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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)      ISBN: 9 78047081 0866 . CHEM1002 [Part 2]. A/Prof Adam Bridgeman (Series 1) Dr Feike Dijkstra (Series 2) Weeks 8 – 13

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  1. Unless otherwise stated, all images in this file have been reproduced from: Blackman, Bottle, Schmid, Mocerino and Wille,Chemistry, 2007 (John Wiley)     ISBN: 9 78047081 0866

  2. CHEM1002 [Part 2] A/Prof Adam Bridgeman (Series 1) Dr FeikeDijkstra (Series 2) Weeks 8 – 13 Office Hours: Monday 2-3, Friday 1-2 Room: 543a e-mail:adam.bridgeman@sydney.edu.au e-mail:feike.dijkstra@sydney.edu.au

  3. Summary of Last Lecture • 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

  4. Chemical Kinetics II • Lecture 16 • Chemical Kinetics • Rate of Reaction • Rate Laws • Reaction Order • Blackman Chapter 14, Sections 14.1 - 14.3 • Lecture 17 • Half lives • The Temperature Dependence of Reaction Rates • Catalysis • Blackman Chapter 14, Sections 14.4 - 14.6

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

  6. The half life of a reaction is the time required for the concentration to fall to half its initial value. [A] 1 = 2 [A]0 [A] ln = -kt [A]0 Half Life (t1/2) • For a first order reaction, • so t1/2 = ln 2 / k

  7. Half Life 2 N2O5 = 4NO2 + O2 t1/2 = 24 min constant – 1st order

  8. Molecules must collide to react collision frequency increases with temperature molecules must be correctly orientated molecules must have sufficient energy to react Collision theory • 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(Ea)

  9. Energy in Chemical Reactions transition state (highest energy point) Ea forward DH (forward) (exothermic)

  10. 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. Multistep Reactions • NO + Br2 NOBr2 fast equilibrium • NOBr2 + NO  2NOBr slow Ea (1) < Ea(2) so step 2is rate determining

  12. k = Ae-Ea/RT Describes the temperature dependence of the rate Ea 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 Arrhenius Equation

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

  14. Catalysts • A catalyst increases the rate of a chemical reaction without itself being changed • A catalyst provides an alternative reactionpathway of lower activation energy with catalyst

  15. Catalysts • Do not effect how favourable reaction is • Do not effect the position of the equilibrium • Do not change Keq • Do effect how fast the reaction is with catalyst

  16. B A Enzyme = Biological Catalyst • 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 A catalytic surface

  17. 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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