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LECTURE 4 AOSC 637 Atmospheric Chemistry

LECTURE 4 AOSC 637 Atmospheric Chemistry. 1 Finlayson-Pitts and Pitts Ch. 5 (Kinetics & Lab Techniques) Wayne Chapt. 3 (Kinetics) Wark & Warner Chapt. 8 & 9 Seinfeld and Pandis Ch. 4 Kinetics; Ch. 9 Thermodynamics. Outline Thermodynamics Free energy Kinetics

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LECTURE 4 AOSC 637 Atmospheric Chemistry

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  1. LECTURE 4AOSC 637Atmospheric Chemistry 1Finlayson-Pitts and Pitts Ch. 5 (Kinetics & Lab Techniques) Wayne Chapt. 3 (Kinetics) Wark & Warner Chapt. 8 & 9 Seinfeld and Pandis Ch. 4 Kinetics; Ch. 9 Thermodynamics. Outline Thermodynamics Free energy Kinetics Rates, rate constants and order of rxns Lifetime & half-life © R. R. Dickerson 2011 1

  2. FREE ENERGY We have a problem, neither energy (E) nor enthalpy (H) is the "criterion of feasibility". Chemical systems generally tend toward the minimum in E and H, but not always. Everyday experience tells us that water evaporates at room temperature, but this is uphill in terms of the total energy, E. © R. R. Dickerson 2011 2

  3. Example 1 H2O(l) H2O(g) P = 10 torr, T = 25°C E = +9.9 kcal/mole The enthalpy, H, is also positive, about 10 kcal/mole, and PdV is too small to have an impact (R =1.99 cal/mole K; V2/V1 ~ 1000). © R. R. Dickerson 2011 3

  4. Example 2. The formation of nitric oxide from nitrogen and oxygen occurs at combustion temperature. We know that H >>0 at room temperature, but what about at combustion temperature? We can calculate H as a function of temperature with heat capacities, Cp, found in tables. Remember that R = 1.99 cal/moleK and dH = Cp dT N2 + O2 2NO © R. R. Dickerson 2011 4

  5. N2 + O2 2NO At room temperature: H298 = + 43.14 kcal/mole The reaction is not favored, but combustion and lightning heat the air, and dH = CpdT © R. R. Dickerson 2011 5

  6. What is ∆H at 1500K? H1500 = H298 + Integral from 298 to 1500 Cp dT We can approximate Cp with a Taylor expansion.  Cp = 2Cp (NO) - Cp (N2) - Cp(O2) Cp (O2)/R = 3.0673 + 1.6371x10-3 T - 5.118x10-7 T 2 Cp (N2)/R = 3.2454 + 0.7108x10-3 T - 0.406-7 T 2 Cp (NO)/R = 3.5326 - 0.186x10-3 T - 12.81x10-7 T2 - 0.547x10-9 T3 Cp /R = 0.7525 - 2.7199x10-3 T + 26.5448x10-7 T2 - 1.094x10-9 T3 © R. R. Dickerson 2011 6

  7. What is H1500 ? © R. R. Dickerson 2011 7

  8. But we know that high temperature combustion and lightning produce large quantities of NO! Therefore, these reactions must be driven by some force other than internal energy or heat. Note: The independence of H with T will be useful. © R. R. Dickerson 2011 8

  9. Entropy and the Second Law of Thermodynamics DFN: dS = đQ/T For a reversible reaction the change in entropy, S, is a function of state of system only. To get a feel for what entropy is, let us derive an expression for the entropy of an ideal gas. General Relations: dE = đQ - PdV dS = đQ/T dS = dE/T + PdV/T © R. R. Dickerson 2011 9

  10. For one mole of an ideal gas, P/T = R/V At constant volume, dE = Cv dT Thus Integrating S = Cv ln(T) + R ln(V) + So Where So is the residual entropy. This equation lets you calculate entropy for an ideal gas at a known T and V. © R. R. Dickerson 2011 10

  11. Gibbs Free Energy dS = đQ/T For an irreversible reaction: dS ≥đQ/T At constant T & P: dE = đQ – PdV But VdP = 0 dE – TdS + pdV ≤ 0 Because dP and dT are zero we can add VdP and SdT to the equation. dE – TdS – SdT + PdV – VdP ≤ 0 d(E + PV – TS) ≤ 0 © R. R. Dickerson 2011 11

  12. We define G as (E + PV - TS) or (H - TS)dG ≡ dH – TdS∆G ≡ ∆H – T∆S ∆G for a reaction is the Gibbs free energy, and it is the criterion of feasibility. G tends toward the lowest value. If ∆G is greater than zero then the reaction will not proceed spontaneously; the reactants are favored over the products. © R. R. Dickerson 2011 12

  13. Gibbs Free Energy, ∆G, andEquilibrium constants Keq Consider the isothermal expansion of an ideal gas. dG = VdP From the ideal gas law, dG = (nRT/P)dP Integrating both sides, © R. R. Dickerson 2011 13

  14. Now let’s apply this to a reaction.aA + bB ↔ cC + dDwhere the small letters represent coefficients. © R. R. Dickerson 2011 14

  15. If a + b is not the same as c + d, we can get into trouble trying to take the log of an expression with units. (Note error in Hobbs’ book.) For this type of reaction, the Gibbs free energy is the sum of the G for the chemical reaction and the G for the change in pressure. Assuming that the reactants start at 1.0 atm and go to an equilibrium pressure and assuming that the products finish at 1.0 atm. The units will always cancel. aA(PA = 1)  aA(PA) GA = aRTln (PA/1) © R. R. Dickerson 2011 15

  16. For the products: cC(PC)  cC(PC = 1) G C = cRTln(1/PC) = −cRTln(PC) Go = Grxn + GA + GB + GC + GD I = Grxn + aRTln(PA/1) + bRTln(PB/1) + cRTln(1/PC) + dRTln(1/PD) © R. R. Dickerson 2011 16

  17. Combining the log terms, For the equilibrium partial pressures where Grxn = 0, © R. R. Dickerson 2011 17

  18. If you remember that each of the partial pressures was a ratio with the initial or final pressure taken as 1.0 and that the ln(1) = 0 are left out you can see that Keq is always dimensionless. © R. R. Dickerson 2011 18

  19. Kinetics Thermodynamics tells us if a reaction can proceed and gives equilibrium concentration. Kinetics tells us how fast reactions proceed. If thermodynamics alone controlled the atmosphere it would be dissolved in the oceans as nitrates - we would be warm puddles of carbonated water. Reaction Rates and Order of Reactions 1. FIRST ORDER A  PRODUCTS dA/dt = -k[A] © R. R. Dickerson 2011 19

  20. The red line describes first order loss with a rate constant of 1 min-1 The blue line describes the rate of formation of the product. minutes © R. R. Dickerson 2011 20

  21. Examples: Radioisotope decay, thermal decomposition and photolysis. A → Products 222Rn → 218Po +  N2O5 = NO2 + NO3 NO2 + hn → NO + O Radon is important source of indoor air pollution, and N2O5 is nitric acid anhydride, important in air pollution nighttime chemistry. The rate equations take the form: d[prod.]/dt = -k[A] = -d[react.]/dt For example: d[Po]/dt = kRn [Rn] = -d[Rn]/dt Where k is the first order rate constant and k has units of time-1 such as s-1, min-1, yr-1. We usually express concentration, [Rn], in molecules cm-3 and k in s-1. © R. R. Dickerson 2011 21

  22. Also d[NO2] /dt = k [N2O5] And d[N2O5]/dt = -k [N2O5] Separating Integrating If we define the starting time as zero: Rate constants at 298 K are:kRn = 0.182 days-1 kN2O5 = 0.26 s-1 © R. R. Dickerson 2011 22

  23. “j-Values”: Definition NO2 + h NO + O ( < 424 nm) Actinic flux (photons cm-2 s-1 nm-1) absorption cross section (cm2 molec-1) photolysis quantum yield (molec. photon-1) © R. R. Dickerson 2011 23

  24. International Photolysis Frequency Measurement and Modeling Intercomparison (IPMMI) NCAR Marshall Field Site, 39°N 105°W, elevation: 1.8 km; June 15–19, 1998 Objectives: j [NO2 NO + O], j [O3  O2 + O(1D)], spectral actinic flux. Measurements by 21 researchers from around the world. Photolysis Frequency of NO2: Measurement and Modeling During the International Photolysis Frequency Measurement and Modeling Intercomparison (IPMMI), R. E. Shetter, W. Swartz, et al., J. Geophys. Res., 108(D16), 10.1029/2002JD002932,2003.

  25. UMD jNO2Actinometer Schematic NO2 + h NO + O

  26. Problem for the student: Show that for 1.00 ppm NO2, 1.00 atm pressure, exposure times of 1.00 s, and j(NO2) values of 10-2 s-1 the errors to: from complicating reactions are less than 1%. 1. O + O2 + M → O3 + M k1 = 6.0 x10-34 cm6 s-1 2. O3 + NO → NO2 + O2 k2 = 1.9×10–14 cm3 s-1 3. O + NO2 → NO +O2 k3 = 1.04×10–11 cm3 s-1 © R. R. Dickerson 2011 26

  27. Trailer #2 UMD Actinometer

  28. UMD Actinometer on top inside quartz photolysis tube

  29. UMD jNO2Actinometer Data

  30. 2. Second Order A + B  PRODUCTS Examples NO + O3 NO2 + O2 HCl + OH  H2O + Cl Examples of the rate equations are as follows: d[NO]/dt = -k[NO][O3] d[Cl]/dt = k[OH][HCl] Units of k are conc-1 time-1. 1/(molecules/cm3) (s-1) = cm3 s-1 Rate constants have the following values: kNO-O3 = 1.9x10-14 cm3 s-1 kHCl-OH = 8.0x10-13 cm3 s-1 © R. R. Dickerson 2011 30

  31. 3. Third Order A + B + C  PRODUCTS d[A]/dt = -k[A][B][C] Examples 2NO + O2 = 2NO2 O + O2 + M = O3 + M† M is any third body (usually N2) needed to dissipate excess energy. From the ideal gas law and Avg's number: Where Mo is the molecular number density at STP in molecules cm-3. Third order rate constants have units of conc-2 time-1. These are usually (cm-3)-2 s-1. kNO-O2 = 2.0 x 10-38 cm6 s-1 kO-O2 = 4.8 x 10-33 cm6 s-1 © R. R. Dickerson 2011 31

  32. Useful idea: For the following reversible reaction: A + B ↔ C + D d[C]/dt = kf [A][B] - kr [C][D] At steady state d[C]/dt = 0, by definition. Thus: © R. R. Dickerson 2011 32

  33. Half-life and Lifetime Definition: Half-life, t1/2, is the time t such that: [A]t / [A]0 = 1/2 Definition of e-folding lifetime or residence time, , comes from kinetics, where k is the first order rate constant with units of time-1. We know that: [A]t/[A]0 = exp(-kt) The lifetime, , is when t = 1/k so  1/k We can link half-life and lifetime: t1/2 = ln(2)/k  0.69/k For radon 222 (222Rn) the lifetime is 5.5 days, but the half-life is only 3.8 days.  = (k[Ā])-1 © R. R. Dickerson 2011 33

  34. For second and third order reactions we can sometime approximate first order conditions – or use pseudo first order kinetics. A + B → Prod. If [B] >> [A] Then k[B] is approximately constant. We call this pseudo first order rate constant k’. © R. R. Dickerson 2011 34

  35. For second and third order reactions we can sometime approximate first order conditions – or use pseudo first order kinetics. For example: NO + O3 NO2 + O2 k = 1.9 x 10-14 cm3 s-1 Assume: [O3] >> [NO] and d[O3]/dt ~ 0.0 Let: mean [O3] = 50 ppb (a reasonable value for air near the surface). CONCLUSION: any NO injected into such an atmosphere (by a car for example) will quickly turn into NO2 , if there are no other reactions that play a role. We will call k[O3] the pseudo first order rate constant. © R. R. Dickerson 2011 35

  36. For third order reactions we must assume that two components are constant. For example: O + O2 + M  O3 + M k = 4.8x10-33 cm6 s-1 ASSUME: d[O2]/dt = d[M]/dt = 0.0 We know that [O2] = 0.21 and that [M] ~ [O2] + [N2] ~ 1.00. At RTP P02 = 0.21 atm and PM ~ 1.0 atm. Therefore the lifetime of O atoms is 1/k[O2][M]M02 where M0 is the conversion to molecules per cm3. = 1.6x10-6 s, short indeed! © R. R. Dickerson 2011 36

  37. Example 2 Same reaction at stratospheric temperature and pressure. P30km ~ P0 exp(-30/7) = 0.014 atm = 2.1x10-3 s This is still short, but it is a thousand times longer than in the troposphere! The pressure dependence has a major impact on the formation and destruction of tropospheric and stratospheric ozone. Problem for the student: Compare the rate of loss of O atoms to reaction with O2 vs. NO2 in the troposphere where [NO2] ~ 10-8and in the stratosphere where [NO2] ~ 10-6. © R. R. Dickerson 2011 37

  38. If two of the reactants in a third order reaction are the same, we can derive a useful expression for the rate of loss of the reactant. A + A + B  PROD For a great excess of B: d[A]/dt = -(2k[B])[A]2 [A]-2 d[A] = -(2k[B])dt -[A]t-1 + [A]0-1 = -(2k[B])t [A]t-1 = 2k[B]t + [A]0-1 Now we can calculate the concentration at any time t in terms of the initial concentration and the rate constant k. © R. R. Dickerson 2011 38

  39. The method works for the self reaction of nitric oxide: NO + NO + O2→ 2NO2 and shows that this reaction can ruin NO in N2 calibration standards if any air gets into the cylinder. d[O2]/dt = -k[NO]2[O2] d[NO]/dt = -2k[NO]2[O2] = 2d[NO2]/dt k = 3.3x10-39 exp (530/T) cm6 molecules-2 s-1 In the atmosphere it is only important for highly concentrated exhaust gases. © R. R. Dickerson 2011 39

  40. Lecture 4 Summary • Changes in enthalpy andentropy, DH and DS, are nearly independent of temperature. • Gibbs free energy provides the criterion of feasibility. • The residence time (lifetime), t, is the inverse of the first order rate constant, k. • If second or third order reactions can be approximated as first order then lifetimes can be estimated. • For reversible reactions, kf/kr = Keq © R. R. Dickerson 2011

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