Introduction

In an earlier course you learned that the Gibb's free energy is a fundamental criterion of spontaneity for a chemical reaction. We can write

DG^o = DH^o - TDS^o

For the sample reaction

n_AA + n_BB Û n_yY + n_ZZ

where n_AA, n_BB etc. are the stoichiometric coefficients. The equilibrium constant for this reaction can be written as:

At equilibrium the free energy change for the reaction is:

DG^o = -RT ln K_eq

The criterion for a spontaneous reaction is:

DG^o < 0

If the reverse reaction is spontaneous then:

DG^o > 0.

Using the properties of logarithms we can write:

DG^o = -n_yRT ln[Y] + -n_ZRT ln[Z] -(-n_ART ln[A] + -n_BRT ln[B])

This expression breaks the equilibrium constant into terms that depend on the individual concentrations of the reactions and products. The individual terms are known as the chemical potential, m_Y^o m_Z^o etc., for products Y and Z and reactants A and B, respectively. The chemical potential is:

m_Y^o = - RT ln[Y]_eq

m_Z^o = - RT ln[Z]_eq

m_A^o = - RT ln[A]_eq

m_B^o = - RT ln[B]_eq

The free energy can be recast as:

DG^o = n_ym_Y^o + n_Zm_Z^o - n_Am_A^o - n_Bm_B^o

For a non-equilibrium condition:

m_Y = - RT ln[Y]

m_Z = - RT ln[Z]

m_A = - RT ln[A]

m_B = - RT ln[B]

or

m_Y = m_Y^o + RT(ln[Y]/[Y]_eq)

m_Z = m_Z^o + RT(ln[Z]/[Z]_eq) etc.

and

DG^o = DG + RT lnQ

where Q is the reaction quotient.

The microscopic and macroscopic views

This macroscopic view of chemical equilibrium is a central topic of Physical Chemistry. Consequently, there is a microscopic view that underlies the macroscopic view. The microscopic view starts with molecular energy levels. These are:

Electronic

Vibrational

Rotational

Spin

The macroscopic properties are determined by averaging over the microscopic energy levels.

Fundamental question: What is the relationship between temperature and molecule energy levels?

We know that DG^o depends on temperature. What is the connection between the temperature and the entropy?

To answer these questions we turn to the Boltzmann factor. (M&S 17-1). Let us consider a molecule with two energy levels.

The two level system

We define a molecule with two energy states and then make an analogy with macroscopic systems. Think of population of E₂ as the products and population of E₁ as reactants. We write down a molecular equilibrium constant between these two levels as the ratio of the populations of the levels. This ratio is called R.

R = N₂/N₁ à DE₂₁ = - k_BT lnR.

Here k_B is to a molecule, what the ideal gas constant R is to a mole. In fact,

R = N_Ak_B. Thus the units of R are J/mol-K while those of k_B are J/molecule-K

or just J/K. So,

R = exp{-E₂/ k_BT }/exp{-E₁/ k_BT}

N₂ µ exp{-E₂/ k_BT } and N₁ µ exp{-E₁/ k_BT }

We could normalize the population of states 1 and 2 by dividing by the total population N₁ + N₂ to obtain the probabilities:

p₁ = N₁/(N₁ + N₂) = N₁/Q

p₂ = N₂/(N₁ + N₂) = N₂/Q

We have given the total population a name. We call it the partition function Q. In the two state system:

Q = exp{-E₂/k_BT} + exp{-E₁/k_BT}

Thus, far we have started with the assumption of a thermal population (equilibrium) between the levels and arrived at explicit probability distributions for each state. The probability distributions can be used to calculate average properties. This type of reasoning provides a connection between microscopic properties of molecules and macroscopic quantities. Before we consider explicitly how this averaging is to take place, we need to consider the nature of the energy levels. Thermal averaging tends for important when the spacing of the energy levels is around thermal energy

E_thermal = k_BT

or less. If the energy level spacing is big compared to k_BT then changes in population of levels due to temperature changes are not important. A concrete example are the electronic energy levels in the hydrogen atom

E_i = -R/n_i²

where n_i is the principal quantum number of the ith state and R is the Rydberg constant. The spacing of the energy levels is given by

DE_j1 = R(1 - 1/n_j²)

Given the R = 107,696 cm^-1 it is relatively large relative to thermal energy. It is useful to carry around conversions for k_B so that you can translate k_BT into meaningful units.

k_B = 1.380 x 10^-23 J/K

Is MKS units, but not very useful for energy comparisons with spectroscopy.

Spectroscopists express transition energy in units of cm^-1.

To convert from cm^-1 to Hz or s^-1 just multiply by the speed of light.

n = cn (n is in Hz and n is in cm^-1)

To convert from nm to cm^-1 use the following relation:

cm^-1 = 10⁷/l(nm).

In cm^-1 k_B = 0.697 cm^-1/K. At ambient temperature T = 300 K, k_BT = 207 cm^-1.

Returning to electronic transition of the H atom we can see that the energy level spacing for transitions among levels is roughly a factor of 500 times larger than thermal energy, i.e. R >> k_BT.

In the next lecture we will consider vibrational and rotational energy levels. Here we treat translational energy levels. In some ways this is trickiest of the energy levels schemes. The energy level spacing is so small that you can treat translation using classical, rather than quantum physics. However, the quantum treatment should also be valid. This is in accord with the correspondence principle that states that quantum mechanics and classical mechanics should give the same number for systems that have sufficiently closely spaced quantum energy levels or sufficiently high quantum numbers. We can treat translation in both ways:

1. Quantum treatment - particle in a box

2. Classical treatment - kinetic energy

Quantum approach - Particle in a box

The particle in a box is a classic problem in introductory quantum mechanics. It also has wide application as a starting point in theories of spectroscopy and statistical mechanics of molecules.

Based on the idea that a particle of gas can be treated using the QM particl in a box method, we can estimate the spacing of energy levels and the average quantum number. Here we picture a cubic box of length a on a side. The quantized energy per molecule is:

Assuming that a thermal weighting will fill the levels sequentially, the average level will be roughly equal in energy to (3/2)k_BT. Thus, we can consider an average quantum number <n_j> where:

for the particle j. Thus we can equate the average energies

and solve for <n_j>,

For the sake of a concrete example, let's assume that a volume is filled with molecular oxygen, O₂. The molar mass of O₂ is »32 a.m.u. To convert to MKS units we multiply by 1.672 x 10^-27 kg/a.m.u. We use k_B = 1.38 x 10^-23 J/K and a temperature of 300 K to obtain the result <n_j> » 4.5 x 10¹⁰. This seems like a huge value for the quantum number. To be sure that this is a reasonable number, we check by calculating the energy level spacing that these values give for the molecules in the gas phase.

We consider the transition from the average quantum number <n> to the next level up <n+1>. We have dropped the subscript j and we consider that the expressions apply to all molecules in an ensemble. These values enter energy expression as square so the transition energy is:

The factor of h² assures that this energy will be small. Even for quantum numbers of 4.5 x 10¹⁰ the energies are E_n+1,n 9.32 x 10^-31 J! This model calculation shows that the energy spacing of translation energy levels is very close. The spacing forms a continuum because there are so many levels. For this reason translation can be treated very accurately using a totally classical approach.

Classical approach - the kinetic theory of gases

Imagine ballistic particles with a classical kinetic energy of (1/2) mu² where u is the velocity. If consider one direction of space for simplicity, say the x-direction then the kinetic energy of a particle traveling in that direction is given by:

The thermal weighting factor can be expressed in the same terms as the two state problem above. The probability that each possible kinetic energy state will be observed is:

The denominator is the important normalization factor. The probability must be 1 for averages over all velocities. The integral in the denominator is a Gaussian integral. Gaussians are written in two forms.

In some ways it is easiest to see the solution for the G(u) function. We can equate:

And plug into the normalized Gaussian distribution, G(u) to obtain the Maxwell-Boltzmann distribution

If you want to see how to get the normalization constant and how to calculate Gaussian integrals go to the tutorial on Gaussians.

Showing that the integral of H(u) is:

We can define the quantity I₀(a).

This allows for a simple trick. Integrals of Gaussian times even polynomials can be obtained by taking derivatives of I₀(a) with respect to a.

This integral allows one to calculate the mean square velocity.

Using the relation

Once normalization is properly accounted for the mean-square velocity is:

Thus, the kinetic energy along each direction is:

We assume that all three directions of space are equivalent. The total kinetic of a particle is the sum of the energies along each of the three individual directions.

The total kinetic energy is the summ of these:

This leads to the expression:

The molar equivalents of these quantities can be obtained using the molar mass M and gas constant R.

R = N_Ak_B = 8.31 J/mol-K

M = N_Am

For a mole of gas the average kinetic energy is 3RT/2.