JEE Main20206 Sep 2020Morning ShiftChemistryThermodynamics (C)Actual
The variation of equilibrium constant with temperature is given below : Temperature EquilibriumConstant T 1 = 25 ° C K 1 = 10 T 2 = 100 ° C K 2 = 100 The values of ΔH ° , ΔG ° at T 1 and Δ G ° at T 2 (in kJ mol - 1 ) respectively, are close to [use R = 8 . 314 JK - 1 mol - 1 ]
Options
- A28 . 4 , - 7 . 14  and  - 5 . 71
- B0 . 64 , - 7 . 14  and  - 5 . 71
- C28 . 4 , - 5 . 71  and  - 14 . 29
- D0 . 64 , - 5 . 71   and   - 14 . 29
Correct answer
A. 28 . 4 , - 7 . 14  and  - 5 . 71
Step-by-step solution
   T 1 = 323 K    T 2 = 373 K k 1 = 10    k 2 = 100 log k 2 k 1 = Δ H 2 . 303 R 1 T 1 - 1 T 2 log 100 10 = Δ H 2 . 303 × 8 . 314 1 298 - 1 373 log 10 = Δ H 2 . 303 × 8 . 314 75 298 × 373 Δ H = 2 . 303 × 8 . 314 × 298 × 373 75 = 28 . 4 KJ At T 1 = 25 ° C = 298 K , K 1 = 10 Δ G = - 2 . 303 RT 1 logk  1 = - 2 . 303 × 8 . 314 × 298 × log ( 10 ) = - 2 . 303 × 8 . 314 × 298 × 1 = - 5 . 7 KJ At T 2