1. Tying up the loose ends in chemical kinetics

2. Reaction mechanisms Here is a sample reaction mechanism Step 1ClO - + H 2 O  HOCl + OH - Step 2Br - + HOCl  HOBr + Cl - Step 3OH - + HOBr  H 2 O + BrO - There are several questions they will ask you about these reaction mechanisms 1. What is the overall reaction 2. Identify “intermediates” and “catalysts”

3. Reaction mechanisms The overall reaction Step 1ClO - + H 2 O  HOCl + OH - Step 2Br - + HOCl  HOBr + Cl - Step 3OH - + HOBr  H 2 O + BrO - This is easy, just add up the reactions like you did when you were working on Hess’s law. Cross out all the chemicals that appear on both sides and then add up.

4. Reaction mechanisms The overall reaction Step 1ClO - + H 2 O  HOCl + OH - Step 2Br - + HOCl  HOBr + Cl - Step 3OH - + HOBr  H 2 O + BrO - ClO- + Br-  Cl- + BrO- This is the overall reaction. Not so bad. You don’t even need to flip any equations, or multiply the coefficients. All you do is cancel and add.

5. Reaction mechanisms “Intermediates” and “catalysts” Step 1ClO - + H 2 O  HOCl + OH - Step 2Br - + HOCl  HOBr + Cl - Step 3OH - + HOBr  H 2 O + BrO - ClO- + Br-  Cl- + BrO- The “intermediates” are crossed out in black. They are all the chemicals that cancel out that are not present in on the reactant side of the first step or the product side of the last step.

6. Reaction mechanisms “Intermediates” and “catalysts” Step 1ClO - + H 2 O  HOCl + OH - Step 2Br - + HOCl  HOBr + Cl - Step 3OH - + HOBr  H 2 O + BrO - ClO- + Br-  Cl- + BrO- In the reaction above H2O is a “catalyst”. Water is present on the reactant side of the first step and on the product side of the last step. This makes H 2 O a catalyst.

7. Reaction mechanisms “Intermediates” and “catalysts” Step 1H 2 O 2 + I -  OI - +H 2 O Step 2H 2 O 2 + OI -  I - + H 2 O + O 2 1. Write the overall reaction 2. Indentify both the catalyst and the intermediate.

8. Free energy diagram As we discussed earlier in thermodynamics, not all spontaneous reaction occur at observable rates. A reaction that occurs slowly is likely to have a high activation energy (E a ). What is this called again? Is it positive or negative?

9. Free energy diagram What is the effect of a catalyst? A catalyst lowers the activation energy! In this class (i.e. for the AP exam) there are two ways to increase the rate of a reaction. 1. Add a catalyst 2. Increase the temperature.

10. How to make it faster If the temperature of a non-reversible reaction is increased then the reaction rate will increase. The higher temperature increases the average kinetic energy of the molecules. Therefore more molecules have sufficient energy to overcome the activation energy (transition state energy).

12. Catalyst A catalyst lowers the activation energy! If the activation energy is lowered then less energy is required to go from reactants to products. Therefore the reaction will proceed more quickly.

13. What you will be asked They will try to trick you!!! (the AP people that is) Here are the tricks 1) The will get you to say ΔG is negative and that the reaction is spontaneous. They will then ask you to comment on the speed of the reaction. ANSWER: Rate of reaction is only affected by temperature and activation energy. Therefore the rate of the reaction is completely unaffected by any thermodynamic data you could possibly present me.

14. What you will be asked Here’s trick number 2 2) You will be told that concentration, pressure, or volume has been change or that an inert extra chemical has been added to your system. ANSWER: Rate of reaction is only affected by temperature and activation energy. Therefore the rate of the reaction is completely unaffected by anything else.

15. What you will be asked Here’s trick number 3 3) You will be told that anything other than a temperature change or the addition of a catalyst has occured. ANSWER: Rate of reaction is only affected by temperature and activation energy. Therefore the rate of the reaction is completely unaffected. If temperature goes ↑ then k goes ↑, therefore rate ↑. If temperature goes down then k goes down, therefore rate decrease. If a catalyst is added then activation energy goes down, therefore the rate of reaction increases

16. Order of reaction from mechanism This is really not hard at all. I’ll show you. Mechanism Step 1H 2 O 2 + I -  HOI + OH - (slow) Step 2HOI+ I -  I 2 + OH - (fast) Step 32OH - + H 3 O +  4H 2 O(fast) First, write the overall reaction H 2 O 2 + 2I - + 2H 3 O +  I 2 + 4H 2 O

17. Order of reaction from mechanism This is really not hard at all. I’ll show you. Mechanism Step 1H 2 O 2 + I -  HOI + OH - (slow) Step 2HOI+ I -  I 2 + OH - (fast) Step 32OH - + H 3 O +  4H 2 O(fast) Next, draw a line under the slow step H 2 O 2 + 2I - + 2H 3 O +  I 2 + 4H 2 O

18. Order of reaction from mechanism This is really not hard at all. I’ll show you. Mechanism Step 1H 2 O 2 + I -  HOI + OH - (slow) Step 2HOI+ I -  I 2 + OH - (fast) Step 32OH - + H 3 O +  4H 2 O(fast) Next, draw a line under the slow step H 2 O 2 + 2I - + 2H 3 O +  I 2 + 4H 2 O

19. Order of reaction from mechanism This is really not hard at all. I’ll show you. Mechanism Step 1H 2 O 2 + I -  HOI + OH - (slow) Step 2HOI+ I -  I 2 + OH - (fast) Step 32OH - + H 3 O +  4H 2 O(fast) Last, add up all the reactants up to the slow step only H 2 O 2 + I - This makes: Rate = k [H 2 O 2 ][I - ]

20. Order of reaction from mechanism What if the location of the slow step changes Mechanism Step 1H 2 O 2 + I -  HOI + OH - (fast) Step 2HOI+ I -  I 2 + OH - (slow) Step 32OH - + H 3 O +  4H 2 O(fast) First, write the overall reaction. H 2 O 2 + 2I - + 2H 3 O +  I 2 + 4H 2 O

21. Order of reaction from mechanism This is really not hard at all. I’ll show you. Mechanism Step 1H 2 O 2 + I -  HOI + OH - (fast) Step 2HOI+ I -  I 2 + OH - (slow) Step 32OH - + H 3 O +  4H 2 O(fast) Next, draw a line under the slow step H 2 O 2 + 2I - + 2H 3 O +  I 2 + 4H 2 O

22. Order of reaction from mechanism This is really not hard at all. I’ll show you. Mechanism Step 1H 2 O 2 + I -  HOI + OH - (fast) Step 2HOI+ I -  I 2 + OH - (slow) Step 32OH - + H 3 O +  4H 2 O(fast) Next, draw a line under the slow step H 2 O 2 + 2I - + 2H 3 O +  I 2 + 4H 2 O

23. Order of reaction from mechanism This is really not hard at all. I’ll show you. Mechanism Step 1H 2 O 2 + I -  HOI + OH - (slow) Step 2HOI+ I -  I 2 + OH - (fast) Step 32OH - + H 3 O +  4H 2 O(fast) Last, add up all the reactants up to the slow step only H 2 O 2 + 2I - This makes: Rate = k [H 2 O 2 ][I - ] 2 The coefficients become the exponents

24. You work it What if the location of the slow step changes Mechanism Step 1H 2 O 2 + I -  HOI + OH - (fast) Step 2HOI+ I -  I 2 + OH - (fast) Step 32OH - + H 3 O +  4H 2 O(slow) Limitless practice, just click the circle below