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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:

A

483 K, spontaneous

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B

443 K, non-spontaneous

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C

443 K, spontaneous

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D

463 K, non-spontaneous

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Solution

## The correct option is D 463 K, non-spontaneous 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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