The document provides information about an upcoming exam for a general chemistry course. It includes details like the date, time, location, and topics to study. The reading assignment covers reaction mechanisms and includes topics like molecularity, rate laws, elementary steps, and examples of unimolecular and bimolecular reactions.

2. General Chemistry II CHEM 152 Week 4

3. Week 4 Reading Assignment Chapter 13 – Section 13.6 (mechanisms)

5. Isomerization Simulation Mechanism and Order A(blue) -> B(red) An isomerization reaction occurs in a single step (mechanism) . This is a “ unimolecular ” process. What is the order of this reaction?

6. Reaction Mechanisms UNIMOLECULAR : the structure of a single particle rearranges to produce a different particle or particles. Reaction Order? Rate= k[A] 1 st -Order Isomerization

8. Dimerization simulation This is a “bimolecular” process. Mechanism and Order 2 A(blue) -> A 2 (red) What is the order of the reaction?

10. Reaction Mechanisms BIMOLECULAR: Two particles collide and rearrange into products Reaction Order? RATE= k[A][B] 2 nd Order

11. Elementary Step Molecularity Rate Law Unimolecular Bimolecular Bimolecular Termolecular Rate = k [A] Rate = k [A] 2 Rate = k [A][B] Rate = k [A] 2 [B] Rate Laws for General Elementary Steps A product 2A product A + B product 2A + B product

12. PLAN: Determining Molecularity and Rate Laws for Elementary Steps The following two reactions are proposed as elementary steps in the mechanism of an overall reaction: (a) Write the overall balanced equation. (b) Determine the molecularity of each step. (a) The overall equation is the sum of the steps. (b) The molecularity is the sum of the reactant particles in the step. (c) Write the rate law for each step. (1) NO 2 Cl( g ) NO 2 ( g ) + Cl ( g ) (2) NO 2 Cl( g ) + Cl ( g ) NO 2 ( g ) + Cl 2 ( g )

13. SOLUTION: rate 2 = k 2 [NO 2 Cl][Cl] 2NO 2 Cl( g ) 2NO 2 ( g ) + Cl 2 ( g ) (1) NO 2 Cl( g ) NO 2 ( g ) + Cl ( g ) (2) NO 2 Cl( g ) + Cl ( g ) NO 2 ( g ) + Cl 2 ( g ) (a) Step(1) is unimolecular. Step(2) is bimolecular. (b) rate 1 = k 1 [NO 2 Cl] (c)

14. The Rate-Determining Step of a Reaction Mechanism The overall rate of a reaction is related to the rate of the slowest, or rate-determining step. Correlating the Mechanism with the Rate Law The elementary steps must add up to the overall equation. The elementary steps must be physically reasonable. The mechanism must correlated with the rate law.

15. The mechanism for the gas-phase reaction 2NO + Cl 2 -> 2NOCl is suggested to be: (1) NO + NO -> N 2 O 2 (slow) (2) N 2 O 2 + Cl 2 -> 2NOCl (fast) Derive the rate law from this mechanism

16. Consider the following mechanism for the oxidation of bromide ions by hydrogen peroxide in aqueous acid solution. (1) H + + H 2 O 2  H 3 O 2 + (fast) (2) H 3 O 2 + + Br ¯ -> HOBr + H 2 O (slow) (3) HOBr + H + + Br ¯ -> Br 2 + H 2 O (fast) What is the overall reaction equation and rate law? Why do we not want intermediates in the rate law if it can be avoided?

17. What would be a reasonable rate law for a reaction with the following mechanism? (1) NO(g) + Br 2 (g)  NOBr 2 (g) fast (2) NOBr 2 (g) + NO(g)  2NOBr(g) slow What is the overall reaction equation and rate law?

18. NO 2 +F 2  NO 2 F + F F + NO 2  NO 2 F RATE LAW FOR THE REACTION? Reaction energy diagram for the two-step NO 2 -F 2 reaction

19. NO 2 +F 2  NO 2 F + F F + NO 2  NO 2 F 2NO 2 + F 2 -> 2NO 2 F Slow Rate = k[NO 2 ][F 2 ] = Rate overall Reaction energy diagram for the two-step NO 2 -F 2 reaction

20. Consider the reaction O 3 + NO  O 2 + NO 2 where the rate law was found experimentally to be rate = k[O 3 ][NO] Which of the following mechanisms is consistent with the rate law? (1) O 3 + NO  O + NO 3 (slow) O + O 3  2O 2 (fast) NO 3 + NO  2NO 2 (fast) (2) O 3  O 2 + O (fast) NO + O  NO 2 (slow) (3) O 3 + NO  O 2 + NO 2

21. S N 2 Mechanism Kinetics CH 3 I + OH –  CH 3 OH + I – find: Rate = k [CH 3 I][OH – ],  bimolecular Both CH 3 I and OH – involved in rate-limiting step reactivity: R-I > R-Br > R-Cl >> R-F C-X bond breaking involved in rate-limiting step  single-step mechanism: CH 3 I + OH – CH 3 OH + I – [HO---CH 3 ---I] –

22. S N 2 Mechanism

23. S N 2 Mechanism

24. Group Activity: Molecules can be designed so that they undergo this S N 2 faster or slower – If you want the S N 2 reaction to happen – you choose a molecule which can easily undergo the reaction . If you want another reaction to occur other than the S N 2 reaction – you choose a molecule which is harder to have the S N 2 reaction occur. How would you alter the following molecule so that the S N 2 reaction is LESS LIKELY to occur? (same conc./T)

25. Group Activity: Draw 2 reaction coordinate diagrams to express the steric effect difference.

26. S N 2 Mechanism Steric effects Compound Rel. Rate Methyl C H 3 Br 150 1º RX CH 3 C H 2 Br 1 2º RX (CH 3 ) 2 C HBr 0.008 3º RX (CH 3 ) 3 C Br ~0 increasing steric hindrance

27. S N 2 Mechanism Steric effects minimal steric hindrance maximum steric hindrance

28. Summary Activity: An important reaction in the formation of smog is the reaction between ozone and NO NO(g) + O 3 (g)  NO 2 (g) + O 2 (g) Experimentally, the [NO] was monitored and the ln[NO] vs time was fairly linear. In another experiment, the O 3 concentration was tripled – and the rate also went up by a factor of 3. The rate constant for the reaction is 80. L/(mol s) at 25  C. If this reaction were to occur in a single step, would the rate law be consistent with the observed order of the reaction? What is the rate of the reaction at 25  C when [NO] = 3.0 x 10 -6 M and [O 3 ] = 5.0 x 10 -9 M?

29. Is the rate law consistent with a single step?

30. What is the rate of the reaction at 25 ºC when [NO] = 3.0 x 10 -6 M and [O 3 ] = 5.0 x 10 -9 M? Enter scientific notation in the format: #.#E# rate = ___________ M/s