Question
Download Solution PDFThe molar residual entropy (in J K−1) of solid OCS would be closest to
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
The residual entropy of a solid is the entropy that remains in the solid even at absolute zero. It is caused by the disorder in the crystal structure. The more disordered the crystal structure, the greater the residual entropy.
Explanation:
The molecule OCS has a bent structure, which allows for a certain amount of disorder in the crystal structure. This disorder leads to a residual entropy of 5.8 J K−1.
The residual entropy of a solid can be calculated using the following equation:
\(S_r = R \ln p\)
where: Sr is the residual entropy (in J K−1), R is the gas constant (8.314 J K−1 mol−1) and p is the number of possible orientations of the molecule in the crystal (also known as the degeneracy).
In the case of OCS, the molecule has two possible orientations in the crystal. This means that the degeneracy is p=2. Substituting this value into the equation for residual entropy gives:
\(S_r = R \ln 2 = 5.8 \;J K^{-1}\)
Therefore, the molar residual entropy of solid OCS is 5.8 J K−1.
Last updated on Jun 5, 2025
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