2. Chapter 14 & 16 Chemical Equilibrium and reaction rates

3. 16.1 Rates of Rxn: Collision Theory Amount of time for a reaction to take place can vary greatly - the measurement of this time is in the form of a rate -measurement of the speed of any change that occurs within an interval of time. ex- iron rusting may be expressed as 0.50 mol/year

4. Reaction Rate The change in concentration of a reactant or product per unit of time

5. Collision Theory Reaction rate depends on the collisions between reacting particles. Successful collisions occur if the particles... –collide with each other –have the correct orientation –have enough kinetic energy to break bonds CO + NO 2  CO 2 + NO Ineffective collision Effective collision N O O O C N O O C O

6. Collision Theory Collision Theory-atoms, ions and molecules can react to form products when they collide, provided they have enough kinetic energy. This minimum amount of KE in order to react is called the activation energy (E a )

7. Collision Theory Activation Energy –depends on reactants –low E a = fast rxn rate EaEa

8. Endothermic Reactions

9. Exothermic Reactions

10. Factors Affecting Rxn Rate 5 mph “fender bender” 50 mph “high-speed crash” 1. Temperature- usually increasing temperature increases the reaction rate. This increases the # of particles having enough KE

11. Factors Affecting Rxn Rate 2. Concentration –increased concentration = increased rxn rate –more opportunities for collisions (collision frequency increases)

12. Factors Affecting Rxn Rate 3. Particle Size –smaller particle size increases the rate. –This increases the amount of total surface area exposed therefore increasing the collision frequency

13. Factors Affecting Rxn Rate 4. Catalyst* –substance that increases rxn rate without being consumed in the rxn –lowers the activation energy *inhibitor- a substance that interferes with the catalyst. This can slow down the rate or even stop it completely.

14. Catalysis Catalyst: A substance that speeds up a reaction without being consumed Enzyme: A large molecule (usually a protein) that catalyzes biological reactions. Homogeneous catalyst: Present in the same phase as the reacting molecules. Heterogeneous catalyst: Present in a different phase than the reacting molecules.

15. Exothermic Reaction with a Catalyst

16. Endothermic Reaction with a Catalyst

17. A catalyst can speed up the rate of a given chemical reaction by A increasing the equilibrium constant in favor of products. B lowering the activation energy required for the reaction to occur. C raising the temperature at which the reaction occurs. D increasing the pressure of reactants, thus favoring products.

18. Which reaction diagram shows the effect of using the appropriate catalyst in a chemical reaction?

19. H 2 O 2 hydrogen peroxide, naturally breaks down into H 2 O and O 2 over time. MnO 2, manganese dioxide, can be used to lower the energy of activation needed for this reaction to take place and, thus, increase the rate of reaction. What type of substance is MnO 2 ? A a catalyst B an enhancer C an inhibitor D a reactant

20. 14.1 Chemical Equilibrium Reversible Reactions: A chemical reaction in which the products can react to re-form the reactants Chemical Equilibrium: When the rate of the forward reaction equals the rate of the reverse reaction the concentration of products and reactants remains unchanged 2HgO(s)  2Hg(l) + O2(g) Arrows going both directions (  ) indicates equilibrium in a chemical equation

21. Rate Comparison for H 2 (g) + I 2 (g) 2HI(g) Equal amounts of H 2 and I 2, [HI] goes up. Decline in the rate of the forward system, most reaction has been carried out The greater number of HI, reverse reaction starts to occur and reverse reaction increases in rate. At some point, rate of forward and reverse becomes equal

22. Chemical Equilibria are Dynamic In some cases, the equilibrium has a higher concentration of products than reactants. This type of equilibrium is also shown by using two arrows. The forward reaction has a longer arrow to show that the products are favored.

23. 14.2 Systems at Equilibrium Equilibrium Constants (K eq ) is the ratio of product concentration to reactant concentration aA + bB  cC + dD K eq = [C] c [D] d K eq = [product] [A] a [B] b [reactant] –K eq < 1, reactants favored at equal. –K eq > 1, products favored at equal. –Does not include pure solids and liquids

24. Write the equilibrium expressions, K c, for the following reaction: 2 NO 2 (g)  2 NO(g) + O 2 (g) K eq ; Equilibrium constant (No Units) K c (concentration); Used for solutions (aq) or gases. K p (Pressure); Used for gases only. Other Equilibrium Expressions

25. Write a equilibrium expression for PCl 5(g)  PCl 3(g) + Cl 2(g) For the above reaction, the equilibrium constant is 35. If the concentrations of PCl 5 and PCl 3 are 0.015 M and 0.78 M respectively. What is the concentration of Cl 2 ? [Cl 2 ] = 0.67 M

26. An aqueous solution of carbonic acid reacts to reach equilibrium as described below. The solution contains the following solution concentrations: carbonic acid, 3.3 × 10 −2 M; bicarbonate ion, 1.19 × 10 −4 M hydronium ion, 1.19 × 10 −4 M. Determine the K eq. Which is more favorable? Forward or backward? Backward, LOW Keq

27. For the system involving N 2 O 4 and NO 2 at equilibrium at a temperature of 100C, the product concentration of N 2 O 4 is 4.0x10 -2 M and the reactant concentration of NO 2 is 1.4x10 -1 M. What is the K eq value for this reaction? 2 NO 2  N 2 O 4 Keq= 2.0

28. Calculating equilibrium concentrations from initial concentrations: Steps in calculating equilibrium concentrations: (1) Write equilibrium expression (2) Calculate change in concentration of each substance (3)Calculate equilibrium concentrations Given the following reaction: A  B + C. The initial concentration of A is 0.260 M and at equilibrium, [C] is 0.160 M. Calculate the K eq. Initial (M) Change (M) Equilibrium (M) A  B + CK eq = ? = 0.256 0.26000  x + x  0.160+ 0.160 0.160 0.100

29. Given the following reaction: H 2(g) + I 2(g) ↔ 2 HI (g). At 25°C, 0.100 mol of H 2 and 0.100 mol of I 2 react in a 1.00 L container. At equilibrium, the [H 2 ] is 0.080 M. Calculate K eq at this temperature. Initial (M) Change (M) Equilibrium (M) 0.100 0  x + 2x  0.020 + 0.040 0.0400.080 H 2 + I 2  2 HIM = mol / L = 0.25

30. K eq for the equilibrium below is 1.8 × 10 −5 at a temperature of 25°C. Calculate when [NH 3 ] = 6.82 × 10 −3. Initial (M) Change (M) Equilibrium (M) 6.82 x 10 -3 00  x + x xx6.82x10 -3 - x Keq = ? NH 3  NH 4 + + OH- Disregard, to make the calculation easier! x 2 = (1.8  10 −5 )  (6.82  10 −3 ) = 1.2 × 10 −7

31. 14.3 Le Chatelier’s Principle When a system at equilibrium is placed under stress, the system will undergo a change in such a way as to relieve that stress.

32. When you take something away from a system at equilibrium, the system shifts in such a way as to replace what you’ve taken away. Le Chatelier Translated: if a stress is applied to a system in a dynamic equilibrium, the system changes to relieve the stress

33. Factors Affecting Equilibrium: Le Chatelier’s Principle 2 SO 2(g) + O 2(g)  2SO 3(g) + heat –Concentration: Increasing O 2, will cause the shift  (away from O 2 ) –Temperature: Increasing T will cause the shift  (away from heat) –Pressure: Increase in P will shift  (favors side with less gas molecules)

34. LeChatelier Example #1 A closed container of ice and water at equilibrium. The temperature is raised. Ice + Energy (ΔH)  Water The equilibrium of the system shifts to the _______ to use up the added energy. right

35. LeChatelier Example #2 A closed container of N 2 O 4 and NO 2 at equilibrium. NO 2 is added to the container. N 2 O 4 (g) + Energy (ΔH)  2 NO 2 (g) The equilibrium of the system shifts to the _______ to use up the added NO 2. left

36. LeChatelier Example #3 A closed container of water and its vapor at equilibrium. Vapor is removed from the system. water + Energy (ΔH)  vapor The equilibrium of the system shifts to the _______ to replace the vapor. right

37. LeChatelier Example #4 A closed container of N 2 O 4 and NO 2 at equilibrium. The pressure is increased. N 2 O 4 (g) + Energy (ΔH)  2 NO 2 (g) The equilibrium of the system shifts to the _______ to lower the pressure, because there are fewer moles of gas on that side of the equation. left

38. 6CO 2(g) + 6H 2 O (l) C 6 H 12 O 6(s) + 6O 2(g) ∆H = 2820 kJ. Some CO 2(g) is added and causes... Shift to right so get more C 6 H 12 O 6(s). Temperature is raised... Shift to right, more C 6 H 12 O 6(s) forms. Volume is decreased There are equal moles of gaseous reactants and products, so no change.

39. 6CO 2(g) + 6H 2 O (l) C 6 H 12 O 6(s) + 6O 2(g) ∆H = 2820 kJ Some O 2(g) is removed... Shift to right, more C 6 H 12 O 6(s) forms. Some C 6 H 12 O 6(s) is removed... No change (glucose is a solid so is not in the equilibrium expression). No more C 6 H 12 O 6(s). A catalyst is added... No effect on equilibrium, only allows it be attained more quickly. No more C 6 H 12 O 6(s). Some H 2 O is removed... No change, water is a liquid. No additional C 6 H 12 O 6(s) forms.

40. When a reaction is at equilibrium and more reactant is added, which of the following changes is the immediate result? –A The reverse reaction rate remains the same. –B The forward reaction rate increases. –C The reverse reaction rate decreases. –D The forward reaction rate remains the same.

41. In which of the following reactions involving gases would the forward reaction be favored by an increase in pressure? –A A + B  AB –B A + B  C + D –C 2A + B  C + 2D –D AC  A + C

42. 4HCl(g) + O 2(g)  2H2O(l)+ 2Cl2(g) + 113kJ Which action will drive the reaction to the right? –A heating the equilibrium mixture –B adding water to the system –C decreasing the oxygen concentration –D increasing the system’s pressure

43. NO2(g) + CO(g)  NO(g) + CO2(g) The reaction shown above occurs inside a closed flask. What action will shift the reaction to the left? –A pumping CO gas into the closed flask –B raising the total pressure inside the flask –C increasing the NO concentration in the flask

44. 16.2 How can rate reaction be explained? The rate law describes the way in which reactant concentration affects reaction rate. –A rate law is the expression that shows how the rate of formation of product depends on the concentration. (all other than solvent)

45. Rate law equation Consider the equation A + B  C + D rate = k [reactant] x x = Order of reaction can not be determined from the equation. MUST be calculated Rate of reaction = k [A] x [B] y k = rate constant x = order of reaction with respect to A y = order of reaction with respect to B x + y = overall order of reaction

46. To calculate the rate of reaction, Ratio of rates a two different concentrations of A

47. Determine the reaction order and write the rate law equation 2HI(g) → H 2 (g) + I 2 (g) 4 = 2 x x = 2 Rate of reaction = k [HI] x Rate = k [HI] 2

48. The initial rate of decomposition of acetaldehyde, CH 3 CHO, CH 3 CHO --> CH 4 + CO at 600  C was measured at a series of concentrations with the following results: a) Determine the reaction order. (b) Write the rate law. Experiments1234 [CH 3 CHO] (M)0.100.200.300.40 rate (mol/L s)0.0850.340.761.4 4 = 2 x x = 2 Rate of reaction = k [CH 3 CHO] x Rate = k [CH 3 CHO] 2 (The power of the [ ] of a reactant) Choose any two experiments