1. CHEMICAL KINETICS Goal of kinetics experiment is to measure concentration of a species at particular time during a rxn so a rate law can be determined Rate of Rxn: describes how fast reactants used up & pdts formed rates are obtained from concen. vs fct of time Chem Kinetics: 1)study of rates, 2) factors that affect rxn rates, & 3) mechanisms (steps) by which rxns occur From a chem eqn, rate can be determined by following the  concen of any subst that is quantitatively detected

2. 4 factors that affect chem rxns 1) nature of reactants 2) concen of reactants 3) temp 4) catalyst present Write rate law for rxn to describe how rate depends on concen. Order of rxn cannot be deduced from chemical eqn. of rxn Rate law expressions - calculate rate of rxn from rate constant & reactant concen - convert into eqn to determine concen of reactants @ any time Rate Law is deduced experimentally from how its rate varies w/ concen Order of rxn cannot be deduced from chemical eqn. of rxn

3. exponents x & y - usually integers - value of x is the order of rxn w/ respect to A - y?? Values for k, x, & y have no relation to coeff of balanced chem eqn., remember, must be determined experimentally For rxn: A + B -----> pdts general form: rate = k[A] x [B] y Exponent Define 0 1 2 rate not depend on [reacts] rate is directly proportional to [reacts] rate is directly proportional to square of concen; [reacts] 2 overall order of rxn = x + y Sum of orders of reacts

4. order in NO: overall order: - order in rate law may not match coeff. in balanced eqn - no way to predict rxn orders overall from balanced eqn - orders must be determined experimentally Examples of observed rate laws for following rxns 3NO (g) ------> N 2 O (g) + NO 2 (g) rate = k[NO] 2 2NO 2 (g) + F 2 (g) ------> 2NO 2 F (g) rate = k[NO 2 ][F 2 ] order in NO 2 : order in F 2 : overall order: 2 nd 1 st 2 nd quick summary

5. rate law can be determined by 2 methods: 1) Method of Initial Rates (if time) 2) using Integrated Rate Eqn

6. ZERO ORDER Has a rate which is independent of concentration of reactant(s), therefore, increasing concen. of rxning species not speed up rate Rate is: Rate is a CONSTANT A -----> pdts integration gives eqn called integrated zero-order rate law [A] = -kt + [A] o Initial concentr concentr of chemical @ particular time

7. [A] = -kt + [A] o eqn line: y = mx + b time, t [A] calculate k from plot of graph; straight line plot of [A] vs time, t ; slope = -k Determine units:

8. half-life describes time needed for half of reactant to be depleted

9. FIRST ORDER Depends on concentration of only 1 reactant, if other reactants present but each will be zero-order eqn for first-order reaction A -----> pdts 1 st order rate constant, units of 1/time integration gives eqn called integrated first-order rate law ln[A] = -kt + ln[A] o rate is: know: eqn line: y = m x + b

10. calculate k from plot of graph; plot of ln[A] vs time, t ; gives straight line slope = -k time, t [A] time, t ln[A] Determine units:

11. half-life describes time needed for half of reactant to be depleted

12. SECOND ORDER A. A ------ > pdts depends on concentration of 2 nd -order reactant second-order rate law integrated in the form: [A] O @ t = 0 & [A] @ t: B. or, A + B = pdts two 1 st -order reactants: rate is: ln r = ln k + 2 ln[A] Another way to represents rate laws, take ln of both sides:

13. Plot 1/[A] vs time, t ; slope = 2 nd -order rate constant; +k eqn line: y = m x + b time, t 1/[A]

14. half-life for 2 nd order dependent on one 2 nd order reactant: Determine units:

15. FIRST ORDER REACTION 2 N 2 O 5 (aq) --------> 4 NO 2 (aq) + O 2 (g) ln[N 2 O 5 ] 1/[N 2 O 5 ], M -1 DATA Time, s [ N 2 O 5 ], M 0 0.0365 600 0.0274 1200 0.0206 1800 0.0157 2400 0.0117 3000 0.00860 3600 0.00640 - 3.310 - 3.597 - 3.882 - 4.154 - 4.448 - 4.756 - 5.051 27.4 36.5 48.5 63.7 85.5 116 156

16. time,s [N 2 O 5 ] time,s ln[N 2 O 5 ] time,s 1/[N 2 O 5 ] rate = k[N 2 O 5 ]

17. time,s ln[N 2 O 5 ] slope =

18. SECOND ORDER REACTION 2 NO 2 (g) --------> 2 NO (g) + O 2 (g) ln[N 2 O 5 ] 1/[N 2 O 5 ], M -1 DATA Time, s [ N 2 O 5 ], M 0 60 120 180 240 300 360 - 4.605 - 4.986 - 5.263 - 5.477 - 5.655 - 5.806 - 5.937 100 146 193 239 286 332 379 0.0100 0.00683 0.00518 0.00418 0.0350 0.00301 0.00264

19. time,s [NO 2 ] time,s ln[NO 2 ] time,s 1/[NO 2 ] rate = k[NO 2 ] 2

20. time,s 1/[NO 2 ] slope =

21. ZERO ORDER REACTION Can occur if: 1) rate limited by [catalyst] 2) photochemical rxn if rate determined by light intensity 3) most often occur when subst as a metal surface or enzyme required for rxn to occur 2 N 2 O (g) --------> 2 N 2 (g) + O 2 (g) N2ON2O N2ON2O N2ON2O N2ON2O N2ON2O N2ON2O N2ON2O N2ON2O N2ON2O N2ON2O N2ON2O N2ON2O N2ON2O N2ON2O N2ON2O N2ON2O N2ON2O N2ON2O N2ON2O N2ON2O N2ON2O N2ON2O N2ON2O N2ON2O N2ON2O N2ON2O N2ON2O N2ON2O N2ON2O N2ON2O N2ON2O N2ON2O Pt metal surface

22. N2ON2O N2ON2O N2ON2O N2ON2O N2ON2O N2ON2O N2ON2O N2ON2O N2ON2O N2ON2O N2ON2O N2ON2O N2ON2O N2ON2O N2ON2O N2ON2O N2ON2O N2ON2O N2ON2O N2ON2O N2ON2O N2ON2O N2ON2O N2ON2O N2ON2O N2ON2O N2ON2O N2ON2O N2ON2O N2ON2O N2ON2O N2ON2O Describe what is happening Rxn occurs on a hot Pt surface, when surface completely covered w/ N 2 O molecules, an increase of [N 2 O] has no effect on rate, since only N 2 O molecules on the surface are reacting. Therefore, the rate is constant because rsn is controlled by what happens on Pt surface rather than total [N 2 O].

23. time,s [N 2 O] rate = k[N 2 O] 0

24. Determine: UNITS HALF-LIFE 0 ORDER 1 ST ORDER 2 ND ORDER

25. Summary for reaction orders 0, 1, 2, & n Zero-Order First-Order Second-Order nth-Order Rate Law Integrated Rate Law Units of Rate Constant (k) Linear Plot to determine k y-intercept Half-life [A] = [A] O - kt [A] = [A] O e -kt ln[A] = ln[A] O - kt [A] vs t -k ln[A] vs t -k [A] O ln[A] O

26. NOTES Rate Rxn - describe rate rxn must determine concen of react/pdt at various times as rxn proceeds - devising methods is challenge for chemists -spectroscopic method: if 1 subst colored measure inc/dec in intensity of color 4 Factors: help control rates

27. Comparing the 2 experiments, [B] is  ed by factor of: 1 1.0 * 10 -2 M 1.0 * 10 -2 M 1.5 * 10 -6 M. s -1 2 1.0 * 10 -2 M 2.0 * 10 -2 M 3.0 * 10 -6 M. s -1 3 2.0 * 10 -2 M 1.0 * 10 -2 M 6.0 * 10 -6 M. s -1 M ETHOD OF INITIAL RATES describing same rxn in each experiment, same rate law, form: rate = k[A] x [B] y Notice, [A] O same in #1 & #2, what would affect the rxn rate? deduce rate law from experimental rate data Experiment [A] O [B] O initial rate  es in rxn rate due to diff initial concen of B

28. rate  es by factor of: What order is rxn order in [B]? rate ratio = ([B]) y Exponent y deduced from: 2.0 = (2.0) y solving, y = 1 rate = k[A] x [B] 1 Experiments 1 & 3 show [B] O same but [A] O different [A] is  ed by factor of: rate  es by factor of:

29. What order is rxn order in [A]? rate ratio = ([A]) x Exponent x deduced from: 4.0 = (2.0) x solving, x = 2 rate = k[A] 2 [B] 1 Rate constant, k, substitute data from any set of 3 sets into rate-law expression or, rate = 1.5 M -2. s -1 [A] 2 [B] rate 1 = k[A] 1 2 [B] 1 1