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Which species will have the largest rotational partition function at a given temperature: \(\mathrm{H}_{2}, \mathrm{HD},\) or \(\mathrm{D}_{2} ?\) Which of these species will have the largest translational partition function assuming that volume and temperature are identical? When evaluating the rotational partition functions, you can assume that the high-temperature limit is valid.

Short Answer

Expert verified
At a given temperature, D₂ will have the largest rotational partition function since it has the smallest \(\theta_{rot}\), while H₂ will have the largest translational partition function since it has the smallest mass.

Step by step solution

01

Find the characteristic rotational temperatures for each species

First, we need to find the characteristic rotational temperature, \(\theta_{rot}\), for all three species. This can be usually found from a data table or using the formula: \(\theta_{rot} = \frac{hcB}{k}\), where \(h\) is Planck's constant, \(c\) is the speed of light, \(B\) is the rotational constant, and \(k\) is the Boltzmann constant. The rotational constants for the species are: - H₂: \(B = 60.853 \space cm^{-1}\) - HD: \(B = 44.363 \space cm^{-1}\) - D₂: \(B = 30.427 \space cm^{-1}\) Plug these values into the equation to get the rotational temperatures for each species.
02

Find the rotational partition functions for each species

Next, use the formula for the high-temperature limit of the rotational partition function and the \(\theta_{rot}\) values we found in Step 1: \(q_{rot} = \frac{T}{\theta_{rot}}\) We can now perform the calculation for each species to determine which has the largest rotational partition function at a given temperature.
03

Compare the translational partition functions for each species

Since the volume and temperature are identical for each species, we can use the ratio of their masses to compare their translational partition functions. The equation is: \(q_{trans} = (\frac{2\pi mkT}{h^2})^{3/2}V\) The masses for the species are: - H₂: \(m = 2 \space amu\) - HD: \(m = 3 \space amu\) - D₂: \(m = 4 \space amu\) Given that the other factors in the equation are the same for all species, we can simply compare their masses to determine which species will have the largest translational partition function.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Physical Chemistry
Physical chemistry is a branch of chemistry focused on the study of how matter behaves on a molecular and atomic level, and how chemical reactions occur. It blends principles of physics and chemistry to understand the physical properties of molecules, the forces that act upon them, and the energy changes that occur during chemical reactions. One of the key concepts in physical chemistry is the partition function, which is an essential tool for studying systems in statistical mechanics and thermodynamics. It provides valuable insights into the distribution of molecular energy states and allows us to calculate important thermodynamic properties like entropy, free energy, and heat capacity. The calculations of different partition functions, such as translational, rotational, vibrational, and electronic, help construct a comprehensive picture of the molecular energy landscape.
Translational Partition Function
The translational partition function describes the statistical distribution of particles in different translational energy states in a given volume and temperature. Translational motion refers to the movement of the entire particle from one place to another. For an ideal gas, the translational partition function, denoted as qtrans, is related to the mass m of the particles, the temperature T, the volume V, and fundamental constants like Planck's constant h and Boltzmann's constant k. The formula is expressed as:

\(q_{trans} = (\frac{2\pi mkT}{h^2})^{3/2}V\)

This function plays a crucial role in the determination of the overall energy distribution of a gas. Since each type of motion (translational, rotational, vibrational) contributes to the total energy, understanding each individual partition function is necessary for a full thermodynamic assessment of a system.
Characteristic Rotational Temperature
Every molecule has a characteristic rotational temperature, denoted as θrot. It is a temperature that correlates with the energy required for a molecule to occupy its first excited rotational energy level. The concept of a characteristic rotational temperature allows us to compare the rotational energy levels of different molecules. The formula to calculate it is:

\(\theta_{rot} = \frac{hcB}{k}\)

Here, h is Planck's constant, c is the speed of light, B is the rotational constant specific to each molecule, and k is Boltzmann's constant. The values of these constants are standardized, but the rotational constant B varies between molecules, as it is inversely related to the moment of inertia. Larger values of B correspond to larger rotational constants and higher characteristic rotational temperatures. This reflects how molecules with lower moments of inertia (commonly lighter molecules) require less energy to rotate, and therefore their energy levels are closer together.
High-Temperature Limit
In the context of rotational partition functions, the 'high-temperature limit' approximates the behavior of a system when the temperature is so high that a significant number of rotational states are populated. Under this condition, it is assumed that energy changes for rotation are small compared to the thermal energy (kT), allowing for a simplification of the rotational partition function, qrot. The simplified formula used is:

\(q_{rot} = \frac{T}{\theta_{rot}}\)

This equation illustrates that as the temperature T increases, the rotational partition function also increases. It is important to note that this high-temperature limit gives an approximate value that's accurate enough for practical uses when temperature is much greater than the characteristic rotational temperature of the molecule. When using this approximation, we can more easily compare and determine the impacts of rotational energy on the thermodynamic properties of molecules at sufficiently high temperatures.

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Most popular questions from this chapter

A general expression for the classical Hamiltonian is \(H=\alpha p_{i}^{2}+H^{\prime}\) where \(p_{i}\) is the momentum along one dimension for particle \(i, \alpha\) is a constant, and \(H^{\prime}\) are the remaining terms in the Hamiltonian. Substituting this into the equipartition theorem yields \(q=\frac{1}{h^{3 N}} \iint e^{-\beta\left(\alpha p_{i}^{2}+H^{\prime}\right)} d p^{3 N} d x^{3 N}\) a. Starting with this expression, isolate the term involving \(p_{i}\) and determine its contribution to \(q\) b. Given that the average energy \(\langle\varepsilon\rangle\) is related to the partition function as follows, \(\langle\varepsilon\rangle=\frac{-1}{q}\left(\frac{\delta q}{\delta \beta}\right)\) evaluate the expression the contribution from \(p_{i} .\) Is your result consistent with the equipartiton theorem?

NO is a well-known example of a molecular system in which excited electronic energy levels are readily accessible at room temperature. Both the ground and excited electronic states are doubly degenerate and are separated by \(121.1 \mathrm{cm}^{-1}\) a. Evaluate the electronic partition function for this molecule at \(298 \mathrm{K}\) b. Determine the temperature at which \(q_{E}=3\)

In using statistical mechanics to describe the thermodynamic properties of molecules, high-frequency vibrations are generally not of importance under standard thermodynamic conditions since they are not populated to a significant extent. For example, for many hydrocarbons the \(\mathrm{C}-\mathrm{H}\) stretch vibrational degrees of freedom are neglected. Using cyclohexane as an example, the IR-absorption spectrum reveals that the \(C-H\) stretch transition are located at \(\sim 2850 \mathrm{cm}^{-1}\). a. What is the value of the vibrational partition function for a mode of this frequency at \(298 \mathrm{K} ?\) b. At what temperature will this partition function reach a value of \(1.1 ?\)

Hydrogen isocyanide, HNC, is the tautomer of hydrogen cyanide (HCN). HNC is of interest as an intermediate species in a variety of chemical processes in interstellar space \((\mathrm{T}=2.75 \mathrm{K})\). a. For HCN the vibrational frequencies are \(2041 \mathrm{cm}^{-1}\) (CN stretch), \(712 \mathrm{cm}^{-1}\) (bend, doubly degenerate), and \(3669 \mathrm{cm}^{-1}\) (CH stretch). The rotational constant is \(1.477 \mathrm{cm}^{-1}\) Calculate the rotational and vibrational partition functions for \(\mathrm{HCN}\) in interstellar space. Before calculating the vibrational partition function, is there an approximation you can make that will simplify this calculation? b. Perform the same calculations for HNC which has vibrational frequencies of \(2024 \mathrm{cm}^{-1}(\mathrm{NC} \text { stretch }), 464 \mathrm{cm}^{-1}\) (bend, doubly degenerate), and \(3653 \mathrm{cm}^{-1} \mathrm{v}\) (NH stretch). The rotational constant is \(1.512 \mathrm{cm}^{-1}\). c. The presence of HNC in space was first established by Snyder and Buhl (Bulletin of the American Astronomical Society \(3[1971]: 388\) ) through the microwave emission of the \(\mathrm{J}=1\) to 0 transition of \(\mathrm{HNC}\) at \(90.665 \mathrm{MHz}\) Considering your values for the rotational partition functions, can you rationalize why this transition would be observed? Why not the \(\mathrm{J}=20-19\) transition?

For IF \(\left(\widetilde{\nu}=610 . \mathrm{cm}^{-1}\right)\) calculate the vibrational partition function and populations in the first three vibrational energy levels for \(T=300\). and \(3000 .\) K. Repeat this calculation for \(\operatorname{IBr}\left(\widetilde{\nu}=269 \mathrm{cm}^{-1}\right) .\) Compare the probabilities for IF and IBr. Can you explain the differences between the probabilities of these molecules?

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