1. www.cengage.com/chemistry/cracolice Mark S. Cracolice Edward I. Peters Mark S. Cracolice The University of Montana Chapter 18 Chemical Equilibrium

2. Collision Theory of Reactions Collision Theory of Gas-phase Reactions A chemical reaction can occur only when two molecules collide with a kinetic energy at least equal to certain energy E a, called activation energy of the reaction. The success of a collision also depends the relative orientation of molecules. This direction-dependence is called the steric requirement of the reaction.

3. Collision Theory of Reactions a) Sufficient energy proper orientation b) Proper orientation not sufficient energy c) Sufficient energy poor orientation Chemical reaction is the overall effect of collisions between reacting molecules

4. Collision Theory of Reactions A conversion of kinetic energy to potential energy occurs during formation of an intermediate complex that can either go on to form products or fall apart into the unchanged reactants. This can be shown by a graph that traces the energy of the system before, during, and after the collision.

5. Energy Changes During a Reaction

6. Transition State Complex In the transition state complex, the original bonds have weakened, whereas the new bonds are only partially formed. Activation Energy E a : The difference between the energy of the transition state complex and the reactant energy.

7. Rate of a Chemical Reaction Three important factors influence the speed of chemical reactions: Temperature The higher the temperature, the faster the rate of reaction. Catalysis A catalyst increases the rate of reaction. Concentration of Reactants The greater the concentration, the greater the rate of reaction.

8. Effect of Temperature on the Distribution of Energy

9. Effect of Temperature on Reaction Rates Kinetic energy distribution curves at two temperatures explain the effect of temperature on reaction rates. E a, the activation energy, is the same at both temperatures. Only the fraction of the particles in the sample represented by the area beneath the curve to the right of E a is able to react. The fraction of molecules that is able to react increases rapidly as the temperature is raised.

10. Effect of Catalyst on the Rate of a Chemical Reaction: Change in Activation Energy A catalyst speeds up a reaction by providing a new pathway that has a lower activation energy.

11. Effect of Change in Activation Energy on the Rate of a Chemical Reaction

12. The fraction of molecules that collide with kinetic energy that is at least equal to the activation energy, E a, is bigger in a catalyzed reaction because the activation energy barrier is lowered. Since E a ‘ < E a The catalyzed reaction rate is faster.

13. Effect of Concentration on Reaction Rate Reaction rate depends on the frequency of effective collisions: The more particles there are in a given volume, the more frequently collisions will occur and the more rapidly the reaction will take place.

14. Development of Equilibrium For a reversible reaction in a closed system, the equilibrium is established when the forward reaction rate is equal to the reverse reaction rate.

15. Development of Equilibrium If the system is not in equilibrium, the concentration of the species in the faster reaction will decrease, and thus the reaction will become slower; the concentration of the species in the slower reaction will increase, and thus the reaction will become faster. Opposite rates will eventually become equal, and an equilibrium will be established.

16. The Equilibrium Constant Consider the reaction H 2 (g) + I 2 (g) 2 HI(g) At equilibrium the following ratio is a constant K is called equilibrium constant

17. The Equilibrium Constant For the general equilibrium a A + b B c C + d D When writing an equilibrium constant expression, use only the concentrations of gases, (g), or dissolved substances, (aq). Do not include solids, (s), or liquids, (l).

18. The Equilibrium Constant Equilibrium Constant, K For any equilibrium at a given temperature, the ratio of the product of the concentrations of the species on the right side of the equilibrium equation, each raised to a power equal to its coefficient in the equation, to the corresponding product of the concentrations on the left side of the equation, each raised to a power equal to its coefficient in the equation, is a constant. The equilibrium constant is both equation-dependent and temperature-dependent.

19. Significance of the Value of K Example: Is the forward reaction favored, the reverse reaction favored, or are appreciable quantities of all species present at equilibrium in the following reaction? HC 2 H 3 O 2 (aq) H + (aq) + C 2 H 3 O 2 – (aq) K = 1.8 × 10 –5. Solution: Since K is very small, the reverse reaction is favored.

20. Significance of the Value of K Consider the general reaction: Reactants Products If the equilibrium constant is very large (K > 100), [Products] > [Reactants], so the forward reaction is favored. If the equilibrium constant is very small (K < 0.01), [Products] < [Reactants], so the reverse reaction is favored. If the equilibrium constant is neither larger nor small, [Products] ≈ [Reactants], so appreciable quantities of all species are present at equilibrium.

21. Le Chatelier’s Principle If a system is in equilibrium, any change imposed on the system tends to shift the equilibrium in a direction that tends to counteract the initial change. Le Chatelier’s principle only suggests an outcome; it does not provide an explanation.

22. Le Chatelier’s Principle The Pressure (Volume) Effect A gas-phase equilibrium responds to compression-a reduction in volume of the reaction vessel. If a gaseous equilibrium is compressed, the equilibrium will be shifted in the direction of formation of fewer molecules, thus minimizes the increase in pressure. If the system is expanded, the shift will be in the direction of formation of more molecules. 3 H 2 (g) + N 2 (g) ↔ 2 NH 3 (g) More molecules less molecules

23. Le Chatelier’s Principle The Pressure (Volume) Effect 3 H 2 (g) + N 2 (g) ↔ 2 NH 3 (g) If the pressure increases, the equilibrium will be shifted to the right (less molecules). If the pressure decreases, the shift will be to the left. To increase the yield of ammonia, industrial process uses pressures of 250 atm or higher.

24. Le Chatelier’s Principle The Temperature Effect If a reaction is exothermic, the reverse reaction is endothermic. If the temperature increases, the equilibrium will be shifted to the direction of consuming heat (endothermic, to the left for NH 3 reaction lelow). If the temperature decreases, the shift will be in direction of producing heat (exothermic, to the right). 3 H 2 (g) + N 2 (g) ↔ 2 NH 3 (g) + 92kJ

25. Le Chatelier’s Principle The Temperature Effect The left tube at 25 0 C contains very little brown gas compared to the tube on the right at 80 0 C

26. Le Chatelier’s Principle The Concentration Effect Let us consider the reaction equilibrium: 3H 2 (g) + N 2 (g) ↔ 2 NH 3 (g) If H 2 is added to the reaction chamber, the shift will be in the forward direction to counteract the increase in the number of hydrogen molecules thus producing more NH 3. If H 2 is removed, the equilibrium will shift to the reverse direction to increase the H 2 concentration.

27. Le Chatelier’s Principle The Concentration Effect Let us consider the reaction equilibrium: 3H 2 (g) + N 2 (g) ↔ 2 NH 3 (g) If H 2 is added (increase of [H 2 ]), the shift will be in the forward direction to increase in the concentration [NH 3 ] and decrease the concentrations [H 2 ] and [N 2 ]. Adding an inert gas has no effect on the equilibrium, although the total pressure increases.

28. Solubility Equilibria The equation for dissolving AgCl, a low-solubility compound, is AgCl (s) ↔ Ag + ( aq) + Cl - (aq) This equilibrium is characterized by the solubility product constant K sp K sp = [Ag + ] [Cl - ]

29. Solubility Equilibria Calculation of solubility product from the solubility: The chloride ion concentration of a saturated solution of silver chloride is 1.3 x 10 -5 M. Calculate the solubility product for silver chloride. K sp = [Ag + ] [Cl - ] In saturated solution of pure silver chloride the concentration of [Ag + ] and [Cl - ] are equal. Therefore K sp = [Ag + ] [Cl - ] =(1.3 x 10 -5 ) x(1.3 x 10 -5 ) = 1.7 x 10 -10

30. Solubility Equilibria Solubility and Solubility Product. For compounds of similar structure, the smaller the solubility product, the smaller the solubility. For example the solubility of silver bromide ( K sp = 5.2 x 10 -13 ) is lower than the solubility of silver chloride. Common Ion Effect Suppose that a soluble chloride, such as NaCl were to be added to the saturated solution of silver chloride. According to the Le Chatelier’s principle the equilibrium would shift the equilibrium in the reverse direction, reducing the solubility of silver chloride..

31. Ionization Equilibria: Weak acid Weak acids ionize only slightly when dissolved in water. For a general weak acid HA, HA(aq) H + (aq) + A – (aq) Major species: HA(aq) Minor Species: H + (aq) + A – (aq)

32. Ionization Equilibria: Weak acid The ionization of a weak acid is usually so small that it is negligible compared with the initial concentration of the acid. We assume that all ionization concentrations are negligible when subtracted from the initial concentration. In other words, the initial concentration of the weak acid is also the final concentration after the acid ionizes.

33. Ionization Equilibria: Weak acid Example: Find the pH of 0.1 M nitrous acid. K a = 4.5 × 10 –4. Solution: HNO 2 (aq) H + (aq) + NO 2 – (aq) Let x = [H + ] = [NO 2 – ]; [HNO 2 ] = 0.1 M

34. Ionization Equilibria: Weak acid Find the pH of 0.1 M nitrous acid. K a = 4.5 × 10 –4. Solution: x 2 = (0.1) (4.5 × 10 –4 ) x = [H + ] = 7 × 10 –3 pH = – log [H + ] = – log (7 × 10 –3 ) = 2.2

35. Ionization Equilibria: Buffer Solution Buffer Solution A solution that resists changes in pH because it contains relatively high concentrations of both a weak acid and a weak base. The acid reacts with any added OH – ; The base reacts with any added H +.

36. Ionization Equilibria: Buffer Solution Determine the pH of a solution that is 0.25 M in HAc and 0.35 M in NaAc. K a = 1.8 × 10 –5. NaAc(aq)  Na + (aq) + Ac – (aq) HAc(aq) H + (aq) + Ac – (aq) pH = – log [H + ] = – log (1.3 × 10 –5 ) = 4.89

37. Ionization Equilibria: Buffer Solution Example: Determine the acid-to-base concentration ratio that will yield a buffer solution with a pH of 4.50 if the acid has K a = 1.0 × 10 –5. Solution: HA(aq) H + (aq) + A – (aq) [H + ] = antilog (–pH) = antilog (–4.50) = 3.2 × 10 –5 M

38. Homework Homework: 25, 27, 29, 31, 33, 35, 37, 41, 43, 47, 61, 63, 71, 75