NEETChemistryThermodynamics (C)
Match List I with List II: List I List II (A) H ₂ O(l) H ₂ O(s) (I) S > 0 , n_ g > 0 (B) I ₂ (s) I ₂ (g) (II) S (C) N ₂ (g) + 3 H ₂ (g) 2 NH ₃ (g) (III) S > 0 , physical change (D) PCl ₅ (g) PCl ₃ (g) + Cl ₂ (g) (IV) S Choose the correct answer from the options given below:
Options
- A(A)-(III), (B)-(IV), (C)-(I), (D)-(II)
- B(A)-(IV), (B)-(III), (C)-(I), (D)-(II)
- C(A)-(IV), (B)-(III), (C)-(II), (D)-(I)
- D(A)-(III), (B)-(IV), (C)-(II), (D)-(I)
Correct answer
C. (A)-(IV), (B)-(III), (C)-(II), (D)-(I)
Step-by-step solution
(A) H ₂ O(l) H ₂ O(s) represents the freezing of water. It is a physical change where a less ordered liquid converts into a highly ordered solid, so S (B) I ₂ (s) I ₂ (g) represents the sublimation of iodine. It is a physical change where a solid converts into a highly disordered gas, so S > 0 . Thus, (B) matches (III). (C) N ₂ (g) + 3 H ₂ (g) 2 NH ₃ (g) is a chemical reaction. The change in the number of moles of gas is n_ g = 2 - (1 + 3) = -2 . Since n_ g (D) PCl ₅ (g) PCl ₃ (g) + Cl ₂ (g) is a chemical reaction.