NTA Abhyas JEE Main2020ChemistryThermodynamics (C)Practice
Match the column I with column II and mark the appropriate choice. Column I Column II (p) H 2 g + B r 2 g → 2 H B r g (i) Δ H = Δ U - 2 R T (q) P C l 5 g → P C l 3 g + C l 2 g (ii) Δ H = Δ U + 3 R T (r) N 2 g + 3 H 2 g → 2 N H 3 g (iii) Δ H = Δ U (s) 2 N 2 O 5 g → 4 N O 2 g + O 2 g (iv) Δ H = Δ U + R T
Options
- A(p)-(iii), (q)-(i), (r)-(ii), (s)-(iv)
- B(p)-(iii), (q)-(iv), (r)-(i), (s)-(ii)
- C(p)-(ii), (q)-(i), (r)-(iv), (s)-(iii)
- D(p)-(iv), (q)-(ii), (r)-(i), (s)-(iii)
Correct answer
B. (p)-(iii), (q)-(iv), (r)-(i), (s)-(ii)
Step-by-step solution
(p) Δ n g = 2 - 2 = 0 ; hence Δ H = Δ U (q) Δ n g = 2 - 1 = 1 ; hence Δ H = Δ U + R T (r) Δ n g = 2 - 4 = - 2 ; hence Δ H = Δ U - 2 R T (s) Δ n g = 5 - 2 = 3 ; hence Δ H = Δ U + 3 R T