Statistical Thermodynamics and Rate Theories/Rotational energy

Rotational Energy

Molecules that consist of more than one atom have rotational angular momentum. The rotational motion which is generated by these molecules is purely kinetic as each molecule rotates around the molecules centre of mass. A good approximation which makes interpreting the rotational energy levels easier is the rigid rotor approximation which is mentioned below. Essentially this approximation assumes that the molecule has a fixed bond length and it does not vibrate while it undergoes rotation about its centre of mass. This approximation makes it easy to calculate the bond length by simply knowing the rotational energy spacings of the molecule observed in a spectroscopic experiment.

The rotational kinetic energy of an object depends on its moment of inertia, I,

$I=\Sigma _{i}m_{i}r_{i}^{2}$

where $m_{i}$ is the mass of the atom $i$, $r_{i}$ is the distance of that atom from the centre of mass of the molecule. Any molecule has three perpendicular axes of rotation each of which intersect at the centre of mass for that molecule, the moment of inertia is the rotational motion found at the center of mass of a molecule. Each of these axes of rotation could be distant from one another, or they could all be equal which leads to different types of rigid rotors. A large moment of inertia is basically based on either larger atomic masses and/or longer bonds.

Types of Rigid Rotor

Type	Moments of Inertia	Examples
Linear Rotors	$I_{1}=I_{2}=I_{3}=0$	N2, CO2, C2H2
Spherical Rotors	$I_{1}=I_{2}=I_{3}\neq 0$	CH4, SF6
Symmetric Rotors	$I_{1}=I_{2}\neq I_{3}$	NH3, CH3Cl
Asymmetric Rotors	$I_{1}\neq I_{2}\neq I_{3}$	H2O, H2CO, Most Molecules

Linear Rotors

In this course we are mostly concerned with linear rotors, using these systems simplifies the math. If we are using a linear rigid rotor, the bond length is constant due to the rigid rotor approximation, and the centre of mass is constant, therefore the moment of inertia can be simplifies using the reduced mass as follows:

$I=\Sigma \mu r_{e}^{2}$

here, $\mu$ is the reduced mass of the linear rotor where ${\frac {m_{1}m_{2}}{m_{1}+m_{2}}}$, and $r_{e}$ is the bond length of the diatomic linear rotor. The reduced mass the inertial mass of two atom system following Newtonian Mechanics. This can then be used to solve the Schrödinger equation which is given in example 2 below. By solving this equation we find that there are two quantum numbers which determine the energy levels of a rotating rigid rotor, M and J both of which are described below in example 2.

Zero Point Energy of Linear Rotors

Unlike the particles in a box model, which has a non zero energy minima due to the fact that the lowest energy state corresponds to the n=1 state, the linear rotor can have an energy of zero in the ground state corresponding to a molecule which is not rotating in space. This is the case since $J=0,1,2...$ when $J=0$ the energy of the system is equal to zero. Since the orientation in space is not known for a linear rotor it obeys the Heisenberg uncertainty principle which states that we cannot simultaneously determine the momentum and the position of the particle. Where each energy level can exist using following equation

$E_{J}={\frac {\hbar }{\mu r_{e}^{2}}}\times J(J+1)$

where J is the quantum number, $\mu$ is the reduced mass, r_e² is the length of the bond from the center of mass and ħ is reduced Planck's constant where Planck's constant is divided by 2π. From inserting J=0 in this equation above, we see $\epsilon _{0}=0$.

Degeneracy of Rotational States

The rotational levels of a diatomic only depend on the quantum number J (orbital angular momentum quantum number) and does not have any dependence on the quantum number M. However, as the quantum number J increases, the degeneracy of the states increases since there are more possible values of M, which is the magnetic quantum number which are as follows

$J=0,1,2...$

$M=-J,...,0,...,+J$

Therefore, as the quantum number J increases, the degeneracy of the states also increases. The degeneracy of the states can be calculated using the following equation:

$g_{J}=2J+1$

Degeneracy occurs when there are accessible states present at the same energy level, for example, when J=1, M_J= -1,0,+1 meaning there are 3 different degeneracies at quantum number J=1

Rotational Spectroscopy of Linear Rotors

Linear rotors which contain a permanent dipole are able to be observed through rotational-vibrational spectroscopy. These molecules exhibit fine structure of lines which are all separated by a constant value. This value can be used to determine the bond length of the diatomic in question but using the following equations.

$v=2B(J+1)$ and $B={\frac {h}{8\pi ^{2}cI}}$

where $v$ is the spacing of the rotational lines, $B$ is the rotational constant, $h$ is planks constant which has a value of $6.626\times 10^{-34}Js$, $c$ is the speed of light and $I$ is the moment of inertia. Once we know the value of $B$ we can then use this value to calculate the bond length of the diatomic which is being observed.

Example 1

Calculation of the bond length of HCl from the rotational constant.

Rotational Angular Momentum and the Rigid Rotor Approximation

Rotational energy is purely kinetic, which is a result of angular momentum in the molecule (How quickly the molecule is spinning in space). It can be assumed that the molecule is rigid for an ideal gas, this is because the change in the bond length for a vibration is small in comparison to the total length of the bond, this is called the Rigid Rotor Approximation.

Linear Rotors

The rotation of an object depends on its moment of inertia $I=\Sigma _{i}m_{i}r_{i}^{2}$ where $m_{i}$ is the mass of atom i, and $r_{i}$ is the distance of atom i to the axis of rotation. This expression works well if the diatomic molecule consists of only the same element (e.g. $O_{2},N_{2}$). For diatomic molecules which do not consist of the same elements (e.g. $HCl,NO,NaF$) this expression can be more work intensive then needed. For diatomic molecules which do not comprise of the same elements the expression $I=\mu r_{e}^{2}$ is significantly less work intensive to use in comparison to the previously stated equation. Where $r_{e}^{2}$ is the bond length of the diatomic molecule and $\mu$ is the reduced mass of the molecule, $\mu$ is given by the equation $\mu ={\frac {m_{1}m_{2}}{m_{1}+m_{2}}}$ where $m_{1}$ is the atomic mass of atom 1, and $m_{2}$ is the atomic mass of atom 2.

Rotational Spacings

Example of a pure rotational absorption spectrum

The spacings of rotational transitions are constant, and are commonly referred to as a rotational constants ${\bar {B}}$. For linear rotors the spacing is $2{\bar {B}}$ between each rotational transition as can be seen in the image to the right.

Utilizing the equation for the rotational constant ${\bar {v}}=2{\bar {B}}(J+1)$ we can find the value for ${\bar {B}}$ and then use it in the equation ${\bar {B}}={\frac {h}{8\pi ^{2}cI}}$ and solve for the moment of inertia $I$, where $h$ is Planck's constant, and $c$ is the speed of light. Once the moment of inertia is calculated it is simply the matter of solving for the diatomic bond length $r_{e}$ which can be found by $r_{e}={\sqrt {\frac {I}{\mu }}}$.

Sample Problem

Calculate the Bond Length of HCl from the Rotational Constant.

Knowing HCl has a rotational constant value of 10.59341 cm⁻¹, the Planck's constant is 6.626 × 10⁻³⁴J s, and the speed of light being 2.998 × 10¹⁰ cm s⁻¹. We then solve for the moment of inertia such that:

${\bar {B}}={\frac {h}{8\pi ^{2}cI}}$

$I={\frac {h}{8\pi ^{2}c{\bar {B}}}}$

$I={\frac {6.626\times 10^{-34}Js}{8\pi ^{2}(2.998\times 10^{10}cm/s)(10.59341cm^{-1})}}$

$I=2.642401\times 10^{-47}Js^{2}$

$I=2.642401\times 10^{-47}(kgm^{2}/s^{2})s^{2}$

$I=2.642401\times 10^{-47}kgm^{2}$

The next step is to calculate the reduced mass $\mu$. Knowing that the mass of hydrogen is 1.00794 u, and the mass of chlorine is 35.4527 u.

$\mu ={\frac {m_{1}m_{2}}{m_{1}+m_{2}}}$

$\mu ={\frac {(1.00794u)\times (35.4527u)}{(1.00794u)+(35.4527u)}}$

$\mu ={\frac {35.73419u^{2}}{36.46064u}}$

$\mu =0.9800758u$

$\mu =0.9800758u\times {\frac {1.6605\times 10^{-27}kg}{1u}}$

$\mu =1.627454\times 10^{-27}kg$

Now knowing all of the variable it is just a matter of solving for the bond length $r_{e}$.

$r_{e}={\sqrt {\frac {I}{\mu }}}$

$r_{e}={\sqrt {\frac {2.642401\times 10^{-47}kgm^{2}}{1.627454\times 10^{-27}kg}}}$

$r_{e}={\sqrt {1.623641\times 10^{-20}m^{2}}}$

$r_{e}=1.274222\times 10^{-10}m$

$r_{e}=1.274222$ Å

$r_{e}=1.274$ Å

Example 2

Calculation of the energy for the transition of N₂ from the ground to first excited rotational state.

Solution

For N₂, the reduced mass is calculated using an atomic mass of 14.0067 u for nitrogen. Since it is a diatomic molecule, the equation can be simplified to the following where $m$ represents the mass of a nitrogen atom:

$\mu ={\frac {m^{2}}{2m}}$

$\mu ={\frac {m}{2}}$

$\mu ={\frac {14.0067u}{2}}$

$\mu =7.0033u$

$\mu =(7.0033u)\left({\frac {1.660539040\times 10^{-27}kg}{u}}\right)$

$\mu =1.1629\times 10^{-26}kg$

For the rigid rotor, the lowest possible energy state is J=0 so the ground state rotational energy is always zero. The rotational energy levels are only dependent on quantum number J and only the ∆𝐽 = ±1 transitions can occur. In the case of the transition from the ground to the first excited state, the following is a diagram that represents the transition.

The energy for the transition J=1 ← J=0 is calculated as follows:

$\Delta \epsilon ={\frac {\hbar ^{2}}{2\mu r_{e}^{2}}}J(J+1)-{\frac {\hbar ^{2}}{2\mu r_{e}^{2}}}J(J+1)$

$\Delta \epsilon ={\frac {\hbar ^{2}}{2\mu r_{e}^{2}}}1(1+1)-{\frac {\hbar ^{2}}{2\mu r_{e}^{2}}}0(0+1)$

$\Delta \epsilon ={\frac {\hbar ^{2}}{\mu r_{e}^{2}}}$

$\Delta \epsilon ={\frac {(1.0545718Js)^{2}}{(1.1629\times 10^{-26}kg)(1.098\times 10^{-10}m)^{2}}}$

$\Delta \epsilon =7.9320\times 10^{-23}J$