1. © R. R. Dickerson 201111 LECTURE 4 AOSC 637 Atmospheric 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

2. © R. R. Dickerson 201122 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.

3. © R. R. Dickerson 201133 Example 1 H 2 O ( l )  H 2 O (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; V 2 /V 1 ~ 1000).

4. © R. R. Dickerson 201144 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, C p, found in tables. Remember that R = 1.99 cal/moleK and dH = C p dT N 2 + O 2  2NO

5. © R. R. Dickerson 201155 N 2 + O 2  2NO At room temperature:  H 298 = + 43.14 kcal/mole The reaction is not favored, but combustion and lightning heat the air, and dH = C p dT

6. © R. R. Dickerson 201166 What is ∆H at 1500K?  H 1500 =  H 298 + Integral from 298 to 1500  C p dT We can approximate C p with a Taylor expansion.  C p = 2C p (NO) - C p (N 2 ) - C p (O 2 ) C p (O 2 )/R = 3.0673 + 1.6371x10 -3 T - 5.118x10 -7 T 2 C p (N 2 )/R = 3.2454 + 0.7108x10 -3 T - 0.406 -7 T 2 C p (NO)/R = 3.5326 - 0.186x10 -3 T - 12.81x10 -7 T 2 - 0.547x10 -9 T 3  C p /R = 0.7525 - 2.7199x10 -3 T + 26.5448x10 -7 T 2 - 1.094x10 -9 T 3

7. © R. R. Dickerson 201177 What is  H 1500 ?

8. © R. R. Dickerson 201188 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.

9. © R. R. Dickerson 201199 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

10. © R. R. Dickerson 201110 For one mole of an ideal gas, P/T = R/V At constant volume, dE = C v dT Thus Integrating S = C v ln(T) + R ln(V) + S o Where S o is the residual entropy. This equation lets you calculate entropy for an ideal gas at a known T and V.

11. © R. R. Dickerson 201111 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

12. © R. R. Dickerson 201112 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.

13. © R. R. Dickerson 201113 Gibbs Free Energy, ∆G, and Equilibrium constants K eq Consider the isothermal expansion of an ideal gas. dG = VdP From the ideal gas law, dG = (nRT/P)dP Integrating both sides,

14. © R. R. Dickerson 201114 Now let’s apply this to a reaction. aA + bB ↔ cC + dD where the small letters represent coefficients.

15. © R. R. Dickerson 201115 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(P A = 1)  aA(P A )  G A = aRTln (P A /1)

16. © R. R. Dickerson 201116 For the products: cC(P C )  cC(P C = 1)  G C = cRTln(1/P C ) = −cRTln(P C )  G o =  G rxn +  G A +  G B +  G C +  G D I =  G rxn + aRTln(P A /1) + bRTln(P B /1) + cRTln(1/P C ) + dRTln(1/P D )

17. © R. R. Dickerson 201117 Combining the log terms, For the equilibrium partial pressures where  G rxn = 0,

18. © R. R. Dickerson 201118 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 K eq is always dimensionless.

19. © R. R. Dickerson 201119 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]

20. © R. R. Dickerson 201120 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

21. © R. R. Dickerson 201121 Examples: Radioisotope decay, thermal decomposition and photolysis. A → Products 222 Rn → 218 Po +  N 2 O 5 = NO 2 + NO 3 NO 2 + h → NO + O Radon is important source of indoor air pollution, and N 2 O 5 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 = k Rn [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.

22. © R. R. Dickerson 201122 Also d[NO 2 ] /dt = k [N 2 O 5 ] And d[N 2 O 5 ]/dt = -k [N 2 O 5 ] Separating Integrating If we define the starting time as zero: Rate constants at 298 K are: k Rn = 0.182 days -1 k N2O5 = 0.26 s -1

23. © R. R. Dickerson 201123 “j-Values”: Definition NO 2 + h  NO + O ( < 424 nm ) Actinic flux (photons cm -2 s -1 nm -1 ) absorption cross section (cm 2 molec -1 ) photolysis quantum yield (molec. photon -1 )

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 [NO 2  NO + O], j [O 3  O 2 + O( 1 D)], spectral actinic flux. Measurements by 21 researchers from around the world. Photolysis Frequency of NO 2 : 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 j NO2 Actinometer Schematic NO 2 + h  NO + O

26. © R. R. Dickerson 201126 Problem for the student: Show that for 1.00 ppm NO 2, 1.00 atm pressure, exposure times of 1.00 s, and j(NO 2 ) values of 10 -2 s -1 the errors to: from complicating reactions are less than 1%. 1. O + O 2 + M → O 3 + M k 1 = 6.0 x10 -34 cm 6 s -1 2. O 3 + NO → NO 2 + O 2 k 2 = 1.9×10 –14 cm 3 s -1 3. O + NO 2 → NO +O 2 k 3 = 1.04×10 – 11 cm 3 s -1

27. Trailer #2 UMD Actinometer

28. inside on top quartz photolysis tube

29. UMD j NO2 Actinometer Data

30. © R. R. Dickerson 201130 2. Second Order A + B  PRODUCTS Examples NO + O 3  NO 2 + O 2 HCl + OH  H 2 O + Cl Examples of the rate equations are as follows: d[NO]/dt = -k[NO][O 3 ] d[Cl]/dt = k[OH][HCl] Units of k are conc -1 time -1. 1/(molecules/cm 3 ) (s -1 ) = cm 3 s -1 Rate constants have the following values: k NO-O3 = 1.9x10 -14 cm 3 s -1 k HCl-OH = 8.0x10 -13 cm 3 s -1

31. © R. R. Dickerson 201131 3. Third Order A + B + C  PRODUCTS d[A]/dt = -k[A][B][C] Examples 2NO + O 2 = 2NO 2 O + O 2 + M = O 3 + M † M is any third body (usually N 2 ) needed to dissipate excess energy. From the ideal gas law and Avg's number: Where M o 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. k NO-O2 = 2.0 x 10 -38 cm 6 s -1 k O-O2 = 4.8 x 10 -33 cm 6 s -1

32. © R. R. Dickerson 201132 Useful idea: For the following reversible reaction: A + B ↔ C + D d[C]/dt = k f [A][B] - k r [C][D] At steady state d[C]/dt = 0, by definition. Thus:

33. © R. R. Dickerson 201133 Half-life and Lifetime Definition: Half-life, t 1/2, 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: t 1/2 = ln(2)/k  0.69/k For radon 222 ( 222 Rn) the lifetime is 5.5 days, but the half-life is only 3.8 days.  = (k[Ā]) -1

34. © R. R. Dickerson 201134 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’.

35. © R. R. Dickerson 201135 For second and third order reactions we can sometime approximate first order conditions – or use pseudo first order kinetics. For example: NO + O 3  NO 2 + O 2 k = 1.9 x 10 -14 cm 3 s -1 Assume: [O 3 ] >> [NO] and d[O 3 ]/dt ~ 0.0 Let: mean [O 3 ] = 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 NO 2, if there are no other reactions that play a role. We will call k[O 3 ] the pseudo first order rate constant.

36. © R. R. Dickerson 201136 For third order reactions we must assume that two components are constant. For example: O + O 2 + M  O 3 + M k = 4.8x10 -33 cm 6 s -1 ASSUME: d[O 2 ]/dt = d[M]/dt = 0.0 We know that [O 2 ] = 0.21 and that [M] ~ [O 2 ] + [N 2 ] ~ 1.00. At RTP P 02 = 0.21 atm and P M ~ 1.0 atm. Therefore the lifetime of O atoms is 1/k[O 2 ][M]M 0 2 where M 0 is the conversion to molecules per cm 3. = 1.6x10 -6 s, short indeed!

37. © R. R. Dickerson 201137 Example 2 Same reaction at stratospheric temperature and pressure. P 30km ~ P 0 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 O 2 vs. NO 2 in the troposphere where [NO 2 ] ~ 10 -8 and in the stratosphere where [NO 2 ] ~ 10 -6.

38. © R. R. Dickerson 201138 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.

39. © R. R. Dickerson 201139 The method works for the self reaction of nitric oxide: NO + NO + O 2 → 2NO 2 and shows that this reaction can ruin NO in N 2 calibration standards if any air gets into the cylinder. d[O 2 ]/dt = -k[NO] 2 [O 2 ] d[NO]/dt = -2k[NO] 2 [O 2 ] = 2d[NO 2 ]/dt k = 3.3x10 -39 exp (530/T) cm 6 molecules -2 s -1 In the atmosphere it is only important for highly concentrated exhaust gases.

40. © R. R. Dickerson 201140 Lecture 4 Summary Changes in enthalpy and  entropy,  H and  S, 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, k f /k r = K eq