20 B Week VI Chapters 18 Chemical Kinetics

1. • The Rates of Chemical Reactions A-->B Rate=d[A]/dt= -d[B]/dt. units molL-1s-1 A is the Reactant and B is the product of the reaction • Rate Laws at early times =k[A]n k is the rate constant and n is the reaction order • Elementary reactions: single step reactions Unimolecular(1st), Bimolecular(2nd), Termolecular(3rd) • The Arrhenius equation for the rate constant k(T)=Aexp(-Ea/RT) • Ea the activation energy the A-factor • Reaction path diagrams from reactants to product. 20 B Week VI Chapters 18 Chemical Kinetics

2. B + surface(wall) Collision Rate Ch 9.7) Zwall= {(1/4)<u>A} [B] Average over incident velocities <v>=(1/4)<u> A = surface area B + surface condensation Rate Rate= --d[B]/dt ~ Zwall x (probability of sticking to the surface) --d[B]/dt ~ {(1/4)<u>A} [B] exp{-Ea/kT} - d[B]/dt = k[B] k~{(1/4)<u>A} exp{-Ea/kT} Rate constant (molecule/surface property) Ea the activation energy

3. A + B Collision Rate ZAB=[A][B]{(√2)<u>(πd2)}(N0)2 A + B Reaction Rate Reaction Rate= -d[A]/dt=-d[B]/dt ~ ZAB p exp{-Ea/kT} -d[A]/dt=-d[B]/dt =[A][B]{(√2)<u>(πd2)}(N0)p exp{-Ea/kT} -d[A]/dt=-d[B]/dt = k[A][B]

4. Collision Rate for one molecule B moving a gas of A Molecules ZB= DN/Dt = (NA /V) DV/Dt = [A](<urel>Dt)(πd2) /Dt ZB=[A]√2<u>πd2) s = πd2 collision cross-section πd2 D Fig. 9-20, p. 426

5. A + B Collision Rate A and B collision rate per unit volume [B]ZB={B][A] √2<u>πd2 ZAB=[A][B]{(√2)<u>(πd2)}(N0)2 Reaction Rate= -d[A]/dt=-d[B]/dt = ZAB p exp{-Ea/kT} -d[A]/dt=-d[B]/dt =[A][B]{(√2)<u>(πd2)}(N0)p exp{-Ea/kT} -d[A]/dt=-d[B]/dt = k[A][B] p = the Steric factor( fraction of molecules with the correct orientation to react) exp{-Ea/kT} = Boltzmann factor, the fraction of molecules with KE ≥ Ea and therefore to react!

6. The Rate Constant k k={(√2)<u>(πd2)}(N0)p exp{-Ea/kT} k = (collision rate {cross-section for a πd2N0} ) (p= fraction with correct orientation) (fraction with energy ≥Ea ) k=A exp{-Ea/kT} The Arrhenius Form and the A-factor A= {(√2)<u>(πd2)}(N0)p This is the A-factor for Elementary Bimolecular Reactions only

7. HCl potential curve H2 potential curve Energy H--H--Cl 0 H2 + Cl H + HCl H HCl H2 Cl

8. Reaction Path Diagram HD + Cl H + DCl Energy “Transition State” reactants before this point. products after it [H--D--Cl]# Activated Complex Ear Eaf H-D + Cl H + D-Cl Rxn Path

9. The Boltzmann Factor- exp(-Ea/RT) Fraction of Molecules That will react at Temperature T

10. Unimolecular A Products Rate = k[A] 1st Order Cis  Trans Isomerization

11. Transition State the Highest Point along reaction path potential energy curve(surface) Energy 180° 90°  C–C Bond length(Re) 0°

12. [transition State]# Cis Trans Isomerization reaction =180° Reactant “Cis” Product “Trans”

13. Let [C2H4] =[A] Rate=-d[A]/dt= k[A] d[A]/[A]=-kdt dln[A]= d[A]/[A]=-kdt ln[A] - ln[A]0 = -kt [A]=[A]0exp{-kt} [A]=[A]0/2 -In(2)=-kt1/2 t1/2= ln2/k Half-Life

14. Let [C2H4] =[A] Rate=-d[A]/dt= k[A] d[A]/[A]=-kdt dln[A]= d[A]/[A]=-kdt ln[A] - ln[A]0 = -kt [A]=[A]0exp{-kt}

15. [A]0 Rate=-d[A]/dt= k[A] d[A]/[A]=-kdt dln[A]= d[A]/[A]=-kdt ln[A] - ln[A]0 = -kt [A]=[A]0exp{-kt} ln[A] = ln[A]0 - kt ln[A]

16. 2nd Order reactions Rate = k[A]2 Bimolecular are 2nd order reactions A + A  Products 2NO2 2NO + O2 let [NO2]=[A] Rate=-(1/2)d[A]/dt = k[A]2 d[A]/[A]2=- d(1/[A])= - 2kdt 1/[A] = 1/[A]0 + 2kt [A]=[A]0/2 (1/[A]0)= 2kt1/2 t1/2= 1/2k[A]0

17. Kinetics of Approaching Equilibrium Equilibrium: Evaporation rate equal the condensation Rate  Equilibrium C2H4(l)  C2H4(g) Rate Law Rate=k1=const, , Order zero • C2H4(l)  C2H4(g) Evap • C2H4(g)  C2H4(l) Cond Rate Law Rate=k2[C2H4(g)] The order n =1 Fig. 10-16, p. 459

18. Fig. 18-3, p. 838