Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

In general, the high-temperature limit for the rotational partition function is appropriate for almost all molecules at temperatures above the boiling point. Hydrogen is an exception to this generality because the moment of inertia is small due to the small mass of H. Given this, other molecules with \(\mathrm{H}\) may also represent exceptions to this general rule. For example, methane \(\left(\mathrm{CH}_{4}\right)\) has relatively modest moments of inertia \(\left(I_{A}=I_{B}=I_{C}=5.31 \times 10^{-40} \mathrm{g} \mathrm{cm}^{2}\right)\). a. determine \(B_{A}, B_{B},\) and \(B_{C}\) for this molecule. b. Use the answer from part (a) to determine the rotational partition function. Is the high-temperature limit valid? and has a relatively low boiling point of \(T=112 \mathrm{K}\)

Short Answer

Expert verified
The rotational constants for methane are: \(B_A = B_B = B_C = B \approx 1.56\ \mathrm{cm^{-1}}\). The high-temperature limit of the rotational partition function is not valid for methane, as it predicts a partition function of around 4.56, which is much smaller than that of other molecules at temperatures above their boiling point.

Step by step solution

01

a. Determine the rotational constants B_A, B_B, and B_C for methane.

In order to determine the rotational constants B_A, B_B, and B_C for methane, we will use the formula: \(B = \frac{h}{8\pi^2Ic}\), where \(h\) is the Planck constant, \(I\) is the moment of inertia, and \(c\) is the speed of light in vacuum. For methane, the moments of inertia are equal, \(I_A = I_B = I_C = I = 5.31 \times 10^{-40}\ \mathrm{g\ cm^2}\). Applying the formula, we get \(B_A = B_B = B_C = B = \frac{h}{8\pi^2Ic}\) Plug in the values of constants \(h\) and \(c\): \(B = \frac{6.626 \times 10^{-34}\ \mathrm{J\ s}}{8\pi^2 (5.31 \times 10^{-40}\ \mathrm{g\ cm^2})(2.998 \times 10^{10}\ \mathrm{cm\ s^{-1}})}\) Now, after calculating, we get: \(B_A = B_B = B_C = B \approx 1.56\ \mathrm{cm^{-1}}\)
02

b. Determine the rotational partition function and validate the high-temperature limit.

We'll use the high-temperature limit approximation for the rotational partition function: \(q_\text{rot} \approx T^{3/2}(\frac{2\pi I_{\mathrm{rms}}}{h})^{3/2}\) We have \(I_{\mathrm{rms}} = \sqrt{\frac{I_A^2 + I_B^2 + I_C^2}{3}} = \sqrt{\frac{3I^2}{3}} = I = 5.31 \times 10^{-40}\ \mathrm{g\ cm^2}\). Hence, the expression becomes: \(q_\text{rot} \approx T^{3/2}(\frac{2\pi I}{h})^{3/2}\) Now, we plug in the values given in the problem: \(q_\text{rot} \approx (112\ \mathrm{K})^{3/2}(\frac{2\pi (5.31 \times 10^{-40}\ \mathrm{g\ cm^2})}{6.626 \times 10^{-34}\ \mathrm{J\ s}})^{3/2}\) After calculating, we get: \(q_\text{rot} \approx 4.56\) The high-temperature limit of the rotational partition function is not valid for methane, as it predicts a partition function of around 4.56, which is much smaller than that of other molecules at temperatures above their boiling point. This is due to the small moments of inertia of the hydrogen atoms in methane, as mentioned in the problem statement.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Rotational Constants
Rotational constants, denoted typically as B, are critical for understanding a molecule's rotational motion. They are directly related to the molecule's moment of inertia, and they help us describe the energy levels associated with its rotation. In quantum mechanics, these constants are used to quantify the energy levels of a diatomic or polyatomic molecule within the rigid rotor approximation.

The formula to calculate the rotational constant is given by
\[ B = \frac{h}{8\pi^2Ic} \]
where B represents the rotational constant in units of inverse centimeters (cm-1), h is Planck's constant, I is the moment of inertia, and c is the speed of light in a vacuum. For symmetrical molecules like methane (CH4), the rotational constants are the same for all axes (BA = BB = BC). Understanding rotational constants provides us with essential information about a molecule's rotational spectra and is invaluable for interpreting infrared and microwave spectroscopy data.
High-Temperature Limit
In the realm of statistical mechanics, the high-temperature limit refers to an approximation used to simplify calculations of a molecule's rotational partition function when the temperature is significantly high. This approximation assumes that because of the elevated temperature, the energy spacing between rotational levels is so small that they can effectively be treated as a continuum.

The high-temperature limit approximation for the rotational partition function qrot is given by:
\[ q_\text{rot} \approx T^{3/2}\left(\frac{2\pi I_{\mathrm{rms}}}{h}\right)^{3/2} \]
where T is the temperature, Irms is the root mean square moment of inertia, and h is Planck's constant. This approximation is usually valid for temperatures above the boiling point of the molecule and fails when the moment of inertia is small, as in the case of hydrogen within molecules. The validity of this approximation can affect the accuracy with which we predict properties like heat capacity and spectral lines.
Moment of Inertia
The moment of inertia, often symbolized as I, is a fundamental concept in physics that measures the extent to which an object resists rotational acceleration about a particular axis. It depends not only on the mass of the object but also on how that mass is distributed with respect to the axis of rotation. For a given molecule, the moment of inertia is crucial in determining its rotational kinetic energy and rotational constants.

In the context of molecular rotations, the moment of inertia is calculated as I = m*r2, where m is the mass and r is the distance from the axis of rotation. For polyatomic molecules like methane (CH4), calculating the moment of inertia involves considering the geometry and mass distribution of all atoms. The smaller the moment of inertia, the higher the rotational constant, which means a molecule with negligible mass, like hydrogen, will have considerably different rotational spectra and partition functions when compared to heavier molecules.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Determine the total molecular partition function for \(\mathrm{I}_{2}\) confined to a volume of \(1000 . \mathrm{cm}^{3}\) at \(298 \mathrm{K}\). Other information you will find useful is that \(B=0.0374 \mathrm{cm}^{-1},\left(\widetilde{\nu}=208 \mathrm{cm}^{-1}\right)\) and the ground electronic state is nondegenerate.

Evaluate the translational partition function for \(^{35} \mathrm{Cl}_{2}\) confined to a volume of \(1 \mathrm{L}\) at \(298 \mathrm{K}\). How does your answer change if the gas is \(^{37} \mathrm{Cl}_{2} ?\) (Hint: Can you reduce the ratio of translational partition functions to an expression involving mass only?)

Calculate the rotational partition function for \(^{35} \mathrm{Cl}_{2}\left(\mathrm{B}=0.244 \mathrm{cm}^{-1}\right)\) at \(298 \mathrm{K}\).

The effect of symmetry on the rotational partition function for \(\mathrm{H}_{2}\) was evaluated by recognizing that each hydrogen is a spin \(1 / 2\) particle and is, therefore, a fermion. However, this development is not limited to fermions, but is also applicable to bosons. Consider \(\mathrm{CO}_{2}\) in which rotation by \(180^{\circ}\) results in the interchange of two spin 0 particles. a. Because the overall wave function describing the interchange of two bosons must be symmetric with respect to exchange, to what \(J\) levels is the summation limited in evaluating \(q_{R}\) for \(\mathrm{CO}_{2} ?\) b. The rotational constant for \(\mathrm{CO}_{2}\) is \(0.390 \mathrm{cm}^{-1}\). Calculate \(q_{R}\) at \(298 \mathrm{K} .\) Do you have to evaluate \(q_{R}\) by summation of the allowed rotational energy levels? Why or why not?

Rhodopsin is a biological pigment that serves as the primary photoreceptor in vision (Science \(266[1994]: 422)\) The chromophore in rhodopsin is retinal, and the absorption spectrum of this species is centered at roughly 500 nm. Using this information, determine the value of \(q_{E}\) for retinal. Do you expect thermal excitation to result in a significant excitedstate population of retinal?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free