Thermodynamics · Question 2

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

1. P → 2; Q → 3; R → 5; S → 4
2. P → 4; Q → 3; R → 5; S → 1
3. P → 2; Q → 5; R → 1; S → 4
4. P → 2; Q → 5; R → 1; S → 3

Answer

(3) P → 2; Q → 5; R → 1; S → 4

Solution

(P) Physisorption: exothermic, and the gas loses freedom on the surface. H<0, S<0 → (2)

(Q) Diamond → graphite: H = -1.9 kJ mol-1, and graphite is less ordered (higher S ). H<0, S>0 → (5)

(R) Denaturation of protein: H-bonds break (endothermic) and the chain unfolds into a random coil. H>0, S>0 → (1)

(S) Propene → cyclopropane: the strained ring has higher enthalpy, and ring closure reduces freedom. H>0, S<0 → (4)

Answer: (C)