Statistical thermodynamics (overview)

Statistical thermodynamics is a theory that connects the microscopic behaviour of atoms and molecules with the macroscopic laws of thermodynamics.

It uses the principles of probability and statistics to predict the average behaviour of a large number of particles, allowing us to understand how the properties of individual atoms and molecules give rise to observable thermodynamic quantities such as temperature, pressure and entropy. Rather than tracking every particle individually, it focuses on the distribution of particles among various energy states and uses this information to compute macroscopic properties to explain the laws of thermodynamics.

Statistical thermodynamics has a wide range of applications in chemistry, including explaining and predicting chemical equilibrium and the transition state theory; deriving expressions for heat capacity, internal energy and entropy based on molecular motions; and understanding phenomena like melting, boiling, and magnetisation by analysing how microscopic interactions change with temperature and pressure.

To fully comprehend the theory, one must first understand the concept of a canonical ensemble.

 

Next article: The canonical ensemble
Content page of statistical thermodynamics
Content page of advanced chemistry
Main content page

Molecular partition function

The molecular partition function  is a mathematical expression that describes how a non-interacting particle in a system is distributed among different energy levels at a given temperature.

It naturally emerges when analysing the canonical partition function of a system composed of non-interacting, distinguishable particles. Under these conditions, the total partition function factorises into a product of single-particle partition functions, allowing the definition of the molecular partition function as the canonical partition function of an individual particle.

Mathematically, it is given by

where is the -th energy level of the particle and is the Boltzmann constant.

It is also equivalent to the denominator of the Boltzmann distribution, which gives the probability of finding a particle in the -th energy state at thermal equilibrium:

 

Question

What are non-interacting molecules?

Answer

Non-interacting molecules behave independently and do not interact with one another through forces such as chemical bonding, electrostatic attraction/repulsion or van der Waals forces. The energy levels of each molecule are determined solely by its own internal structure (electronic, vibrational, rotational and translational states), and not altered by the presence of nearby molecules. Ideal gases, or very dilute gases where interactions are negligible, can be considered systems containing non-interacting molecules.

 

We know that is a function of temperature , but why is it called a partition function? Consider the case when is very high. In this limit, each term in the sum approaches 1. Conversely, when , only the term with (the ground state) is equal to 1. This means that tells us how many energy states are accessible to a molecule at a given temperature. When it appears in the Boltzmann distribution, it then reveals how the total probability is partitioned among the available energy levels.

Although the molecular partition function is defined for a single molecule, it can be used to describe the behaviour of a collection of non-interacting molecules in a system. This is because, when the molecules do not interact, the probability that any one molecule occupies a particular energy state given by eq50, also represents the fraction of all molecules found in that state at thermal equilibrium. This situation is analogous to rolling a four-sided biased die, where each face has a different probability of appearing. In this analogy, the probability of a particular outcome corresponds to the probability of a molecule occupying a specific energy state. Since each independent roll follows the same probability distribution (e.g. 40% for 1, 30% for 2, 17% for 3 and 13% for 4), the likelihood of any outcome is the same for a single roll or for many rolls. Likewise, at equilibrium, the probability that a single molecule is in a given energy state is equal to the fraction of the total number of molecules in that state.

To elaborate on this concept, consider an equally spaced array of energy levels (0, , , …, ) like those of a harmonic oscillator, with

Let . Since for (see this article for details), eq252 and eq251 become

and

respectively, where .

Most diatomic molecules are in their vibrational ground state at room temperature. For example, if and , the state populations (according to eq254) are , and . At higher temperatures, e.g. , more vibrational levels become populated, with , , , , , , and . These results are illustrated in the diagram below.

Notably, a higher value of corresponds to a larger number of significantly populated states. Therefore, the molecular partition function provides a measure of how many energy states are thermally accessible at a given temperature and, together with the Boltzmann distribution, partitions the total probability among the available energy levels.

 

Question

If the molecular partition function is related to the Boltzmann distribution, which is based on distinguishable particles, is it applicable to distinguishable or indistinguishable particles since it can also be used to describe a collection of particles?

Answer

The molecular partition function is defined for a single non-interacting molecule and does not itself assume distinguishability or indistinguishability. When extended to a system of non-interacting molecules, the probability that a molecule occupies a particular state  remains , which is the same fraction of molecules in that state regardless of whether the molecules are considered distinguishable or indistinguishable. This is because each molecule independently follows the same single-particle probability distribution. The distinction between distinguishable and indistinguishable particles only affects the form of the canonical partition function , not the calculation of these fractions.

 

The molecular partition function can also be expressed as:

where refers to the number of states having the same energy (degenerate states), and the sum is now over degenerate states, not individual states.

 

Question

Use an example to show that eq250 and eq255 are equivalent.

Answer

 

Energy Level,

Degeneracy, Contributing States

0

2

A, B

3

C, D, E

1

F

 

Using eq250,

Using eq255,

 

Finally, the total energy of a molecule in state  can be approximated as:

where

is the translational energy of the molecule in state
is the rotational energy of the molecule in state
is the vibrational energy of the molecule in state
is the electronic energy of the molecule in state

The validity of eq256 is based on the following assumptions:

    1. The four types of energy are independent of each other. For instance, coupling between vibrational and rotational motion is small and can be neglected.
    2. The Born-Oppenheimer approximation is applied, which assumes that electronic motion can be treated independently of nuclear motion (translation, rotation, vibration) because electrons move much faster than nuclei. This results in the total wavefunction of the molecule being approximated as the product of four wavefunctions: .
  1. Substituting eq256 into eq250 gives

Since the sum is over all states , we have

This factorisation of the molecular partition function is extremely useful in statistical mechanics because it allows us to treat each mode of motion — translational, rotational, vibrational, and electronic — independently. Instead of summing over all possible combinations of quantum states for the entire molecule (which can be computationally intractable), we can compute simpler, separate partition functions for each type of motion. This greatly simplifies calculations of thermodynamic properties such as internal energy, entropy and heat capacity. It also helps to explain the transition state theory.

 

Next article: translational Molecular partition function
Previous article: Canonical partition function for a system of non-interacting particles
Content page of statistical thermodynamics
Content page of advanced chemistry
Main content page

Thermodynamic functions involving the molecular partition function

Thermodynamic functions involving the molecular partition function describe the macroscopic thermodynamic properties of solids and ideal gases composed of non-interacting particles in the canonical ensemble.

These functions are derived using thermodynamic functions involving the canonical partition function :

Thermodynamic Property Function Derivation
Pressure eq146
Internal energy

eq148

eq157

Entropy eq159
Helmholtz energy eq159a
Enthalpy eq159b
Constant-volume heat capacity eq159c
Gibbs energy eq159d
Chemical potential eq159e

For example, substituting eq166 into eq157 yields the internal energy for a system of  non-interacting, distinguishable particles (such as those in a solid):

Here, is the total single-particle partition function (i.e. eq257), with contributions from translational, rotational, vibrational and electronic degrees of freedom. Eq320 assumes that all components of , including , are defined using absolute energy levels — for example, the vibrational molecular partition function includes the zero-point energy (i.e. eq290). However, if is instead defined relative to the zero-point energy (i.e. eq292), then we need to add back the zero-point energy term giving:

 

Question

Show how eq320 is applicable to a system in which particles can occupy two states with absolute energies 0 and .

Answer

From eq250, . Therefore,

As , , which implies that all particles are in the ground state. As , . This is also physically intuitive. At very high temperatures, thermal energy far exceeds the energy gap . The two energy states are equally populated, and the average energy per particle is . The total internal energy is therefore  times this average energy.

 

The above Q&A demonstrates that although the molecular partition function is defined for a single particle, thermodynamic functions derived from it describe the macroscopic properties of a system. In the case of an ideal gas (composed of non-interacting, indistinguishable particles), substituting eq167 into eq157 also results in eq320. Similarly, for non-interacting, indistinguishable particles, each of the expressions for pressure , enthalpy and constant-volume heat capacity in terms of the molecular partition function is the same. They are given by:

 

Question

Compare the expressions for pressure derived from the canonical partition function and the molecular partition function.

Answer

Comparing eq322 and eq146,

Aspect

Dependence

Canonical (full system) partition function

Molecular (single-particle) partition function

Assumed interactions

Valid for both interacting and non-interacting particles

Assumes non-interacting, indistinguishable particles

Application Real gases

Ideal gases

 

However, the statistical thermodynamic expressions for entropy , as well as for Helmholtz energy, Gibbs energy and chemical potential, differ between distinguishable and indistinguishable particles due to the Gibbs paradox. For an ideal gas, substituting eq167 into eq159 results in:

It follows that the Helmholtz energy expression for an ideal gas is:

For Gibbs energy, we have . Hence,

 

Question

Show that .

Answer

Substituting into eq326 yields , where . For an ideal gas, , where is the number of moles and , with  being the Avogadro constant. So,

where .

Using Stirling’s approximation and ,

where is the molar molecular partition function.

Dividing eq328 through by gives:

 

Substituting eq167 into eq159e gives the expression for the chemical potential of an ideal gas:

 

Next article: Gibbs paradox
Previous article: MEan Molecular energies of non-interacting molecules
Content page of statistical thermodynamics
Content page of advanced chemistry
Main content page

Statistical thermodynamic derivation of the ideal gas law

The ideal gas law is derived from the statistical thermodynamic expression for pressure using the canonical partition function of non-interacting, indistinguishable particles.

Substituting eq257 into eq322 gives:

Since only the translational molecular partition function is dependent on , we have

Substituting eq266 into eq350 yields:

With , where is the Avogadro constant and is the number of moles, we have the ideal gas law:

where the universal gas constant .

This derivation demonstrates how macroscopic thermodynamic laws are consistent with the microscopic principles of quantum mechanics, as described by statistical thermodynamics. It reinforces the concept that, in an ideal gas, the particles are non-interacting point masses whose energy is solely due to translational motion.

 

Next article: Internal energy in statistical thermodynamics
Previous article: Gibbs paradox
Content page of statistical thermodynamics
Content page of advanced chemistry
Main content page

Equilibrium constant in statistical thermodynamics

The equilibrium constant in statistical thermodynamics relates to the ratio of molecular partition functions for products and reactants. It reflects how molecular energy levels and their populations determine macroscopic equilibrium.

Consider the reaction

where is the stoichiometric coefficient of the species in the reversible reaction. Note that the stoichiometric coefficients for reactants are negative by convention, while those for products are positive.

The reaction Gibbs energy for the reaction is:

Substituting eq329 into eq380 gives,

where the reaction zero-point energy change is .

Eq381 can be rewritten as:

Substituting eq8 into eq382 and rearranging yields:

 

Question

Show that eq383 is consistent with the classical thermodyamic equilibrium constant .

Answer

As is derived using (see this article), we need to show that the statistical thermodynamics expression for the chemical potential of an ideal gas can be transformed into . Substituting eq257 into eq330 yields:

Substituting eq266 into eq384 gives:

where .

Substituting the ideal gas law into eq385 gives:

or equivalently,

From eq386, we can define the standard chemical potential as . Then,

where is the activity of an ideal gas.

 

 

Previous article: Internal energy in statistical thermodynamics
Content page of statistical thermodynamics
Content page of advanced chemistry
Main content page

Internal energy in statistical thermodynamics

The internal energy of a thermodynamic system consisting of non-interacting particles can be represented statistically using the molecular partition function .

It is equivalent to the product of the mean molecular energy and the number of non-interacting particles in the system. From eq320, where , we have:

As mentioned in an earlier article, is the total molecular partition function (i.e. eq257), with contributions from translational, rotational, vibrational and electronic degrees of freedom. Eq360 assumes that all components of , including , are defined using absolute energy levels — for example, the vibrational molecular partition function includes the zero-point energy (i.e. eq290). However, if is instead defined relative to the zero-point energy (i.e. eq292), then we need to add back the zero-point energy term giving:

where .

 

Monatomic ideal gas

For a monatomic ideal gas, each atom is a point mass with no rotational or vibrational degrees of freedom. This implies that there is only a single rotational ground state and a single vibrational ground state, each with energy equal to zero. Therefore, . Furthermore, as explained in the derivation of eq297, for an atom not subjected to extreme temperatures. Hence, eq361 simplifies to:

Substituting eq266 into eq362 and differentiating gives:

Comparing eq363 with eq305, . Substituting , where is the Avogadro constant and is the number of moles, into eq363 yields:

where the universal gas constant is .

 

Diatomic ideal gas

For a diatomic ideal gas, is again given by eq363. is derived by substituting eq275 into and differentiating to give:

Comparing eq364 with eq309, .

For , we substitute eq293 into and differentiating to give:

where .

At high but not extreme temperatures, , which allows us to expand as a Taylor series () to give:

Comparing eq366 with eq313, .

Similar to an atom,  for a diatomic molecule (see derivation of eq297). Hence,

which is consistent with eq316.

The total internal energy of a system of a diatomic ideal gas not subjected to extreme temperatures is therefore:

 

Polyatomic ideal gas

The translational component of the internal energy of a polyatomic ideal gas is again:

As mentioned in this article, each normal mode of vibration of a polyatmomic molecule behaves approximately like a separate harmonic oscillator, with the total vibrational partition function given by eq294:

where for non-linear molecules,  for linear molecules, and is the number of atoms per molecule.

Substituting into of of and differentiating gives:

It follows that at high but not extreme temperatures,

for a linear rotor is derived by substituting eq275 into and differentiating to yield:

for a non-linear rotor (including spherical, symmetric and asymmetric rotors), is derived by substituting either eq280, eq286 or eq287 into and differentiating to give:

Once again, for a polyatomic molecule not subjected to extreme temperatures. Therefore, the total internal energy of a system of a polyatomic ideal gas not subjected to extreme temperatures is:

 

Next article: Equilibrium constant in statistical thermodynamics
Previous article: Statistical thermodynamic derivation of the ideal gas law
Content page of statistical thermodynamics
Content page of advanced chemistry
Main content page

Effect of nuclear statistics on rotational states

The study of nuclear statistics plays a crucial role in understanding the rotational states of atomic nuclei, particularly in systems composed of identical particles, for example, homonuclear diatomic molecules.

Nuclear statistics dictate which rotational states are populated for molecules with identical nuclei. This effect arises due to a postulate of quantum mechanics, which states that the total wavefunction of a molecule must be either symmetric or antisymmetric with respect to the exchange of identical nuclei, depending on whether the nuclei are bosons or fermions. Such a statistical constraint directly influences the observed spectroscopic transitions.

To understand how nuclear statistics constrain rotational states, we refer to the total wavefunction for a rotating molecule, which can be approximated as a product of its component wavefunctions:

where

is the nuclear spin wavefunction
is the rotational wavefunction
is the vibrational wavefunction
is the electronic wavefunction

Consider the total ground state wavefunction of 16O2, where each nucleus is a boson. There are two unpaired electrons occupying separate degenerate antibonding molecular orbitals, while the remaining (core) electrons occupy doubly filled orbitals in singlet states.

The total two-electron wavefunction describing each doubly occupied molecular orbital (MO) is:

The spin wavefunction is antisymmetric under electron exchange. As required by the Pauli exclusion principle, the spatial component must be symmetric with respect to electron exchange, and hence the form . Under a nuclear exchange, , which is independent of nuclear coordinates, is invariant. The spatial part , being a product of two identical one-electron functions, also remains unchanged because any sign change on the individual orbital under a symmetry operation is cancelled out. Therefore, is symmetric with respect to nuclear exchange.

The molecular orbitals of the two unpaired electrons (triplet state) are formed from the atomic orbitals 2px and 2py of the oxygen atoms. Since the triplet spin wavefunction is symmetric under electron exchange, the associated spatial part of the wavefunction must be antisymmetric, in accordance with the Pauli exclusion principle:

When subject to a rotation operation about the principal rotation axis perpendicular to the molecular -axis (see above diagram),

It follows that the valence electronic wavefunction is antisymmetric with respect to this symmetry operation, which corresponds to nuclear exchange in the molecular frame. Therefore, the total ground state electronic wavefunction of 16O2, approximated as the product of the core and valence contributions, is antisymmetric under nuclear exchange.

With regard to eq97, the total vibrational ground state wavefunction of 16O2 is totally symmetric under all symmetry operations of the molecular point group, since these operations leave the squares of the normal coordinates invariant.

The nuclear spin wavefunction for 16O2 is given by (see this article for details), where each 16O nucleus has zero spin. Since both nuclear spin states are identical, i.e. , swapping the subscripts results in the exact same wavefunction. Therefore, the nuclear spin wavefunction is symmetric under nuclear exchange.

Rotational wavefunctions are represented by spherical harmonics , where the symmetry under nuclear exchange depends on the value of . These wavefunctions are symmetric for even and antisymmetric for odd , i.e. they change sign by under an inversion. Since the total wavefunction of 16O2 must be symmetric, half of the possible rotational states are forbidden by nuclear statistics. In other words, 16O2 can only occupy odd rotational states in its ground electronic configuration.

 

Question

Prove that the rotational wavefunction of a diatomic molecule is symmetric for even and antisymmetric for odd .

Answer

With reference to eq400, the unnormalised rotational wavefunction of a diatomic molecule is:

where .

The general and most comprehensive way to analyse the symmetry of a rotational wavefunction is to subject it to an inversion. In spherical coordinates, the inversion operator transforms a position vector from to , where the polar angle is reflected about the equatorial plane and the azimuthal angle is rotated by . Since , it follows that

Letting and using the chain rule gives , and hence . Furthermore, , where . So, . Therefore,

Since is always even, the wavefunction is symmetric with respect to inversion for even and antisymmetric for odd .

 

While nuclear statistics do not affect the value of the rotational energy levels, they can result in diminished spectral intensities or even “missing” peaks, particularly in Raman spectroscopy. However, this restriction does not apply to heteronuclear diatomic molecules, since the nuclei are not identical and there is no exchange symmetry requirement.

 

Question

How do nuclear statistics constrain rotational energy levels of ground state symmetrical linear polyatomic molecules like CO2?

Answer

The principles for analysing nuclear spin statistics in CO2 are similar to those for O2. , and must be symmetric under nuclear exchange, which involves two oxygen nuclei in a closed-shell molecule. The ground state vibrational wavefunction of CO2 is also given by eq97, which is symmetric under nuclear exchange. Therefore, CO2 can only occupy even rotational states in its ground electronic and vibrational configuration.

 

Next article: rotational selection rules
Previous article: Molecular rotational energies
Content page of rotational spectroscopy
Content page of advanced chemistry
Main content page

Particle in a box

A particle in a box is a mathematical model used to illustrate the key principles of quantum mechanics in a simplified system involving the particle moving freely within the box.

1-D box

Consider the case of a free particle of mass moving between and , where , along the -axis of a one-dimensional box. Since a free particle is under the influence of no potential energy, we set in eq44 to give

where the boundary conditions impose the restriction such that the probability of finding the particle outside the box is zero.

The general non-trivial solution of eq45a is , where and are non-zero constants. Applying the first boundary condition , must be zero, which results in the specific solution . Imposing the second boundary condition , we have . This implies that , where

 

Question

Can be zero or negative?

Answer

If , then , and for all . This would mean that the probability of finding the particle anywhere within the box is zero, which is not a physically meaningful solution. Negative values of would give the same probability density as positive values of . Therefore, including negative integers for would not yield any new, physically distinct solutions. The set of positive integers is sufficient to describe all possible states of the particle in the box.

 

It follows that the wavefunction for a particle in a 1-D box is

Solving eq45a using eq45b yields:

Eq45c shows that the energy of the particle is quantised, with being the quantum number.

 

Question

Normalise the wavefunction in eq45b.

Answer

Setting , with , and noting that , gives

Therefore, and .

 

3-D box

For a three-dimensional box of lengths , and , the Hamiltonian is given by eq45. Similarly, we set , resulting in:

Since the motion of the particle in the box along each axis is independent, we can approximate the wavefunction as a product of three functions, each depending on one of the independent coordinates , and :

Using the separation of variables method, we substitute eq45e into eq45d and divide through by eq45e to yield:

Since , and  are independent variables, the value of each term on RHS of eq45f varies independently, each producing a constant corresponding to the energy in that dimension, with the sum equaling , i.e. . In other words, eq45f can be solved by evaluating the following three equations separately:

The respective boundary conditions are

Therefore, the expressions for the three wavefunctions can be derived using the same logic as for the 1-D case, giving:

with

and

Since the energy of a particle moving in a 1-D (or 3-D) box is purely kinetic, eq45c (or eq45h) describes the translational energy of that particle in the box. In general, the wavefunction and energy equations of a particle in 1-D and 3-D boxes have applications in testing the validity of the position and momentum operators, as well as in perturbation theory and statistical thermodynamics.

 

Next article: one-particle, time-independent schrodinger equation in spherical coordinates
Previous article: one-particle, time-independent schrodinger equation
Content page of quantum mechanics
Content page of advanced chemistry
Main content page

Centrifugal distortion

Centrifugal distortion refers to the deformation of a molecule’s structure that occurs due to the rotational motion of the molecule.

As a molecule rotates, the centrifugal force, which acts outward from the axis of rotation, causes the molecule’s bond lengths and bond angles to stretch and distort, particularly at higher rotational speeds. This effect is most pronounced in larger, heavier molecules, where the rotational energy is sufficient to cause a noticeable shift in the geometry. Centrifugal distortion is an important factor in molecular spectroscopy, as it can influence the rotational energy levels of a molecule, leading to shifts in the observed spectral lines, and must be accounted for when interpreting high-resolution rotational spectra.

What is centrifugal force, and how is it different from centripetal force?

Centripetal force and centrifugal force are both associated with circular motion, but they are described in different frames of reference. Centripetal force is described in an inertial frame of reference or in a frame where the object is experiencing actual physical forces that are measurable and observable. In the inertial frame, the force that keeps the object in circular motion is real and necessary to counteract the object’s tendency to move in a straight line (due to inertia).

Consider a particle moving around a circle (see diagram above) of radius with an instantaneous velocity at and at , where . If is small, the arc length is approximately equal to and hence to . Since the ratio of corresponding sides of two similar triangles is always equal, . Dividing both sides of this equation by and rearranging (noting that and ) gives:

Substituting (see this article for details) into eq85 yields:

Therefore the centripetal force acting on the rotating particle of reduced mass is

Centrifugal force, on the other hand, is experienced in a non-inertia frame of reference (rotating frame). It appears to act outward on the rotating particle. However, it does not correspond to any real physical force but is instead a fictitious force introduced to account for the observed effects of acceleration in the rotating frame. Since the magnitude of centrifugal force is the same as that of the centripetal force, the magnitude of the centrifugal force experienced by a diatomic molecule of reduced mass and bond length rotating at angular velocity is .

At equilibrium, the centrifugal force (or the centripetal force) equals the restoring force given by Hooke’s law:

where is the force constant and is the equilibrium bond length.

For small displacements, the motion expressed by eq88 is simple harmonic, with the following classical Hamiltonian for the system:

where the kinetic energy term is given by the classical form of eq4a and the potential energy term is derived from this article.

Rearranging eq88 results in , and therefore

For small displacements, , which implies from that , and hence . So, we can ignore the last term in eq90 as a first approximation, giving:

Substituting (see this article for details) into RHS of eq91 gives:

Substituting eq92 and eq88 into eq89 yields:

Since for small displacements, eq93 becomes:

Using the quantum mechanics postulate that “to every observable quantity in classical mechanics there corresponds a linear, Hermitian operator in quantum mechanics”, and replacing the classical angular momentum with the quantum mechanical operator gives

The corresponding eigenvalues of eq94 (derived using orbital angular momentum ladder operators) are:

or

where is the rotational constant and is the centrifugal distortion constant.

However, rotational energies are typically reported as wavenumbers in molecular spectroscopy. Substituting and into eq95 yields , which is a function of . Therefore,

 

Question

Can centrifugal distortion be observed from an inertia frame?

Answer

Yes, centrifugal distortion can indeed be observed from an inertial frame, though it’s important to clarify that there are no “fictitious forces” like centrifugal force in an inertial frame. However, centrifugal distortion is not a fictitious effect. It is a real, physical phenomenon that arises due to the rotation of the molecule, and this can be observed in an inertial reference frame.

In a non-rotating molecule, the bonds between atoms are at their equilibrium lengths, representing the most stable configuration where the bond is neither stretched nor compressed. When the molecule begins to rotate, each atom wants to move in a straight line (due to its inertia) tangentially away from the center of rotation. To keep the atoms moving in a circular path, the chemical bond must provide an inward pulling force (the centripetal force).

Just like a spring, a chemical bond can only exert a restoring force when it is displaced from its equilibrium position. To generate a small centripetal force (restoring force) during low-speed rotation, the bond needs to stretch only a little. For faster rotation, the bond must stretch more to provide a larger centripetal force. This stretching of the bond beyond its equilibrium length is the physical phenomenon of centrifugal distortion, which can be observed and measured in an inertia frame.

 

Previous article: Microwave spectra
Content page of rotational spectroscopy
Content page of advanced chemistry
Main content page

Microwave spectra

Microwave spectra represent the electromagnetic radiation absorbed or emitted by molecules when they undergo transitions between discrete rotational energy levels. These spectra are a key component in the study of molecular structure and dynamics. They provide valuable insights about molecular properties, such as bond lengths and moments of inertia.

An ideal, pure absorption rotational spectrum of a rigid linear molecule, generated in the microwave frequency range of 3 to 600 GHz (1 to 200 cm-1), typically exhibits a series of absorption lines that correspond to the rotational transitions between discrete rotational energy levels (see diagram below). In the absence of an external magnetic field, these transitions follow the selection rule , meaning that a molecule can only transition from one rotational level to the next higher level.

Question

What is the difference between an absorption spectrum and an emission spectrum?

Answer

An external radiation is required to produce an absorption spectrum. The detector measures the amount of energy absorbed by the sample at each frequency, corresponding to transitions with .

In an emission spectrum, the sample itself acts as the source. It must first be excited to higher-energy rotational states, for example by heating or applying an electrical discharge. The detector then measures the amount of energy emitted by the sample, corresponding to transitions with .

Both types of spectra can be obtained using the same spectrometer, with different settings. Suppose the spectrometer is configured to record the absorption spectrum of a sample. If two rotational states are populated under the experimental conditions, the sample can both absorb and emit radiation at the same frequency, since both processes involve transitions with between the same pair of quantised energy levels. However, the conditions for measuring an absorption spectrum are typically set to ensure that net absorption occurs.

 

The rotational energy levels are given by eq44 or eq45 and the frequencies of the transitions are directly related to the rotational constant , and as such, the spacing between the spectral lines provides information about the moment of inertia and the molecular structure. Using the definition , we have

Therefore, the first peak, obtained by substituting into eq80, lies at the point along the horizontal axis. Similarly, the second and third peaks are at and respectively. In other words, the peaks in an ideal rotational spectrum (without centrifugal distortion) of a rigid linear rotor are equally spaced at .

 

Question

Calculate the moment of inertia and bond length of HCl if the line spacing in the rotational spectrum is 21.2 cm-1?

Answer

Since the spacing is , we have cm-1. Using the formula gives kg m2. Furthermore, . Therefore, the bond length is m.

 

The intensity of each transition depends on the population of the initial level , which is governed by the Boltzmann distribution, with higher- states being less populated at lower temperatures:

where

is the Boltzmann constant.
is the number of particles in the energy state .
is the total number of particles in the system.

Since represents the fraction of particles in the state ,  the intensity of a spectral line is proportional to , where the value corresponds to a specific energy level. However, there are degenerate states associated with each value of  in a rigid linear molecule. This degeneracy increases the statistical weight of higher levels, redistributing the population towards more degenerate states, which become thermally accessible at typical laboratory temperatures. In other words, at a given temperature, more particles will occupy a higher state with greater degeneracy at equilibrium than they would if the state were non-degenerate. Therefore, when a sample is exposed to an external microwave field, the intensity of each allowed transition in the pure rotational spectrum is proportional to the population of the initial state, which is the product of  and . This gives:

where is a proportionality constant and is given by eq45.

The factor in eq81 increases linearly with , while the exponential term decreases exponentially. As a result, the graph of versus forms a skewed bell curve (see diagram above), with a maximum found by treating as a continuous variable and taking the derivative of with respect to :

Setting , and noting that for any finite , gives

Since must be an integer, the maximum intensity occurs at the value of closest to the result of eq82.

 

Question

Is the intensity of each transition dependent on the electric dipole moment?

Answer

Although the transition probability between adjacent rotational states is proportional to , which depends on the magnitude of the electric dipole moment, the sample molecules in the waveguide of a typical microwave spectrometer are randomly oriented. As a result, the effect transition dipole moment does not vary significantly from one transition to the next for a given molecule. Therefore, while the absolute intensity depends on the dipole moment, it is often treated as a constant factor when comparing transitions within the same molecule.

 

For a symmetric rotor, and is also given by eq80. However, the spectrum is not just a single series of equally spaced lines like that of a linear rotor. Instead, each value of is associated with a series of lines with equal spacing of . In other words, a symmetric top spectrum consists of multiple superimposed series of equally spaced lines, each corresponding to a different value of . As a result, the individual lines may not be as clearly resolved and may appear as a denser pattern compared to that of a linear molecule. Nevertheless, the overall intensity envelope of the spectrum still resembles a skewed bell curve, with the maximum intensity occurring at the value where the population is highest.

In practice, the peaks in a real rotational spectrum are not vertical lines but have finite width and shape, appearing as broadened curves. This broadening arises, under typical laboratory conditions, from several effects:

    • Doppler broadening, due to the thermal motion of molecules, causes a spread in observed frequencies as molecules move towards or away from the detector.
    • Instrumental broadening results from the finite resolution of the spectrometer itself.

Beyond broadening, other phenomena affect the detailed structure and spacing of the spectral lines:

    • Centrifugal distortion causes deviations from the ideal rigid rotor model. As rotational speed increases with higher levels, the molecular bond stretches slightly, increasing the moment of inertia and decreasing the rotational constant . This leads to uneven spacing between lines, especially at higher , and is accounted for using a correction term.
    • Isotopic substitution alters the moment of inertia due to the change in atomic mass, thereby changing the rotational constant . Different isotopologues of the same molecule produce distinct sets of rotational lines, each with slightly different spacings. If multiple isotopes are present in the sample, the resulting spectrum may show clusters or duplications of lines, corresponding to each isotopologue.

Together, these effects lead to a rotational spectrum that is richer and more complex than the idealised, evenly spaced series of lines predicted by the rigid rotor model.

 

Question

Do large molecules, such as polymers, enzymes and DNA, undergo rotational transitions when radiated by microwaves?

Answer

Yes, large molecules can undergo rotational transitions when exposed to microwave radiation. However, such molecules are flexible and do not behave like rigid rotors, which invalidates some of the assumptions used in microwave rotational spectroscopy. If we assume that still applies, the rotational energy levels will be very closely spaced because the moment of inertia is large, and becomes very small. As a result, individual rotational transitions for large molecules often overlap, leading to broad, unresolved bands or a complete loss of identifiable peaks.

Furthermore, instead of undergoing uniform rotation, different parts of these complex molecules can move independently in response to the microwave field, including side-chain rotations and wobbling. These movements cause the molecules to collide with neighbouring molecules, converting the absorbed microwave energy into kinetic energy and heat. This general heating effect may denature the large molecules, leading to further changes in their structure and, consequently, variations in the observed spectra.

 

Next article: Centrifugal distortion
Previous article: Microwave spectroscopy (instrumentation)
Content page of rotational spectroscopy
Content page of advanced chemistry
Main content page
Mono Quiz