Question

The simplest polyatomic molecular ion is \(\mathrm{H}_{3}^{+},\) which can be thought of as molecular hydrogen with an additional proton. Infrared spectroscopic studies of interstellar space have identified this species in the atmosphere of Jupiter and other interstellar bodies. The rotational- vibration spectrum of the \(\nu_{2}\) band of \(\mathrm{H}_{3}^{+}\) was first measured in the laboratory by \(\mathrm{T}\). Oka in \(1980[\text { Plyss. Rev: Lett. } 45(1980): 531] .\) The spectrum consists of a series of transitions extending from 2450 to \(2950 \mathrm{cm}^{-1}\) Employing an average value of \(2700 . \mathrm{cm}^{-1}\) for the energy of the first excited rotational- vibrational state of this molecule relative to the ground state, what temperature is required for \(10 \%\) of the molecules to populate this excited state?

Short answer

The temperature required for 10% of the \(\mathrm{H}_{3}^{+}\) molecules to populate the first excited rotational-vibrational state is approximately 517.5 K.

Step-by-step solution

Convert energy from wavenumbers to joules

To begin, we need to convert the given energy from wavenumbers (cm⁻¹) to joules (J). We can do this by using the conversion factor, where 1 cm⁻¹ is equal to \(1.9863\times10^{-23}\) J: \(E = 2700 \,\mathrm{cm}^{-1} \cdot 1.9863\times10^{-23} \, \mathrm{J}/\,\mathrm{cm}^{-1} = 5.363\times10^{-20}\) J.

Boltzmann distribution formula

We will use the Boltzmann distribution formula to find the required temperature. The formula relates the population ratio of molecules in two energy states to their energy difference and temperature: \(\frac{N_2}{N_1} = \frac{g_2}{g_1}\exp{\left(-\frac{ΔE}{k_B T}\right)}\), where: \(N_1\) and \(N_2\) are the number of molecules in the ground state and the first excited state, respectively. \(g_1\) and \(g_2\) are the statistical weights of the ground state and first excited state, respectively. (In our case, they are equal, so they will cancel out) ΔE is the energy difference between the two states. (In our case, it is the energy of the first excited state) \(k_B\) is the Boltzmann constant (\(1.380649\times10^{-23}\) J/K) T is the temperature in Kelvin.

Calculate the temperature

We are given that 10% of the molecules need to be in the excited state: \(\frac{N_2}{N_1} = \frac{1}{9}\), because for every 1 molecule in the excited state, there are 9 molecules in the ground state. Since \(g_1 = g_2\), we can remove the statistical weights from the formula: \(\frac{1}{9} = \exp{\left(-\frac{ΔE}{k_B T}\right)}\) Next, we solve for temperature: \(T = \frac{ΔE}{k_B \ln{\left(\frac{9}{1}\right)}}\) Now, plug in the values of ΔE and \(k_B\) we obtained in Step 1 and the known value of the Boltzmann constant, respectively: \(T = \frac{5.363 \times 10^{-20}\, \mathrm{J}}{1.380649\times10^{-23}\, \mathrm{J/K} \ln{\left(9\right)}}\) Calculate the temperature: \(T \approx 517.5\) K. Thus, a temperature of approximately 517.5 K is required for 10% of the \(\mathrm{H}_{3}^{+}\) molecules to populate the first excited rotational-vibrational state.

Key concepts

Understanding Boltzmann Distribution

The concept of Boltzmann distribution is a fundamental principle in statistical mechanics. It describes the distribution of particles, such as atoms or molecules, among various energy states in thermal equilibrium.

The key equation for the Boltzmann distribution is: \[\frac{N_2}{N_1} = \frac{g_2}{g_1} \times e^{-\frac{\Delta E}{k_B T}}\],where:\begin{itemize}\item \(N_1\) and \(N_2\) represent the number of particles in two different energy states.\item \(g_1\) and \(g_2\) are their respective statistical weights,\item \(\Delta E\) is the energy difference between these states,\item \(k_B\) is the Boltzmann constant, and\item \(T\) is the absolute temperature in Kelvin.\end{itemize}In simpler terms, the formula gives us insight into how the number of molecules in one energy state relates to the number at another, based on the temperature of the system. Higher temperatures enable a greater number of particles to reach higher energy states, as seen in our exercise where we find the temperature for a specific population ratio.

Infrared Spectroscopy & Molecular Vibrations

Infrared spectroscopy is an analytical technique used to identify and study compounds and their molecular structures. It involves passing infrared light through a sample and measuring the absorbed frequencies. Different bond vibrations absorb different frequencies of the infrared, creating a spectrum.Molecules absorb infrared radiation at frequencies that match the energy needed to change their vibrational states. This includes both stretching and bending vibrations of the chemical bonds. The spectrum resulting from this absorption can be used to identify the functional groups within the molecule and thus, determine its molecular identity and structure.

In studying interstellar molecular ions like \(\mathrm{H}_{3}^{+}\), infrared spectroscopy allows scientists to understand the molecular composition of celestial bodies by analyzing the spectrum of light emitted or absorbed by them.

Interstellar Molecular Ions

Interstellar molecular ions, such as \(\mathrm{H}_{3}^{+}\), play a crucial role in the chemistry of the universe. They are highly reactive and can be found in the interstellar medium, which is the matter that exists in the space between the stars. These ions influence the creation and destruction of other chemical species and can affect the evolution of galaxies.

\(\mathrm{H}_{3}^{+}\) is of particular interest because it is believed to be a fundamental component in the sequence of chemical reactions that produce complex molecules in space. The discovery and study of \(\mathrm{H}_{3}^{+}\)'s rotational-vibration spectrum provide insights into the physical conditions prevalent in space, such as temperature and density, contributing to our understanding of the cosmic environment.

Statistical Weights and Energy States

Statistical weights, represented as \(g\) in the Boltzmann formula, are an essential factor when determining the population of particles in different energy states. The statistical weight corresponds to the number of ways a particular energy state can be populated. In quantum mechanics, this is often related to the degeneracy of the energy level, which is determined by quantum numbers. For simple cases, a non-degenerate level has a statistical weight of 1, but more complex situations, such as those involving electron spins or molecular rotations, can result in higher statistical weights.

For the \(\mathrm{H}_{3}^{+}\) ion, assuming equal statistical weights for both the ground and the first excited state simplifies our calculations, as they cancel each other out in the Boltzmann formula. However, understanding how these weights contribute is key in more complex molecules where this may not be the case.