Download presentation
Presentation is loading. Please wait.
1
Kinetics Chapter 14
2
Kinetics Study of the speed at which a chemical reaction occurs
Reaction Rates Also study of how the reaction occurs Reaction Mechanism
3
Review from Regents Collision Theory Reaction Rate
a reaction is more likely to occur if reactant particles collide with proper energy and orientation Reaction Rate How fast the reaction proceeds Activation Energy (EA) Minimum energy that colliding particles must have in order to react
4
Proper Orientation
5
Review from Regents When a bond is formed energy is released
Bonded atoms are more stable together than free atoms Energy must be added or absorbed in order to break a bond It takes energy to pull atoms apart, since they are more stable together
6
Potential Energy Diagrams
Reaction Process Energy EA Energy Reaction Process EA ΔH PEProducts PEReactants ΔH PEProducts PEReactants
7
Potential Energy Diagrams
Transition State Reaction Process Energy An activated complex is the name of the species at the transition state
8
Potential Energy Diagrams
9
Factors Affecting Reaction Rates
Concentration increasing the number of particles in a given volume (concentration) increases the reaction rate Increasing the number of reactant particles increases the frequency of collisions, and therefore effective collisions.
10
Factors Affecting Reaction Rates
Temperature Increasing temperature increases the reaction rate Increasing the temperature increase the frequency of collisions. Increasing the temperature also increases the number of reactant particles with high enough energy to overcome the activation energy barrier.
11
Factors Affecting Reaction Rates
Temperature
12
Factors Affecting Reaction Rates
Surface Area increasing surface area increases reaction rate Increasing the surface area increases the number of reactant particles that can interact and collide.
13
Factors Affecting Reaction Rates
Catalyst Allows the reaction to proceed through an alternate pathway with lower activation energy, increasing the reaction rate Catalysts are not used up during a reaction Present before and after reaction
14
Factors Affecting Reaction Rates
Catalyst Can hold the reactants in order to break bonds (enzymes)
15
Factors Affecting Reaction Rates
Catalyst
16
Homework Read Chapter 14 Sections 1, 5, and 7
Answer Questions: 1, 2, 7, 13 For Fri Finish POGIL Packet Read Chapter 14 Sections 2, 3
18
Reaction Rate Speed at which a reaction proceeds
Measured in molarity per second (M/s) Most reactions slow down over time because the concentration of reactants decreases. Rate changes over time
19
Instantaneous Reaction Rate
20
Reaction Rates 𝐴 ⟶ 𝐵 𝑟𝑎𝑡𝑒= Δ[𝐵] Δ𝑡 =− Δ[𝐴] Δ𝑡
![Reaction Rates 𝐴 ⟶ 𝐵 𝑟𝑎𝑡𝑒= Δ[𝐵] Δ𝑡 =− Δ[𝐴] Δ𝑡](http://slideplayer.com./slide/16470022/96/images/20/Reaction+Rates+%F0%9D%90%B4+%E2%9F%B6+%F0%9D%90%B5+%F0%9D%91%9F%F0%9D%91%8E%F0%9D%91%A1%F0%9D%91%92%3D+%CE%94%5B%F0%9D%90%B5%5D+%CE%94%F0%9D%91%A1+%3D%E2%88%92+%CE%94%5B%F0%9D%90%B4%5D+%CE%94%F0%9D%91%A1.jpg "Reaction Rates 𝐴 ⟶ 𝐵 𝑟𝑎𝑡𝑒= Δ[𝐵] Δ𝑡 =− Δ[𝐴] Δ𝑡")
21
Reaction Rates 𝑎𝐴+ 𝑏𝐵 ⟶𝑐𝐶+ 𝑑𝐷
In General….. 𝑎𝐴+ 𝑏𝐵 ⟶𝑐𝐶+ 𝑑𝐷 𝑟𝑎𝑡𝑒=− 1 𝑎 Δ[𝐴] Δ𝑡 =− 1 𝑏 Δ[𝐵] Δ𝑡 = 1 𝑐 Δ[𝐶] Δ𝑡 = 1 𝑑 Δ[𝐷] Δ𝑡
22
Rate Law 𝐴 ⟶ 𝐵 𝑅𝑎𝑡𝑒 = 𝑘 [𝐴] 𝑛
Equation relating concentration of reactants to the rate without time 𝐴 ⟶ 𝐵 𝑅𝑎𝑡𝑒 = 𝑘 [𝐴] 𝑛 k is the rate constant, a number which has units (temperature dependent) n is the order, usually a whole number (0, 1, 2)
![Rate Law 𝐴 ⟶ 𝐵 𝑅𝑎𝑡𝑒 = 𝑘 [𝐴] 𝑛](http://slideplayer.com./slide/16470022/96/images/22/Rate+Law+%F0%9D%90%B4+%E2%9F%B6+%F0%9D%90%B5+%F0%9D%91%85%F0%9D%91%8E%F0%9D%91%A1%F0%9D%91%92+%3D+%F0%9D%91%98+%5B%F0%9D%90%B4%5D+%F0%9D%91%9B.jpg "Equation relating concentration of reactants to the rate without time. 𝐴 ⟶ 𝐵. 𝑅𝑎𝑡𝑒 = 𝑘 [𝐴] 𝑛. k is the rate constant, a number which has units. (temperature dependent) n is the order, usually a whole number (0, 1, 2)")
23
Reaction Order, n 𝐴 ⟶ 𝐵 𝑅𝑎𝑡𝑒 = 𝑘 [𝐴] 𝑛
Usually a whole number (0, 1, 2) Determines how the rate is dependent on the reactant concentration 𝐴 ⟶ 𝐵 𝑅𝑎𝑡𝑒 = 𝑘 [𝐴] 𝑛
![Reaction Order, n 𝐴 ⟶ 𝐵 𝑅𝑎𝑡𝑒 = 𝑘 [𝐴] 𝑛](http://slideplayer.com./slide/16470022/96/images/23/Reaction+Order%2C+n+%F0%9D%90%B4+%E2%9F%B6+%F0%9D%90%B5+%F0%9D%91%85%F0%9D%91%8E%F0%9D%91%A1%F0%9D%91%92+%3D+%F0%9D%91%98+%5B%F0%9D%90%B4%5D+%F0%9D%91%9B.jpg "Usually a whole number (0, 1, 2) Determines how the rate is dependent on the reactant concentration. 𝐴 ⟶ 𝐵. 𝑅𝑎𝑡𝑒 = 𝑘 [𝐴] 𝑛.")
24
Reaction Orders 𝑅𝑎𝑡𝑒 = 𝑘 [𝐴] 0 𝑅𝑎𝑡𝑒 = 𝑘 [𝐴] 1 𝑅𝑎𝑡𝑒 = 𝑘 [𝐴] 2
Zero Order Rate is independent of the reactant concentration First Order Rate is directly proportional to the reactant concentration Second Order Rate is directly proportional to the square of the reactant concentration 𝑅𝑎𝑡𝑒 = 𝑘 [𝐴] 0 𝑅𝑎𝑡𝑒 = 𝑘 [𝐴] 1 𝑅𝑎𝑡𝑒 = 𝑘 [𝐴] 2
![Reaction Orders 𝑅𝑎𝑡𝑒 = 𝑘 [𝐴] 0 𝑅𝑎𝑡𝑒 = 𝑘 [𝐴] 1 𝑅𝑎𝑡𝑒 = 𝑘 [𝐴] 2](http://slideplayer.com./slide/16470022/96/images/24/Reaction+Orders+%F0%9D%91%85%F0%9D%91%8E%F0%9D%91%A1%F0%9D%91%92+%3D+%F0%9D%91%98+%5B%F0%9D%90%B4%5D+0+%F0%9D%91%85%F0%9D%91%8E%F0%9D%91%A1%F0%9D%91%92+%3D+%F0%9D%91%98+%5B%F0%9D%90%B4%5D+1+%F0%9D%91%85%F0%9D%91%8E%F0%9D%91%A1%F0%9D%91%92+%3D+%F0%9D%91%98+%5B%F0%9D%90%B4%5D+2.jpg "Zero Order. Rate is independent of the reactant concentration. First Order. Rate is directly proportional to the reactant concentration. Second Order. Rate is directly proportional to the square of the reactant concentration. 𝑅𝑎𝑡𝑒 = 𝑘 [𝐴] 0. 𝑅𝑎𝑡𝑒 = 𝑘 [𝐴] 1. 𝑅𝑎𝑡𝑒 = 𝑘 [𝐴] 2.")
25
Rate Law 𝑎𝐴+ 𝑏𝐵 ⟶𝑐𝐶+ 𝑑𝐷 𝑅𝑎𝑡𝑒 = 𝑘 [𝐴] 𝑚 [𝐵] 𝑛
For multi reactant reactions… 𝑎𝐴+ 𝑏𝐵 ⟶𝑐𝐶+ 𝑑𝐷 𝑅𝑎𝑡𝑒 = 𝑘 [𝐴] 𝑚 [𝐵] 𝑛
![Rate Law 𝑎𝐴+ 𝑏𝐵 ⟶𝑐𝐶+ 𝑑𝐷 𝑅𝑎𝑡𝑒 = 𝑘 [𝐴] 𝑚 [𝐵] 𝑛](http://slideplayer.com./slide/16470022/96/images/25/Rate+Law+%F0%9D%91%8E%F0%9D%90%B4%2B+%F0%9D%91%8F%F0%9D%90%B5+%E2%9F%B6%F0%9D%91%90%F0%9D%90%B6%2B+%F0%9D%91%91%F0%9D%90%B7+%F0%9D%91%85%F0%9D%91%8E%F0%9D%91%A1%F0%9D%91%92+%3D+%F0%9D%91%98+%5B%F0%9D%90%B4%5D+%F0%9D%91%9A+%5B%F0%9D%90%B5%5D+%F0%9D%91%9B.jpg "For multi reactant reactions… 𝑎𝐴+ 𝑏𝐵 ⟶𝑐𝐶+ 𝑑𝐷. 𝑅𝑎𝑡𝑒 = 𝑘 [𝐴] 𝑚 [𝐵] 𝑛.")
26
Differential Rate Laws
Rate laws in this form are known as differential rate laws 𝑅𝑎𝑡𝑒 = 𝑘 [𝐴] 𝑚 [𝐵] 𝑛
27
Reaction Order Reaction order is specified by each individual reactant and by the overall reaction The overall reaction order is the sum of the exponents in the rate law
28
Reaction Order 𝑁𝐻 4 + (𝑎𝑞)+ 𝑁𝑂 2 − (𝑎𝑞)⟶ 𝑁 2 (𝑔)+ 2 𝐻 2 𝑂(𝑙)
For example, First-order in 𝑁𝐻 4 + First-order in [ 𝑁𝑂 2 − ] Overall Second-order reaction 𝑁𝐻 4 + (𝑎𝑞)+ 𝑁𝑂 2 − (𝑎𝑞)⟶ 𝑁 2 (𝑔)+ 2 𝐻 2 𝑂(𝑙) 𝑅𝑎𝑡𝑒=𝑘 𝑁𝐻 4 + [ 𝑁𝑂 2 − ]
29
Example
30
Homework Re-Read Chapter 14 Sections 2, 3
Answer Questions: 3, 4, 20, 24,
32
Integrated Rate Laws 𝑙𝑛 [𝐴] 𝑡 [𝐴] 0 = −𝑘𝑡 𝑅𝑎𝑡𝑒 = 𝑘 [𝐴] 1
Integrating the differential rate law equations turns them into equations relating the concentration to time 𝑙𝑛 [𝐴] 𝑡 [𝐴] 0 = −𝑘𝑡 𝑅𝑎𝑡𝑒 = 𝑘 [𝐴] 1
![Integrated Rate Laws 𝑙𝑛 [𝐴] 𝑡 [𝐴] 0 = −𝑘𝑡 𝑅𝑎𝑡𝑒 = 𝑘 [𝐴] 1](http://slideplayer.com./slide/16470022/96/images/32/Integrated+Rate+Laws+%F0%9D%91%99%F0%9D%91%9B+%5B%F0%9D%90%B4%5D+%F0%9D%91%A1+%5B%F0%9D%90%B4%5D+0+%3D+%E2%88%92%F0%9D%91%98%F0%9D%91%A1+%F0%9D%91%85%F0%9D%91%8E%F0%9D%91%A1%F0%9D%91%92+%3D+%F0%9D%91%98+%5B%F0%9D%90%B4%5D+1.jpg "Integrating the differential rate law equations turns them into equations relating the concentration to time. 𝑙𝑛 [𝐴] 𝑡 [𝐴] 0 = −𝑘𝑡. 𝑅𝑎𝑡𝑒 = 𝑘 [𝐴] 1.")
33
Integrated Rate Laws 𝑙𝑛 [𝐴] 𝑡 [𝐴] 0 = −𝑘𝑡 𝑙𝑛 [𝐴] 𝑡 −𝑙𝑛 [𝐴] 0 = −𝑘𝑡
First Order 𝑙𝑛 [𝐴] 𝑡 [𝐴] 0 = −𝑘𝑡 𝑙𝑛 [𝐴] 𝑡 −𝑙𝑛 [𝐴] 0 = −𝑘𝑡 𝑙𝑛 [𝐴] 𝑡 = −𝑘𝑡+𝑙𝑛 [𝐴] 0 𝑦 = 𝑚𝑥 + 𝑏
![Integrated Rate Laws 𝑙𝑛 [𝐴] 𝑡 [𝐴] 0 = −𝑘𝑡 𝑙𝑛 [𝐴] 𝑡 −𝑙𝑛 [𝐴] 0 = −𝑘𝑡](http://slideplayer.com./slide/16470022/96/images/33/Integrated+Rate+Laws+%F0%9D%91%99%F0%9D%91%9B+%5B%F0%9D%90%B4%5D+%F0%9D%91%A1+%5B%F0%9D%90%B4%5D+0+%3D+%E2%88%92%F0%9D%91%98%F0%9D%91%A1+%F0%9D%91%99%F0%9D%91%9B+%5B%F0%9D%90%B4%5D+%F0%9D%91%A1+%E2%88%92%F0%9D%91%99%F0%9D%91%9B+%5B%F0%9D%90%B4%5D+0+%3D+%E2%88%92%F0%9D%91%98%F0%9D%91%A1.jpg "First Order. 𝑙𝑛 [𝐴] 𝑡 [𝐴] 0 = −𝑘𝑡. 𝑙𝑛 [𝐴] 𝑡 −𝑙𝑛 [𝐴] 0 = −𝑘𝑡. 𝑙𝑛 [𝐴] 𝑡 = −𝑘𝑡+𝑙𝑛 [𝐴] 0. 𝑦 = 𝑚𝑥 + 𝑏.")
34
Integrated Rate Laws
35
Integrated Rate Laws Zero Order [𝐴] 𝑡 = −𝑘𝑡+ [𝐴] 0
![Integrated Rate Laws Zero Order [𝐴] 𝑡 = −𝑘𝑡+ [𝐴] 0](http://slideplayer.com./slide/16470022/96/images/35/Integrated+Rate+Laws+Zero+Order+%5B%F0%9D%90%B4%5D+%F0%9D%91%A1+%3D+%E2%88%92%F0%9D%91%98%F0%9D%91%A1%2B+%5B%F0%9D%90%B4%5D+0.jpg "Integrated Rate Laws Zero Order [𝐴] 𝑡 = −𝑘𝑡+ [𝐴] 0")
36
Integrated Rate Laws 1 [𝐴] 𝑡 = 𝑘𝑡+ 1 [𝐴] 0 Second Order
![Integrated Rate Laws 1 [𝐴] 𝑡 = 𝑘𝑡+ 1 [𝐴] 0 Second Order](http://slideplayer.com./slide/16470022/96/images/36/Integrated+Rate+Laws+1+%5B%F0%9D%90%B4%5D+%F0%9D%91%A1+%3D+%F0%9D%91%98%F0%9D%91%A1%2B+1+%5B%F0%9D%90%B4%5D+0+Second+Order.jpg "Integrated Rate Laws 1 [𝐴] 𝑡 = 𝑘𝑡+ 1 [𝐴] 0 Second Order")
37
Integrated Rate Laws
38
Half Life, t1/2 Amount of time for half of a reactant to react At t1/2 [𝐴] 𝑡 = 0.5 [𝐴] 0
![Half Life, t1/2 Amount of time for half of a reactant to react At t1/2 [𝐴] 𝑡 = 0.5 [𝐴] 0](http://slideplayer.com./slide/16470022/96/images/38/Half+Life%2C+t1%2F2+Amount+of+time+for+half+of+a+reactant+to+react+At+t1%2F2+%5B%F0%9D%90%B4%5D+%F0%9D%91%A1+%3D+0.5+%5B%F0%9D%90%B4%5D+0.jpg "Half Life, t1/2 Amount of time for half of a reactant to react At t1/2 [𝐴] 𝑡 = 0.5 [𝐴] 0")
39
Half Life, t1/2 𝑙𝑛 [𝐴] 𝑡 [𝐴] 0 = −𝑘𝑡 𝑙𝑛0.5= −𝑘 𝑡 1/2 −0.693= −𝑘 𝑡 1/2
First Order 𝑙𝑛 [𝐴] 𝑡 [𝐴] 0 = −𝑘𝑡 𝑙𝑛0.5= −𝑘 𝑡 1/2 −0.693= −𝑘 𝑡 1/2 𝑙𝑛 0.5[𝐴] 0 [𝐴] 0 = −𝑘 𝑡 1/2 0.693 𝑘 = 𝑡 1/2
![Half Life, t1/2 𝑙𝑛 [𝐴] 𝑡 [𝐴] 0 = −𝑘𝑡 𝑙𝑛0.5= −𝑘 𝑡 1/2 −0.693= −𝑘 𝑡 1/2](http://slideplayer.com./slide/16470022/96/images/39/Half+Life%2C+t1%2F2+%F0%9D%91%99%F0%9D%91%9B+%5B%F0%9D%90%B4%5D+%F0%9D%91%A1+%5B%F0%9D%90%B4%5D+0+%3D+%E2%88%92%F0%9D%91%98%F0%9D%91%A1+%F0%9D%91%99%F0%9D%91%9B0.5%3D+%E2%88%92%F0%9D%91%98+%F0%9D%91%A1+1%2F2+%E2%88%920.693%3D+%E2%88%92%F0%9D%91%98+%F0%9D%91%A1+1%2F2.jpg "First Order. 𝑙𝑛 [𝐴] 𝑡 [𝐴] 0 = −𝑘𝑡. 𝑙𝑛0.5= −𝑘 𝑡 1/2. −0.693= −𝑘 𝑡 1/2. 𝑙𝑛 0.5[𝐴] 0 [𝐴] 0 = −𝑘 𝑡 1/ 𝑘 = 𝑡 1/2.")
40
Half Life, t1/2
41
Half Life, t1/2 [𝐴] 0 2𝑘 = 𝑡 1/2 1 𝑘 [𝐴] 0 = 𝑡 1/2 Zero Order
Second Order [𝐴] 0 2𝑘 = 𝑡 1/2 1 𝑘 [𝐴] 0 = 𝑡 1/2
![Half Life, t1/2 [𝐴] 0 2𝑘 = 𝑡 1/2 1 𝑘 [𝐴] 0 = 𝑡 1/2 Zero Order](http://slideplayer.com./slide/16470022/96/images/41/Half+Life%2C+t1%2F2+%5B%F0%9D%90%B4%5D+0+2%F0%9D%91%98+%3D+%F0%9D%91%A1+1%2F2+1+%F0%9D%91%98+%5B%F0%9D%90%B4%5D+0+%3D+%F0%9D%91%A1+1%2F2+Zero+Order.jpg "Second Order. [𝐴] 0 2𝑘 = 𝑡 1/2. 1 𝑘 [𝐴] 0 = 𝑡 1/2.")
43
Reaction Mechanisms Chemical equations used to represent a chemical reaction usually represent the overall chemical reaction and not the series of individual steps involved with the reaction Many reactions occur through a series of separate steps A reaction mechanism is the process by which a reaction happens.
44
Reaction Mechanisms Some reactions happen through one reaction step
Elementary reaction or elementary step Most reactions happen through multiple steps multiple elementary reactions The rate law for an overall chemical reaction is dependent on the reaction mechanism
45
Reaction Mechanisms In multi-step reactions, usually one of the steps is slower than the other steps This slower step limits the speed of the reaction Rate limiting or rate determining step The rate law for the overall reaction is dependent on the rate limiting step
46
Molecularity Elementary reactions have a set rate law based on the molecularity of the reaction Number of reactant molecules in the step Unimolecular – one reactant molecule Bimolecular – two reactant molecules Termolecular – three reactant molecules Very rare
47
Molecularity
48
Reaction Mechanisms 3 Possibilities for a chemical reaction
Reaction is one elementary step Reaction is a multistep reaction First step is the rate limiting step (slow) First step is not the rate limiting step (slow)
49
Slow First Step 𝑁𝑂 2 𝑔 +𝐶𝑂 𝑔 ⟶ 𝑁𝑂 𝑔 + 𝐶𝑂 2 (𝑔)
Rate law has been experimentally determined to be: rate = k[NO2]2 𝑁𝑂 2 𝑔 +𝐶𝑂 𝑔 ⟶ 𝑁𝑂 𝑔 + 𝐶𝑂 2 (𝑔)
50
Slow First Step 𝑁𝑂 2 + 𝑁𝑂 2 ⟶ 𝑁𝑂+ 𝑁𝑂 3 𝑁𝑂 3 +𝐶𝑂 ⟶ 𝑁𝑂 2 + 𝐶𝑂 2
Proposed Mechanism Step 1: Step 2: For step 1: rate = k[NO2]2 CO is not involved in the slow step (rate determining) NO3 is produced in one step and used in another step Intermediate 𝑁𝑂 2 + 𝑁𝑂 2 ⟶ 𝑁𝑂+ 𝑁𝑂 3 (Slow) 𝑁𝑂 3 +𝐶𝑂 ⟶ 𝑁𝑂 2 + 𝐶𝑂 2 (Fast)
51
Reaction Mechanism Intermediate
A species that is produced in one step and used in another Does not appear in overall reaction Different from a catalyst
52
Fast First Step 2𝑁𝑂 𝑔 + 𝐵𝑟 2 𝑔 ⟶ 2𝑁𝑂𝐵𝑟 𝑔
Rate law has been experimentally determined to be: rate = k[NO]2[Br] 2𝑁𝑂 𝑔 + 𝐵𝑟 2 𝑔 ⟶ 2𝑁𝑂𝐵𝑟 𝑔
53
Fast First Step 𝑁𝑂+ 𝐵𝑟 2 ⇄ 𝑁𝑂𝐵𝑟 2 𝑁𝑂𝐵𝑟 2 +𝑁𝑂 ⟶2𝑁𝑂𝐵𝑟
Proposed Mechanism Step 1: Step 2: Rate Law of Slow Step: Can’t have intermediates in rate law 𝑁𝑂+ 𝐵𝑟 2 ⇄ 𝑁𝑂𝐵𝑟 2 (Fast) 𝑁𝑂𝐵𝑟 2 +𝑁𝑂 ⟶2𝑁𝑂𝐵𝑟 (Slow) 𝑟𝑎𝑡𝑒= 𝑘 2 𝑁𝑂𝐵𝑟 2 [𝑁𝑂]
54
Fast First Step 𝑁𝑂+ 𝐵𝑟 2 ⇄ 𝑁𝑂𝐵𝑟 2 𝑁𝑂𝐵𝑟 2 +𝑁𝑂 ⟶2𝑁𝑂𝐵𝑟
Proposed Mechanism Step 1: Step 2: Assume first step is in equilibrium 𝑁𝑂+ 𝐵𝑟 2 ⇄ 𝑁𝑂𝐵𝑟 2 (Fast) 𝑁𝑂𝐵𝑟 2 +𝑁𝑂 ⟶2𝑁𝑂𝐵𝑟 (Slow) 𝑟𝑎𝑡𝑒 𝑓𝑜𝑟𝑤𝑎𝑟𝑑 = 𝑟𝑎𝑡𝑒 𝑟𝑒𝑣𝑒𝑟𝑠𝑒 𝑘 1 𝑁𝑂 [ 𝐵𝑟 2 ]= 𝑘 −1 [ 𝑁𝑂𝐵𝑟 2 ]
55
Fast First Step 𝑁𝑂+ 𝐵𝑟 2 ⇄ 𝑁𝑂𝐵𝑟 2 𝑁𝑂𝐵𝑟 2 +𝑁𝑂 ⟶2𝑁𝑂𝐵𝑟
Proposed Mechanism Step 1: Step 2: Assume first step is in equilibrium 𝑁𝑂+ 𝐵𝑟 2 ⇄ 𝑁𝑂𝐵𝑟 2 (Fast) 𝑁𝑂𝐵𝑟 2 +𝑁𝑂 ⟶2𝑁𝑂𝐵𝑟 (Slow) 𝑟𝑎𝑡𝑒 𝑓𝑜𝑟𝑤𝑎𝑟𝑑 = 𝑟𝑎𝑡𝑒 𝑟𝑒𝑣𝑒𝑟𝑠𝑒 𝑘 1 𝑘 −1 𝑁𝑂 [ 𝐵𝑟 2 ]= [ 𝑁𝑂𝐵𝑟 2 ]
56
Fast First Step 𝑁𝑂+ 𝐵𝑟 2 ⇄ 𝑁𝑂𝐵𝑟 2 𝑁𝑂𝐵𝑟 2 +𝑁𝑂 ⟶2𝑁𝑂𝐵𝑟
Proposed Mechanism Step 1: Step 2: Rate Law of Slow Step: 𝑁𝑂+ 𝐵𝑟 2 ⇄ 𝑁𝑂𝐵𝑟 2 (Fast) 𝑁𝑂𝐵𝑟 2 +𝑁𝑂 ⟶2𝑁𝑂𝐵𝑟 (Slow) 𝑟𝑎𝑡𝑒= 𝑘 2 𝑁𝑂𝐵𝑟 2 [𝑁𝑂] 𝑘 1 𝑘 −1 𝑁𝑂 [ 𝐵𝑟 2 ]= [ 𝑁𝑂𝐵𝑟 2 ]
57
Fast First Step 𝑁𝑂+ 𝐵𝑟 2 ⇄ 𝑁𝑂𝐵𝑟 2 𝑁𝑂𝐵𝑟 2 +𝑁𝑂 ⟶2𝑁𝑂𝐵𝑟
Proposed Mechanism Step 1: Step 2: Rate Law of Slow Step: 𝑁𝑂+ 𝐵𝑟 2 ⇄ 𝑁𝑂𝐵𝑟 2 (Fast) 𝑁𝑂𝐵𝑟 2 +𝑁𝑂 ⟶2𝑁𝑂𝐵𝑟 (Slow) 𝑟𝑎𝑡𝑒= 𝑘 2 𝑘 1 𝑘 −1 𝑁𝑂 [ 𝐵𝑟 2 ][𝑁𝑂]
58
Fast First Step 𝑁𝑂+ 𝐵𝑟 2 ⇄ 𝑁𝑂𝐵𝑟 2 𝑁𝑂𝐵𝑟 2 +𝑁𝑂 ⟶2𝑁𝑂𝐵𝑟
Proposed Mechanism Step 1: Step 2: Rate Law of Slow Step: 𝑁𝑂+ 𝐵𝑟 2 ⇄ 𝑁𝑂𝐵𝑟 2 (Fast) 𝑁𝑂𝐵𝑟 2 +𝑁𝑂 ⟶2𝑁𝑂𝐵𝑟 (Slow) 𝑟𝑎𝑡𝑒=𝑘 𝑁𝑂 2 [ 𝐵𝑟 2 ]
59
Reaction Mechanism A proposed mechanism can be validated if two criteria are met All elementary steps sum to match overall reaction Rate law predicted by mechanism is consistent with experimentally determined rate law
60
Reaction Mechanism Why is one step slow and another fast?
61
Homework Re-Read Chapter 14 Section 6 Answer Questions: 9, 65, 68, 96
Similar presentations
![Reaction Rate Change in concentration of a reactant or product per unit time. [A] means concentration of A in mol/L; A is the reactant or product being.](/17/5277620/big_thumb.jpg "Reaction Rate Change in concentration of a reactant or product per unit time. [A] means concentration of A in mol/L; A is the reactant or product being.>")
