Question

At \(0 \mathrm{~K},\) the entropy of carbon monoxide crystal is not zero but has a value of \(4.2 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol},\) called the residual entropy. According to the third law of thermodynamics, this means that the crystal does not have a perfect arrangement of the CO molecules. (a) What would be the residual entropy if the arrangement were totally random? (b) Comment on the difference between the result in part (a) and \(4.2 \mathrm{~J} / \mathrm{K} \cdot\) mol. (Hint: Assume that each CO molecule has two choices for orientation, and use Equation 18.1 to calculate the residual entropy.)

Short answer

The calculated entropy is 5.76 J/K·mol; the real one (4.2 J/K·mol) indicates some order.

Step-by-step solution

Understand the Concept of Entropy

Entropy is a measure of disorder or randomness. The third law of thermodynamics indicates that a perfect crystal should have an entropy of zero because there is only one way to arrange its atoms. However, some crystals, like CO, can have multiple orientations in a lattice, increasing the entropy even at absolute zero.

Identify Equation 18.1

Equation 18.1, relevant to this problem, describes the entropy of a system in terms of the number of possible configurations (W). The equation is given by Boltzmann's formula: \[ S = k_B imes ext{ln}(W) \] where \(k_B\) is Boltzmann's constant, \(1.38 \times 10^{-23} \text{ J/K}\), and \(W\) is the number of arrangements.

Calculate Residual Entropy with Random Arrangement

If each CO molecule can be randomly oriented in two ways, the number of arrangements \(W\) for one mole (\(N_A = 6.022 \times 10^{23}\)) of CO molecules is \(W = 2^{N_A}\). Using Boltzmann's formula:\[ S = k_B imes ext{ln}(2^{N_A}) \]\[ S = k_B imes N_A imes ext{ln}(2) \]Substitute the values:\[ S = (1.38 \times 10^{-23}) \times (6.022 \times 10^{23}) \times ext{ln}(2) \]\[ S ≈ 5.76 \text{ J/K/mol} \]

Compare the Calculated Residual Entropy to Given Value

The calculated residual entropy if the arrangement of CO molecules is totally random is \(5.76 \text{ J/K/mol}\). The given residual entropy at 0 K is \(4.2 \text{ J/K/mol}\). The difference suggests that the arrangement of molecules in the crystal is not entirely random, indicating some order or preference likely driven by intermolecular interactions.

Key concepts

Residual Entropy

Residual entropy refers to the entropy present in a crystal at absolute zero (0 K) where, ideally, its entropy should be zero according to the Third Law of Thermodynamics. This law states that a perfect crystal with a perfectly ordered structure should lack any disorder, leading to zero entropy. However, certain substances like carbon monoxide (CO) display what is termed residual entropy. This phenomenon occurs because the molecules within the crystal aren't perfectly ordered. Instead, they maintain a level of randomness or disorder.
For the CO crystal discussed in the exercise, this residual entropy is observed to be 4.2 J/K∙mol. It indicates an inherent chaotic arrangement of molecules that even at the lowest possible temperatures, total order isn't achieved.
Understanding residual entropy is vital because it distinguishes substances whose molecular structures refuse to reach perfect orderliness, giving them non-zero entropy even at 0 K.

Boltzmann's Formula

Ludwig Boltzmann introduced a groundbreaking formula that links microscopic molecular configurations to macroscopic properties like entropy. The formula is crucial in statistical mechanics and thermodynamics, expressed as:
\[ S = k_B \times \text{ln}(W) \]
In this equation, \( S \) represents entropy, \( k_B \) is Boltzmann's constant (1.38 × 10^{-23} J/K), and \( W \) denotes the number of possible microstates or configurations of the system.
Applying Boltzmann's formula to calculate entropy essentially involves determining how many ways the molecules within a substance can be arranged. In the case where each CO molecule can be oriented in two ways, calculating \( W \) for a mole gives an immense number of configurations, considering Avogadro's number (\( N_A = 6.022 \times 10^{23} \)).
This formula allows us to quantify the degree of disorder and introduce precise calculations for the entropy of systems with variable microstates, such as our CO crystal.

Entropy Calculation

Calculating entropy not only helps us understand a system's disorder but also tells us about how distributed the energy states are within that system. Using Boltzmann's formula to calculate the residual entropy for a random arrangement of CO molecules involves several key steps.
First, the number of possible configurations, \( W \), is determined by the freedom of orientation. For a CO molecule, each molecule can be oriented in two ways. Considering a whole mole, this leads to \( W = 2^{N_A} \).
Subsequently, inserting \( W \) into Boltzmann's equation results in:
\[ S = k_B \times N_A \times \text{ln}(2) \]
Plugging in the values, we perform the calculation:
\[ S = (1.38 \times 10^{-23}) \times (6.022 \times 10^{23}) \times \text{ln}(2) \]
Gives us an entropy value of approximately 5.76 J/K/mol. This predicted value for a totally random CO arrangement shows a contrast to the observed 4.2 J/K·mol, suggesting partial order in the system.
Such calculations are a cornerstone of thermodynamics, explaining the seeming paradox of disorder at absolute zero and predicting material behaviors.