AP Chemistry Notes

1. AP Chemistry Notes Chapter 14 Kinetics

2. Kinetics speed of chemical reactions • Factors affecting chemical reactions A. concentrations of reactants B. temperature 10ºC↑=2X rate C. presence of a catalyst D. surface area↑, particle size↯

3. 14.1 Reaction Rates A. average rate= Δmol/liter Δt B. instantaneous rate= at any point during reaction

4. p. 577 (now table 14.1)C4H9Cl + HOH  C4H9OH + HCl

6. C. Stoichiometry[ ]= concentration M 2HI → H2 + I2 Rate= -½ Δ[HI] = Δ[H2]= Δ[I2] Δt Δt Δt aA + bB → cC + dD Rate= -1Δ[A] = -1Δ[B] = +1Δ[C] = +1Δ[D] aΔt bΔt cΔt dΔt

7. Sample 14.2 2O3→ 3O2 -½Δ[O3] = 1/3 Δ[O2] Δt Δt 6.0 x 10-5 m/s x 2 mol O3 = 4.0 x 10-5 m/s 3 mol O2

8. 14.2 Dependence of Rate on Concentration • Rate Law Rate=k[A][B] • k=rate constant – can be changed by changes in temperature • Reaction Order Rate=k[A]1[B]2 1st order with respect to A 2nd order with respect to B 3rd order overall

9. Basic form for a rate law equation Zeroth order Rate = k[A]0 1 = k (10)0 1 = k (20)0 1 = k (30)0 • Rate = k[A]x[B]y

10. 1st order • Rate = k [A]1 • 10 = k (10)1 • 20 = k (20)1 • 30 = k (30)1 • 2nd order • Rate = k [A]2 • 100 = k (10)2 • 400 = k (20)2 • 900 = k (30)2

11. D. Units of k Rate = k[A] M/s=kM k=M/s=sec-1 M 2nd order: Rate=k[A]2M=kM2 s k=M/s= 1 = l , M2 s-M s-mol

12. E. Using Initial Rates to Determine Rate Laws a.) Rate = k[A]2 [B]0 = k[A]2 b.) 4.0x 10-5 = k(.1)2 k= 4.0x10-3M-1S-1 c.) Rate= (4.01x10-3)(.05)2 = 1.0x10-5M/s NOT BASED ON STOICHIOMETRY!!

13. 14.3 Δ of Conc. w/ Time • First Order Reactions A. ln [A]t = -kt ln[A]0 = kt [A]0 [A]t ln[A]t – ln [A]0 = -kt B. ln[ A]t = -kt + ln [A]0 y=mx+b 1. ln[A] vs. time should be a straight line

14. ½ life, t ½ t ½ = .693 = ln 2 k k concentration has no effect on t1/2 II 2nd Order Reactions A. 1 [A]t B. t1/2 = 1 k[A]0

15. First order vs second order

16. Zeroth order graph [M] time

17. 14.4 Temperature and Rate • Collision Model A + B  C + D • Activation NRG- min. amt. NRG needed to start a rxn 1. activated complex > highest energy 2. lowered by catalyst 3. increased by inhibitors C. Orientation Factor

19. D. Arrhenius Equations 1. Fraction of molecules possessing Ea or greater 2. # of collisions 3. fraction of coll. That have app. orientation 4. k= Ae –Ea/RT 5. ln k1= Ea (1 – 1) k2 R T2 T1 k= rate constant Ea = act. NRG R= 8.31 J/ mol-K T= temp in Kelvin

20. Svante August Arrhenius • Nobel Prize for chemistry in 1903 • Born in Vik, Sweden

22. 14.5 I. Reaction Mechanisms- step by step process by which a reaction occurs II. Elementary Steps- single step rxn A. Molecularity B. Unimolecular- one molecule is involved in reaction C. bimolecular- 2 molecules in reaction D. termolecular- 3 molecules ( rare) A + B + C - D

24. III. Multistep Mechanisms NO2 + NO2 NO3 + NO (slow) NO3 + CO NO2 + CO2(fast) Overall = NO2 + CO  NO + CO2 • NO3 is intermediate- made in one step, used up in another. NOT overall reaction B. Rate- determining step – the slowest step. C. catalyst-starts in first step, made in last. NOT in overall rxn

27. Energy Profile Diagram ΔH = negative #

28. ΔH = positive #

29. 1. Rate = k [A]x[B]y A + B  C (slow) C + D  E ( fast) A + B +D  E Rate = k [A]1[B]1

30. 2. Rate = k[A]x[B]y A +A  A2 ( slow) A2 + B  C ( fast) 2A + B  C Rate = k [A]2[B]0 OR rate = k[A]2

31. 3. Rate = k[A]x[B]y A  X (slow) X + B  C (fast ) C + B  D ( fast) A + 2B  D Rate = k[A]1[B]0 OR rate = k[A]1

32. 4. Rate = k[A]x[X]y A +X  C (fast) C +X  D + E (slow) D + Y  A + B( fast) 2X + Y  E +B Rate = k[X]2[Y]0 OR rate = k[X]2