Unit 5 How do we predict chemical change?

1. Comparing the relative stability of different substances M1. Analyzing Structure Determining the directionality and extent of a chemical reaction. M2. Comparing Free Energies Analyzing the factors that affect reaction rate. M3. Measuring Rates Identifying the steps that determine reaction rates. M4. Understanding Mechanism Unit 5How do we predict chemical change? The central goal of this unit is to help you identify and apply the different factors that help predict the likelihood of chemical reactions.

2. Unit 5 How do we predict chemical change? Module 4: Understanding Mechanism Central goal: To use reaction mechanisms to make predictions about reaction rate and vice versa.

3. TransformationHow do I change it? The Challenge Imagine that you were interested in understanding why certain types of substances and processes appeared in our planet. How can we use reaction mechanisms to make predictions? How can we deduce reaction mechanisms based on reaction outcomes?

4. CO2(g) + 2H2O(g) Most reaction mechanisms involve several steps. However, some steps play a more central role than others in determining the overall rate of reaction. CH4(g) + O2(g) Reaction Pathways Our ability to predict the most likely outcomes of a chemical reaction improves considerably when we understand the mechanism that leads from reactants to possible products.

5. A + A  B + C Bimolecular Unimolecular B  C + E 2A 2C + E B is an intermediate Reaction Steps In the mechanistic model, the overall reaction is viewed as the result of multiple elementary reactions or steps occurring simultaneously in the system. For example, the overall reaction: 2A 2C + E may involve two elementary steps: Detecting intermediates is an important means of investigating reaction mechanisms.

6. A + A  B + C Slow Fast B  C + E 2A 2C + E Overall Rate Order In the mechanistic model, the overall rate of the reaction is an “emergent property” of the rates of the individual steps. For many reactions, one step is slow enough to limit the rate of the overall reaction: Overall Rate = k [A][A] = k [A]2

7. Mirror Images Consider the following problem of central relevance for our understanding of the origin of life: L D Most amino acids found on Earth appear in only one of two possible mirror-image forms, called enantiomers or optical isomers. Non-superimposable These isomers have most of the same properties, but react differently with L and D isomers of other “chiral” substances.

8. Identify the chiral carbons in this molecule: Let′s think! L-Glucose Chiral Centers Molecular chirality is commonly caused by the presence of carbon atoms in a molecule attached to four different groups: Chiral carbon Non-Chiral carbons

9. Chirality is of central importance for many biological functions. Different enantiomers interact differently with the chiral molecules (proteins, DNA) in our body. Left or Right Handed Many biologically active molecules are chiral, including the naturally occurring amino acids, which tend to be “left-handed” (L). How this preferred chirality emerged on our planet?

10. H+ Mechanism NH2 1. Unimolecular Step DG forward backward 2. Bimolecular Step C H2N NH2 H3C COOH H3C C H H C CH3 HOOC COOH DGorxn= 0 L D Ea= 124 kJ/mol Reaction Coordinate Racemization Pure samples of L or D amino acids eventually convert into a mixture of both forms. This process is called racemization. How do we explain it?

11. Mechanism The racemization process can be explained by this two-step mechanism: L C- + H+ Unimolecular (Slow) Rate = k[L] H+ + C-  D Bimolecular (Fast) Rate = k[C-][H+] where H+ and C- are intermediates. The rate of the overall reaction is determined by the slowest step (Rate Determining Step), thus: L DRate = k [L] (First Order Reaction) The same ideas apply to the backward reaction D L.

12. Follow the evolution of the system. When does the process “stop”? t (s) L-Ala (M) D-Ala (M) 0 1 0 1 2 3 Let’s Think According to this mechanism, the interconversion between the L and D forms should occur with the same probability. Imagine that you start with 1 M solution of L-Ala and 30% of it transforms to D-Ala every second. The same percentage of D-Ala in the system transforms to L-Ala in that time.

13. t (s) L-Ala (M) D-Ala (M) 0 1 0 LD 0.3 DL  0 1 0.7 0.3 LD 0.7 x 0.3 = 0.21 DL  0.3 x 0.3 = 0.09 2 0.58 0.42 LD 0.58 x 0.3 = 0.174 DL  0.42 x 0.3 = 0.126 3 0.532 0.468 LD 0.532 x 0.3 = 0.1596 DL 0.468 x 0.3= 0.1404 4 0.5128 0.4872 Equilibrium? Etc. Let’s Think L D40% D  L40%

14. LD Kinetics and Equilibrium Chemical equilibrium is reached when the rate of the forward reactions is equal to the rate of the backward process. L D Rate = kf[L] D L Rate = kb[D] kf[L]eq = kb[D]eq In this case, kf= kb. Thus, Kc= 1. The reaction keeps going at equilibrium but [L] and [D] remain constant.

15. Imagine that once amino acids are linked into proteins, the % of an L amino acid converted to the D form is 20% every second, versus 80% conversion from D to L in the same period of time. What would be [D]/[L] at equilibrium? Let′s think! Amino Acid Chirality The origin of biologically active amino acids’ chirality is an unresolved problem in science. Many hypothesis have been suggested. L Dkf D  Lkb Kc = ¼ kb = 4 kf

16. + 2 2 2 H2N 3 3 NH2 2 H3C C H H C CH3 2 + H2O 2 HOOC COOH 3 3 3 Let’s Think Consider the following data derived through our work in this unit: DecompositionEa = 177 kJ/mol DimerizationEa = 88.7 kJ/mol RacemizationEa = 124 kJ/mol Build a hypothesis about what could have prevented amino acid racemization on primitive Earth.

17. As we know, these “catalysts” act by either reducing the activation energy Ea or changing the reaction mechanism. Changing Mechanism It has been suggested that one crucial step in the origin of life was the synthesis of substances that could speed up the rate of certain chemical reactions.

18. (S) (P) (E) (ES) (E) E + SES Fast ES E + P Slow Mechanism Enzymes The substances that catalyze biological processes are called enzymes. One common model to explain their behavior is the “lock and key” model.

19. If we assume that equilibrium is reached in the first step: E + SES Fast ES E + P Slow Enzyme Kinetics DGo ES E+S The rate law for this process is determined by the second step: Rate = k2 [ES] E + P This expression is not very useful given that we cannot easily follow [ES] as a function of time. Rate = k2 Kc [E][S]

20. Let’s Think For a given concentration of enzyme [E]o, the reaction rate is first order in [S], but only at low concentrations of the substrate. At high concentrations: Rate ~ k [S]0 = k (constant) (Zeroth order process) Use the lock and key model to explain this behavior.

21. E + SES Fast ES EP Slow EPE + P Fast Let’s Think The "lock and key" model has proven inaccurate. The induced fit model is the most currently accepted Draw an energy profile for this mechanism and analyze whether the associated rate law needs to be modified.

22. Induced Fit Model The same rate law: Rate = k2 Kc [E][S]

23. [B] Why this shape? t Auto-Catalysis Processes in which the reaction is catalyzed by its own products are called auto-catalytic and may have played a central role in the origin of life. A + B 2 B Rate = k [A] [B] The presence of auto-catalytic steps in some reaction mechanisms may explain the appearance of metabolic cycles.

24. Write the overall reaction; • Identify the auto-catalytic steps and the intermediate species; • Predict the structure of the plots [X] and [Y] vs. t as the reaction proceeds. [X] t Let’s Think Consider this mechanism: A + X 2 X X + Y  2 Y Y  B Hint: Think of A as grain, X as ducks, Y as wolves, and B as “dead” wolves.

25. Oscillating Reactions A + X 2 X X + Y  2 Y Y  B [X] [Y]

26. Let′s apply! Assess what you know

27. Let′s apply! 2 2 + H2O 2 3 3 3 Analyze Consider the dimerization of amino acids (Aa): which we found to be a second order reactionRate = k [Aa]2 Several possible mechanisms have been proposed for this type of reaction.

28. Let′s apply! Analyze 2Aa Aa-Aa + H2O DGo Is there a way to determine which of these mechanisms is more plausible? Could you propose a different mechanism that leads to the same experimental rate law Rate=k[Aa]2? Aa-Aa+ H2O 2Aa 2Aa Aa-Aa* Aa-Aa*  Aa-Aa + H2O DGo Aa-Aa* Aa-Aa+ H2O 2Aa

29. Let′s apply! Another possibility: 2AaAa-Aa* Fast Aa-Aa*  Aa-Aa + H2O Slow DGo Aa-Aa* Aa-Aa+ H2O 2Aa Analyze Both mechanisms lead to Rate = k[Aa]2. We would have to experimentally confirm the existence of the intermediate. Rate = k2 [Aa-Aa*] Kc = [Aa-Aa*]/[Aa]2 [Aa-Aa*] = Kc [Aa]2 Rate = k2 Kc [Aa]2

30. Discuss with a partner one thing you do not fully understand about the content of this Module.

31. A + B C + D Bimolecular Rate = k1[A][B] C  B + E Unimolecular Rate = k2[C] A  D + E Understanding Mechanism Summary Reaction mechanisms allow us to understand reaction kinetics and make predictions about most likely outcomes. Most chemical processes can be thought as occurring in a sequence of elementary steps: The rate law is determined by the slowest step.

32. Catalysts The reaction mechanism can be altered by the presence of substances that help create alternative reaction paths. Catalysts” act by either reducing the activation energy Ea or changing the reaction mechanism.

33. Are You Ready?

34. The Quest for Ammonia Ammonia, NH3, is one of the most important industrial chemical substances. It is widely used in the production of fertilizers, pharmaceuticals, refrigerants, explosives, and cleaning agents. It ranks as one of the 10 top chemicals substances produced annually in the world.

35. The Synthesis Ammonia is mainly produced via this simple chemical reaction: 1/2 N2(g) + 3/2 H2(g) NH3(g) Let′s think! Compare the energetic and entropic stabilityof reactants and products. Make a prediction of the signs of DHorxn and DSorxnfor this process. Energy: DHorxn < 0 A-A bonds  A-B bonds Entropy: DSorxn < 0 4mol gas 3mol gasmixture  one compound

36. Directionality Use these data to calculate DGorxn. Write Kp for this reaction and calculate its value at25oC. Let′s think! DGrxn= DHrxn–TDSrxn =-45.9 – 298.15*0.0991 = -16.4 kJ

37. Let′s think! Is the synthesis of ammonia thermodynamically favored at low or high temperatures? Estimate the temperature at which the directionality switches (K = 1)? The reaction is favored at low temperatures. K = 1 DGrxn =0 DGrxn= -45.9 –T*0.0991 = 0 T = 463 K

38. Ep 1/2 N23/2 H2 325 kJ NH3 45.9 KJ Reaction Coordinate Energy Profile Use the following information to build the energy profile for the reaction: Let′s think!

39. Reaction Conditions Given its high Ea, the reaction is normally done at high T (~500oC) and P (~200 atm). Discuss how the increased T will affect the: Rate (Calculate k500/k25)Extent (Calculate K500/K25) of the reaction. Let′s think!

40. Reaction Conditions

41. Reaction Conditions High temperature increases reaction rate, but decreases reaction extent. That is why the reaction is carried out at high P too. Discuss why high P favors the product side in this process: N2(g) + 3H2(g) 2NH3(g) Let′s think! The collision rate is higher in the side with more particles. The forward reaction is favored.

42. For example: T = 500oC P = 200 atm Fe catalyst The rate does only depends on the concentration of N2(g). What is the reaction order with respect to N2(g)? What is the value of the rate constant under these conditions? Let′s think! Catalysts and Reaction Order The synthesis of NH3 is carried out in the presence of catalysts. The order of the reaction depends on the composition and structure of this catalyst.

43. Rate = k [N2(g)] ln(C)= -kt + ln(Co) First order with respect to [N2(g)] k = 1.72 s-1 Reaction Order

44. Let′s think! What other mechanistic steps are involved in the synthesis of NH3? Which step can be expected to be the slowest given that Rate = k[N2]? Reaction Mechanism In the presence of a solid catalyst, the reaction takes place on the surface of the solid. N2(g)  N2(ad) N2(ad)  2N(ad) N(ad) +H(ad)  NH(ad)

45. Reaction Mechanism H2(g)   H2(ad) N2(g)  N2(ad) N2(ad)  2N(ad) H2(ad)  2H(ad) N(ad) +H(ad)  NH(ad) NH(ad) + H(ad)  NH2(ad) NH2(ad) + H(ad)  NH3(ad) NH3(ad)   NH3(g) or N2(ad)  2N(ad) Slowest: N2(g)  N2(ad) Rate = k’ [N2(ad)] Kc = [N2(ad)]/[N2(g)] Rate = k [N2(g)] Rate = k’ Kc [N2(g)]

46. Activation Energy The presence of the catalyst reduces Ea considerably. Use the following data to derive Eafor an Fe-based catalyst. How many times faster is the reaction at T = 500oC in the presence of the catalyst? (Ea= 325kJ/mol without it) Let′s think!

47. Activation Energy Ea= 9743R Ea= 81 kJ/mol