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Mechanisms, Catalysts Intermediates and k

Mechanisms, Catalysts Intermediates and k. Tying up the loose ends in chemical kinetics. Reaction mechanisms. Here is a sample reaction mechanism Step 1 ClO - + H 2 O  HOCl + OH - Step 2 Br - + HOCl  HOBr + Cl - Step 3 OH - + HOBr  H 2 O + BrO -

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Mechanisms, Catalysts Intermediates and k

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  1. Mechanisms, CatalystsIntermediates and k Tying up the loose ends in chemical kinetics

  2. Reaction mechanisms Here is a sample reaction mechanism Step 1 ClO- + H2O  HOCl + OH- Step 2 Br- + HOCl HOBr + Cl- Step 3 OH- + HOBr  H2O + BrO- There are several questions they will ask you about these reaction mechanisms • What is the overall reaction • Identify “intermediates” and “catalysts”

  3. Reaction mechanisms The overall reaction Step 1 ClO- + H2O  HOCl + OH- Step 2 Br- + HOCl HOBr + Cl- Step 3 OH- + HOBr  H2O + 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 1 ClO- + H2O  HOCl + OH- Step 2 Br- + HOCl HOBr + Cl- Step 3 OH- + HOBr  H2O + 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 1 ClO- + H2O  HOCl + OH- Step 2 Br- + HOCl HOBr + Cl- Step 3 OH- + HOBr  H2O + 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 1 ClO- + H2O  HOCl + OH- Step 2 Br- + HOCl HOBr + Cl- Step 3 OH- + HOBr  H2O + 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 H2O a catalyst.

  7. Reaction mechanisms “Intermediates” and “catalysts” Step 1 H2O2 + I- OI- +H2O Step 2 H2O2 + OI-  I- + H2O + O2 • Write the overall reaction • 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 (Ea). • 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. • Add a catalyst • 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).

  11. 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.

  12. 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.

  13. 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.

  14. 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

  15. Order of reaction from mechanism • This is really not hard at all. I’ll show you. Mechanism Step 1 H2O2 + I- HOI + OH- (slow) Step 2 HOI + I-  I2 + OH- (fast) Step 3 2OH- + H3O+  4H2O (fast) • First, write the overall reaction H2O2 + 2I- + 2H3O+  I2 + 4H2O

  16. Order of reaction from mechanism • This is really not hard at all. I’ll show you. Mechanism Step 1 H2O2 + I- HOI + OH- (slow) Step 2 HOI + I-  I2 + OH- (fast) Step 3 2OH- + H3O+  4H2O (fast) • Next, draw a line under the slow step H2O2 + 2I- + 2H3O+  I2 + 4H2O

  17. Order of reaction from mechanism • This is really not hard at all. I’ll show you. Mechanism Step 1 H2O2 + I- HOI + OH- (slow) Step 2 HOI + I-  I2 + OH- (fast) Step 3 2OH- + H3O+  4H2O (fast) • Next, draw a line under the slow step H2O2 + 2I- + 2H3O+  I2 + 4H2O

  18. Order of reaction from mechanism • This is really not hard at all. I’ll show you. Mechanism Step 1 H2O2 + I- HOI + OH- (slow) Step 2 HOI + I-  I2 + OH- (fast) Step 3 2OH- + H3O+  4H2O (fast) • Last, add up all the reactants up to the slow step only H2O2 + I- • This makes: Rate = k [H2O2][I-]

  19. Order of reaction from mechanism • What if the location of the slow step changes Mechanism Step 1 H2O2 + I- HOI + OH- (fast) Step 2 HOI + I-  I2 + OH- (slow) Step 3 2OH- + H3O+  4H2O (fast) • First, write the overall reaction. H2O2 + 2I- + 2H3O+  I2 + 4H2O

  20. Order of reaction from mechanism • This is really not hard at all. I’ll show you. Mechanism Step 1 H2O2 + I- HOI + OH- (fast) Step 2 HOI + I-  I2 + OH- (slow) Step 3 2OH- + H3O+  4H2O (fast) • Next, draw a line under the slow step H2O2 + 2I- + 2H3O+  I2 + 4H2O

  21. Order of reaction from mechanism • This is really not hard at all. I’ll show you. Mechanism Step 1 H2O2 + I- HOI + OH- (fast) Step 2 HOI + I-  I2 + OH- (slow) Step 3 2OH- + H3O+  4H2O (fast) • Next, draw a line under the slow step H2O2 + 2I- + 2H3O+  I2 + 4H2O

  22. Order of reaction from mechanism • This is really not hard at all. I’ll show you. Mechanism Step 1 H2O2 + I- HOI + OH- (slow) Step 2 HOI + I-  I2 + OH- (fast) Step 3 2OH- + H3O+  4H2O (fast) • Last, add up all the reactants up to the slow step only H2O2 + 2I- • This makes: Rate = k [H2O2][I-]2 • The coefficients become the exponents

  23. You work it • What if the location of the slow step changes Mechanism Step 1 H2O2 + I- HOI + OH- (fast) Step 2 HOI + I-  I2 + OH- (fast) Step 3 2OH- + H3O+  4H2O (slow) Limitless practice, just click the circle below

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