Question

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 \( W = k^N \)):

• A. 5.76 J K\(^{-1}\) mol\(^{-1}\)
• B. 7.76 J K\(^{-1}\) mol\(^{-1}\)
• C. 9.76 J K\(^{-1}\) mol\(^{-1}\)
• D. 11.76 J K\(^{-1}\) mol\(^{-1}\)

Hint:

The entropy change for a system with multiple arrangements can be calculated using the Boltzmann formula \( S = k_B \ln W \).

Answer 1

The Correct Option is A

The entropy change due to a change in the number of available states is given by: \[ S = k_B \ln W \] For two possible arrangements (CO and OC), the thermodynamic probability \( W \) is: \[ W = 2 \text{(since there are two arrangements)} \] Thus, the entropy is: \[ S = k_B \ln 2 \] Where \( k_B = 1.38 \times 10^{-23} \, \text{J/K} \) is Boltzmann's constant. For 1 mol of CO: \[ S = 1.38 \times 10^{-23} \times \ln 2 \times N_A \] Where \( N_A = 6.022 \times 10^{23} \) (Avogadro's number). \[ S \approx 5.76 \, \text{J K}^{-1} \, \text{mol}^{-1} \] Final Answer: \[ \boxed{5.76 \, \text{J K}^{-1} \, \text{mol}^{-1}} \]