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



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