Upcoming SlideShare
×

# New chm 152 unit 2 power points sp13

524 views
267 views

Published on

0 Likes
Statistics
Notes
• Full Name
Comment goes here.

Are you sure you want to Yes No
• Be the first to comment

• Be the first to like this

Views
Total views
524
On SlideShare
0
From Embeds
0
Number of Embeds
2
Actions
Shares
0
8
0
Likes
0
Embeds 0
No embeds

No notes for slide

### New chm 152 unit 2 power points sp13

1. 1. CHEMICAL EQUILIBRIUM It is a dynamic state of things. Chapters 4.7 & 17 - Silberberg
2. 2. Goals & Objectives      See the following Learning Objectives on page 773. Understand These Concepts: 17.1-7, 9, 11-16. Master These Skills: 17.1-4, 6-17
3. 3. The Equilibrium State All reactions are reversible and under suitable conditions will reach a state of equilibrium. At equilibrium, the concentrations of products and reactants no longer change because the rates of the forward and reverse reactions are equal. At equilibrium: rateforward = ratereverse Chemical equilibrium is a dynamic state because reactions continue to occur, but because they occur at the same rate, no net change is observed on the macroscopic level.
4. 4. Basic Concepts   Most chemical reactions do not go to completion. Reversible reactions do not go to completion and can occur in either direction.
5. 5. Basic Concepts    Reversible reactions may be represented in general terms as the following: aA(g) + bB(g) = cC(g) + dD(g) Chemical equilibrium exists when the forward and reverse reactions are occuring at exactly the same rate and therefore there is no net change in the concentration of reactants and products.
6. 6. Basic Concepts    For the generalized reaction A(g) + B(g) = C(g) + D(g) the reaction may be represented graphically
7. 7. Figure 17.3 The change in Q during the N2O4-NO2 reaction.
8. 8. Basic Concepts   It makes no difference if we start with the “reactants” or the “products” of a reversible reaction, because equilibrium can be established from either direction. Once established, the equilibrium is “dynamic”--it continues with no net change.
9. 9. The Equilibrium Constant     The general form for the equilibrium constant for the reaction aA(g) + bB(g) = cC(g) + dD(g) Kc = [C]c[D]d [A]a[B]b
10. 10. The Equilibrium Constant     Write equilibrium constant expressions for the following reactions at 500oC. 1. 2. 3. + PCl5(g) = PCl3(g) + Cl2(g) H2(g) + I2(g) = 2HI(g) 4NH3(g) + 5O2(g) = 4NO(g) 6H2O(g)
11. 11. The Equilibrium Constant  The thermodynamic definition of the equilibrium constant involves activities rather than concentrations.    For pure liquids and solids, the activity is taken to be 1. For components of ideal solutions, the activity of each component is taken to be its molar concentration. For mixtures of ideal gases, the activity of each component is taken to be its partial pressure in atmospheres.
12. 12. The Equilibrium Constant  Because of the use of activities, equilibrium constants have no units.
13. 13. Exercises   One liter of the following reaction system at a high temperature was found to contain 0.172 moles of phosporous trichloride, 0.086 moles of chlorine and 0.028 moles of phosporous pentachloride at equilibrium. Determine the value of Kc for the reaction. PCl5 (g) = PCl3 (g) + Cl2(g)
14. 14. Exercises   The decomposition of PCl5 was studied at another temperature. One mole of PCl5 was introduced into an evacuated 1.00L container at the new temperature. At equilibrium, 0.60 moles of PCl3 were present in the container. Calculate the equilibrium constant at this temperature. PCl5 (g) = PCl3 (g) + Cl2(g)
15. 15. Exercises   At a given temperature, 0.80 moles of N2 and 0.90 moles of H2 were placed in an evacuated 1.00L container. At equilibrium, 0.20 moles of NH3 were present. Determine the value for Kc for the reaction. N2(g) + 3H2(g) = 2NH3(g)
16. 16. The Reaction Quotient, Q   The reaction quotient, Q, has the same form as the equilibrium constant, Kc, except the concentrations are not necessarily equilibrium concentrations. Comparison of Q with Kc enables the prediction of the direction the reaction will occur to the greater extent when a system is not at equilibrium.
17. 17. The Reaction Quotient, Q   The relationship between Q and Kc When    Q = Kc Q > Kc the Q < Kc the the system is at equilibrium reaction occurs to the left to greater extent reaction occurs to the right to greater extent
18. 18. Determining the Direction of Reaction The value of Q indicates the direction in which a reaction must proceed to reach equilibrium. If Q < K, the reactants must increase and the products decrease; reactants → products until equilibrium is reached. If Q > K, the reactants must decrease and the products increase; products → reactants until equilibrium is reached. If Q = K, the system is at equilibrium and no further net change takes place.
19. 19. Figure 17.5 Reaction direction and the relative sizes of Q and K. Q>K Q<K Q=K
20. 20. Solving Equilibrium Problems If equilibrium quantities are given, we simply substitute these into the expression for Kc to calculate its value. If only some equilibrium quantities are given, we use a reaction table to calculate them and find Kc. A • • • • reaction table shows the balanced equation, the initial quantities of reactants and products, the changes in these quantities during the reaction, and the equilibrium quantities.
21. 21. Figure 17.6 Steps in solving equilibrium problems. PRELIMINARY SETTING UP 1. Write the balanced equation. 2. Write the reaction quotient, Q. 3. Convert all amounts into the correct units (M or atm). WORKING ON THE REACTION TABLE 4. When reaction direction is not known, compare Q with K. 5. Construct a reaction table.  Check the sign of x, the change in the concentration (or pressure).
22. 22. Figure 17.6 continued SOLVING FOR x AND EQUILIBRIUM QUANTITIES 6. Substitute the quantities into Q. 7. To simplify the math, assume that x is negligible: ([A]init – x = [A]eq ≈ [A]init) 8. Solve for x.  Check that assumption is justified (<5% error). If not, solve quadratic equation for x. 9. Find the equilibrium quantities.  Check to see that calculated values give the known K.
23. 23. Exercises  The equilibrium constant for the following reaction is 49 at 450oC. If 0.22 moles of I2, 0.22 moles of H2 and 0.66 moles of HI were placed in an evacuated 1.00L container, determine if the system is at equilibrium. If not, in which direction will the reaction occur to the greater extent to achieve equlibrium.
24. 24. Exercises(cont)  H2(g) + I2(g) = 2HI(g)
25. 25. Uses of the Equilibrium Constant  The equilibrium constant, Kc, is 3.00 for the following reaction at a given temperature. If 1.00 moles of SO2 and 1.00 moles of NO2 are put into an evacuated 2.00L container and allowed to reach equilibrium, determine the concentration of each compound at equilibrium.
26. 26. Uses of the Equilibrium Constant(cont.)  SO2(g) + NO2(g) = SO3(g) + NO(g)
27. 27. Uses of the Equilibrium Constant(cont.)   The equilibrium constant is 49 for the following reaction at 450oC. If 1.00 mole of HI is placed in an evacuated 1.00L container and allowed to reach equilibrium, determine the equilibrium concentration of all species. H2(g) + I2(g) = 2HI(g)
28. 28. Le Châtelier’s Principle When a chemical system at equilibrium is disturbed, it reattains equilibrium by undergoing a net reaction that reduces the effect of the disturbance. A system is disturbed when a change in conditions forces it temporarily out of equilibrium. The system responds to a disturbance by a shift in the equilibrium position. A shift to the left is a net reaction from product to reactant. A shift to the right is a net reaction from reactant to product.
29. 29. Le Châtelier’s Principle When a chemical system at equilibrium is disturbed, it reattains equilibrium by undergoing a net reaction that reduces the effect of the disturbance. A system is disturbed when a change in conditions forces it temporarily out of equilibrium. The system responds to a disturbance by a shift in the equilibrium position. A shift to the left is a net reaction from product to reactant. A shift to the right is a net reaction from reactant to product.
30. 30. Factors That Affect Equilibrium  LeChatelier’s Principle   If a change of conditions(stress) is applied to a system in equilibrium, the system responds in the way that best tends to reduce the stress by reaching a new state of equilibrium. Changes in concentrations, pressure, and temperature are considered stresses to the system.
31. 31. Figure 17.7 The effect of a change in concentration.
32. 32. 1. Changes in concentration  Consider the following reaction at equilibrium   H2(g) + I2(g) = 2HI(g) Determine the equilibrium shift if     H2 is added H2 is removed HI is removed I2 is added
33. 33. 2. Changes in volume  If the volume of the container is decreased for a system at equilibrium, the concentrations of all gases(but not liquids or solids) will increase. If the balanced equation has more moles of gas on the reactant side than on the product side, the forward reaction is favored. An increase in volume has the opposite effect.
34. 34. 2. Changes in volume  Consider the following reaction at equilibrium at a constant temperature:   2 NO2(g) = N2O4(g) Determine the equilibrium shift if    the volume is decreased(pressure increased) the volume is increased(pressure decreased) the pressure is increased at constant volume by the addition of an inert gas
35. 35. The Effect of a Change in Temperature To determine the effect of a change in temperature on equilibrium, heat is considered a component of the system. Heat is a product in an exothermic reaction ( H rxn < 0). Heat is a reactant in an endothermic reaction ( H rxn > 0). An increase in temperature adds heat, which favors the endothermic reaction. A decrease in temperature removes heat, which favors the exothermic reaction.
36. 36. 3. Changes in temperature   Increasing the temperature always favors the reaction that consumes heat, and vice-versa. Consider the following reaction at equilibrium at a given temperature   2SO2(g) + O2(g) = 2SO3(g) + 198kJ Determine the equilibrium shift if   T is increased T is decreased
37. 37. 4. Introduction of a catalyst  Addition of a catalyst increases the rate at which equilibrium is achieved but has no effect on the final equilibrium. The same equilibrium is achieved but in a shorter time.
38. 38. Changes in the Value of Kc    Changes in concentration, volume, and the addition of a catalyst do not change the value of Kc. Changes in temperature do effect the value of Kc. Kc will increase if the forward reaction is endothermic and decrease if the forward reaction is exothermic. See table 17.4.
39. 39. Table 17.4 Effects of Various Disturbances on a System at Equilibrium
40. 40. Exercises  Given the following reaction at equilibrium in a closed container at 500oC. Indicate the direction the equilibrium would shift(left, right, no shift) and the effect on the value of Kc(increase, decrease, no change) for each of the following changes.
41. 41. Exercises(cont.)        N2(g) + 3H2(g) =2NH3(g) H=-92kJ Position Kc Increase T Decrease V Add N2 Remove NH3 Add a catalyst
42. 42. Exercises  Given the following reaction at equilibrium in a closed container at 500oC. Indicate the direction the equilibrium would shift(left, right, no shift) and the effect on the value of Kc(increase, decrease, no change) for each of the following changes.
43. 43. Exercises        H2(g) + I2(g) = 2HI (g) H= +25kJ Position Kc Increase T Increase V Add HI Remove H2 Add a catalyst
44. 44. K and Q for hetereogeneous equilibrium A hetereogeneous equilibrium involves reactants and/or products in different phases. A pure solid or liquid always has the same “concentration”, i.e., the same number of moles per liter of solid or liquid. The expressions for Q and K include only species whose concentrations change as the reaction approaches equilbrium. Pure solids and liquids are omitted from the expression for Q or K.. For the above reaction, Q = [CO2]
45. 45. Heterogeneous Equilibria  involve two or more phases or states of matter. The activities of solids and liquids are unity. Write Kc expressions for each of the lowing equilibria. CaCO3(s) = CaO(s) + CO2(g) SO2(g) + H2O(l) = H2SO3(aq)  CaF2(s)    = Ca2+(aq) + 2F-1(aq)