Chemical Thermodynamics Question 378
Question: A certain reaction is non spontaneous at 298K. The entropy change during the reaction is $ 121J{{K}^{-1}} $ . Is the reaction is endothermic or exothermic- The minimum value of $ \Delta H $ for the reaction is
Options:
A) endothermic, $ \Delta H=36.06kJ $
B) exothermic, $ \Delta H=-36.06kJ $
C) endothermic, $ \Delta H=60.12kJ $
D) exothermic, $ \Delta H=-60.12kJ $
Show Answer
Answer:
Correct Answer: A
Solution:
- For non spontaneous reaction $ \Delta G=+ve $
$ \Delta G=\Delta H-T\Delta S $ and $ \Delta S=121J{{K}^{-1}} $ For $ \Delta G=+ve $
$ \Delta H $ has to be positive. Hence the reaction is endothermic. The minimum value of $ \Delta H $ can be obtained by putting $ \Delta G=0 $
$ \Delta H=T\Delta S=298\times 121J $
$ =36.06kJ $
