Copyright © R. R. Dickerson Lecture 5 AOSC/CHEM 637 Atmospheric Chemistry R. Dickerson OUTLINE KINETICS Activation Energy Kinetic Theory of Gases Calc. Rate Constants w/Collision Theory Finlayson-Pitts (2000) Ch. 5 Seinfeld and Pandis (2006) Ch. 3 1

Copyright © R. R. Dickerson Kinetics continued Activation Energy The energy hill that reactants must climb in order to produce products; a barrier to thermodynamic equilibrium; for a second order reaction: ENERGY DIAGRAM C + D A + B AB † ΔH Ea A + B → AB † AB † + M → AB + M † AB † → C + D (Transition state or activated complex) (Sometimes there can be quenching) (Reaction) 2

Copyright © R. R. Dickerson ACTIVATION ENERGY Remember the Van't Hoff (or Gibbs-Helmholtz) equation. dlnK eq /dT =  H/(RT 2 ) This suggests: dlnk/dT = E a /(RT 2 ) Which is the Arrhenius expression where E a is the activation energy. If we integrate both sides: ln(k) = (-E a /R) 1/T + ln(A) Where ln(A) is the constant of integration. Rearranging: k = A e (-Ea/RT) This is the Arrhenius Equation in which A is the pre-exponential factor, also called the Arrhenius factor, and exp(-E a /RT) is the Boltzmann factor. 3

Copyright © R. R. Dickerson KINETIC THEORY OF GASES When molecules in the gas phase collide they sometimes rearrange their chemical bonds to form new molecules. The rate of formation of the new molecules is determined by the fraction of molecules with sufficient energy to overcome the activation energy barrier. POSTULATES OF CHEMICAL KINETICS 1. Pressure is the result of molecular collisions. 2. Collisions are elastic, i.e. no change in kinetic energy. 3. Volume of the molecules << volume occupied by gas. 4. Kinetic energy proportional to T and independent of gas, i.e., the same for all gases. 4

Copyright © R. R. Dickerson Boltzmann Distribution N 1 /N 0 = e {-(E1 - E0 )/kT} (Also called Maxwell distribution for ideal gases) WHERE: N 1 = number of particles (molecules) with energy E 1 N 0 = number of particles (molecules) with energy E 0 M = molecular weight dN/N 0 = M/kT exp [{ -Mc -2 }/2kT] c dc WHERE: c 2 = V 2 + U 2 + W 2 SEE: Lavenda, "Brownian Motion," Sci. Amer., 252(2), 70-85,

Copyright © R. R. Dickerson

7 CALCULATING RATE CONSTANTS FROM COLLISION THEORY From thermodynamics and Arrhenius: k = A exp(-E a /RT) A is a function of diameter, temperature, and mass; its maximum possible value is the frequency of collisions. WHERE: k = Boltzmann const. = 1.38x erg/K T = Abs. Temp. (K) d = Diameter of molecules.  = reduced mass = {M 1 x M 2 } / {M 1 + M 2 } A has units of (molecules cm -3 ) -1 s -1 or cm 3 s -1 7

Copyright © R. R. Dickerson Example: Collisions Between Nitrogen Molecules d N2 = 3.2x10 -8 cm M N2 = 28/6.023x10 23 g For N 2 + N 2 This is a good estimate for the maximum rate constant for any reaction. Note A is proportional to d 2,  , T 1/2. One would expect the Arrhenius factor to have a T 1/2 factor, but this is usually swamped out by the exponential temperature dependence of activation energy. (Remember ergs are g cm 2 s -2 ). 8

Copyright © R. R. Dickerson Example calculation of a rate constant O 3 + NO ↔ NO 2 + O 2 ENERGY DIAGRAM ↑ NO 2 + O 2 O 3 + NO O=O-O ∙ ∙ ∙ N=O ΔH EaEa O=O-O ∙ ∙ ∙ N=O is the activate complex E a is the activation energy, unknown. ΔH is the enthalpy of the reaction, known from thermodynamics. 9

Copyright © R. R. Dickerson Rate const for the forward reaction: k f = A exp(-E a /RT) We need the enthalpy: ∆ H o rxn =  ∆ H f o prod. -  ∆ H f o react. ∆ H f = { } = kcal/mole ∆ H r = - ∆ H f = kcal/mole We need the pre-exponential factors A f and A r DIAMETERS d(NO) = 0.40 nm d(O 3 ) = 0.46 nm d(NO 2 ) = 0.46 nm d(O 2 ) = nm 10

Copyright © R. R. Dickerson REDUCED MASSES  f = 18.5/6.023x10 23 g  r = 18.9/6.023x10 23 g A f < Forward collision rate = 3.4x s -1 cm 3 A r < Reverse collision rate = 2.6x s -1 cm 3 Now we need an estimate of activation energy, E a E ar  47.5 kcal/mole E af We know nothing! This is very slow! STUDENTS: Calculate the lifetime of NO 2 with respect to conversion to NO at the typical oxygen content of the atmosphere. 11

Copyright © R. R. Dickerson To get at k f lets look to thermodynamics. k f /k r = K eq = exp(-  G/RT)  G o = = kcal/mole The products are heavily favored. k f = K eq x k r But we knew that much from the collision rate already. The measured rate constant for this reaction is: 12

Copyright © R. R. Dickerson The measured "A" is 113 times smaller than the maximum "A". Why? Not every collision with sufficient energy results in a reaction. The molecules must have the proper orientation. STERIC FACTOR: A(collisional)/A(actual) = 113 Only one collision in 113 has the proper orientation. Now lets try to calculate a better value for k r. Assume same steric factor. E ar = E af +  H r = k r = 3.0x exp(-25500/T) cm 3 s -1 = 2.0x cm 3 s -1 at 298 K Thermodynamics says: k r = k f /K eq = 3.2x cm 3 s -1 13

Copyright © R. R. Dickerson The agreement is not too bad, less than a factor of two difference! The thermodynamic value is more likely correct. We cannot measure the reverse rate constant because it is too slow. For example if we took a 1 atm mixture of 50% NO 2 and 50% O 2 at equilibrium (square brackets represent partial pressure) the ozone and nitric oxide concentrations would be much too small to measure. Pretty small, so we can say that all the NO and O 3 react away. 14

Copyright © R. R. Dickerson KJ/mole +286 KJ/mole +428 KJ/mole Stedman’s Kinetic Kidney Enthalpy ↑ O 2 → 2O  H = +498 KJ/mole k = A e (-Ea/RT)

Copyright © R. R. Dickerson Lecture 5 Summary For simple bimolecular (second order) reactions, the Arrhenius expression describes the temperature dependence of the rate constants. Collision theory gives some insight into the limits of rate constants. The enthalpy of a reaction provides insight into the activation energy. Molecular oxygen, O 2, is not generally an important oxidant in atmospheric reactions. The radicals OH and HO 2 are the “hit men” of atmospheric chemistry.