Microscopic energies

In our attempt to understand the molecular connections to thermodynamics We need to first get a handle on the molecular level energies.  To that end, we will develop a brief understanding of molecular dynamics and energies.

the Schrödinger equation

We have seen from first year that there are functions called wave functions that are used to describe the sizes and shapes of atoms and molecules.  The particular wave functions you were introduced to for atoms were simple spherical harmonics.  in Molecules, you eventually learned that these atomic wave functions (orbitals) were combined in some way to create a new manifold of states that described the energies of the system.  What you didn't ever learn was how we can determine the proper mix of these basis functions to come up with good overall functions that described the molecule and it's energies.  Erwin Schrödinger, from Austria, developed an equation to model the wave function of any system. 

the time-independent Schrödinger equation for a particle of mass m moving in one dimension with energy E is

where the symbol

is called h-bar.  This equation has two terms on the left-hand side.  The factor V(x) is the potential energy of the particle at position x and the first term is related to the kinetic energy of the particle.

the wave function itself ψ  is understood in Quantum mechanics to contain all the information there is to know about the motion of the particle.  If we use the Born interpretation we can get the probability density by taking the square modulus, |ψ |² = ψ *ψ ,  of the wave function and by multiplying by dx, we get the probability of finding the particle in the infinitesimal region between x and dx.  Thus, there is never a negative probability of finding the particle (which would not make sense anyway).  There is at minimum a zero probability, in cases where the square modulus is zero.

Additionally, it turns out that we can multiply the wave functions by a constant such that the integral of this function is one.

If this integral is one then the wave functions are deemed to be normalized.   This makes it useful for comparing wave functions of different types. additionally, if we consider a wave function as a "container" for an electron then the probability of finding the electron in it's container should be 1.  Additionally, when we take linear combinations of wave functions (as in determining MOs, we can ensure that the final wave functions are also normalized).

in three Dimensions, the normalization equation would be

or, in spherical polar coordinates

Translational motion

if we consider a particle moving freely with no external forces (potential is zero) then the Schrodinger equation is

The Schrodinger equation is often written as

where Ĥ, called the Hamiltonian, is an operator that when applied to the wave function returns the energies of the system and the regenerated wave function.  In this case (a freely moving particle in a zero potential space), the Hamiltonian is

The Schrödinger equation is a differential equation that has solutions of the form
 

where k is the index identifying the eigenfunctions and eigenvalues of , respectively. 

Fig. 9.1  A particle in a one-dimensional region with impenetrable walls. Its potential energy is zero between x = 0 and x = L, and rises abruptly to infinity as soon as it touches the walls.

Particle in a box

Consider a particle confined in a 1-dimensional box from x = 0 to x = L.  We can use the same solutions to the Hamiltonian as we derived for the free particle except that there are boundary conditions (the sides of the box) beyond which the potential is infinity (so that the particle can never get there).  within the box, the potential is always zero, just like in the free particle equation.

We can also use the expansion of e^ikx = cos kx + i sin kx and e^-ikx = cos kx - i sin kx to rewrite the wave function as

ψ _k = A e^ikx + B  e^-ikx = A( cos kx + i sin kx ) + B( cos kx - i sin kx )

    = (A+B) cos kx + (A-B) i sin kx   by renaming C = A+B and D = A-B, we get

ψ(k,x) = C cos kx + D i sin kx 

boundary conditions dictate that y(k,0) = 0 and ψ(k,L) = 0  for all values of k.

These boundary conditions mean that only certain wave functions will work.  These turn out to be a set of quantized wave functions of the form
 

2.7

with energies of

2.8

In this case, the factor (1/L)^1/2 is the normalization factor for the wave function.

The allowed energy levels for a particle in a box. Note that the energy levels increase as n2, and that their separation increases as the quantum number increases.

  Note that while the original Hamiltonian has ħ, but because of the boundary conditions, the solution for the energies has only h. 

Example:

Consider the conjugated p system, butadiene CH₂=CH-CH=CH₂.  the p electrons in this distributed MO can travel freely from one end to the other end of the (almost) linear molecule.  Thus, we can use the particle in a box concept to figure out the energies of the electrons in this system.

What is the lowest energy absorption that we might expect in this molecule?

Since there are 4 p-electrons in the system, we would expect them to occupy (according to the Pauli exclusion principle) the lowest two energy levels.  Thus, the lowest energy transition will be between n=2 and n=3.  The length of the "box" that these electrons are trapped in is the sum of the bond lengths if we use 140 nm as a rough average then L=3*150nm = 450nm

The energies are given in equation 2.8 for each value of n so we can calculate the n=2 and n=3 energy levels

 

E2 =	22 × (6.626×10–34 J s)2	= 1.19 ×10–18
	8 × (9.110×10–31 kg) × (450 × 10–9 m)2

 

E3 =	32 × (6.626×10–34 J s)2	= 2.68 ×10–18
	8 × (9.110×10–31 kg) × (450 × 10–9 m)2

Thus, the transition energy ΔE = E₃ – E₂  = 1.49×10^–18 J.  Since we know that ΔE = hc/λ,  we can calculate the wavelength of light that would be absorbed by this transition.

So butadiene will be colorless to our eyes since it only absorbs in the UV region.

Molecules generally can move in 3 dimensions so we need to generalize the formula we derived for a one-dimensional box to three dimensions.

In a 2-d box, we would have motion in a plane, say, the xy plane and the Schrödinger equation would become

2.9

Using the separation of variables technique for solving these equations, we can divide the wave function into two functions

ψ(x,y) = X(x)Y(y)

where the solutions for each dimension are simply the one-dimensional solutions.

2.10

after some work, we see the eigenfunctions and eigenvalues of the Schrödinger equation in 2-d are

2.11

similarly, we can extend this to three dimensions to get

2.12

This last equation is interesting in that it allows us to explore how degeneracies can occur.  Consider, for a moment a particle trapped in a cube where all dimensions L₁, L₂, and L₃ are the same, L.  now, the equation for the energy becomes

Clearly, there will be combinations of quantum numbers that will give the same energies, even though they will refer to different physical states.  for example, consider the set of numbers (1,1,2), (1,2,1) and (2,1,1)  clearly these three different sets of values for n₁, n₂, n₃, will give energy values that are identical to each other. If the system was not fully symmetric, say one length was different from the other two then there would still be some degeneracies, just not as many.  So it is easy to see that symmetry and degeneracy is related.

OK, so let's get back to our question of molecular energies for motion.

Consider the H₂ molecule confined in a 1dm³ cubic box.  What are the energy levels?

m = 2.008g/mol * 1/6.023×10²³

m = 3.334×10^–27 kg

 

The lowest energy level is the level   (1,1,1)	E = 4.938×10–39 J (note: it's not zero – the molecule never stops)
the next lowest level is triply degenerate with (1,1,2), (1,2,1), and (2,1,1)	E = 9.877×10–39 J
next is also triply degenerate (1,2,2), (2,1,2) and (2,2,1)	E = 1.482 ×10–38 J

Clearly, these energies are very small and hence so are their differences.  Generally, we ignore the quantization effects for translational energies.

Vibrational energies

Since any particle with more than one atom in it will have vibrations, we need to include these energies in any consideration of molecular energies.  We will use the the harmonic oscillator to model such vibrational motions.

Any particle that experiences a restoring force proportional to it's displacement will undergo harmonic motion (oscillations).

where F is the restoring force, k is the "spring" constant and x is the displacement from centre. the "stiffer the spring", the larger the value of k; in a molecule, this could be related to the bond strength.

Force is related to potential energy (see appendix 3 of Atkins) via the equation

so we can now include a potential energy term in the Schrödinger equation such that our particle in a box does not move in a zero potential field anymore but is constantly experiencing a potential energy related to the restoring force.

2.16

The energies (eigenvalues) derived from this equation are

2.17

where omega w =  (k/m)^1/2  is an angular frequency (2p × freq.) of vibration.  Notice that the energies we get from equation 2.17 have a gap that does not depend on the value of the vibrational quantum number v (v can be 0,1,2,3,...).

2.18

Also note that even at the lowest vibrational level, there is energy (vibrations never stop).  This is called the zero point energy for the harmonic oscillator.

Consider a diatomic molecule, H₂.  if we look at the molecular (IR) spectroscopy absorption for H₂, we see an absorption band at 8.7428×10^–20 J.  This is the H-H vibrational frequency and clearly it is much much larger than the energy differences we calculated for translation of the H₂ molecule.  These quantization effects are measurable.

 

Rotational energies

Molecules can rotate along any axis that does not have spherical symmetry.  So, a spherical particle (single atom) has no noticeable rotation.  A diatomic molecule (or any perfectly linear molecule) can rotate about two linearly independent axes (both perpendicular to the bond axis) and a non-linear molecule can rotate about all three Cartesian axes.

for a diatomic molecule, the energies are quantized and can be calculated via

E = l(l+1)	2	2.19
	2I

where the moment of inertia I = Sm_ir_i² or, using reduced mass  I = m r², where r is now the bond length and m is the reduced mass.

for Hydrogen, ²/2l = 1.2088×10^–21J so this is smaller than vibrations but still measurable.

Electronic energies

Electronic wave functions describe the possible positions of the electrons in the molecule.  They can only be solved exactly (analytically) for one-electron species like H and Li⁺.  We can use computers to solve the Schrödinger equation with approximate (numerical) methods.  For example, since the Schrödinger equation has the wave function on both sides, we take the output wave function and feed it back into the input for a second run and keep repeating until the wave function doesn't change anymore.  This is (loosely) what is done in a technique called the self-consistent field (SCF) method of calculating Molecular orbitals (SCF-MO).

For the H atom, we can find the energy levels for n = 1,2,3,...

En =	–m  e4	2.20
	32 p2 eo2  2  n2

where m is the mass of the electron,  e is the charge on the electron, ε_o is the permittivity of free space and n is the quantum number.  we should also remember that for every quantum number, there are certain degeneracies, depending on the the allowed values of the other quantum numbers l, m_l and m_s.  By substituting in the parameters, we can solve for the energies

E₃ = –2.42×10^–18J     s,p,d orbitals  9-fold degenerate
E₂ = –5.45×10^–18J     s,p orbitals     4-fold degenerate
E₁ = –2.18×10^–17J    s orbitals only  

2.1 Electronic, Vibrational and Rotational energy levels for the hydrogen molecule.  Note the scales are only approximate but clearly, the difference between the electronic levels is very large compared to both the vibrational levels and the rotational levels

This time it is easy to see that the energy gaps between the electronic levels are huge compared to the vibrational and rotational levels.

At room temperature,  The large energy gap between the first electronic level and the next nearest level practically guarantees that all hydrogen molecules (and most other molecules) exist only in their electronic ground state.

It is, however, more likely that many molecules exist in different excited vibrational or rotational states.

k_BT = 1.381×10^–23J/K * 298K

        = 4.12×10^–21J (= 0.026 eV)

So, Thermal energy is just enough to excite some vibrational levels and many rotational levels in H₂ molecules.

Putting it all together

The total energy for a system is simply the sum of all the individual energy sources.

For an atomic system, it is

For a molecular system, it is

Molecules can be made to jump between different energy levels, electronic, vibrational and rotational, by applying different frequencies of light.  If the energy of the light matches the energy gap of the energy levels (among other criteria ... selection rules) then we may get an absorption pattern (spectrum) that can be used to uniquely identify the molecule.

Energy in large systems

The energy equations we have looked at up to now were not necessarily restricted to single molecular systems, even though our particular discussion was for single particle systems.  From this point in this discussion, we need to distinguish clearly the energy states of individual particles e_i versus energies of ensembles of many particles E. 

Consider a sample of N atoms of an atomic ideal gas in a cube of dimension L.  There will be a set of energy levels ε_i , populated by n_i atoms each where the first energy level is artificially set a value of zero and the levels increase in energy as i = 0,1,2,3...  The total energy of any given energy level is

E_i(N,V) = n_iε_i,                                                                                                                                             2.21

where i runs from 0 to the total number of states, and V is the volume of the container.  Obviously, the sum of all the n_i values S_in_i =  N .  The total energy of the system is obviously given by

E(N,V) = S_in_iε_i.                                                                                                                                          2.22

We need some method of understanding how the energy is distributed among the molecules, in more precise terms, we want to determine the values of n_i.  It seems obvious that some molecules will have high translational kinetic energy while others have quite low translational kinetic energy.  For a given molecule, Statistical Mechanics gives us the Boltzmann distribution which shows that the population n_i of the i^th energy level drops exponentially as the energy gets higher.

Another way of writing this equation that may be more succinct is

where q is the molecular partition function

and the parameter β is called the Boltzmann temperature and is given by the equation

Fig. 16.6  (Atkins) The populations of the energy levels of a harmonic oscillator at different temperatures, and the corresponding values of the partition function.  Note that β = 1/kT and ε is the spacing of the energy levels

The Boltzmann distribution tells us the probability that the i^th energy level of a molecule will be populated.  We see that that probability is highest for the lowest levels and reduces exponentially as the energy of the level i gets higher.  The probability for some upper energy being populated goes up as the temperature increases and goes down as the energy level itself goes up.  Said another way, high energy states don't get populated as readily as low energy states.  The molecular partition function can be thought of as a measure of the number of states that are accessible at the particular temperature.  Thus, as T → 0, the value of q reduces to the number of energy states at i = 0, i.e., the degeneracy of the ground state.  As T → 0, the value of q approaches the number of states in the system.

We can also look at the relationship in a different light.  By using the total energy of the given state E_i. rather than the energy level of the molecules, we get

where Q is now the Canonical Partition Function,

• For a system of identical distinguishable particles, . 

  Q = qN

  2.29a

  
  This kind of system could be a crystalline solid, where each particle can be identified by its location.
  

• For a system of identical indistinguishable particles,

  Q = qN/N!.

  2.29b

  
  This might be a gas, for example, where the motion of the particles makes it impossible to keep track of the molecules.

Fig. 16.15  The energy density of states is the number of states in an energy range divided by the width of the range.

Fig. 16.16  To construct the form of the distribution of members of the canonical ensemble in terms of their energies, we multiply the probability that any one is in a state of given energy, eqn 2.27, by the number of states corresponding to that energy (a steeply rising function). The product is a sharply peaked function at the mean energy, which shows that almost all the members of the ensemble have that energy.

Equation 2.27 shows us that at the energy E_i gets higher, the probability of finding the system with that energy is lower.  If  E_i  is much smaller than kT then there is lots of energy around and so the level is easily accessible.  If the reverse is true and E_i is much larger than kT then p_i will be very small and the system is less likely to have this energy.

Now, we can eliminate the electronic energy levels from our consideration since only the ground state is ever populated at normal temperatures because of the large gaps in energy.  Note that the exponent term involves energy factors that are a sum of translational and electronic energies.  So, the exponent of the sum is the same as the product of the exponents

by similar logic then, the partition function that comes from the sum of translational and electronic energies can be rewritten simply as the product of two partition functions

now, since the value of ε_i for all values of i > 0 are very much larger than k_BT, then all the exponent values in the sum will be zero and the electronic partition function reduces to the first term only.  Since the numerator can be similarly treated can rewrite the Boltzmann distribution function as

clearly, the electronic terms cancel and our new Boltzmann distribution function involves only the translational terms.
 

so for an atomic system, we will restrict ourselves to translation kinetic energy terms.

Example:  3 neon atoms in a 1 dm³ box.

Each Ne atom can have energy levels given by equation 2.13

Thus, the lowest energy (1,1,1) is 4.914×10^–40 J

The next lowest level is                  9.828×10^–40 J

The next lowest level is                  1.474×10^–39 J

level (10,10,10) is                           4.914A×10^–38 J

Level (25,25,25) is                          3.071×10^–37 J       (~14,000^th level)

Since this system has three atoms, it's energy is E(N,V) = E(3,1dm³) = ε₁+ ε₂+ ε₃. 

Lowest energy state for the system is 4.914×10^–40 J × 3 = 1.474×10^–39J

Next Lowest is 2×4.914×10^–40 J + 9.828×10^–40 J = 1.9656×10^–39 J

a higher level would be 3×3.071×10^–37 J  = 9.213×10^–37J.

Now consider the probability of finding the system at these energies for various temperatures

T = 10^–16 K so k_BT = 1.3806×10^–39J.

T = 5 K so k_BT = 6.9032×10^–23J.

Clearly, even at very low temperatures, many many levels are probable.

Once all the probabilities are known, we can calculate the total internal energy

take the partial derivative of the partition function

So, we can write the expression for the energy of the system as

So, if we can find the correct value for Q for our system, we can find it's internal energy.  Very powerful function.

Often, determining the molecular partition function is easier as it's a simple sum over states so we can use the relationships 2.32a and 2.32b to calculate the canonical partition function.

In general, for an ideal gas,

or,

now, substituting back into 2.33 gives us
 

This is the same result we found for the kinetic energy of an ideal gas from the kinetic molecular theory. 

If we had started this derivation using a molecular ideal gas, rather than the atomic gas, we would have needed to include the vibrational partition function and the rotational partition function as well.  In that case, the final equation to calculate the internal energy of the gas would be
 

Where the factors v_rot  and v_vib are the degrees of freedom of rotation and of vibration multiplied by the rotational and vibrational partition function, respectively.

In the case of rotations, since the value of k_BT is large compared to the spacing between levels, we can assume that all the rotational levels are accessible so we get

For vibrational levels, the spacing is larger than k_BT so we can assume that most levels are not accessible so then  v_vib is the number of active vibrational levels.   as a simple rule if

but if

anything in between is uncertain.  We need to consider each of the 3n-6 (non-linear) or 3n-5 (linear) vibrations to determine which are active.

If we were to try to calculate the contributions for real gases, we would also need to include terms that take into account the intermolecular interactions.  (not in this course :)