Reaction Mechanisms Consider the following reaction: 4 HBr (g) +O 2(g) →2H 2 O (g) +2Br 2(g) There are five reactants… Typically more than 3 molecules.

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Presentation transcript:

Reaction Mechanisms

Consider the following reaction: 4 HBr (g) +O 2(g) →2H 2 O (g) +2Br 2(g) There are five reactants… Typically more than 3 molecules colliding at any one time is very improbable. So reaction having 5 reactants in the bal. equation must occur in more than one step. Even a reaction containing 2 or 3 molecules can occur in more than one step. The actual sequence of steps which make up an overall reaction is called a reaction mechanism!

Mechanism 1. HBr (g) +O 2(g) → HOOBr (g) slow HBr (g) + HOOBr (g) → 2HOBr (g) fast 2HOBr (g) +2HBr (g) →2H 2 O (g) +2Br 2(g) fast PE Reaction Path

Mechanism 1 Cancel out identical formulas to get the overall equation. HBr (g) +O 2(g) → HOOBr (g) slow HBr (g) + HOOBr (g) → 2HOBr (g) fast 2HOBr (g) +2HBr (g) →2H 2 O (g) +2Br 2(g) fast PE Reaction Path

Mechanism 1 Cancel out identical formulas to get the overall equation. HBr (g) +O 2(g) → HOOBr (g) slow HBr (g) + HOOBr (g) → 2HOBr (g) fast 2HOBr (g) +2HBr (g) →2H 2 O (g) +2Br 2(g) fast PE Reaction Path

Mechanism 1 Cancel out identical formulas to get the overall equation. HBr (g) +O 2(g) → HOOBr (g) slow HBr (g) + HOOBr (g) → 2HOBr (g) fast 2HOBr (g) +2HBr (g) →2H 2 O (g) +2Br 2(g) fast PE Reaction Path

Mechanism 1 Intermediates are produced and then consumed and cross out from right (first) and left (second). HBr (g) +O 2(g) → HOOBr (g) slow HBr (g) + HOOBr (g) → 2HOBr (g) fast 2HOBr (g) +2HBr (g) →2H 2 O (g) +2Br 2(g) fast PE Reaction Path

Mechanism 1 What is left is the overall reaction. HBr (g) +O 2(g) → HOOBr (g) slow HBr (g) + HOOBr (g) → 2HOBr (g) fast 2HOBr (g) +2HBr (g) →2H 2 O (g) +2Br 2(g) fast PE Reaction Path

Mechanism 1 What is left is the overall reaction. HBr (g) +O 2(g) → HOOBr (g) slow HBr (g) + HOOBr (g) → 2HOBr (g) fast 2HOBr (g) +2HBr (g) →2H 2 O (g) +2Br 2(g) fast 4HBr (g) +O 2(g) →2H 2 O (g) +2Br 2(g) PE Reaction Path

Mechanism 1 Step 1, which is the slow step, is called the rate determining step and has the highest activation energy. HBr (g) +O 2(g) → HOOBr (g) slow HBr (g) + HOOBr (g) → 2HOBr (g) fast 2HOBr (g) +2HBr (g) →2H 2 O (g) +2Br 2(g) fast 4HBr (g) +O 2(g) →2H 2 O (g) +2Br 2(g) PE Reaction Path

Mechanism 1 HBr (g) +O 2(g) → HOOBr (g) slow HBr (g) + HOOBr (g) → 2HOBr (g) fast 2HOBr (g) +2HBr (g) →2H 2 O (g) +2Br 2(g) fast 4HBr (g) +O 2(g) →2H 2 O (g) +2Br 2(g) A potential energy diagram for this reaction might look like this. PE Reaction Path

Mechanism 1 HBr (g) +O 2(g) → HOOBr (g) slow HBr (g) + HOOBr (g) → 2HOBr (g) fast 2HOBr (g) +2HBr (g) →2H 2 O (g) +2Br 2(g) fast 4HBr (g) +O 2(g) →2H 2 O (g) +2Br 2(g) A potential energy diagram for this reaction might look like this. PE Reaction Path

Mechanism 1 HBr (g) +O 2(g) → HOOBr (g) slow HBr (g) + HOOBr (g) → 2HOBr (g) fast 2HOBr (g) +2HBr (g) →2H 2 O (g) +2Br 2(g) fast 4HBr (g) +O 2(g) →2H 2 O (g) +2Br 2(g) highest A potential energy diagram for this Mechanism has three humps The slow step has the highest Ea PE Reaction Path Ea Ea(rev)

The Rate Determining Step The slowest step in the reaction mechanism is called the rate- determining step. It has the highest Ea. To increase the rate, you must increase the rate of this step. Increasing the rate of a fast step will not increase the rate of the overall reaction.

Identifying a Catalyst in a Mechanism How to find a catalyst in a reaction mechanism?? A catalyst is used in one step; it speeds up the reaction, and then is regenerated in a later step. Therefore, a catalyst will cancel out but will be first found on the left side (with reactants) in an earlier step then on the right side.

Identifying a Catalyst in a Mechanism A catalyst is not consumed in the reaction. It is used in one step; it speeds up the reaction, and then is regenerated in a later step. A catalyst will cancel out but will be on the left side in an earlier step then on the right side. C→ →C

Identifying a Catalyst in a Mechanism A catalyst is not consumed in the reaction. It is used in one step; it speeds up the reaction, and then is regenerated in a later step. A catalyst will cancel out but will be on the left side in an earlier step then on the right side. C→ →C

Identifying a Catalyst in a Mechanism A catalyst is not consumed in the reaction. It is used in one step; it speeds up the reaction, and then is regenerated in a later step. A catalyst will cancel out but will be on the left side in an earlier step then on the right side. C→ →C An intermediate is produced and then consumed. It will be on the right side in an earlier step then on the left side.

Identifying a Catalyst in a Mechanism A catalyst is not consumed in the reaction. It is used in one step; it speeds up the reaction, and then is regenerated in a later step. A catalyst will cancel out but will be on the left side in an earlier step then on the right side. C→ →C An intermediate is produced and then consumed. It will be on the right side in an earlier step then on the left side. →I I→

Identifying a Catalyst in a Mechanism A catalyst is not consumed in the reaction. It is used in one step; it speeds up the reaction, and then is regenerated in a later step. A catalyst will cancel out but will be on the left side in an earlier step then on the right side. C→ →C An intermediate is produced and then consumed. It will be on the right side in an earlier step then on the left side. →I I→

Mechanism 2 1.A+B→C 2.C+D→CD 3.CD+E→ABE+D Overall Equation: Catalyst: Intermediate:

Mechanism 2 1.A+B→C 2.C+D→CD 3.CD+E→ABE+D Overall Equation: Catalyst: Intermediate:

Mechanism 2 1.A+B→C 2.C+D→CD 3.CD+E→ABE+D Overall Equation: Catalyst: Intermediate: C

Mechanism 2 1.A+B→C 2.C+D→CD 3.CD+E→ABE+D Overall Equation: Catalyst: Intermediate: C

Mechanism 2 1.A+B→C 2.C+D→CD 3.CD+E→ABE+D Overall Equation: Catalyst: Intermediate: CCD

Mechanism 2 1.A+B→C 2.C+D→CD 3.CD+E→ABE+D Overall Equation: Catalyst: Intermediate: CCD

Mechanism 2 1.A+B→C 2.C+D→CD 3.CD+E→ABE+D Overall Equation: Catalyst:D Intermediate: CCD

Mechanism 2 1.A+B→C 2.C+D→CD 3.CD+E→ABE+D Overall Equation:A +B + E →ABE Catalyst:D Intermediate: CCD

Mechanism 3 Step 1Br 2 → 2Br Step 2Br + OCl 2 → BrOCl + Cl Step 3Br + Cl→ BrCl Overall Reaction: Intermediates: PE Reaction Path

Mechanism 3 Step 1Br 2 → 2Br Step 2Br + OCl 2 → BrOCl + Cl Step 3Br + Cl→ BrCl Overall Reaction: Intermediates: PE Reaction Path

Mechanism 3 Step 1Br 2 → 2Br Step 2Br + OCl 2 → BrOCl + Cl Step 3Br + Cl→ BrCl Overall Reaction:Br 2 + OCl 2 → BrOCl + BrCl Intermediates: PE Reaction Path

Mechanism 3 Step 1Br 2 → 2Br Step 2Br + OCl 2 → BrOCl + Cl Step 3Br + Cl→ BrCl Overall Reaction:Br 2 + OCl 2 → BrOCl + BrCl Intermediates:BrCl PE Reaction Path

Mechanism 3 Step 1Br 2 → 2Br Step 2Br + OCl 2 → BrOCl + Cl Step 3Br + Cl→ BrCl Overall Reaction:Br 2 + OCl 2 → BrOCl + BrCl Intermediates:BrCl ΔH= Ea (forward) = Ea (reverse) = The enthalpy of Br The enthalpy of BrCl PE Reaction Path

Mechanism 3 Step 1Br 2 → 2Br Step 2Br + OCl 2 → BrOCl + Cl Step 3Br + Cl→ BrCl Overall Reaction:Br 2 + OCl 2 → BrOCl + BrCl Intermediates:BrCl ΔH=200 kJ Ea (forward) = Ea (reverse) = The enthalpy of Br The enthalpy of BrCl PE Reaction Path

Mechanism 3 Step 1Br 2 → 2Br Step 2Br + OCl 2 → BrOCl + Cl Step 3Br + Cl→ BrCl Overall Reaction:Br 2 + OCl 2 → BrOCl + BrCl Intermediates:BrCl ΔH=200 kJ Ea (forward) = Ea (reverse) = The enthalpy of Br The enthalpy of BrCl PE Reaction Path

Mechanism 3 Step 1Br 2 → 2Br Step 2Br + OCl 2 → BrOCl + Cl Step 3Br + Cl→ BrCl Overall Reaction:Br 2 + OCl 2 → BrOCl + BrCl Intermediates:BrCl ΔH=200 kJ Ea (forward) =600 kJ Ea (reverse) = The enthalpy of Br The enthalpy of BrCl PE Reaction Path

Mechanism 3 Step 1Br 2 → 2Br Step 2Br + OCl 2 → BrOCl + Cl Step 3Br + Cl→ BrCl Overall Reaction:Br 2 + OCl 2 → BrOCl + BrCl Intermediates:BrCl ΔH=200 kJ Ea (forward) =600 kJ Ea (reverse) =400 kJ The enthalpy of Br The enthalpy of BrCl PE Reaction Path

Mechanism 3 Step 1Br 2 → 2Br Step 2Br + OCl 2 → BrOCl + Cl Step 3Br + Cl→ BrCl Overall Reaction:Br 2 + OCl 2 → BrOCl + BrCl Intermediates:BrCl ΔH=200 kJ Ea (forward) =600 kJ Ea (reverse) =400 kJ The enthalpy of Br300 kJ The enthalpy of BrCl PE Reaction Path

Mechanism 3 Step 1Br 2 → 2Br Step 2Br + OCl 2 → BrOCl + Cl Step 3Br + Cl→ BrCl Overall Reaction:Br 2 + OCl 2 → BrOCl + BrCl Intermediates:BrCl ΔH=200 kJ Ea (forward) =600 kJ Ea (reverse) =400 kJ The enthalpy of Br300 kJ The enthalpy of BrCl 400 kJ PE Reaction Path

Mechanism 4 The following mechanism shows how chlorine-containing fluorocarbons destroy the ozone layer. Ozone is O 3 and O 1.CFCl 3 →CFCl 2 +Cl 2.Cl+O 3 →ClO+O 2 3.ClO+O→Cl+O 2 Overall Reaction: Intermediates:

Mechanism 4 The following mechanism shows how chlorine-containing fluorocarbons destroy the ozone layer. Ozone is O 3 and O 1.CFCl 3 →CFCl 2 +Cl 2.Cl+O 3 →ClO+O 2 3.ClO+O→Cl+O 2 Overall Reaction: Intermediates:

Mechanism 4 The following mechanism shows how chlorine-containing fluorocarbons destroy the ozone layer. Ozone is O 3 and O 1.CFCl 3 →CFCl 2 +Cl 2.Cl+O 3 →ClO+O 2 3.ClO+O→Cl+O 2 Overall Reaction:CFCl 3 + O 3 + O → CFCl 2 +2O 2 + Cl Intermediates:

Mechanism 4 The following mechanism shows how chlorine-containing fluorocarbons destroy the ozone layer. Ozone is O 3 and O 1.CFCl 3 →CFCl 2 +Cl 2.Cl+O 3 →ClO+O 2 3.ClO+O→Cl+O 2 Overall Reaction:CFCl 3 + O 3 + O → CFCl 2 +2O 2 + Cl Intermediates:ClClO Formula of activated complex in step 2:

Determine the Missing Step of the Mechanism Mechanism 5 A+B→C C+D→B+AD Overall:A+D+E→ADE

Determine the Missing Step of the Mechanism Mechanism 5 A+B→C C+D→B+AD Overall:A+D+E→ADE

Determine the Missing Step of the Mechanism Mechanism 5 A+B→C C+D→B+AD Overall:A+D+E→ADE Put in step 3 so that you get the overall reaction

Determine the Missing Step of the Mechanism Mechanism 5 A+B→C C+D→B+AD AD Overall:A+D+E→ADE Put in AD to cancel

Determine the Missing Step of the Mechanism Mechanism 5 A+B→C C+D→B+AD AD+→ Overall:A+D+E→ADE Put in AD to cancel

Determine the Missing Step of the Mechanism Mechanism 5 A+B→C C+D→B+AD AD+E→ Overall:A+D+E→ADE Put in E

Determine the Missing Step of the Mechanism Mechanism 5 A+B→C C+D→B+AD AD+E→ADE Overall:A+D+E→ADE Put in ADE

Determine the Missing Step of the Mechanism Mechanism 6 C+D→CD CD+E→ABE+ D Overall:A+B+E→ABE

Determine the Missing Step of the Mechanism Mechanism 6 C+D→CD CD+E→ABE+ D Overall:A+B+E→ABE

Determine the Missing Step of the Mechanism Mechanism 6 C+D→CD CD+E→ABE+ D Overall:A+B+E→ABE

Determine the Missing Step of the Mechanism Mechanism 6 C+D→CD CD+E→ABE+ D Overall:A+B+E→ABE

Determine the Missing Step of the Mechanism Mechanism 6 →C C+D→CD CD+E→ABE+ D Overall:A+B+E→ABE

Determine the Missing Step of the Mechanism Mechanism 6 →C C+D→CD CD+E→ABE+ D Overall:A+B+E→ABE

Determine the Missing Step of the Mechanism Mechanism 6 A+B→C C+D→CD CD+E→ABE+ D Overall:A+B+E→ABE

Determine the Missing Step of the Mechanism Mechanism 7 CFCl 3 →CFCl 2 +Cl ClO+O→Cl+ O 2 Overall: O + CFCl 3 + O 3 →CFCl 2 + 2O 2 +Cl

Determine the Missing Step of the Mechanism Mechanism 7 CFCl 3 →CFCl 2 +Cl ClO+O→Cl+ O 2 Overall: O + CFCl 3 + O 3 →CFCl 2 + 2O 2 +Cl

Determine the Missing Step of the Mechanism Mechanism 7 CFCl 3 →CFCl 2 +Cl Cl ClO+O→Cl+ O 2 Overall: O + CFCl 3 + O 3 →CFCl 2 + 2O 2 +Cl

Determine the Missing Step of the Mechanism Mechanism 7 CFCl 3 →CFCl 2 +Cl Cl ClO+O→Cl+ O 2 Overall: O + CFCl 3 + O 3 →CFCl 2 + 2O 2 +Cl

Determine the Missing Step of the Mechanism Mechanism 7 CFCl 3 →CFCl 2 +Cl Cl + O 3 ClO+O→Cl+ O 2 Overall: O + CFCl 3 + O 3 →CFCl 2 + 2O 2 +Cl

Determine the Missing Step of the Mechanism Mechanism 7 CFCl 3 →CFCl 2 +Cl Cl + O 3 →ClO+ O 2 ClO+O→Cl+ O 2 Overall: O + CFCl 3 + O 3 →CFCl 2 + 2O 2 +Cl

Determine the Missing Step of the Mechanism Mechanism 7 CFCl 3 →CFCl 2 +Cl Cl + O 3 →ClO+ O 2 ClO+O→Cl+ O 2 Overall: O + CFCl 3 + O 3 →CFCl 2 + 2O 2 +Cl

Determine the Missing Step of the Mechanism Mechanism 7 CFCl 3 →CFCl 2 +Cl Cl + O 3 →ClO+ O 2 ClO+O→Cl+ O 2 Overall: O + CFCl 3 + O 3 →CFCl 2 + 2O 2 +Cl

Determine the Missing Step of the Mechanism Mechanism 7 CFCl 3 →CFCl 2 +Cl Cl + O 3 →ClO+ O 2 ClO+O→Cl+ O 2 Overall: O + CFCl 3 + O 3 →CFCl 2 + 2O 2 +Cl

Determine the Missing Step of the Mechanism Mechanism 7 CFCl 3 →CFCl 2 +Cl Cl + O 3 →ClO+ O 2 ClO+O→Cl+ O 2 Overall: O + CFCl 3 + O 3 →CFCl 2 + 2O 2 +Cl

Reaction Intermediate Notes: An intermediate doesn’t accumulate (like a product) because as soon as it is formed, it gets used up again. Intermediates are not necessarily unstable (in other circumstances, they may last a while). An activated complex is very unstable and short-lived. It doesn’t usually obey bonding “rules.”

HOMEWORK Hebden Textbook Page 28 Questions #47-53

Reaction Mechanism Review Questions

Provincial Exam Questions

PE Diagram for a Reaction Mechanism AC (Step 1) AC (Step 2) AC (Step 3) HOOBr PE HBr + O 2 HOBr H 2 O + Br 2 Reaction Proceeds Label this diagram: RDS, Ea (Overall Rxn), Ea (Step 2), ∆H.

PE Diagram for a Reaction Mechanism Notes: each “hump” is a step the higher the hump (greater Ea), the slower the step the highest hump (greatest Ea) is for the RDS AC’s are found at top of humps, intermediates in middle “valleys”, products in the final “valley” the Ea for the forward overall rxn is vertical distance from reactants to top of highest hump

PE Diagram for a Reaction Mechanism The Ea for the overall forward reaction is the difference in energy between the reactants and the top of the highest peak.

PE Diagram for a Reaction Mechanism Example: Given the following Potential Energy Diagram for a reaction mechanism: PE Reaction Proceeds

PE Diagram for a Reaction Mechanism 1. This mechanism has steps 2.Ea for overall rxn = ______kJ 3.Ea (reverse rxn) = kJ 4. The overall rxn is thermic 5.  H = kJ 6.  H for reverse rxn = kJ 7.Ea (step1)= 8. RDS for forward rxn is step ________

PE Diagram for a Reaction Mechanism If asked for Ea for a particular step, you use: Ea= PE (activated complex for that step)-PE (reactants for THAT PARTICULAR step) If asked for Ea for overall reaction, use: Ea= PE (highest energy activated complex)-PE (original reactantsd) Rate determining step is simply the one with the highest numerical value of Ea you find.

Catalysts Catalyst: an introduced substance which produces an alternate mechanism with a lower activation energy.

Provincial Exam Questions

HOMEWORK Hebden Textbook Page 30 Questions Page34 Exercises 56-61

WE ARE DONE UNIT 1!