1. Reaction Rates and Equilibrium Chapter 17

2. Collision Theory or Model Molecules react by colliding with each other with enough energy and proper orientation to break bonds, rearrange and form new bonds Explains why increasing concentration of reactants (solutions) increases rate of reaction (more particles collide) Explains why increasing temperature increases reaction rate (increases the energy (speed)) of the particles

3. Collision Theory

5. Collision Theory or Model Why does increasing concentration or temperature increase the rate of reaction? More particles have collisions with enough energy and proper orientation to reach activation energy

6. Collision Theory or Model Increasing pressure of gases increases rate of reaction. Why? It increases the number of collisions that occur by reducing the volume.

7. Collision Theory or Model Increasing surface area increases rate of heterogeneous reaction. Why? More places for reactions to occur (locations for collisions)

8. Catalyst and Reaction Rate Catalyst is not used up in the reaction Lowers the activation energy of the reaction Provides a “shortcut” for reaction Enzymes are examples of catalysts

9. Activation Energy

10. Homogeneous Reactions Reactions that involve only one phase or state of matter. Ex. N 2 (g) + 3H 2 (g)  2NH 3 (g)

11. Heterogeneous Reactions Reactions involving more than one phase or state of matter. Ex. Mg(s) + 2HCl(aq)  MgCl 2 (aq) + H 2 (g)

12. Equilibrium When forward and backward reactions are occurring at the same rate, the system is said to be at equilibrium. Formation of reactants and products occur at the same rate, but the quantities are not necessarily the same.

13. Assignment Read Chapter 17.1 Answer Section17.1 Review Questions 1-7

15. Answers – page 604 1. Molecules must collide for reactions to occur. 2. A catalyst provides a new pathway for the reaction at a lower activation energy. 3. At lower temperatures, molecules have less kinetic energy. Therefore, fewer molecules will have the necessary activation energy for the reaction making it slower.

16. Answers con’t. 4. Grinding increases surface area which increases locations for collisions to occur. More collisions increases the rate. 5. 2NaCl(s) + H 2 SO 4 (aq)  Na 2 SO 4 (aq) + 2HCl (aq) 2H 2 (g) + O 2 (g)  2H 2 O(g) 6. The rates of the forward and reverse reactions are equal at equlibrium. 7. Changes are occurring at the molecular level.

17. LeChatelier’s Principle When a stress is applied to a system, it will react in such a fashion as to counter or offset the stress. Most reactions are reversible so stress may favor formation of either reactant(s) or product(s). It depends on the type of “stress”.

18. LeChatelier’s Principle What kind of stressors may be applied to a system? Addition or removal of a reactant Addition or removal of a product Increase or decrease temperature Increase or decrease pressure Pure solids and liquids do not affect equilibrium. Why?

19. Pressure Increasing pressure tends to favor the side of the reaction with the fewest number of gas molecules. Decreasing pressure tends to favor the side of the reaction with the largest number of gas molecules.

20. If more of a reactant is added, which way do you think the equilibrium will shift- towards the reactants or towards the product? Why? If the reaction is exothermic and the temperature is increased, which way will the equilibrium shift? Why? LeChatelier’s Principle

21. Equilibrium Constant Equilibrium is expressed in terms of concentrations of the products and reactants From the balanced chemical equation, the coefficients become exponents (superscripts) If the actual concentrations (moles per liter) of the reactants and products is known, K can be calculated Remember, pure solids and liquids ( such as water) do not affect equilibrium K = [Prod] x [ucts] y [Reac] x [tants] y Ex. 2O 3 (g)  3O 2 (g) K = [O 2 ] 3 [O 3 ] 2

22. Assignment Read 17.2 Complete Practice Problem 17.3 (yellow box) a-d, bottom of page 610 Complete Section 17.2 Section Review (page611) 1-5

24. Answers p.610 A. K = [O 2 ] 3 B. K = [NO 2 ] [H 2 O] 2 C. K = __1__ [ CO 2 ] D. K = __1__ [SO 3 ]

25. Answers p. 611 1. K= __[NO 2 ] 2 __ [NO] 2 [O 2 ] 2. K = [0.141] 2 [0.000104] [0.000201] 3 K = 2.35 X 10 13 3. Equilibrium position is a set of equilibrium concentrations. The equilibrium constant is a specific ratio of these concentrations. Ex..5 = 1/2, 2/4

26. Answers con’t. p. 611 4. Their concentrations do not change. 5. a) K = ___1___ [NH 3 ] [HCl] b) K = [H 2 O] 2 [H 2 ] 2 [O 2 ] c) K = __1__ [H 2 ] 2 [O 2 ]