Foundations of Chemical Kinetics Lecture 5: The Boltzmann distribution Marc R. Roussel Department of Chemistry and Biochemistry
Ensembles ◮ Even without quantum mechanics, matter has some inherently probabilistic qualities. ◮ In every collision between molecules, energy and momentum are exchanged. ◮ Even if we knew the positions and momenta of every particle in a container at some time t , our ability to predict the energies, positions and momenta of particles after a few collisions would be essentially zero, except in a statistical sense. ◮ We treat such systems using ensembles, which are large, imaginary sets (approaching infinite size) of copies of a system of interest which have some variables in common and which have been similarly prepared.
The Boltzmann distribution ◮ Imagine an ensemble of systems all of which are held at the same temperature or an ensemble of non-interacting molecules in a container at a fixed temperature T . = ⇒ canonical ensemble ◮ According to Boltzmann, the probability that one of these systems (molecules) has energy ǫ i is given by � � − ǫ i P ( ǫ i ) = g i exp / Q k B T where g i is the degeneracy of energy ǫ i , i.e. the number of different microscopic states that give the same energy, k B is Boltzmann’s constant, Q is a constant chosen such that � i P ( ǫ i ) = 1.
The partition function ◮ The normalization constant Q is called the partition function. ◮ We must have � i P ( ǫ i ) = 1, so � − ǫ i � � � P ( ǫ i ) = g i exp / Q = 1 k B T i i ∴ 1 � − ǫ i � � g i exp = 1 Q k B T i � � − ǫ i � ∴ Q = g i exp k B T i
Meaning of the partition function � − ǫ i � � Q = g i exp k B T i ◮ Suppose that, at some temperature of interest, k B T ≫ ǫ i for i ≤ n and k B T ≪ ǫ i for i > n . ◮ For states well below k B T , exp( − ǫ i / k B T ) ≈ 1. ◮ For states well above k B T , exp( − ǫ i / k B T ) ≈ 0. ◮ Thus, n � Q ≈ g i . i =1 ◮ Roughly speaking, the partition function counts the number of states accessible at temperature T .
Example: The partition function for a two-level system ◮ Suppose that ǫ 1 = 0, g 1 = 1 and ǫ 2 = 1 × 10 − 19 J, g 2 = 3. � � − ǫ 2 Q = 1 + 3 exp = 1 + 3 exp( − 7243 K / T ) k B T 4 3.5 3 2.5 Q 2 1.5 1 100 1000 10000 100000 1e+06 T /K
A minor rewrite of the partition function We can drop the degeneracy if we sum over quantum states (distinct sets of quantum numbers) rather than over energy levels: � � − ǫ i � Q = exp k B T i
Separability of partition functions ◮ Suppose that the molecular energy can be decomposed into a sum of independent contributions, i.e. ǫ = ǫ (1) + ǫ (2) + . . . ◮ Then, using the second form of the partition function, we have � − ǫ i + ǫ j + ǫ k + . . . � � � � Q = · · · exp k B T i j k � − ǫ i � � − ǫ j � � − ǫ k � � � � = · · · exp exp exp · · · k B T k B T k B T i j k � − ǫ i � � � − ǫ j � � � − ǫ k � � = exp exp exp · · · k B T k B T k B T i j k = Q (1) Q (2) Q (3) . . .
The translational partition function ◮ Consider some molecules in a rectangular container in the gas phase. We can treat their translational degrees of freedom using a particle-in-a-box treatment. In three dimensions, we just include three particle-in-a-box terms, one for each dimension and with its own quantum number. ◮ For an oxygen molecule with an average kinetic energy at room temperature in a 1 cm box of 2 × 10 − 21 J, the translational (particle-in-a-box) quantum number is n x ≈ 4 × 10 8 . x ≈ 9 × 10 − 30 J ◮ E n x +1 − E n x = (2 n + 1) h 2 / 8 mL 2
The translational partition function (continued) ◮ Since the energy levels are very close together, Q x = � n x exp( − ǫ n x / k B T ) is a very accurate Riemann sum for the corresponding integral (with ∆ n x = 1). ◮ Thus, � ∞ n 2 x h 2 � � Q x ≈ exp dn x − 8 mL 2 x k B T 0 ◮ This is a known integral: Q x = L x � 2 π mk B T h ◮ The translational partition function is therefore Q tr = Q x Q y Q z = L x L y L z (2 π mk B T ) 3 / 2 = V h 3 (2 π mk B T ) 3 / 2 h 3
The vibrational partition function ◮ For a harmonic oscillator, E v i = � ω ( i ) v i + 1 � � . 0 2 We are allowed to set our zero of energy wherever we want. In particular, we could set it so that E 0 = 0, i.e. use E v i = � ω ( i ) 0 v i : � − � ω ( i ) � ∞ 0 v i � Q i = exp k B T v i =0 �� v i � � − � ω ( i ) ∞ � 0 = exp k B T v i =0 �� − 1 � � − � ω ( i ) 0 = 1 − exp k B T
Rotational partition function ◮ The treatment of rotation in quantum mechanics is complex, with several cases to consider. ◮ The most general case (a nonlinear molecule with no special symmetries) leads to the following partition function: � 1 / 2 � 8 π 2 I b k B T � 1 / 2 � 8 π 2 I c k B T � 1 / 2 � 8 π 2 I a k B T Q rot = π 1 / 2 h 2 h 2 h 2 ◮ I a , I b and I c are the moments of inertia of the molecule around its three principal rotational axes.
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