Foundations of Chemical Kinetics Lecture 6: Further studies of the Boltzmann distribution Marc R. Roussel Department of Chemistry and Biochemistry
The Boltzmann distribution in classical mechanics ◮ One of the main differences between classical and quantum mechanics is that energy is not quantized in the former theory. ◮ Accordingly, in classical mechanics, it doesn’t make sense to ask for the probability that a molecule has exactly energy ǫ . The answer to this question is zero. ◮ The correct question to ask is what is the probability that a molecule has energy between two specified limits, i.e. what is P ( ǫ low ≤ ǫ ≤ ǫ high ) for some specified values of the two limits?
The Boltzmann distribution in classical mechanics (continued) ◮ Define the density of states g ( ǫ ) such that g ( ǫ ) d ǫ is the number of states between energies ǫ and ǫ + d ǫ . ◮ Let ǫ o be the result of an observation of ǫ . Then, � � ǫ P ( ǫ ≤ ǫ o ≤ ǫ + d ǫ ) = g ( ǫ ) exp − d ǫ/ Q k B T or � ǫ high � � P ( ǫ low ≤ ǫ o ≤ ǫ high ) = 1 ǫ g ( ǫ ) exp − d ǫ Q k B T ǫ low
Classical partition function ◮ Let A be the set of allowed energies. This will normally be an interval of energies, possibly semi-infinite. ◮ Using very similar reasoning to that used to obtain the quantum partition function, we get � � ǫ � Q = g ( ǫ ) exp − d ǫ k B T A
Example: harmonic oscillator density of states A “calculation” based on the quantum-mechanical energy ◮ For the quantum-mechanical harmonic oscillator, the energy levels are equally spaced by � ω 0 . ◮ Provided ∆ ǫ ≫ � ω 0 , the number (∆ G ) of energy levels in an interval of size ∆ ǫ is therefore ∆ G ≈ ∆ ǫ � ω 0 ◮ By definition, the density of states is g ( ǫ ) = dG / d ǫ . Rearrange the last equation and take a “physicist’s limit” ∆ ǫ → 0: ∆ G 1 ∆ ǫ = � ω 0 ∆ G ∆ ǫ = dG 1 ∴ g ( ǫ ) = lim d ǫ = � ω 0 ∆ ǫ → 0
The Arrhenius equation V x ) ( ε a reactant product x specific rate of reaction = (probability that ǫ > ǫ a ) × (specific rate of crossing if ǫ > ǫ a )
The Arrhenius equation (continued) ◮ If the barrier is high enough, there will be many states below E a , so energy can be treated approximately as a continuous variable. � ∞ ǫ a g ( ǫ ) exp( − ǫ/ k B T ) d ǫ P ( ǫ > ǫ a ) = � ∞ 0 g ( ǫ ) exp( − ǫ/ k B T ) d ǫ ◮ Suppose that there is one, roughly constant, density of states below the top of the barrier and another above, i.e. that � g b for ǫ < ǫ a g ( ǫ ) = for ǫ > ǫ a g a
The Arrhenius equation (continued) ◮ Then � ∞ g a ǫ a exp( − ǫ/ k B T ) d ǫ P ( ǫ > ǫ a ) = � ǫ a � ∞ g b 0 exp( − ǫ/ k B T ) d ǫ + g a ǫ a exp( − ǫ/ k B T ) d ǫ ◮ For a high barrier, very few states above the barrier will be populated compared to the number of states in the reactant well. Thus, � ∞ g a ǫ a exp( − ǫ/ k B T ) d ǫ P ( ǫ > ǫ a ) ≈ � ǫ a g b 0 exp( − ǫ/ k B T ) d ǫ ◮ For the same reason, we make only a small error by extending the range of integration in the denominator to infinity: � ∞ g a ǫ a exp( − ǫ/ k B T ) d ǫ P ( ǫ > ǫ a ) ≈ � ∞ g b 0 exp( − ǫ/ k B T ) d ǫ
The Arrhenius equation (continued) � ∞ g a ǫ a exp( − ǫ/ k B T ) d ǫ P ( ǫ > ǫ a ) ≈ � ∞ g b 0 exp( − ǫ/ k B T ) d ǫ = − g a k B T exp( − ǫ/ k B T ) | ∞ ǫ a − g b k B T exp( − ǫ/ k B T ) | ∞ 0 = g a exp( − ǫ a / k B T ) = g a exp( − E a / RT ) g b g b ◮ Suppose that the specific rate at which molecules with sufficient energy cross the barrier is r . Then k = r g a exp( − E a / RT ) = A exp( − E a / RT ) g b
The distribution of velocities ◮ The Boltzmann distribution ought to apply to kinetic energy: K = 1 2 mv 2 = m v 2 x + v 2 y + v 2 � � z 2 ◮ Sum of terms: apply Boltzmann distribution to each term independently. � − mv 2 p ( v x ) dv x = 1 � x Q exp dv x 2 k B T � ∞ � − mv 2 � x ∴ Q = exp dv x 2 k B T −∞ � = 2 π k B T / m � − mv 2 � m � x ∴ p ( v x ) dv x = 2 π k B T exp dv x 2 k B T
The distribution of velocities (continued) p ( v x , v y , v z ) dv x dv y dv z = � 3 / 2 � � − m ( v 2 x + v 2 y + v 2 z ) � m exp dv x dv y dv z 2 π k B T 2 k B T This gives the probability density at a point in the velocity space, i.e. near particular values of ( v x , v y , v z ).
The distribution of molecular speeds ◮ Speed v related to velocity components by v 2 = v 2 x + v 2 y + v 2 z ◮ The same speed is obtained at every point on a sphere satisfying this equation. ◮ dv x dv y dv z is a volume element. Integrated over the surface of a sphere of radius v , it gives 4 π v 2 dv . ◮ Distribution of speeds: � 3 / 2 � − mv 2 � � m p ( v ) dv = 4 π v 2 exp dv 2 π k B T 2 k B T or � 3 / 2 � − Mv 2 � � M p ( v ) dv = 4 π v 2 exp dv 2 π RT 2 RT
The distribution of molecular speeds 0.0035 T = 100 K T = 200 K 0.003 T = 300 K 0.0025 p ( v )/s m -1 0.002 0.0015 0.001 0.0005 0 0 200 400 600 800 1000 v /m s -1
Average speed ◮ Given a probability density p ( x ), the average of f ( x ) is given by � f ( x ) p ( x ) dx A where A is the region over which x is defined. ◮ For example, the average speed is � ∞ v = ¯ v p ( v ) dv 0 � ∞ � 3 / 2 � − mv 2 � � m 4 π v 3 = exp dv 2 π k B T 2 k B T 0 � 8 k B T = π m
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