Section12_Kinetics_2

Section12_Kinetics_2 - Kinetics (continued) By the end of...

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Page 1 of 17 Kinetics (continued) By the end of this section, you will be able to: (1) Understand activation energies and transition states (2) Manipulate the Arrhenius equation (3) Understand homogeneous and heterogeneous catalysis (4) Describe reaction mechanisms and elementary reactions (5) Deduce rate laws from proposed mechanisms
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Page 2 of 17 Transition States and Activation Energies (MH5, Chapter 11.4 & 11.5) All systems have a tendency to adopt a low enthalpy; this is one of the reasons why many chemical reactions occur “spontaneously”. Consider the combustion of natural gas (mostly CH 4 ): CH 4 + 2 O 2 CO 2 + 2 H 2 O + heat ( Δ H < 0) We use this reaction for heating our houses, obviously it is spontaneous . BUT: A mixture of CH 4 and O 2 is stable, until it is exposed to a spark. Why is this? Consider an energy diagram of this process:
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Page 3 of 17 The spark is required to overcome the activation energy barrier . Once ignited, there is enough heat to keep the process going the gas mixture “continuously re-ignites itself”. Reactants and products of a chemical reaction are always separated by an activation energy barrier, E a . The top of the barrier represents the extremely unstable transition state (“activated complex”). In the transition state, bonds in the reactant are not yet fully broken, and bonds in the product are not yet fully formed. A-B + C A + B-C reactants transition state products Only molecules that have enough energy to overcome the activation energy barrier can react. For example, for A-B + C A + B-C to occur, A-B and C have to collide sufficiently fast . Molecules in a hot gas move faster than in a cold gas. Therefore the reaction rate increases as the temperature is increased.
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Page 4 of 17 MH5, Figure 11.9 This is true for all chemical processes (not only gas- phase processes): Chemical reactions occur faster at higher temperature , because more reactant molecules have enough energy to overcome the activation energy barrier E a . The dependence of the reaction rate on the temperature is given by the Arrhenius equation : RT E a e A k / ! = k = rate constant A = a factor that depends on the reaction of interest (we ʼ re not going to worry about A too much) E a = activation energy R = gas constant (8.31 J K -1 mol -1 ) T = temperature in KELVIN
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Page 5 of 17 If we take the ln of both sides of the Arrhenius equation, we get: T R E A k a 1 ln ln ! " = Therefore, plotting ln k vs. 1/ T gives a straight line
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This note was uploaded on 06/20/2008 for the course CHEM 024b taught by Professor Jones during the Winter '07 term at UWO.

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Section12_Kinetics_2 - Kinetics (continued) By the end of...

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