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Question

The enthalpy change for a given reaction at 298 K is −xcal/mol. If the reaction occurs spontaneously at 298 K, the entropy change at that temperature:

A
Can be negative but numerically larger than x/298calK1mol1
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B
Can be negative but numerically smaller than x/298calK1mol1
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C
Cannot be negative
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D
Cannot be positive
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Solution

The correct option is A Can be negative but numerically larger than x/298calK1mol1
Solution:- (B) Can be negative but numerically smaller than x298cal/Kmol
As we know that,
ΔG=ΔHTΔS
For spontaneity,
ΔG<0
ΔHTΔS<0
Given:- ΔH=xcal/molveT=298K
Therefore,
For ΔG<0ΔS can be negative but the numerical value must be less than x298cal/Kmol.

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