# Statistical Thermodynamics and Rate Theories/Vibrational energy

Molecular vibrations occur when atoms in a molecule are in periodic motion while the molecule as a whole has constant translational and vibrational motion. Vibrational energy is different from translational energy (how a molecule translates through 3-dimensional space along the x,y, and z Cartesian coordinates) and rotational energy (how fast a molecule spins) in that the focus is placed on how a chemical bond behaves with respect to the quantum harmonic oscillator approximation.

## Vibrational Energy Levels

### The Potential Energy Surface of Molecules

Chemical bonds are formed due to electrostatic interactions between protons and electrons, and can occur as nuclear-electron attractions, electron-electron repulsion, and inter-nuclear repulsion. As a bond forms due to the nuclear electron attractions between a proton and electron, the electron density increases in the space between two nuclei. For a diatomic, the potential energy is only a function of the distance between the two atoms, and the strength of these components changes depending on the distance between the atoms. Bonding is generally stronger when two nuclei are close, but this close proximity also increases internuclear and core-core repulsions. Hence, the potential energy is lowest at the equilibrium bond length ${\displaystyle {r_{e}}}$ . Short distances lead to high repulsion, long distances result in no interactions, and middle distances result in stabilized bonding.

### Application of Hooke's Law

Molecules will have some amount of kinetic energy along this degree of freedom and harmonic oscillation will occur along the bond length. A bond between molecules can be approximated as a parabola centred around the equilibrium bond length ${\displaystyle {r_{e}}}$ . Hooke's Law treats the bond as a spring. When a bond is stretched or compressed its potential energy increases, just like the behaviour of a spring under the harmonic oscillation approximation. ${\displaystyle {k}}$  is the constant corresponding to the stiffness of the spring, called the spring constant, and can be correlated to the stiffness of a molecular bond. Higher spring constants correlate to stiffer (stronger) bonds. For example, the value of ${\displaystyle {k}}$  will be larger for a double bond than in a single bond. Hence, a double bond is more stiff.

We can approximate the bond as a parabola centred around the equilibrium bond length ${\displaystyle {r_{e}}}$ :

${\displaystyle {\mathcal {V}}(r)={\frac {1}{2}}k_{bond}\left(r-r_{eq}\right)^{2}}$

### Quantum Harmonic Oscillator Approximation

There are several significant approximations associated with describing chemical bonds as harmonic oscillators:

1. The bond can never dissociate. In reality if we keep pulling two atoms apart the bond will eventually break, but according to the quantum harmonic oscillator approximation the potential energy just continues to increase and bonds are not allowed to break. This can be referred to as anharmonic effects; if bonds were really harmonic no chemical reaction would ever take place.
2. The repulsive wall isn't repulsive enough. In reality, short distances are dominated by repulsive interactions. The low potential energy structures of most important molecules are near the minimum energy geometry.
3. The spacing of the vibrational energy levels are exactly equal. However, in reality they become slightly smaller as quantum number ${\displaystyle {n}}$  increases.

Energy levels of a quantum harmonic oscillator follow:

${\displaystyle E_{n}=h{\nu }\left(n+{1 \over 2}\right)}$

where ${\displaystyle {h}}$  is Planck's Constant (6.626×10−34 J s), ${\displaystyle \nu }$  is the vibrational frequency of the oscillator in s-1 and quantum number ${\displaystyle {n}}$ = 0,1,2...

${\displaystyle \mu ={\cfrac {m_{1}m_{2}}{m_{1}+m_{2}}},\!\,}$

${\displaystyle {v}}$  is related to the spring constant, ${\displaystyle {k}}$ , and reduced mass of a diatomic,${\displaystyle {\mu }}$ .

${\displaystyle v={1 \over {2\pi }}{\sqrt {k \over {\mu }}}\!}$

### Other Important Formulae

Since Planck's constant, ${\displaystyle h}$ , and vibrational frequency, ${\displaystyle \nu }$ , are related by ${\displaystyle E=h\nu }$ , the relation can also be expressed as: ${\displaystyle E={\frac {hc}{\lambda }}.}$

Wavenumbers, also called vibrational frequency, are reported in reciprocal distance and chemists typically use reciprocal centimetres.

${\displaystyle v={\frac {1}{\lambda }}={\frac {E}{hc}}}$

where ${\displaystyle {E}}$  is Energy in Joules, ${\displaystyle {c}}$  is the speed of light (2.9979×108 m/s = 2.9979×1010 cm/s) and, ${\displaystyle {\lambda }}$  is wavelength. This molecular transition energy is usually extremely small and typically follows the order of 1×10−20 joules, J.

### Vibrational Zero Point Energy

The ground state does not have zero energy. This is the vibrational zero-point energy, and is consistent with the Heisenberg Uncertainty principle. If a molecule had zero vibrational energy it would be at rest at the minimum, and therefore we would know its position and momentum completely. Even in the lowest energy state, molecules still vibrate.

${\displaystyle E_{n=0}=h\nu \left(n+{1 \over 2}\right)=h\nu \left(0+{1 \over 2}\right)={1 \over 2}h\nu }$

### Vibrational Absorption Spectra

Vibrational transitions follow the selection rule ${\displaystyle \Delta n=\pm 1[itex],andthereforeallvibrationaltransitionshaveexactlythesameenergy.[itex]\Delta E=h{\nu }\left((n'+1)+{1 \over 2}\right)-h{\nu }\left(n'+{1 \over 2}\right)}$

${\displaystyle \Delta E=h{\nu }\left((n'+1)+{1 \over 2}-(n'+{1 \over 2})\right)}$

${\displaystyle \Delta E=h{\nu }(1)=h{\nu }}$

The vibrational spectrum of diatomics is coupled with rotational transitions. The total selection rule is${\displaystyle \Delta n=\pm 1[itex]and[itex]\Delta J=\pm 1[itex].===VibrationalStatesofPolyatomics===Whilemonatomicgaseousatomshaveavibrationaldegreeoffreedomofzero,polyatomicmoleculeswillhaveseveralvibrationaldegreesoffreedom.Thesevibrationsaretreatedasindependentharmonicoscillators.Polyatomicmoleculesaredividedintotwocategories:Linearandnon-linearmolecules.Forlinearmolecules:[itex]n_{vib}=3N_{atom}-5}$

For non-linear molecules: ${\displaystyle n_{vib}=3N_{atom}-6}$

### Trends in Vibrational Frequencies

• Strong bonds are stiff bonds, which have a large spring constant (${\displaystyle k}$ ).
• Stiff bonds have large vibrational energy level spacings.
• Light atoms give small reduced masses (${\displaystyle {\mu }}$ ).
• Molecules with small reduced masses have large vibrational energy level spacings.

## Vibrational Energy Example 1

The harmonic vibrational frequency of HF in wavenumbers is 4141.3 cm-1. [1]

Using this information, calculate the energy for the transition of HF from the ground to first excited vibrational state.

### Solution

Vibrational energy levels of a quantum harmonic oscillator follow:

${\displaystyle E_{n}=h{v}\left(n+{1 \over 2}\right)}$

However, this relationship requires the vibrational frequency to be in units of s-1. To do this, we multiply the given value of 4141.3 cm-1 by the speed of light, in centimetres.

${\displaystyle v={\widetilde {v}}c}$ ${\displaystyle v=(4141.3cm^{-1})(2.9979\times 10^{10}cm/s)}$

${\displaystyle v=1.2415\times 10^{14}s^{-1}}$

Now that we have the correct units, the ground state vibrational energy can be calculated as follows:

${\displaystyle E_{0}=h{v}\left(n+{1 \over 2}\right)=h{v}\left(0+{1 \over 2}\right)={1 \over 2}h{v}}$
${\displaystyle E_{0}=\left({1 \over 2}\right)(6.626\times 10^{-34}Js)(1.2415\times 10^{14}s^{-1})}$
${\displaystyle E_{0}=4.1131\times 10^{-20}J}$

Next, the first excited state vibrational energy is:

${\displaystyle E_{1}=h{v}\left(n+{1 \over 2}\right)=h{v}\left(1+{1 \over 2}\right)={3 \over 2}h{v}}$
${\displaystyle E_{1}=\left({3 \over 2}\right)(6.626\times 10^{-34}Js)(1.2415\times 10^{14}s^{-1})}$
${\displaystyle E_{1}=1.2339\times 10^{-19}J}$

Finally, the energy for the transition from the ground state (n=0) to the first excited state (n=1) level is:

${\displaystyle \Delta E=E_{1}-E_{0}}$
${\displaystyle \Delta E=1.2339\times 10^{-19}J-4.1131\times 10^{-20}J}$
${\displaystyle \Delta E=8.2259\times 10^{-20}J}$

## Vibrational Energy Example 2

Calculation of the vibrational wavenumber of CO from the force constant and reduced mass.

### Harmonic Oscillator Approximation

As there is some kinetic energy along the vibrational degree of freedom, the bond length undergoes oscillation. The Schrӧdinger equation for the movement of quantum mechanical particles given the potential energy function V(r) is complicated to solve. Instead, the Harmonic Oscillator Approximation is used which proposes that the bond behaves like a spring governed by Hooke's law.

Hooke's law: ${\displaystyle {\mathcal {V}}(r)={\frac {1}{2}}k(r-r_{eq})^{2}}$  where r is the internuclear distance and k is a constant corresponding to the stiffness of the spring

Given this approximation, the following relationship arises

${\displaystyle \nu ={1 \over {2\pi }}{\sqrt {k \over {\mu }}}\!}$

where,

• ${\displaystyle k}$  is the force constant of the bond in ${\displaystyle {\textrm {kg}}\ {\textrm {s}}^{-1}}$
• ${\displaystyle \mu }$  is the reduced mass in kilograms
• ${\displaystyle v}$  is the vibrational frequency of the oscillator in ${\displaystyle s^{-1}}$

The reduced mass ${\displaystyle {\mu }}$  for a heteronuclear diatomic can be calculated as follows

${\displaystyle {\mu }}$  ${\displaystyle ={\cfrac {m_{1}m_{2}}{m_{1}+m_{2}}}}$  where ${\displaystyle m_{1}}$  and ${\displaystyle m_{2}}$  corresponds to the masses of each atom.

### Solution

Firstly, the reduced mass of the carbon monoxide molecule must be calculated given that the atomic masses of carbon and oxygen is ${\displaystyle 12.0107u}$  and ${\displaystyle 15.9994u}$  respectively.

${\displaystyle \mu ={\cfrac {m_{1}m_{2}}{m_{1}+m_{2}}}}$
${\displaystyle \mu ={\cfrac {(12.0107\ u)(15.9994\ u)}{12.0107\ u+15.9994\ u}}}$
${\displaystyle \mu ={\cfrac {192.164\ u^{2}}{28.0101\ u}}}$
${\displaystyle \mu =6.8605\ u}$

Convert into units of kilogram as follows

${\displaystyle \mu =(6.8605\ u)\left({\cfrac {1\ kg}{1000\ g}}\right)\left({\cfrac {1}{6.02214129\times 10^{23}\ mol^{-1}}}\right)}$
${\displaystyle \mu =1.1392\times 10^{-26}\ kg}$

The next step is to derive the vibrational frequency in ${\displaystyle s^{-1}}$  given the value of the reduced mass,${\displaystyle \mu =(1.1392)(10^{-26})kg}$  , and the force constant ${\displaystyle k=1860Nm^{-1}}$  [2]. Note that ${\displaystyle 1\ N\ m^{-1}=1\ kg\ s^{-2}}$

${\displaystyle \nu ={1 \over {2\pi }}{\sqrt {k \over {\mu }}}\!}$
${\displaystyle \nu ={1 \over {2\pi }}{\sqrt {1860kg/s^{2} \over {1.1392\times 10^{-26}\ kg}}}\!}$
${\displaystyle \nu =(0.15915)({\sqrt {(1.6327\times 10^{29})\ s^{-2}}}\!}$
${\displaystyle \nu =(0.15915)(4.0407\times 10^{14})\ s^{-1}\!}$
${\displaystyle \nu =6.4308\times 10^{13}\ s^{-1}\!}$

Convert the vibrational frequency into wavenumbers given the following relationship

${\displaystyle {\tilde {v}}={\cfrac {vibrationalfrequency,v\ (s^{-1})}{\textrm {speedoflight}}}}$

${\displaystyle {\tilde {v}}={\cfrac {6.4308\times 10^{13})\ s^{-1}}{\textrm {speedoflight}}}\!}$

${\displaystyle {\tilde {v}}={\cfrac {6.4308\times 10^{13}s^{-1}}{2.99792458\times \ 10^{10}cm\ s^{-1}}}\!}$

${\displaystyle {\tilde {v}}=2145.1\ cm^{-1}\!}$

## References

1. Atkins, Peter and DePaula, Julio. "Elements of Physical Chemistry". Oxford University Press, 2013, p. 501.
2. HyperPhysics.Quantum Physics: Vibrational Spectra of Diatomic Molecules. http://hyperphysics.phy-astr.gsu.edu/hbase/molecule/vibspe.html (accessed Feb. 2, 2016)