JEE Main Chemistry Thermodynamics 2027: Enthalpy, Entropy & Gibbs Energy

Will a reaction happen by itself? Thermodynamics answers that with one equation, ΔG = ΔH − TΔS. The chapter looks abstract, but its JEE Main questions are mostly clean substitutions once you have the sign convention and the relation between ΔH and ΔU straight.

ΔG below zero means spontaneous. Everything else in the chapter is a way of finding ΔG.

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Chapter at a Glance

Priority reflects past-paper trends, not an official NTA weightage.

What the NTA Syllabus Covers

• System and surroundings, extensive and intensive properties, state functions, entropy, types of processes
• First law: work, heat, internal energy, enthalpy, heat capacity, Hess's law, enthalpies of bond dissociation, combustion, formation, atomisation, sublimation, phase transition, hydration, ionisation and solution
• Second law: spontaneity, ΔS of universe and ΔG of system, ΔG° and equilibrium constant

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Master These Topics

1. First Law and the Chemistry Sign Convention

In Chemistry the first law is written ΔU = q + w, where w is work done on the system. Compare Physics, where it is ΔU = Q − W with W as work by the gas. The two are the same law with different sign conventions, so check which one the question uses.

Enthalpy is H = U + PV, and at constant pressure ΔH = ΔU + Δn_g RT, where Δn_g is the change in moles of gas (products minus reactants).

Worked example: For N₂ + 3H₂ → 2NH₃, Δn_g = 2 − 4 = −2. At 298 K, Δn_g RT = −2 × 8.314 × 10⁻³ × 298 ≈ −4.96 kJ. So ΔH = ΔU − 4.96 kJ.

Trap: ΔH equals ΔU only when Δn_g = 0, or for reactions involving only solids and liquids.

2. Hess's Law

Enthalpy is a state function, so the total change is the same whatever the route.

Worked example: Given C(s) + O₂ → CO₂ with ΔH = −393.5 kJ and CO + ½O₂ → CO₂ with ΔH = −283 kJ, subtract the second from the first to get C(s) + ½O₂ → CO with ΔH = −393.5 − (−283) = −110.5 kJ.

3. Entropy and Gibbs Energy

Entropy S measures disorder, and it rises with temperature, when solids melt or evaporate, and when gas moles increase. Gibbs energy combines both drivers:

ΔG = ΔH − TΔS

Worked example: For ammonia synthesis, ΔH = −92 kJ and ΔS = −199 J/K. At 298 K, ΔG = −92 − 298 × (−0.199) = −92 + 59.3 = −32.7 kJ, so it is spontaneous at room temperature.

4. Link to the Equilibrium Constant

ΔG° = −RT ln K = −2.303 RT log K. At 298 K, 2.303RT ≈ 5.7 kJ/mol.

Worked example: If ΔG° = −5.7 kJ/mol at 298 K, then log K ≈ 1 and K ≈ 10. A more negative ΔG° means a larger K.

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Common Traps to Avoid

• Mixing the Physics sign convention (ΔU = Q − W) with the Chemistry one (ΔU = q + w).
• Assuming ΔH = ΔU for a reaction where gas moles change.
• Forgetting to convert ΔS from J/K to kJ/K when using ΔG = ΔH − TΔS.
• Confusing ΔG (any conditions) with ΔG° (standard state).

60-Second Revision Sheet

• ΔU = q + w; ΔH = ΔU + Δn_g RT
• Hess's law: add or subtract equations and their ΔH values
• ΔG = ΔH − TΔS; spontaneous when ΔG < 0
• ΔG° = −2.303 RT log K

Your Study Plan

1. Day 1: state functions, first law and ΔH versus ΔU.
2. Day 2: Hess's law and bond enthalpy problems.
3. Day 3: entropy, Gibbs energy and temperature of spontaneity.
4. Day 4: ΔG° and K numericals, timed mixed set.

Practice Chemical Thermodynamics Questions Free → (opens in a new tab)

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Continue Your Chemistry Journey

• Previous chapter: Chemical Bonding & Molecular Structure
• Next chapter: Solutions
• All JEE Main Chemistry chapters
• Complete JEE Main Syllabus 2027 guide

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