A dilute gas, composed of a mixture of \(N_{1}\) iodine atoms 1 and \(N_{12}\) iodine molecules \(I_{2}\), is confined to a box of volume \(V=1.0 \mathrm{~m}^{3}\) at temperature \(T=300 \mathrm{~K}\). The rotational temperature of the iodine molecules is \(\theta_{\text {rot }}=0.0537 \mathrm{~K}\) (for simplicity we neglect vibrational modes). (a) Compute the chemical potentials, \(\mu_{1}\) and \(\mu_{12}\), of the iodine atoms and molecules, respectively. (b) The numbers of the iodine atoms and molecules can change via chemical reactions with one another. The condition for chemical equilibrium is \(\mu_{12}=2 \mu_{1}\). Use this condition to find the ratio \(N^{2}{ }_{1} / N_{12}\) when the gas is in equilibrium. (c) Does the inclusion of the rotational degree of freedom increase or decrease the number of \(\bar{I}_{2}\) molecules at chemical equilibrium.

The chemical potential is a significant concept in statistical thermodynamics, particularly when dealing with chemical reactions and phase equilibria. It's a form of potential energy that provides insight into how particles distribute and shift within a system.

To compute the chemical potential for an ideal monoatomic gas such as iodine atoms (\text{I}), we use the equation:

\[ \mu_{1} = -kT \ \ln\left[ \left(\frac{V}{N_{1}} \right)\left(\frac{mkT}{2 \pi \hbar^2} \right)^{3/2} \right] \]

Here, k is the Boltzmann constant, T is the temperature, V is the volume, \(N_{1}\) is the number of iodine atoms, \(m\) is the mass of an iodine atom, and \(\hbar\) is the reduced Planck constant.

For diatomic iodine molecules (\(I_{2}\)), the chemical potential also accounts for the rotational degrees of freedom:

\[ \mu_{12} = -kT \ \ln\left[ \left(\frac{V}{N_{12}} \right)\left(\frac{M k T}{2 \pi \hbar^2} \right)^{3/2}\frac{T}{\theta_{\text{rot}}} \right] \]

In this equation, \(\theta_{\text{rot}}\) is the rotational temperature, and \(M\) is the mass of an iodine molecule. Notice how the term \(\frac{T}{\theta_{\text{rot}}}\) pops up for the diatomic molecules, which will play a crucial role in chemical equilibrium.

In statistical thermodynamics, rotational degrees of freedom refer to the ways in which a molecule can rotate and thus contribute to its overall energy. For diatomic molecules like iodine (\(I_{2}\)), considering rotational degrees of freedom is vital in accurately determining the chemical potential.

Rotational motion adds an extra term in the partition function and subsequently influences the chemical potential. Specifically, the contribution is represented by the factor \(\frac{T}{\theta_{\text{rot}}}\). Here, \(T\) is the temperature, and \(\theta_{\text{rot}}\) is a characteristic rotational temperature specific to the molecule.

This factor lowers the chemical potential, \(\mu_{12}\), making it smaller compared to a calculation that neglects rotational degrees of freedom. As a result, fewer diatomic iodine molecules (\(I_{2}\)) form at equilibrium because the system prefers configurations with higher overall entropy, leading to a larger number of monoatomic iodine (\(I\)) particles.

Always remember, the intricacies of molecular motion, including rotational aspects, are fundamental to grasping the detailed behavior of real gases.

Chemical equilibrium occurs when the forward and reverse reactions happen at the same rate, hence the concentrations of reactants and products remain constant over time. For a system of iodine atoms (\(I\)) and iodine molecules (\(I_{2}\)), the equilibrium condition is formulated by setting the chemical potentials equal: \(\mu_{12} = 2 \mu_{1}\).

This equality ensures that the conversion between iodine atoms and molecules is balanced. By substituting the chemical potential equations derived earlier, we find:

\[ -kT \ \ln\left[ \ \frac{V}{N_{12}} \ \left(\frac{M k T}{2 \pi \hbar^2} \right)^{3/2} \ \frac{T}{\theta_{\text{rot}}} \right] = 2 \ \left( -kT \ \ln\left[ \ \ \frac{V}{N_{1}} \ \left(\frac{m k T}{2 \pi \hbar^2} \right)^{3/2} \right] \right) \]

By simplifying, we derive the relationship for the ratio:\[ \ \frac{N_{1}^{2}}{N_{12}} = \ \ \frac{V}{\left( \ \frac{M k T}{2 \pi \hbar^2} \ \right)^{3/2} \ \ \frac{T}{\theta_{\text{rot}}}} \]

This equation tells us how the presence of rotational degrees of freedom (the term \(\frac{T}{\theta_{\text{rot}}}\)) impacts the equilibrium concentrations. In general, including rotational energy levels results in a greater number of monoatomic iodine (\(I\)) compared to diatomic iodine (\(I_{2}\)).
Thus, understanding chemical equilibrium and incorporating the effects of additional degrees of freedom helps predict the behavior of real gases more accurately. This is especially significant in mixtures of diatomic gases like iodine.