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Question

The enthalpy change for a given reaction at 298K is −x Jmol−1. For the reaction to be spontaneous at 298K, the entropy change at that temperature:

A
can be negative, but numerically greater than x298JK1
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B
can be negative, but numerically smaller than x298JK1
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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 C cannot be negative
The enthalpy change for a given reaction at 298K is −x J/mol. For the reaction to be spontaneous at 298K, the entropy change at that temperature cannot be negative.
For a spontaneous reaction
ΔG=ΔHTΔS<0
We already know that ΔH is negative. If ΔS is negative then ΔG as a whole might turn positive if it is numerically greater than x298.
Hence, the correct option is C.

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