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Unless otherwise stated, all images in this file have been reproduced from: Blackman, Bottle, Schmid, Mocerino and Wille, Chemistry, 2007 (John Wiley)

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Presentation on theme: "Unless otherwise stated, all images in this file have been reproduced from: Blackman, Bottle, Schmid, Mocerino and Wille, Chemistry, 2007 (John Wiley)"— Presentation transcript:

1 Unless otherwise stated, all images in this file have been reproduced from: Blackman, Bottle, Schmid, Mocerino and Wille, Chemistry, 2007 (John Wiley) ISBN:

2 Slide 2/17 e CHEM1002 [Part 2] A/Prof Adam Bridgeman (Series 1) Dr Feike Dijkstra (Series 2) Weeks 8 – 13 Office Hours: Monday 2-3, Friday 1-2 Room: 543a

3 Slide 3/17 e Chemical Kinetics I The rate of change of concentration of a reactant or a product is the rate of the reaction divided by the corresponding stoichiometric coefficient in the chemical reaction The rate law shows how the rate of the reaction depends on the concentration of how reactant The order of the reaction with respect to each reactant is determined from experimental data The order of the reaction with respect to each reactant is not given by the corresponding stoichiometric coefficient in the chemical reaction The rate constant (including its units) is found from experimental data Summary of Last Lecture

4 Slide 4/17 e Lecture 16 Chemical Kinetics Rate of Reaction Rate Laws Reaction Order Blackman Chapter 14, Sections Lecture 17 Half lives The Temperature Dependence of Reaction Rates Catalysis Blackman Chapter 14, Sections Chemical Kinetics II

5 Slide 5/17 e Concentration - Time Relationships For the 1 st order reaction A  B, the rate law is dt = k[A] rate= -d[A] To find how [A] varies with time, this is integrated: [A] 0 = -kt [A] ln concentration at start (t = 0)

6 Slide 6/17 e For a first order reaction,  so Half Life (t 1/2 ) The half life of a reaction is the time required for the concentration to fall to half its initial value. [A] 0 = [A] 2 1 t 1/2 = ln 2 / k [A] 0 = -kt [A] ln

7 Slide 7/17 e Half Life 2 N 2 O 5 = 4NO 2 + O 2 t 1/2 = 24 min constant – 1 st order

8 Slide 8/17 e Collision theory Molecules must collide to react  collision frequency increases with temperature  molecules must be correctly orientated  molecules must have sufficient energy to react Molecules must collide to react  collision frequency increases with temperature  molecules must be correctly orientated  molecules must have sufficient energy to react The minimum energy that molecules must have to react is called the activation energy (E a )

9 Slide 9/17 e Energy in Chemical Reactions E a forward  H (forward) transition state (highest energy point) (exothermic)

10 Slide 10/17 e Multistep Reactions Each elementary step in a reaction has a separate activation energy The step with the largest activation energy is the rate determining step.

11 Slide 11/17 e Multistep Reactions NOBr 2 + NO  2NOBrslow NO + Br 2 NOBr 2 fast equilibrium E a (1) < E a (2) so step 2 is rate determining

12 Slide 12/17 e Arrhenius Equation k = Ae -E a /RT Describes the temperature dependence of the rate E a is the activation energy – the minimum amount of energy that the reacting molecules must possess for the reaction is to be successful A is the pre-exponential factor or the “ A factor ” – depends on the collision frequency and orientation factor

13 Slide 13/17 e Arrhenius Equation k = Ae -E a /RT lnk = lnA – E a /RT For a typical chemical reaction, k doubles for every 10 °C (10 K) increase in temperature

14 Slide 14/17 e Catalysts A catalyst increases the rate of a chemical reaction without itself being changed A catalyst provides an alternative reaction pathway of lower activation energy with catalyst

15 Slide 15/17 e  Do not effect the position of the equilibrium with catalyst  Do not change K eq Catalysts  Do not effect how favourable reaction is Do effect how fast the reaction is

16 Slide 16/17 e The enzyme provides a surface for the reaction This surface stabilizes the transition state, lowering the activation energy The enzyme helps transform the transition state to product B B A catalytic surface A Enzyme = Biological Catalyst

17 Slide 17/17 e Summary: Chemical Kinetics II Learning Outcomes - you should now be able to: Be able to perform calculations using half lives Be able to draw reaction coordinate diagrams Explain why reaction rate increase with temperature Explain what catalysts do and how they do it Embarrass your lecturer with a very high mark in the exam Very Best of Luck...


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