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JEE Main · Chemistry · Class 11
Chemical thermodynamics carries one to two of the 25 JEE Main Chemistry questions in our reading of past papers. Almost all of them test the same three things: signs, the difference between ΔH and ΔU, and whether a reaction is spontaneous at the temperature given.
Quick answer
Questions per paper
1–2
Our analysis of past papers
Marks at stake
4–8
At +4 / −1 per question
Difficulty
Moderate
Sign errors decide the marks
Class
11
Feeds equilibrium and electrochemistry
BrainStride analysis of past papers
ΔU = q + w with w the work done on the system. Heat absorbed by the system is positive; work done by the system in expanding is negative. Fixing this convention once removes most of the errors in the chapter.
U, H, S and G depend only on the state, so their changes are the same by any route. q and w depend on the path — reversible and irreversible expansions between the same states give different work.
H = U + PV, and for reactions involving gases ΔH = ΔU + Δn_g RT. Δn_g is the change in the number of moles of gas only, so a reaction with no gas-phase change has ΔH = ΔU.
q_v = ΔU and q_p = ΔH, which is why bomb calorimetry measures ΔU and open-vessel work measures ΔH. For an ideal gas, C_p − C_v = R per mole.
Because H is a state function, reaction enthalpies add like the equations that produce them: ΔH_rxn = ΣΔH_f(products) − ΣΔH_f(reactants). With bond enthalpies the rule inverts: bonds broken minus bonds formed.
ΔG = ΔH − TΔS at constant T and P. A negative ΔG means spontaneous. When ΔH and ΔS share a sign, temperature decides: an endothermic reaction with positive ΔS becomes spontaneous above T = ΔH/ΔS.
ΔU = q + w
IUPAC convention: w is work done on the system.
w = −p_ext ΔV
Irreversible expansion against constant external pressure.
w = −2.303 nRT log(V₂/V₁)
Ideal gas; ΔU = 0, so q = −w.
ΔH = ΔU + Δn_g RT
Δn_g counts gaseous moles only.
q_v = ΔU · q_p = ΔH
Bomb calorimeter gives ΔU.
C_p − C_v = R
Per mole; γ = C_p/C_v.
ΔH_rxn = Σ ΔH_f(products) − Σ ΔH_f(reactants)
Elements in their standard states have ΔH_f = 0.
ΔH = Σ BE(broken) − Σ BE(formed)
Opposite order to the formation-enthalpy rule.
ΔS = q_rev/T · ΔS_universe > 0 for a spontaneous change
Entropy of the system alone may fall.
ΔG = ΔH − TΔS · ΔG° = −2.303RT log K
ΔG < 0 ⇒ spontaneous at that temperature.
Why it happens
Older books use w as work done by the system, so two conventions circulate.The fix
Use ΔU = q + w with w = −p_extΔV. Expansion increases V, so w is negative and the system loses energy.Why it happens
Δn is read as “change in moles” rather than “change in moles of gas”.The fix
Count gaseous species only. For C(s) + O₂(g) → CO₂(g), Δn_g = 0, so ΔH = ΔU.Why it happens
The degree symbol is easy to lose while copying.The fix
ΔG = ΔG° + 2.303RT log Q. Only when every species is in its standard state does ΔG equal ΔG°.Why it happens
Energy release is taught as the driver of change.The fix
Spontaneity needs ΔG < 0. An exothermic reaction with a large negative ΔS becomes non-spontaneous above T = ΔH/ΔS.Why it happens
Both give ΔH_rxn from tabulated values, but the subtraction runs opposite ways.The fix
Formation: products minus reactants. Bonds: broken minus formed. Write which table you are using before substituting.5 questions
Question 1
For an ideal gas undergoing an isothermal process, which statement is correct?
Question 2
Which of the following is not a state function?
Question 3
For the reaction N₂(g) + 3H₂(g) → 2NH₃(g), the relation between ΔH and ΔU is:
Question 4
A reaction has ΔH = +40 kJ mol⁻¹ and ΔS = +100 J K⁻¹ mol⁻¹. Above roughly what temperature does it become spontaneous?
Question 5
The standard enthalpy of formation of an element in its most stable form is:
A test on one chapter alone tells you little — the exam never asks Thermodynamics in isolation. So this builds a full JEE Main Chemistry paper that leans on it. Weighted like the exam, a 25-question paper gives Thermodynamics 2 questions; marked weak, it gets 3, and the rest of the paper keeps the other chapters honest. Fill the plan from previous-year questions or your module.
Your paper
Press Generate paper plan for a chapter-by-chapter blueprint: questions per chapter, numericals, marks and time — ready to print and fill from your PYQs, modules or the BrainStride question bank.
Chapter weights are BrainStride’s reading of past papers, not official data. For any other chapter mix, use the JEE Main Chemistry test generator.
ΔH and ΔU · 3 marks · moderate
For N₂(g) + 3H₂(g) → 2NH₃(g), ΔH° = −92.4 kJ mol⁻¹ at 298 K. Calculate ΔU°. (R = 8.314 J K⁻¹ mol⁻¹)
Answer: ΔU° ≈ −87.4 kJ mol⁻¹.
Isothermal reversible expansion · 3 marks · moderate
Two moles of an ideal gas expand reversibly and isothermally at 300 K from 5 L to 50 L. Calculate w, ΔU and q.
Answer: w ≈ −11.5 kJ, ΔU = 0, q ≈ +11.5 kJ.
Gibbs energy and spontaneity · 3 marks · easy
A reaction has ΔH = +30 kJ mol⁻¹ and ΔS = +75 J K⁻¹ mol⁻¹. Is it spontaneous at 298 K, and above what temperature does it become spontaneous?
Answer: Not spontaneous at 298 K (ΔG ≈ +7.7 kJ mol⁻¹); spontaneous above about 400 K.
In BrainStride’s analysis of past JEE Main papers, chemical thermodynamics contributes about one to two of the 25 Chemistry questions, and together with electrochemistry forms a two-to-three question block. NTA publishes no official chapter weightage, so use these ranges for planning rather than as a promise.
ΔU is the heat exchanged at constant volume and ΔH at constant pressure. They differ by the work done as gases expand or contract: ΔH = ΔU + Δn_gRT, where Δn_g counts only gaseous moles. When a reaction produces and consumes equal moles of gas, the two are equal.
No. Spontaneity at constant temperature and pressure needs ΔG = ΔH − TΔS to be negative. An exothermic reaction with a strongly negative entropy change stops being spontaneous above T = ΔH/ΔS, which is why many exothermic reactions reverse at high temperature.
Use the IUPAC convention that NCERT follows: ΔU = q + w, where w is the work done on the system. A gas expanding does work on its surroundings, so w is negative. Pick one convention and never switch mid-paper, because both appear in older reference books.
Enthalpy is usually given in kilojoules and entropy in joules per kelvin, and mixing them is the most common error in this chapter. Convert entropy to kilojoules per kelvin before substituting into ΔG = ΔH − TΔS, then check that the magnitude of TΔS is plausible.
The free diagnostic tags every wrong answer as a concept gap, a silly error or time pressure — so you know which one this chapter is for you.