 \mathrm{O}\right) is 3.78 kJ mol−13.78 \mathrm{~kJ} \mathrm{~mol}^{-1}.
  2. (B) Dissolution of gas in water is an endothermic process.
  3. (C) The heat of solution depends on the amount of solvent.
  4. (D)The heat of formation of HCl solution is represented by both (a) and (b).

Answer: (C)

Step-by-step solution

JEE Main 2025 · 4 Apr · Shift 1 · First Law and Internal Energy

One mole of an ideal gas expands isothermally and reversibly from 10dm310 \mathrm{dm}^3 to 20dm320 \mathrm{dm}^3 at 300 K . ΔU,q\Delta \mathrm{U}, \mathrm{q} and work done in the process respectively are

Given: R=8.3 J K−1 mol−1\mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}

ln⁡10=2.3\ln 10=2.3

log⁡2=0.30\log 2=0.30

log⁡3=0.48\log 3=0.48

  1. (A) 0,21.84 kJ,−1.726 J0,21.84 \mathrm{~kJ},-1.726 \mathrm{~J}
  2. (B) 0,21.84 kJ,21.84 kJ0,21.84 \mathrm{~kJ}, 21.84 \mathrm{~kJ}
  3. (C) 0,1.718 kJ,−1.718 kJ0,1.718 \mathrm{~kJ},-1.718 \mathrm{~kJ}
  4. (D) 0,−17.18 kJ,1.718 J0,-17.18 \mathrm{~kJ}, 1.718 \mathrm{~J}

Answer: (C)

Step-by-step solution

JEE Main 2025 · 4 Apr · Shift 1 · Gibbs Free Energy

Let us consider a reversible reaction at temperature, T. In this reaction, both ΔH\Delta \mathrm{H} and ΔS\Delta \mathrm{S} were observed to have positive values. If the equilibrium temperature is Te , then the reaction becomes spontaneous at:

  1. (A) Te>T\mathrm{Te}>\mathrm{T}
  2. (B) T>Te\mathrm{T}>\mathrm{Te}
  3. (C) T=Te\mathrm{T}=\mathrm{Te}
  4. (D) Te=5 T\mathrm{Te}=5 \mathrm{~T}

Answer: (B)

Step-by-step solution

JEE Main 2025 · 29 Jan · Shift 2 · Hess's Law

If C\quad C (diamond )→C) \rightarrow C (graphite) +X kJ mol−1+X \mathrm{~kJ} \mathrm{~mol}^{-1}

C (diamond) +O2( g)→CO2( g)+YkJmol−1+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})+\mathrm{Y} \mathrm{kJ} \mathrm{mol}{ }^{-1}

C (graphite) +O2( g)→CO2( g)+ZkJmol−1+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})+\mathrm{Z} \mathrm{kJ} \mathrm{mol}^{-1}

at constant temperature. Then

  1. (A)

    −X = Y + Z

  2. (B)

    X = Y − Z

  3. (C)

    X = −Y + Z

  4. (D)

    X = Y + Z

Answer: (B)

Step-by-step solution

JEE Main 2025 · 24 Jan · Shift 1 · Entropy and Spontaneity

Standard entropies of X2,Y2\mathrm{X}_2, \mathrm{Y}_2 and XY5\mathrm{XY}_5 are 70, 50 and 110 J K−1 mol−1110 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} respectively. The temperature in Kelvin at which the reaction

12X2+52Y2⇌XY5ΔH⊖=−35 kJ mol−1\frac{1}{2} \mathrm{X}_2+\frac{5}{2} \mathrm{Y}_2 \rightleftharpoons \mathrm{XY}_5 \Delta \mathrm{H}^{\ominus}=-35 \mathrm{~kJ} \mathrm{~mol}^{-1}

will be at equilibrium is __________ (Nearest integer)

Answer: 700

Step-by-step solution

JEE Main 2024 · 4 Apr · Shift 1 · Bond Enthalpy

The enthalpy of formation of ethane (C2H6)(\mathrm{C}_2 \mathrm{H}_6) from ethylene by addition of hydrogen where the bond-energies of C−H,C−C,C=C,H−H\mathrm{C}-\mathrm{H}, \mathrm{C}-\mathrm{C}, \mathrm{C}=\mathrm{C}, \mathrm{H}-\mathrm{H} are 414 kJ,347 kJ,615 kJ414 \mathrm{~kJ}, 347 \mathrm{~kJ}, 615 \mathrm{~kJ} and 435 kJ435 \mathrm{~kJ} respectively is −- __________ kJ\mathrm{kJ}

Answer: 125

Step-by-step solution

All 104 questions are in the free PDF above, grouped by topic.

Frequently asked

How many questions come from Chemical Thermodynamics in JEE Main?

104 questions from 2020 to 2025, about 17 per year. In 2025 it had 15.

Which Chemical Thermodynamics topics are most important for JEE Main?

By past papers: Enthalpy and Heat Capacity (33%), First Law and Internal Energy (22%) and Gibbs Free Energy (16%) of all Chemical Thermodynamics questions.

Is the Chemical Thermodynamics PYQ PDF free?

Yes. The PDF, the answer key and every step-by-step solution on PrepWiser are free, with no login needed to practise.

More Chemistry chapters

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