How much of JEE Main comes from Chemical Thermodynamics?

JEEGenie estimates about 5% of JEE Main marks come from Chemical Thermodynamics, based on recent papers. The share varies from year to year.

Which formulas matter most in Chemical Thermodynamics?

The Chemical Thermodynamics sheet on this page has 19 formula entries, starting with First Law of Thermodynamics, Work Done in Reversible Processes, Enthalpy and Enthalpy Change, Hess's Law, Standard Enthalpy of Formation. Each one comes with when to use it and the trap to avoid.

What topics does Chemical Thermodynamics cover for JEE?

Chemical Thermodynamics covers Enthalpy and Hess's Law, Entropy, Gibbs Free Energy, First Law Applications, Spontaneity, Thermochemistry.

How should I revise Chemical Thermodynamics before the exam?

Learn each formula together with its condition of use, read the trap listed beside it, then practise questions and review every mistake. The practice link on this page opens Chemical Thermodynamics questions on JEEGenie.

How many previous-year questions are there on Chemical Thermodynamics?

JEEGenie has 414 previous-year questions on Chemical Thermodynamics, and this page shows a few with worked explanations.

Is practice on Chemical Thermodynamics free on JEEGenie?

Sign-up is free and includes a 7-day Pro trial. After the trial, free accounts keep a limited daily quota.

Chemical Thermodynamics for JEE Main & Advanced — formulas, what is asked, and traps

JEEGenie estimates about 5% of JEE Main marks come from Chemical Thermodynamics. This page covers 19 formula entries, each with when to use it, what JEE asks and the traps. A common slip: Confusing closed system with isolated system—closed allows energy exchange, isolated does not.

Key points to know

System, Surroundings, and Boundary. The system is the part of the universe under study; surroundings are everything else; the boundary separates them. Types: open (matter & energy exchange), closed (only energy), isolated (neither).
State Functions vs. Path Functions. State functions depend only on the current state (P, V, T, U, H, S, G); path functions depend on the path taken (work W, heat Q).
First Law of Thermodynamics. Energy is conserved: ΔU = q + w. For a system, internal energy change equals heat added plus work done on the system.
Standard Enthalpy of Formation. ΔH_f° is the enthalpy change when 1 mole of compound is formed from its elements in their standard states (1 bar, 298 K). For elements in standard state, ΔH_f° = 0.
Spontaneity and Second Law. A process is spontaneous if total entropy of universe increases (ΔS_univ > 0). Second law: ΔS_univ = ΔS_sys + ΔS_surr ≥ 0.
Gibbs Free Energy. Gibbs free energy G = H - TS. At constant T and P, spontaneity is determined by ΔG < 0. ΔG = ΔH - TΔS.
Third Law of Thermodynamics. The entropy of a perfect crystalline substance is zero at absolute zero (0 K). This allows calculation of absolute entropies.
Carnot Cycle and Efficiency. Carnot cycle is a reversible cycle with two isothermal and two adiabatic steps. Efficiency η = 1 - T_cold/T_hot, maximum for any engine.
Enthalpy of Combustion, Neutralization, and Solution. ΔH_comb: heat released when 1 mol substance burns completely in O2. ΔH_neut: heat released when acid and base neutralize (≈ -57 kJ/mol for strong acid-strong base). ΔH_soln: heat change when 1 mol solute dissolves.
Spontaneity and Temperature: ΔG = ΔH - TΔS. Spontaneity depends on signs of ΔH and ΔS. If both positive, spontaneous at high T; if both negative, spontaneous at low T; if ΔH negative and ΔS positive, always spontaneous.
Clausius-Clapeyron Equation. Relates vapor pressure to temperature: ln(P2/P1) = -ΔH_vap/R (1/T2 - 1/T1). Used for phase transitions.

Formula sheet

Name	Formula	When to use
First Law of Thermodynamics	$Δ U = q + w$	Calculate energy changes in any thermodynamic process.
Work Done in Reversible Processes	$w = - \int_{V_{1}}^{V_{2}} P d V$ ; for isothermal reversible: $w = - n R T \ln (\frac{V_{2}}{V_{1}})$	Calculate work for reversible expansion/compression of ideal gas.
Enthalpy and Enthalpy Change	$H = U + P V$ ; $Δ H = Δ U + Δ (P V)$ ; for ideal gas: $Δ H = n C_{p} Δ T$	Calculate heat absorbed/released at constant pressure (most reactions).
Hess's Law	$Δ H_{reaction} = \sum Δ H_{f}^{\circ} (products) - \sum Δ H_{f}^{\circ} (reactants)$	Calculate ΔH for reactions where direct measurement is difficult.
Standard Enthalpy of Formation	$Δ H_{f}^{\circ}$	Calculate reaction enthalpies using standard formation data.
Spontaneity and Second Law	$Δ S_{univ} = Δ S_{sys} + Δ S_{surr} \geq 0$	Predict spontaneity of a process.
Entropy Change for Ideal Gas	$Δ S = n C_{v} \ln (\frac{T_{2}}{T_{1}}) + n R \ln (\frac{V_{2}}{V_{1}})$ ; also $Δ S = n C_{p} \ln (\frac{T_{2}}{T_{1}}) - n R \ln (\frac{P_{2}}{P_{1}})$	Calculate entropy change for ideal gas undergoing any process.
Gibbs Free Energy	$G = H - T S$ ; $Δ G = Δ H - T Δ S$	Predict spontaneity at constant T and P.
Gibbs Free Energy and Equilibrium	$Δ G^{\circ} = - R T \ln K$ ; $Δ G = Δ G^{\circ} + R T \ln Q$	Calculate equilibrium constant from thermodynamic data or vice versa.
Temperature Dependence of ΔG and K	$\ln (\frac{K_{2}}{K_{1}}) = - \frac{Δ H^{\circ}}{R} (\frac{1}{T_{2}} - \frac{1}{T_{1}})$	Calculate K at different temperatures or find ΔH° from K vs T data.
Third Law of Thermodynamics	$S (0 K) = 0$ for perfect crystal	Calculate absolute entropies of substances at given temperatures.
Standard Entropy Change of Reaction	$Δ S_{rxn}^{\circ} = \sum S^{\circ} (products) - \sum S^{\circ} (reactants)$	Calculate entropy change for a reaction from standard entropy data.
Adiabatic Process for Ideal Gas	$P V^{γ} = constant$ ; $γ = \frac{C_{p}}{C_{v}}$ ; $w = Δ U = n C_{v} (T_{2} - T_{1}) = \frac{n R (T_{2} - T_{1})}{γ - 1}$ (work done by the gas $= \frac{n R (T_{1} - T_{2})}{γ - 1}$ )	Calculate work, temperature, or volume changes in adiabatic processes.
Carnot Cycle and Efficiency	$η = 1 - \frac{T_{c}}{T_{h}}$	Calculate maximum possible efficiency of a heat engine.
Bond Enthalpy and Reaction Enthalpy	$Δ H_{rxn} = \sum Δ H_{bond break} - \sum Δ H_{bond form}$	Estimate reaction enthalpy when formation data is unavailable.
Kirchhoff's Law	$Δ H (T_{2}) = Δ H (T_{1}) + Δ C_{p} (T_{2} - T_{1})$ ; $Δ C_{p} = \sum C_{p} (products) - \sum C_{p} (reactants)$	Calculate ΔH at a different temperature from known ΔH at another temperature.
Spontaneity and Temperature: ΔG = ΔH - TΔS	$Δ G = Δ H - T Δ S$	Determine temperature range for spontaneity.
Trouton's Rule	$Δ S_{vap} = \frac{Δ H_{vap}}{T_{b}} \approx 85 {J mol}^{- 1} K^{- 1}$	Estimate ΔH_vap from boiling point or vice versa.
Clausius-Clapeyron Equation	$\ln (\frac{P_{2}}{P_{1}}) = - \frac{Δ H_{vap}}{R} (\frac{1}{T_{2}} - \frac{1}{T_{1}})$	Calculate vapor pressure at different temperatures or find ΔH_vap.

Open the Chemical Thermodynamics formula page

What JEE asks

System, Surroundings, and Boundary. JEE often asks to classify a given process (e.g., gas in a cylinder with piston) as open/closed/isolated.
State Functions vs. Path Functions. JEE asks: 'Which of the following is a state function?' with options like q, w, ΔH, ΔU.
First Law of Thermodynamics. JEE gives a process (e.g., isothermal expansion) and asks for ΔU, q, or w.
Work Done in Reversible Processes. JEE often asks to compare work done in reversible vs irreversible isothermal expansion.
Enthalpy and Enthalpy Change. JEE asks: 'For a reaction, ΔU = -x kJ, Δn = 2, T = 300 K. Find ΔH.'
Hess's Law. JEE gives multiple reactions with known ΔH and asks for ΔH of a target reaction.
Standard Enthalpy of Formation. JEE provides a table of ΔH_f° and asks for ΔH of a reaction.
Spontaneity and Second Law. JEE asks: 'Which of the following is always spontaneous?' based on ΔH and ΔS signs.
Entropy Change for Ideal Gas. JEE gives a gas expansion and asks for ΔS_sys, ΔS_surr, and ΔS_univ.
Gibbs Free Energy. JEE asks: 'For a reaction, ΔH = +ve, ΔS = +ve. At what temperature is it spontaneous?'
Gibbs Free Energy and Equilibrium. JEE gives ΔG° and asks for K at a given temperature.
Temperature Dependence of ΔG and K. JEE gives K at two temperatures and asks for ΔH° or K at a third temperature.
Third Law of Thermodynamics. JEE asks: 'Which substance has zero entropy at 0 K?' or uses S° values to calculate ΔS° for reactions.
Standard Entropy Change of Reaction. JEE provides S° values and asks for ΔS° of a reaction.
Predicting Sign of ΔS. JEE asks: 'For which of the following reactions is ΔS positive?'
Adiabatic Process for Ideal Gas. JEE gives initial P, V, T and asks for final T or work after adiabatic expansion.
Carnot Cycle and Efficiency. JEE asks: 'A Carnot engine operates between 500 K and 300 K. Find efficiency.'
Bond Enthalpy and Reaction Enthalpy. JEE gives bond enthalpies and asks for ΔH of a reaction.
Enthalpy of Combustion, Neutralization, and Solution. JEE asks: 'Given ΔH_comb of C, H2, and CH4, find ΔH_f of CH4.'
Kirchhoff's Law. JEE gives ΔH at 298 K and heat capacities, asks for ΔH at 500 K.
Spontaneity and Temperature: ΔG = ΔH - TΔS. JEE asks: 'For a reaction with ΔH = +30 kJ, ΔS = +100 J/K, find temperature above which it is spontaneous.'
Trouton's Rule. JEE asks: 'The boiling point of a liquid is 400 K. Estimate ΔH_vap using Trouton's rule.'
Clausius-Clapeyron Equation. JEE gives vapor pressure at two temperatures and asks for ΔH_vap or boiling point at a given pressure.

Common traps

Confusing closed system with isolated system—closed allows energy exchange, isolated does not. System, Surroundings, and Boundary
Thinking heat and work are state functions—they are path functions. State Functions vs. Path Functions
Sign convention: w positive if work done on system, negative if by system; q positive if heat added. First Law of Thermodynamics
Using the isothermal formula for adiabatic processes—different formula applies. Work Done in Reversible Processes
Assuming ΔH = q always; true only at constant pressure. Enthalpy and Enthalpy Change
Forgetting to multiply by stoichiometric coefficients. Hess's Law
Assuming ΔH_f° of O2(g) is not zero; it is zero. Standard Enthalpy of Formation
Thinking ΔS_sys > 0 always means spontaneity; need ΔS_univ > 0. Spontaneity and Second Law
The formula gives ΔS_sys for any process between the given states (S is a state function); for irreversible processes ΔS_surr ≠ -ΔS_sys, so ΔS_univ > 0. Entropy Change for Ideal Gas
Assuming ΔG < 0 always means fast reaction; it only indicates spontaneity, not kinetics. Gibbs Free Energy
Using ΔG° directly for non-standard conditions; must use ΔG = ΔG° + RT ln Q. Gibbs Free Energy and Equilibrium
Assuming ΔH° is constant over temperature range (often valid for small ranges). Temperature Dependence of ΔG and K
Assuming entropy is zero at 0 K for all substances; only perfect crystals. Third Law of Thermodynamics
Assuming all reactions with gas evolution have positive ΔS; check if moles of gas increase. Predicting Sign of ΔS

Topics in Chemical Thermodynamics

Enthalpy and Hess's Law
Entropy
Gibbs Free Energy
First Law Applications
Spontaneity
Thermochemistry

Previous-year questions on Chemical Thermodynamics

JEEGenie has 414 previous-year questions on Chemical Thermodynamics. A few worked examples follow.

Year	Questions
2026	22
2025	32
2024	23
2023	29
2022	25
2021	25
2020	18
2019	43
2018	16
2017	18
2016	11
2015	9
2014	9
2013	15
2012	8
2011	2
2010	7
2009	3
2008	4
2007	8
2006	9
2005	4
2004	7
2003	7
2002	11
2001	4
2000	5
1999	2
1998	2
1997	4
1996	2
1994	1
1993	4
1992	3
1991	2
1990	2
1989	3
1986	2
1985	3
1984	4
1983	2
1982	3
1981	1

Question 1 medium

Consider the following data. (i)

2 Al(s) + 6 HCl(aq) \to {Al}_{2} {Cl}_{6} (aq) + 3 H_{2} (g) + 1200 kJ/mol

(ii)

H_{2} (g) + {Cl}_{2} (g) \to 2 HCl(g) + 164 kJ/mol

(iii)

HCl(g) + aq \to HCl(aq) + 83 kJ/mol

(iv)

{Al}_{2} {Cl}_{6} (s) + aq \to {Al}_{2} {Cl}_{6} (aq) + 663 kJ/mol

The enthalpy of formation of anhydrous solid

{Al}_{2} {Cl}_{6}

is :

$- 648 kJ {mol}^{- 1}$
$- 1350 kJ {mol}^{- 1}$
$- 2002 kJ {mol}^{- 1}$
$- 1527 kJ {mol}^{- 1}$

Answer: D

Explanation. **Principle:** Hess's law. The target is the formation reaction

2 Al(s) + 3 {Cl}_{2} (g) \to {Al}_{2} {Cl}_{6} (s)

. Heat written on the product side means the reaction is exothermic, so the given enthalpies are

Δ H_{1} = - 1200

Δ H_{2} = - 164

Δ H_{3} = - 83

Δ H_{4} = - 663

kJ/mol. **Combining the steps:** (i) as written:

2 Al + 6 HCl(aq) \to {Al}_{2} {Cl}_{6} (aq) + 3 H_{2}

- 12003 \times

(ii):

3 H_{2} + 3 {Cl}_{2} \to 6 HCl(g)

3 (- 164) = - 4926 \times

(iii):

6 HCl(g) + aq \to 6 HCl(aq)

6 (- 83) = - 498

reverse of (iv):

{Al}_{2} {Cl}_{6} (aq) \to {Al}_{2} {Cl}_{6} (s) + aq

+ 663

Adding,

HCl(aq)

HCl(g)

H_{2}

and

{Al}_{2} {Cl}_{6} (aq)

cancel, leaving

2 Al(s) + 3 {Cl}_{2} (g) \to {Al}_{2} {Cl}_{6} (s)

Δ H_{f} = - 1200 - 492 - 498 + 663 = - 1527 {kJ mol}^{- 1}

Answer: (D)

From NTA/IIT papers: JEE Main 2026 · 2 Apr Shift 1

Question 2 hard

Arrange the following in order of magnitude of work done by the system/on the system at constant temperature. (a)

∣ w_{reversible} ∣

for expansion in infinite stages. (b)

∣ w_{irreversible} ∣

for expansion in single stage. (c)

∣ w_{reversible} ∣

for compression in infinite stages. (d)

∣ w_{irreversible} ∣

for compression in single stage. Choose the correct answer from the options given below :

$_{} d > c = a > b$
$c = a > d > b$
$a > c > b > d$
$a > b > c > d$

Answer: A

Explanation. For isothermal processes, magnitude of reversible work is same for expansion and compression:

∣ w_{rev, exp} ∣ = ∣ w_{rev, comp} ∣ = n R T \ln (V_{f} / V_{i})

. For irreversible single-stage expansion against constant external pressure, work done is less than reversible. For irreversible single-stage compression, work done on system is greater than reversible. Thus:

∣ w_{irr, comp} ∣ > ∣ w_{rev, comp} ∣ = ∣ w_{rev, exp} ∣ > ∣ w_{irr, exp} ∣

. So order: d > c = a > b. The answer is option A.

From NTA/IIT papers: JEE Main 2025 · 2 Apr Shift 2

Question 3 easy

The enthalpy of formation of ethane

(C_{2} H_{6})

from ethylene by addition of hydrogen where the bond-energies of

C - H, C - C, C = C, H - H

are

414 k J, 347 k J, 615 k J

and

435 k J

respectively is

-

__________

k J

Answer: 125

Explanation. **Principle.** The reaction enthalpy is the energy needed to break the bonds in the reactants minus the energy released in forming the bonds in the products. Adding

H_{2}

across the double bond breaks one

C = C

and one

H - H

and makes one

C - C

and two new

C - H

bonds. The reaction is

H_{2} C = C H_{2} + H_{2} \to H_{3} C - C H_{3}

Δ H = \sum (bonds broken) - \sum (bonds formed)

**Bonds broken (reactants):** - one

C = C

615

- four

C - H

4 \times 414 = 1656

- one

H - H

435

Total

= 615 + 1656 + 435 = 2706

kJ. **Bonds formed (product):** - one

C - C

347

- six

C - H

6 \times 414 = 2484

Total

= 347 + 2484 = 2831

kJ.

Δ H = 2706 - 2831 = - 125 kJ

The enthalpy change is

- 125

kJ, so the blank is filled with

125

. Answer: 125

From NTA/IIT papers: JEE Main 2024 · 4 Apr Shift 1

Question 4 medium

The value of

\log K

for the reaction

A ⇌ B

298 K

is ___________. (Nearest integer) Given:

Δ H^{\circ} = - 54.07 k J {m o l}^{- 1}

Δ S^{\circ} = 10 J K^{- 1} {m o l}^{- 1}

(Take

2.303 \times 8.314 \times 298 = 5705

)

Answer: 10

Explanation. Use the standard Gibbs energy relation:

Δ G^{\circ} = Δ H^{\circ} - T Δ S^{\circ}

Given:

Δ H^{\circ} = - 54.07 {kJ mol}^{- 1} = - 54070 {J mol}^{- 1}

T = 298 K, Δ S^{\circ} = 10 {J K}^{- 1} {mol}^{- 1}

Δ G^{\circ} = - 54070 - (298) (10)

Δ G^{\circ} = - 54070 - 2980 = - 57050 {J mol}^{- 1}

Now,

Δ G^{\circ} = - 2.303 R T \log K

- 57050 = - (2.303) (8.314) (298) \log K

Using

2.303 \times 8.314 \times 298 = 5705

- 57050 = - 5705 \log K

\log K = \frac{57050}{5705} = 10

Answer: 10

From NTA/IIT papers: JEE Main 2023 · 6 Apr Shift 1

Question 5 hard

2 mol of

H g (g)

is combusted in a fixed volume bomb calorimeter with excess of

O_{2}

at 298 K and 1 atm into

H g O (s)

. During the reaction, temperature increases from 298.0 K to 312.8 K. If heat capacity of the bomb calorimeter and enthalpy of formation of

H g (g)

are 20.00 kJ K

^{- 1}

and 61.32 kJ mol

^{- 1}

at 298 K, respectively, the calculated standard molar enthalpy of formation of

H g O (s)

at 298 K is X kJ mol

^{- 1}

. The value of

∣ X ∣

is ______. [Given: Gas constant R = 8.3 J K

^{- 1}

mol

^{- 1}

]

Answer: 90.39

Explanation. We have the reaction:

2 H g (g) + O_{2} (g) \to 2 H g O (s)

In a bomb calorimeter, the heat released at constant volume is:

q_{V} = - C_{cal} Δ T = - 20.00 \times (312.8 - 298.0) = - 20.00 \times 14.8 = - 296.0 kJ

This is for 2 mol of Hg, so per mole of Hg:

Δ U_{comb, per mol} = \frac{- 296.0}{2} = - 148.0 {kJ mol}^{- 1}

The relation between

Δ H

and

Δ U

for the reaction per mole of Hg is:

Δ H = Δ U + Δ n_{g} R T

Here,

Δ n_{g}

per mole of Hg is: Reactant gases: 1 mol

O_{2}

per 2 mol Hg

\to

per mol Hg,

\frac{1}{2}

mol

O_{2}

consumed, and Hg(g) is also a gas. So per mole of Hg: Gaseous reactants: 1 (Hg) + 0.5 (

O_{2}

) = 1.5 mol gas Gaseous products: 0 Thus

Δ n_{g} = 0 - 1.5 = - 1.5

per mole of Hg.

Δ H_{comb, per mol} = - 148.0 + (- 1.5) (8.3 \times 10^{- 3}) (298)

= - 148.0 - 3.71 = - 151.71 {kJ mol}^{- 1}

This is

Δ H

for:

H g (g) + \frac{1}{2} O_{2} (g) \to H g O (s)

So:

Δ H_{f}^{\circ} (H g O (s)) = Δ H_{comb} + Δ H_{f}^{\circ} (H g (g))

= - 151.71 + 61.32 = - 90.39 {kJ mol}^{- 1}

Thus:

∣ X ∣ = 90.39

90.39

From NTA/IIT papers: JEE Advanced 2022

Frequently asked questions

How much of JEE Main comes from Chemical Thermodynamics?: JEEGenie estimates about 5% of JEE Main marks come from Chemical Thermodynamics, based on recent papers. The share varies from year to year.
Which formulas matter most in Chemical Thermodynamics?: The Chemical Thermodynamics sheet on this page has 19 formula entries, starting with First Law of Thermodynamics, Work Done in Reversible Processes, Enthalpy and Enthalpy Change, Hess's Law, Standard Enthalpy of Formation. Each one comes with when to use it and the trap to avoid.
What topics does Chemical Thermodynamics cover for JEE?: Chemical Thermodynamics covers Enthalpy and Hess's Law, Entropy, Gibbs Free Energy, First Law Applications, Spontaneity, Thermochemistry.
How should I revise Chemical Thermodynamics before the exam?: Learn each formula together with its condition of use, read the trap listed beside it, then practise questions and review every mistake. The practice link on this page opens Chemical Thermodynamics questions on JEEGenie.
How many previous-year questions are there on Chemical Thermodynamics?: JEEGenie has 414 previous-year questions on Chemical Thermodynamics, and this page shows a few with worked explanations.
Is practice on Chemical Thermodynamics free on JEEGenie?: Sign-up is free and includes a 7-day Pro trial. After the trial, free accounts keep a limited daily quota.