Chemical Thermodynamics Question 460
Question: The free energy for a reaction having $ \Delta H=31400,ca; $ . $ \Delta S=32,cal,{{K}^{-1}},mo{{l}^{-1}} $ at $ 1000^{o}C $ is [Orissa JEE 2005]
Options:
A) - 9336 cal
B) - 7386 cal
C) -1936 cal
D) + 9336 cal
Show Answer
Answer:
Correct Answer: A
Solution:
$ \Delta G=\Delta H-T\Delta S=31400-1273\times 32 $
$ =31400-40736=-9336,cal $
BETA
