Reaction mechanisms and catalysts

1. Reaction mechanisms and catalysts 6.4

2. Reaction Mechanisms • The transition state theory is used to explain the process in a reaction that results from an effective collision. • The collision of two particles may result in the formation of a new temporary unstable substance that behaves as a temporary complex in the formation of a more stable product • This high energy unstable state is called an activated complex. • Within the activated complex the bonds are not completely formed or broken which represents a maximum potential energy state called the transition state.

3. Transition state Activated complex Threshold energy Potential Energy (kJ) Ea Products DH Reactants Reaction Progress (s)

4. Reaction Mechanisms • We tend to focus on the start and end of chemical reactions. The pathway from start to finish is often comprised of intermediate or elementary steps. • Elementary steps only involve one, two or possibly three particles in a collision. • Unstable compounds may be produced in these elementary steps and are called reaction intermediates. • The slowest of these elementary steps is the rate determining step. • The sum of these elementary steps represents the overall sequence or reaction mechanism.

5. 4 HBr + O2 2 H2O + 2 Br2 Overall impression of the reaction’s progress Potential Energy (kJ) Actual progression of the reaction through 3 intermediate reactions 4 HBr + O2 2 H2O + 2 Br2 Reaction Progress (s) The overall reaction may be comprised of individual elementary reaction steps. Each involving their own energy changes and rates.

6. 4 HBr + O2 2 H2O + 2 Br2 Potential Energy (kJ) 4 HBr + O2 2 H2O + 2 Br2 Reaction Progress (s) The overall reaction may be comprised of individual elementary reaction steps. Each involving their own energy changes and rates. Activated complexes Reaction intermediates

7. Potential Energy (kJ) 4 HBr + O2 2 H2O + 2 Br2 Reaction Progress (s) HOOBr HOBr • HBr(g) + O2(g) HOOBr(g) (slow) • HOOBr(g) + HBr(g)  2 HOBr(g) (fast) • 2x(HOBr(g) + HBr(g)  H2O(g) + Br2(g)) (fast) 4 HBr(g) + O2(g)  2 H2O(g) + 2 Br2(g)

8. Rate Determining Step Elementary Reaction Steps Potential Energy (kJ) Reaction Intermediates Ea HOOBr HOBr 4 HBr + O2 DH 2 H2O + 2 Br2 Reaction Progress (s) • HBr(g) + O2(g) HOOBr(g) (slow) • HOOBr(g) + HBr(g)  2 HOBr(g) (fast) • 2x(HOBr(g) + HBr(g)  H2O(g) + Br2(g)) (fast) 4 HBr(g) + O2(g)  2 H2O(g) + 2 Br2(g)

9. Rate Law Equations for Elementary Reactions • For elementary reactions, the exponents in the rate law equation are the same as the stoichiometric coefficients for each reactant in the chemical equation.

10. Proposing and Evaluating Mechanisms • When chemists propose a mechanism, they must satisfy the following criteria: • The equation for the elementary steps must combine to give the equation for the overall reaction • The proposed elementary steps must be reasonable • The mechanism must support the experimentally determined rate law.

11. Rate-Determining Step and the Rate law Example: The rate law will depend on the rate-determining reaction.

12. Overall reaction Uncatalyzed reaction path Catalyzed reaction path A catalyst reduces the overall activation energy for the reaction by producing alternative reaction intermediates with lower energy states. Potential Energy (kJ) Uncatalyzed activation energy Catalyzed activation energy Reaction Progress (s)

13. Example: • Reaction of sodium potassium tartrate with H2O2, catalyzed by cobaltous chloride. • CoCl2 is pink in solution. • Contents of beaker turn dark greed, suggesting the formation of a reaction intermediate. • CoCl2 is regenerated when the reaction is over.

14. Explaining & Applying Chemical Kinetics • Increasing the temperature of the system increases the number of particles with sufficient energy to react. ET New particles with sufficient energy to effectively collide in the chemical reaction. Number of particles Particles with sufficient energy to effectively collide in the chemical reaction. Kinetic energy (kJ)

15. Explaining & Applying Chemical Kinetics • An increase in concentration results in more particles with the required threshold energy being added to the system. New particles with sufficient energy to effectively collide in the chemical reaction. Increased concentration ET Number of particles Particles with sufficient energy to effectively collide in the chemical reaction. Kinetic energy (kJ)

16. Explaining & Applying Chemical Kinetics • A catalyst reduces the threshold energy and consequentially, activation energy . New particles with sufficient energy to effectively collide in the chemical reaction. ET Catalyzed ET Numberof particles Particles with sufficient energy to effectively collide in the chemical reaction. Kinetic energy (kJ)

17. Explaining & Applying Chemical Kinetics • A catalyst reduces the activation energy and threshold energy. ET Ep (kJ) ET Catalyzed Reaction Progress

18. Homogeneous Catalysts • Exists in the same phase as the reactant. • Usually catalyze gaseous and aqueous reactions.

19. Heterogeneous Catalysts • Exist in a phase different from the phase of the reaction. • Example: addition of hydrogen to an organic compound that contains C=C bonds: ___________. • Without a catalyst, the reaction is very slow. • Ethylene and hydrogen molecules form bonds with the metal surface. This weakens the bonds of the hydrogen and ethylene • The H-H bonds of the hydrogen molecules break and are somewhat stabilized because of their attraction to the metal. • Hydrogen atoms react with ethylene, forming ethane.