Enzyme Catalysis

1. Enzyme Catalysis • Enzymes have enormous catalytic activity. • The substance whose reaction the enzyme catalyzes is called the substrate. (see Figure 14.21) • Figure 14.22 illustrates the reduction in acivation energy resulting from the formation of an enzyme-substrate complex.

2. Figure 14.22: Potential-energy curves for the reaction of substrate, S, to products, P.

3. Figure 14.17: Comparison of activation energies in the uncatalyzed and catalyzed decompositions of ozone.

4. Figure 17-5a Conformation changes in yeast hexokinase on binding glucose. (a) Space-filling model of a subunit of free hexokinase. (b) Space-filling model of a subunit of free hexokinase in complex with glucose (purple). .

5. Operational Skills • Relating the different ways of expressing reaction rates • Calculating the average reaction rate • Determining the order of reaction from the rate law • Determining the rate law from initial rates • Using the concentration-time equation for first-order reactions • Relating the half-life of a reaction to the rate constant

6. Operational Skills • Using the Arrhenius equation • Writing the overall chemical equation from a mechanism • Determining the molecularity of an elementary reaction • Writing the rate equation for an elementary reaction • Determining the rate law from a mechanism

7. Conceptual Problem 14.43

8. Figure 14.19: Proposed mechanism of a catalytic converter.

9. Figure 14.20: Exhaust gasses pass through the catalytic converter.

10. Three-way catalytic converters • A three-way catalytic converter has three simultaneous tasks: • Reduction of nitrogen oxides to nitrogen and oxygen: • 2NOx → xO2 + N2 • Oxidation of carbon monoxide to carbon dioxide: • 2CO + O2 → 2CO2 • Oxidation of unburnt hydrocarbons (HC) to carbon dioxide • and water: • 2CxHy + (2x+y/2)O2 → 2xCO2 + yH2O

11. The catalyst itself is most often a precious metal. Platinum is the most active catalyst and is widely used. However, it is not suitable for all applications because of unwanted additional reactions and/or cost. Palladium and rhodium are two other precious metals that are used. Platinum and rhodium are used as a reduction catalyst, while platinum and palladium are used as an oxidization catalyst. Cerium, iron, manganese and nickel are also used, though each has its own limitations. Nickel is not legal for use in the European Union (due to nickel hydrate formation). While copper can be used, its use is illegal in North America due to the formation of dioxin. More on catalytic converters http://en.wikipedia.org/wiki/Catalytic_converter# Three-way_catalytic_converters

12. Figure 14.20: Cross-sectional views of some catalytic converters. Photo courtesy of Delphi Automotive Systems.

13. Chapter 15Chemical Equilibrium

14. Chemical Equilibrium • When compounds react, they eventually form a mixture of products and unreacted reactants, in a dynamic equilibrium. • A dynamic equilibrium consists of a forward reaction, in which substances react to give products, and a reverse reaction, in which products react to give the original reactants. • Chemical equilibrium is the state reached by a reaction mixture when the rates of the forward and reverse reactions have become equal.

15. Figure 15.2: Catalytic methanation reaction approaches equilibrium.

16. Chemical Equilibrium • For example, the Haber process for producing ammonia from N2 and H2 does not go to completion. • It establishes an equilibrium state where all three species are present.

17. Suppose we place 1.000 mol N2 and 3.000 mol H2 in a reaction vessel at 450 oC and 10.0 atmospheres of pressure. The reaction is A Problem to Consider • Applying Stoichiometry to an Equilibrium Mixture. • What is the composition of the equilibrium mixture if it contains 0.080 mol NH3?

18. Starting 1.000 3.000 0 Change -x -3x +2x Equilibrium 1.000 - x 3.000 - 3x 2x = 0.080 mol A Problem to Consider • Using the information given, set up the following table. • The equilibrium amount of NH3 was given as 0.080 mol. Therefore, 2x = 0.080 molNH3 (x = 0.040 mol).

19. Starting 1.000 3.000 0 Change -x -3x +2x Equilibrium 1.000 - x 3.000 - 3x 2x = 0.080 mol A Problem to Consider • Using the information given, set up the following table. Equilibrium amount of N2 = 1.000 - 0.040 = 0.960 mol N2 Equilibrium amount of H2 = 3.000 - (3 x 0.040) = 2.880 mol H2 Equilibrium amount of NH3 = 2x = 0.080 mol NH3

20. The Equilibrium Constant • Every reversible system has its own “position of equilibrium” under any given set of conditions. • The ratio of products produced to unreacted reactants for any given reversible reaction remains constant under constant conditions of pressure and temperature. • The numerical value of this ratio is called the equilibrium constant for the given reaction.

21. For the general equation above, the equilibrium-constant expression would be: The Equilibrium Constant • The equilibrium-constant expression for a reaction is obtained by multiplying the concentrations of products, dividing by the concentrations of reactants, and raising each concentration to a power equal to its coefficient in the balanced chemical equation.

22. The Equilibrium Constant • The equilibrium-constant expression for a reaction is obtained by multiplying the concentrations of products, dividing by the concentrations of reactants, and raising each concentration to a power equal to its coefficient in the balanced chemical equation. • The molar concentration of a substance is denoted by writing its formula in square brackets.

23. The Equilibrium Constant • The equilibrium constant, Kc, is the value obtained for the equilibrium-constant expression when equilibrium concentrations are substituted. • A large Kc indicates large concentrations of products at equilibrium. • A small Kc indicates large concentrations of unreacted reactants at equilibrium.

24. Consider the equilibrium established in the Haber process. The Equilibrium Constant • The law of mass action states that the value of the equilibrium constant expression Kc is constant for a particular reaction at a given temperature, whatever equilibrium concentrations are substituted.

25. Note that the stoichiometric coefficients in the balanced equation have become the powers to which the concentrations are raised. The Equilibrium Constant • The equilibrium-constant expression would be

26. 2A + B  2C

27. Consider the decomposition of N2O4, dinitrogen tetroxide. Equilibrium: A Kinetics Argument • If the forward and reverse reaction rates in a system at equilibrium are equal, then it follows that their rate laws would be equal. • If we start with some dinitrogen tetroxide and heat it, it begins to decompose to produce NO2.

28. Models of N2O4 and 2NO2.

29. Obtaining Equilibrium Constants for Reactions • Equilibrium concentrations for a reaction must be obtained experimentally and then substituted into the equilibrium-constant expression in order to calculate Kc.

30. Obtaining Equilibrium Constants for Reactions • Consider the reaction below (see Figure 15.3). • Suppose we started with initial concentrations of CO and H2 of 0.100 M and 0.300 M, respectively.

31. Reactants Products [CO] = 0.0613 M [CH4] = 0.0387 M [H2] = 0.1893 M [H2O] = 0.0387 M Obtaining Equilibrium Constants for Reactions • Consider the reaction below (see Figure 15.3). • When the system finally settled into equilibrium we determined the equilibrium concentrations to be as follows.

32. Obtaining Equilibrium Constants for Reactions • Consider the reaction below (see Figure 15.3). • The equilibrium-constant expression for this reaction is:

33. Obtaining Equilibrium Constants for Reactions • Consider the reaction below (see Figure 15.3). • If we substitute the equilibrium concentrations, we obtain:

34. Obtaining Equilibrium Constants for Reactions • Consider the reaction below (see Figure 15.3). • Regardless of the initial concentrations (whether they be reactants or products), the law of mass action dictates that the reaction will always settle into an equilibrium where the equilibrium-constant expression equals Kc.

35. Obtaining Equilibrium Constants for Reactions • Consider the reaction below (see Figure 15.3). • As an example, let’s repeat the previous experiment, only this time starting with initial concentrations of products: [CH4]initial = 0.1000 M and [H2O]initial = 0.1000 M

36. Reactants Products [CO] = 0.0613 M [CH4] = 0.0387 M [H2] = 0.1893 M [H2O] = 0.0387 M Obtaining Equilibrium Constants for Reactions • Consider the reaction below (see Figure 15.3). • We find that these initial concentrations result in the following equilibrium concentrations.

37. Obtaining Equilibrium Constants for Reactions • Consider the reaction below (see Figure 15.3). • Substituting these values into the equilibrium-constant expression, we obtain the same result. • Whether we start with reactants or products, the system establishes the same ratio.

38. Figure 15.3: Some equilibrium compositions for the methanation reaction.

39. It can be seen from the ideal gas equation that the partial pressure of a gas is proportional to its molarity. The Equilibrium Constant, Kp • In discussing gas-phase equilibria, it is often more convenient to express concentrations in terms of partial pressures rather than molarities (see Figure 15.4).