NTA Abhyas JEE Main2020ChemistryThermodynamics (C)Practice
Standard entropies of x 2 , y 2 and x y 3 are 70, 50 and 60 J K - 1 m o l - 1 respectively. For the reaction 1 2 x 2 + 3 2 y 2 ⇌ x y 3 . Δ H = - 30 k J to be at equilibrium, the temperature should be
Options
- A450 K
- B600 K
- C1200 K
- D300 K
Correct answer
B. 600 K
Step-by-step solution
Δ S = S x y 3 - 1 2 S x 2 - 3 2 S y 2 Δ S = 60 - 1 2 × 70 - 3 2 × 50 = - 50 JK - 1 m o l - 1 At equilibrium Δ G = 0 Δ G = Δ H - T Δ S 0 = - 30 - ( - 50 × 1 0 - 3 × T ) T = 600 K