Chemical Thermodynamics Question 160
Question: The $ \Delta H $ and $ \Delta S $ for a reaction at one atmospheric pressure are +30.558 kJ and $ 0.066,kJ{{k}^{-1}} $ respectively. The temperature at which the free energy change will be zero and below of this temperature the nature of reaction would be [Kerala CET 2005]
Options:
A) 483 K, spontaneous
B) 443 K, non-spontaneous
C) 443 K, spontaneous
D) 463 K, non-spontaneous
E) 463 K, spontaneous
Show Answer
Answer:
Correct Answer: D
Solution:
$ \Delta G=\Delta H-T\Delta S $
$ 0=+30.558-T\times 0.066 $ or $ T=\frac{30.558}{0.066}=463K $ If $ {{(dG)}_{T,P}}=0 $ sign $ ‘=’ $ mean. If is reversible process
