Chapter 121 Chapter 12: Kinetics; Outline 1. Introduction  2. macroscopic determination of rate (experimental) define rate  define rate law, rate constant,

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Presentation transcript:

Chapter 121 Chapter 12: Kinetics; Outline 1. Introduction  2. macroscopic determination of rate (experimental) define rate  define rate law, rate constant, reaction order  rate determination via expt: initial rate  integrated rate laws  first order; half-lives  second order; half-lives  rate & temperature: the Arrhenius equation

Chapter 122 Arrhenius equation: from observation, reaction rates and rate constants increase with temperature examples: food decays faster at higher temperature; cooking; fireflies; from observation, a graph of log(k) vs. 1/T gives a straight line the slope of the log(k) vs. 1/T graph is related to the activation energy

Chapter 123 A graph of ln(k) vs. 1/T gives a straight line with a slope of -E A /R and an intercept of ln(A) where k is rate constant, E A is the activation energy, R is the gas constant in J K -1 mol -1, T is temperature in Kelvin, and A is the “steric factor”

Chapter 124 Example: The rate constant for a certain reaction is measured at four temperatures (see the data below). What is the activation energy for the reaction? What is the value of the rate constant at 400°C?

Chapter 125

6 activation energy = slope ·2.303·(-R) = +1.13x10 5 J mol -1

Chapter molecular view of kinetics mechanisms relationship to rate laws Arrhenius equation 4. Catalysis, Factors affecting rates 3. Molecular View of Kinetics

Chapter 128 Mechanisms A mechanism is a series of elementary reactions, or steps, that describe what happens as a reaction proceeds. Elementary reactions are not overall reactions: overall reactions summarize the products and reactants and give the stoichiometry. Elementary reactions may have intermediates: short-lived (<1 second) species that are formed and then react away as the reaction proceeds. An elementary reaction often involves a collision between two species (a bimolecular step or reaction).

Chapter 129 example: NO 2 +CO  NO+CO 2 rate=k[NO 2 ] 2 [2nd order] NO 2 +NO 2  NO 3 +NO NO 3 +CO  NO 2 +CO 2 possible mechanism:

Chapter 1210 A few comments on the mechanism and its relation to the rate law: NO and NO 3 are intermediates. They don’t appear as reactants or products and they are very reactive (unstable). Each of these steps is bimolecular: involves 2 molecules. The individual reactions add up to give the overall reaction with the correct stoichiometry. The order and the rate equation of an elementary reaction is determined by its stoichiometry step 1: rate(1)=k 1 [NO 2 ] 2 bimolecular step 2: rate(2)=k 2 [NO][NO 3 ]bimolecular

Chapter 1211 If one step is much slower than the others, the rate of that slow step is the rate of the overall reaction step 1: slow(forms two unstable intermediates) step 2: FAST (two unstable, reactive intermediates react) overall rate predicted by the mechanism: k 1 [NO 2 ] 2 rate determined by experiment: k[NO 2 ] 2 Therefore, we say that the proposed mechanism is REASONABLE.

Chapter 1212 example: devise a plausible mechanism consistent with the overall reaction and the overall reaction rates for each of the following Co(CN) 5 H 2 O -2 (aq)+I -1 (aq)  Co(CN) 5 I -3 (aq)+H 2 O(l) rate=k[Co(CN) 5 H 2 O -2 ] 1 (aq) Co(CN) 5 H 2 O -2  Co(CN) H 2 Ok[Co(CN) 5 H 2 O -2 ] Co(CN) I 1-  Co(CN) 5 I -3 k[Co(CN) 5 -2 ][I 1- ] slow fast

Chapter NO 2 (g)+F 2 (g)  2NO 2 F(g)rate=k[NO 2 ][F] [HINT: consider a slow step and a fast step.]

Chapter 1214 Collision TheoryFrom observation, we know 1. reaction rates and rate constants increase with temperature 2. a graph of ln(k) versus 1/T gives a straight line with a negative slope 3. rate constants are relatively slow considering the HUGE number of collisions that occur at room temperature (around in a cm 3 at 1 atm and 298 K) Develop a theory that accounts for these observations

Chapter 1215 reactions occurs as the result of a collision only collisions above some minimum energy (the ACTIVATION ENERGY)will result in product formation only collisions of the correct geometry (orientation) will result in product formation

Chapter 1216 So, we can write an equation based on theory that gives the reaction rate; this equation is based on molecular parameters. rate=(collision rate) x (fraction molecules with E A ) x (fraction molecules with correct geometry) collision rate=Z[A][B] where Z depends on temperature and [A], [B] are concentrations fraction molecules with E A =f fraction molecules with correct geometry=P

Chapter 1217

Chapter 1218 fraction of molecules with enough energy to react: Activation Energy

activation energy lower temperature higher temperature more particles have energy>E A at higher T fraction of particles with energy >E A

Chapter 1220 from before we have line form of Arrhenius equation combining rearranging

Chapter 1221 Catalysis 1. definition: a catalyst is a chemical substance that increases the rate of a chemical reaction but is not consumed in the reaction (does not appear as a reactant or product). catalysts work by 2. providing an alternate, lower energy reaction pathway to product formation or 3. stabilizing the activated complex (transition state) and thus lowering the activation energy

Chapter 1222

Chapter examples 1. HOMOGENEOUS CATALYSIS: ozone depletion Cl· catalyzes the depletion reaction through a series of reactions: Cl· acts as a catalyst in the second two steps

Chapter HETEROGENEOUS CATALYSIS catalytic converter CO(g)+1/2O 2 (g)  CO 2 (g) reactant molecules adsorb (by intermolecular forces or by chemical bonding) to the metal surface in the converter  bonds in reactants weaken  activated complex energy is reduced  reaction goes faster

Chapter enzymes: large molecules that act as catalysts in biological reactions enzymes are specific: they work with only 1 substrate reaction rates are increased by factors of 10 8 to the “turnover number” of number of reaction events per second is large: 10 3 to 10 7 s -1 link to www: