Question:medium

At 0 K, the molecule CO exists in two alternate arrangements (CO and OC) in the solid crystal. The value of the entropy is (where thermodynamic probability \( W = k^N \)):

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The entropy change for a system with multiple arrangements can be calculated using the Boltzmann formula \( S = k_B \ln W \).
Updated On: Feb 10, 2026
  • 5.76 J K\(^{-1}\) mol\(^{-1}\)
  • 7.76 J K\(^{-1}\) mol\(^{-1}\)
  • 9.76 J K\(^{-1}\) mol\(^{-1}\)
  • 11.76 J K\(^{-1}\) mol\(^{-1}\)
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The Correct Option is A

Solution and Explanation

The entropy change associated with a variation in the number of accessible states is defined as:\[S = k_B \ln W\]When there are two possible configurations (CO and OC), the statistical weight \( W \) is:\[W = 2 \text{(representing the two configurations)}\]Consequently, the entropy is calculated as:\[S = k_B \ln 2\]Here, \( k_B = 1.38 \times 10^{-23} \, \text{J/K} \) denotes Boltzmann's constant.For a quantity of 1 mol of CO:\[S = 1.38 \times 10^{-23} \times \ln 2 \times N_A\]Where \( N_A = 6.022 \times 10^{23} \) is Avogadro's number.\[S \approx 5.76 \, \text{J K}^{-1} \, \text{mol}^{-1}\] Final Answer: \[\boxed{5.76 \, \text{J K}^{-1} \, \text{mol}^{-1}}\]
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