Kinetics Pt. 2 Temperature Dependence, Reaction Mechanisms, and Reaction Coordinates

1. Kinetics Pt. 2 Temperature Dependence, Reaction Mechanisms, and Reaction Coordinates

2. Temperature and Rate • The rates of most chemical reactions increase with temperature. • How is this temperature dependence reflected in the rate expression? • Rates increase with temperature because rate constants increase with temperature.An example is the 2nd order reaction: O3+ NO  O2 + NO2 Variation in k with temperature

3. Collision Model • The collision modelmakes sense of this. This model is based on kinetic theory • We’ve seen that the thermal energy of a molecule is converted to kinetic energy in order to facilitate motion, and we’ve seen that the velocity of a molecule increases with T. • The central idea of the collision model is that molecules must collide to react. The more collisions per second, the faster the reaction goes. • This model also rationalizes the concentration dependence. The more molecules present, the more collisions you have.

4. Activation Energy • Of course, there is more to a chemical reaction than just collisions of molecules. • Molecules must have some minimum kinetic energy in order to overcome the energy barrier of reaction. The kinetic energy of molecules is used to stretch, bend, and break bonds in order to cause a reaction. • If the molecules don’t have enough kinetic energy, they simply bounce off one another and nothing happens. • This minimum energy that molecules must have is called the activation energy,(Ea)

5. Ex. O3 + NO  NO2 + O2 1. Effective collision, activation energy absorbed, high energy intermediate forms (slow) 2. Intermediate decomposes and products form (fast)

6. Reaction Coordinate • To the right is a reaction coordinate plot of the reaction. The y-axis represents enthalpy. The x-axis represents the progress of reaction from reactants to products. • The activation energy is represented by the upward bump in the curve. The intermediate is the highest energy state of a reaction. Once the intermediate forms, the reaction is rapid and is energetically “downhill” • This reaction is exothermic because the products have lower enthalpy than the reactants.

7. Arrhenius Equation • Arrhenius noted that reaction-rate data depended on three aspects • (1) the fraction of molecules possessing an energy Ea or greater • (2) the number of collisions per second • (3) the fraction of molecules oriented in the right way for a reaction to occur • These factors are incorporated into the Arrhenius Equation where k is the rate constant, R is the gas constant, T is temperature, Ea is the activation energy, and A is a value called the frequency factor, which is related to (2) and (3) above. • Higher Ea values result in much lower values of k, which slows reaction rate.

8. Arrhenius Equation f = • The energy of molecules in a system is not uniformly distributed. Not every molecule possesses the same energy at a given temperate and time. • The fraction of molecules in a system with enough energy to react (E >Ea) is given by: • A plot of f vs Ek for reactant molecules is shown to the right. At higher values of T, a greater fraction of the curve is beyond Ea.

9. The Arrhenius Equation is a Linear Equation ln k vs 1/T yields a linear plot with slope –Ea/R and a y-intercept of lnA

10. Calculating Changes in k Mathematically • The Arrhenius equation can be rearranged once again to compare k values at different temperatures • Remember:

11. Example • A reaction has an activation energy of 43.5 kJ/mol. The reaction occurs at 298 K with a rate constant of 110 s-1. What will the rate constant be if we increase the temperature to 308 K?

12. Reaction Mechanisms • Consider the following reaction: • In looking at this equation, and applying your understanding of collision theory, you would assume that this reaction proceeds when • NO2 and CO collide • The activation energy is exceeded • The atoms rearrange to form NO and CO2 • Thus, we would expect the rate of the reaction to depend on both [NO2] and [CO]

13. Reaction Mechanisms • However, according to experimental data, the reaction is 2nd order with respect to NO2 and zero order with respect to CO. The rate law is: • This means that the rate of the reaction does not depend on [CO] at all. How can this be? What does this tell us about the reaction mechanism? • This particular reaction must proceed through multiple steps, and the rate-determining stepmust only depend on NO2

14. Reaction Mechanisms • For this particular reaction, there are two steps: • This step-by-step description of the molecular pathway is the reaction mechanism. • The 1st step is substantially slower than the 2nd. Therefore, step (1) acts as the “rate determining step”, and the overall reaction speed can only be as fast as its slowest step.

15. Each step in a reaction mechanism is called an elementary reaction. • An elementary reaction is any reaction that proceeds in a single step. Mechanisms consist of sums of elementary reactions. • Going back to rate laws, each elementary reaction must have a corresponding rate constant • *** For elementary reactions only, you can assume that the rate law depends directly on the number of species present. Thus, the rate law of an elementary reaction follows the stoichiometry. intermediate k1 k2 rate of overall reaction

16. Catalysis • The rates of reactions can be increased by using catalysts. • Catalysts work by changing the mechansim to produce a lower energy intermediate. • The catalyst itself is not consumed in the reaction.

17. uncatalyzed Example: Oxidation of Sulfur Dioxide Pt-Catalyzed 2SO2(g) + O2(g)  2SO3(g) (1) 2SO2(g) ---> 2S(s) + 2O2(g) ΔHo(1) = 593.6 kJ/mol (slow step) (2) 2S(s) + 3O2(g) ---> 2SO3(g) ΔHo (2) = -791.4 kJ/mol 2SO2(g) + O2(g)  2SO3(g) (1) O2(g) ---> 2O ΔHorxn (1) = 498.4 kJ/mol(slow step) (2) SO2(s) + O(g) ---> SO3(g) ΔHorxn (2) = -696.2 kJ/mol catalyzed Pt

18. Example: Oxidation of Sulfur Dioxide Pt-Catalyzed