NTA Abhyas JEE Main2020ChemistryThermodynamics (C)Practice
For a given reaction, Δ H = 35 . 5 kJmo l - 1 and Δ S = 83 . 6 J k - 1 mo l - 1 . The reaction is spontaneous at (Assume that Δ H and Δ S do not vary with tempearature)
Options
- AT > 425 K
- BAll temperatures
- CT > 398 K
- DT < 525 K
Correct answer
A. T > 425 K
Step-by-step solution
As Both Δ H and Δ S are positive so to keep Δ G = - ve High Temp is favourable so that process become spontaneous. Δ G = Δ H - T . Δ S T = Δ H Δ S = 35.5 × 1000 83.6 > 424.6 K T > 425K