Question

In microwave spectroscopy a traditional unit for the rotational constant is the Mc or "mega cycle" equal to \(10^{6} \mathrm{s}^{-1}\) For \(^{14} \mathrm{N}^{14} \mathrm{N}^{16} \mathrm{O}\) the rotational constant is \(12,561.66 \mathrm{Mc}\). a. Convert the above value for the rotational constant from \(\mathrm{Mc}\) to \(\mathrm{cm}^{-1}\). b. Determine the value of the rotational partition function at \(298 \mathrm{K}\).

Short answer

a. The rotational constant in cm⁻¹ is approximately 0.04187 cm⁻¹. b. The rotational partition function value at 298 K is approximately 122.

Step-by-step solution

Convert Mc to Hz

First, we need to convert the given rotational constant value from Mc to Hz (s⁻¹). \(Rotational \; Constant \; (Hz) = 12,561.66 \; Mc × 10^6 \; s^{-1}/Mc\)

Convert Hz to cm⁻¹

Now, we will use the speed of light (c) to convert the rotational constant value from Hz (s⁻¹) to cm⁻¹. The speed of light in cm/s is \(c = 2.998 × 10^{10} \; cm/s\). From the relation \(E = h\nu\), where E is the energy and \(\nu\) is the frequency, we can convert the rotational constant from Hz to cm⁻¹: \(Rotational \; Constant \; (cm^{-1}) = \frac{Rotational \; Constant \; (Hz)}{c}\) \(Rotational \; Constant \; (cm^{-1}) = \frac{12,561.66 \times 10^6 \; s^{-1}}{2.998 \times 10^{10} \; cm/s}\) Calculate the value: \(Rotational \; Constant \; (cm^{-1}) ≈ 0.04187 \; cm^{-1}\)

Calculate the Rotational Partition Function

Now, we have to calculate the value of the rotational partition function (\(q_{rot}\)) at 298 K. The formula for the rotational partition function is: \(q_{rot} = \frac{T}{\theta_{rot}}\) where T is the temperature and \(\theta_{rot}\) is the characteristic rotational temperature. The characteristic rotational temperature relates to the rotational constant (B) by: \(\theta_{rot} = \frac{hcB}{k}\) where h is the Planck's constant (6.626 × 10⁻³⁴ Js), c is the speed of light (2.998 × 10¹⁰ cm/s), and k is the Boltzmann's constant (1.381 × 10⁻²³ J/K). First, let's calculate \(\theta_{rot}\): \(\theta_{rot} ≈ \frac{(6.626 \times 10^{-34} \; Js)(2.998 \times 10^{10} \; cm/s)(0.04187 \; cm^{-1})}{1.381 \times 10^{-23} \; J/K}\) \(\theta_{rot} ≈ 2.4402 \; K\) Now, let's calculate the rotational partition function at 298 K: \(q_{rot} = \frac{298 \; K}{2.4402 \; K}\) \(q_{rot} ≈ 122\) Now that we have completed the steps, we can answer each part of the exercise. a. The value of the rotational constant in cm⁻¹ is approximately 0.04187 cm⁻¹. b. The rotational partition function value at 298 K is approximately 122.

Key concepts

Rotational Constant Conversion

Understanding the rotational constant is crucial in microwave spectroscopy, as it provides information about the energy levels of a molecule’s rotation. In spectroscopy, various units can be used to express this constant, such as mega cycles (Mc) or reciprocal centimeters (\text{cm}^{-1}). Conversion of units is often required to align with the common scientific standards or to make the values more comprehensible.

The conversion from Mc to Hz is straightforward, involving multiplying by the factor of \text{10}^6, since 1 Mc equals \text{10}^6 s^{-1}. Further conversion to \text{cm}^{-1} requires knowledge of the speed of light (c), which highlights the relationship between frequency (\text{Hz}) and wavelength (\text{cm}). With the speed of light at approximately 3.00 \(\times\) \text{10}^{10} \text{cm/s}, the conversion from Hz to \text{cm}^{-1} becomes a matter of dividing the frequency in Hz by the speed of light in \text{cm/s}.

This process allows us to obtain a value for the rotational constant in units of \text{cm}^{-1}, which is widely used in quantum chemistry and molecular physics to predict rotational energy levels and spectra.

Rotational Partition Function

The concept of a rotational partition function (\text{q}_{rot}) is essential in statistical mechanics as it relates to the distribution of energy levels in a rotating molecule. This function plays a critical role in determining the rotational contribution to the thermodynamic properties of gas-phase substances.

The rotational partition function is calculated using the temperature (T) and the characteristic rotational temperature (\text{\(\theta_{rot}\)}), which reflects the spacing between rotational energy levels. This temperature is directly proportional to the rotational constant (B) of a molecule. As shown in the exercise solution, the partition function can be calculated by dividing the actual temperature by the characteristic rotational temperature.

With \text{T} being 298 K for most room temperature calculations, understanding \text{q}_{rot} can aid in predicting how energy is distributed among the rotational states of a molecule and how this affects properties such as heat capacity and entropy. The higher the \text{q}_{rot}, the greater the number of accessible rotational states at a given temperature.

Characteristic Rotational Temperature

The characteristic rotational temperature (\text{\(\theta_{rot}\)}) provides an interesting perspective on the energy associated with the rotation of a molecule. It is represented by the formula \text{\(\theta_{rot} = \frac{hcB}{k}\)}, where \text{h} is Planck’s constant, \text{c} is the speed of light, \text{B} is the rotational constant, and \text{k} is Boltzmann’s constant.

\text{\(\theta_{rot}\)} connects macroscopic measurements of temperature with microscopic rotational energy levels. When the temperature of the system is above \text{\(\theta_{rot}\)}, there will be a significant population of excited rotational levels. Conversely, if the temperature is below \text{\(\theta_{rot}\)}, most of the molecules reside in low rotational energy levels.

This characteristic temperature allows us to simplify the complex interactions within a molecule to a value that can be used comparatively across different substances. For example, knowing \text{\(\theta_{rot}\)} can help in determining the temperature range where specific rotational transitions are likely to be observed in a spectroscopy experiment.