JEE (Advanced) 2026, Paper 1, Chemistry, Section 4 — matching list set, one correct option: +4 if it is chosen, 0 if unanswered, −1 otherwise.
Q.13 List-I contains various physical/chemical processes, and List-II contains combinations of changes in enthalpy (ΔH) and entropy (ΔS). Match each entry in List-I to the appropriate entry in List-II, and choose the correct option.
| List-I | List-II |
|---|---|
| (P) Physisorption | (1) ΔH > 0 and ΔS > 0 |
| (Q) Diamond ⟶ Graphite | (2) ΔH < 0 and ΔS < 0 |
| (R) Denaturation of protein | (3) ΔH < 0 and ΔS = 0 |
| (S) Propene ⟶ Cyclopropane | (4) ΔH > 0 and ΔS < 0 |
| (5) ΔH < 0 and ΔS > 0 |
Four measurements, one bench — the apparatus on the left, the ΔH and ΔS meters on the right, in step. Press START EXPERIMENT: set up → measure P, Q, R, S → the sign plane → test options A–D → answer.
Four processes, two signs each. The question's processes are loaded — press START EXPERIMENT.
Orange waves: heat flowing out of the system (ΔH < 0). Blue waves: heat flowing in (ΔH > 0). The needles swing left for a negative value and right for a positive one; the number is printed under each.
ΔG / |ΔH| = 1 − T·ΔS/ΔH; below zero the process is spontaneous
the measured values, on a log scale so that 1.9 and 1440 both show
not yet examined
not yet examined
not yet examined
not yet examined
Nothing investigated yet.
Pick a process and slide the temperature. The point on the sign plane never moves — ΔH and ΔS keep their signs — but ΔG = ΔH − TΔS can change sign, and with it whether the process goes by itself.
The complete worked solution — every structure, step and calculation.
The experiment has not finished yet. Press START and let it run to the end — or use Reveal anyway in the header if you are revising.
At constant pressure ΔH is the heat taken in. Heat out (bonds or attractions formed): ΔH < 0. Heat in (bonds broken, strain built): ΔH > 0.
Entropy counts the arrangements and motions available. Freedom gained (gas formed, chain unfolded, rotor freed): ΔS > 0; freedom lost: ΔS < 0.
ΔH and ΔS hardly change with temperature, so their signs place a process in one List-II entry. ΔG decides spontaneity at a given T.
An option is correct only if all four of its entries match. One wrong entry rejects it.
Gas molecules are held on the surface by van der Waals forces: an attraction forms, so heat is released (typical enthalpy of physisorption −20 to −40 kJ/mol or less; ≈ −15 kJ/mol for N₂ on charcoal). A molecule that flew freely in three dimensions is now confined to the surface.
Consistent with ΔG = ΔH − TΔS: physisorption is favourable at low temperature and falls off as T rises — exactly what is observed.
Graphite is the standard state of carbon (ΔfH° = 0); diamond lies 1.9 kJ/mol above it. Graphite's weakly held layers vibrate in more low-energy ways than diamond's rigid 3-D network, so its entropy is higher. The change is spontaneous but immeasurably slow. Trap: ΔS = 0 (option A/B's "3") — the two solids have measurably different entropies.
Unfolding breaks many hydrogen bonds and other non-covalent contacts: heat is absorbed (a DSC scan shows an endothermic peak; its area is ΔH, several hundred kJ/mol for a small protein). The unfolded chain can adopt an enormous number of conformations: ΔS > 0.
The ring is strained (60° angles instead of 109.5°), so it lies higher in energy; closing it turns a flexible chain with a freely rotating CH₃ into a rigid ring, so ΔS < 0. With ΔH > 0 and ΔS < 0, ΔG > 0 at every temperature. Trap: "(3) ΔH < 0, ΔS = 0" (option D) — wrong on both counts.
| Option | P | Q | R | S | Verdict |
|---|---|---|---|---|---|
| (A) | 2 ✓ | 3 ✗ | 5 ✗ | 4 ✓ | rejected |
| (B) | 4 ✗ | 3 ✗ | 5 ✗ | 1 ✗ | rejected |
| (C) | 2 ✓ | 5 ✓ | 1 ✓ | 4 ✓ | correct |
| (D) | 2 ✓ | 5 ✓ | 1 ✓ | 3 ✗ | rejected |
For a student opening it for the first time. The whole investigation runs by itself after one press.
Four processes in List-I, five sign combinations in List-II, four options. Which option matches all four?
The question's processes are loaded. You do not need to change anything. The answer is shown above the experiment.
Swap physisorption for chemisorption, or run Q, R or S backwards. The lab says honestly when you have left the question. ↺ RESTORE QUESTION CONDITIONS puts everything back.
Press ▶ START EXPERIMENT under the experiment screen. Nothing else to click (about a minute; SLOW MOTION or 2× changes the pace).
Left: the apparatus — gas sticking, a lattice opening, a protein unfolding in a DSC, a chain closing into a ring. Right: the ΔH and ΔS needles, the Hess ladder or the thermogram, and the List-II entry.
All four processes fly onto a ΔH–ΔS plane; each option is then tested entry by entry. Badges unlock on the way.
In the ΔG–T explorer sweep the temperature and see when each process becomes spontaneous. Then read the Detailed solution.
Press ▶ REPLAY EXPERIMENT, or change a process and press ⚙ RUN WITH CUSTOM VALUES. CLASSROOM MODE enlarges everything for a projector.
| Exam | JEE Advanced 2026, Paper 1, Chemistry, Section 4, Q.13 |
|---|---|
| Chapter | Thermodynamics |
| Topic | Signs of enthalpy and entropy change - adsorption, allotropes, protein denaturation and ring strain |
| Difficulty | Moderate |
| Answer | — run the experiment |