Question
The ΔH and ΔS for a reaction at one atmospheric pressure are +30.558 kJ and 0.066 k J k−1 respectively. The temperature at which the free energy change will be zero and below this temperature the nature of reaction would be:
Answer: Option D
:
D
Now, to get T, let's use
△G = △H−T△S
Here, △H=+30.558KJ
△S=0.066kJk−1
△G=0
⇒0=30.558−T×0.066
⇒T=30.5580.066=463K
Now, at 463 K⇒△G = 0
If we go below 463 K, T△S value will decrease and △G value will become positive.
Therefore, non-spontaneous
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:
D
Now, to get T, let's use
△G = △H−T△S
Here, △H=+30.558KJ
△S=0.066kJk−1
△G=0
⇒0=30.558−T×0.066
⇒T=30.5580.066=463K
Now, at 463 K⇒△G = 0
If we go below 463 K, T△S value will decrease and △G value will become positive.
Therefore, non-spontaneous
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