JEE Advanced 2026 · Paper 1 · Chemistry Q13

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ChemistryList match+4 / -1Question 45 of 48

List-I contains various physical/chemical processes, and List-II contains combinations of changes in enthalpy $(\Delta H)$ and entropy $(\Delta S)$ . Match each entry in List-I to the appropriate entry in List-II, and choose the correct option.

List-I

P
Physisorption
Q
Diamond $\longrightarrow$ Graphite
R
Denaturation of protein
S
Propene $\longrightarrow$ Cyclopropane

List-II

1
$\Delta H>0$ and $\Delta S>0$
2
$\Delta H<0$ and $\Delta S<0$
3
$\Delta H<0$ and $\Delta S=0$
4
$\Delta H>0$ and $\Delta S<0$
5
$\Delta H<0$ and $\Delta S>0$

P → 2; Q → 3; R → 5; S → 4

P → 4; Q → 3; R → 5; S → 1

P → 2; Q → 5; R → 1; S → 4

P → 2; Q → 5; R → 1; S → 3

Answer and worked solution

Official answer: C

(P) Physisorption releases heat and the adsorbed gas loses freedom: $\Delta H<0$ , $\Delta S<0$ → 2.
(Q) Diamond → graphite: graphite is the stable form ( $\Delta H\approx-1.9$ kJ/mol) and has the higher entropy: $\Delta H<0$ , $\Delta S>0$ → 5.
(R) Denaturation of a protein breaks hydrogen bonds (endothermic) and unfolds the ordered chain: $\Delta H>0$ , $\Delta S>0$ → 1.
(S) Propene → cyclopropane: the ring is strained ( $\Delta H>0$ ) and closing a ring lowers entropy ( $\Delta S<0$ ) → 4.

P→2, Q→5, R→1, S→4.

Topics:Entropy First law and enthalpy Amino acids and proteins

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JEE Advanced 2026 · Paper 1, Chemistry question 13