This question asks us to calculate the entropy of Carbon Monoxide (CO) molecules in a solid crystal at absolute zero (0 K). At this temperature, CO molecules can exist in two possible orientations: the standard CO arrangement and the reversed OC arrangement. We need to determine the residual entropy based on these possibilities.

Understanding Residual Entropy at 0 K

At 0 K, ideally, a perfect crystal should have zero entropy, according to the Third Law of Thermodynamics. However, sometimes molecules can get 'stuck' in disordered arrangements even at absolute zero. This remaining entropy is called residual entropy. In the case of CO, the molecule is polar ($C^{\delta+} - O^{\delta-}$), and in the solid state, the molecules might align randomly as CO or OC, leading to disorder.

Applying the Boltzmann Equation for Entropy

The relationship between entropy ($S$), thermodynamic probability ($W$), and the Boltzmann constant ($k_B$) is given by the Boltzmann equation:

$S = k_B \ln W$

Here:

• S is the entropy.
• $k_B$ is the Boltzmann constant, approximately $1.38 \times 10^{-23}$ J K$^{-1}$.
• W is the thermodynamic probability, representing the total number of possible microscopic arrangements (microstates) of the system that correspond to the same macroscopic state.

Calculating Thermodynamic Probability (W) for CO

The question states that the thermodynamic probability is given as $W = K^{N_A}$.

We are also told that at 0 K, CO exists in two alternate arrangements (CO and OC). This means that for each mole of CO, there are $N_A$ molecules (where $N_A$ is Avogadro's number, approximately $6.022 \times 10^{23}$ mol$^{-1}$), and each molecule has 2 possible orientations.

Therefore, the total number of possible arrangements ($W$) for $N_A$ molecules, each having 2 possible states, is:

$W = 2^{N_A}$

Comparing this with the given form $W = K^{N_A}$, we can see that $K = 2$.

Calculating Molar Entropy

Now, we substitute $W = 2^{N_A}$ into the Boltzmann equation:

$S = k_B \ln(2^{N_A})$

Using the properties of logarithms ($\ln(a^b) = b \ln a$), we get:

$S = k_B N_A \ln 2$

We know that the product of the Boltzmann constant ($k_B$) and Avogadro's number ($N_A$) is equal to the ideal gas constant ($R$):

$R = k_B N_A \approx 8.314 \text{ J K}^{-1} \text{ mol}^{-1}$

So, the equation simplifies to:

$S = R \ln 2$

Now, we can calculate the value:

• $R = 8.314$ J K$^{-1}$ mol$^{-1}$
• $\ln 2 \approx 0.693$

$S \approx 8.314 \text{ J K}^{-1} \text{ mol}^{-1} \times 0.693$

$S \approx 5.764 \text{ J K}^{-1} \text{ mol}^{-1}$

Rounding this value gives us 5.76 J K$^{-1}$ mol$^{-1}$.

Conclusion

The calculated entropy value matches the first option provided. This residual entropy arises because the CO molecules in the solid crystal remain disordered due to the two possible orientations (CO and OC) even at 0 K.