Verified Solution • 2025–26 Session moderate

QJEEM19Ch0aeb: JEE Main 2019 Chemistry - Class 11 JEEM19Ch0aeb Solution: A process has H = 200 J mol –1 and S = 40 JK –1 mol –1 . Out of …

A process has Δ\Delta H = 200 J mol–1 and Δ\Delta S = 40 JK–1 mol–1. Out of the values given below, choose the minimum temperature above which the process will be spontaneous :

Multiple Choice Options

  • A 4 K
  • B 20 K
  • C 5 K
  • D 12 K

Step-by-Step Verified Solution

Δ\Delta G = Δ\Delta H −- T Δ\Delta S

T = ΔHΔS=20040=5K{{\Delta H} \over {\Delta S}} = {{200} \over {40}} = 5K

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