dS° = entropy of products - entropy of reactants.
If no. Of molecules on the product side is more than number of molecules on reactant side , then dS° will be positive .
And if number of molecules on the product side is less than number of molecules on reactant side , then dS° will be negative.
Thus only in case of (d)
2IBr(g) -----> I2(s) + Br2(l) , there will be dS° negative.
Although number of molecules on both side is same , but here a gaseous substance is changing into solid and liquid material , and entropy of gases is more than solid and liquid.
Thus entropy change in case of (d) will ne negative.
Option (d) is the correct answer.
Kindly rate the answer.
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