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Chemical Equilibrium

Chemical Equilibrium. AP Chem Unit 13. Chemical Equilibrium. The Equilibrium Condition The Equilibrium Constant Equilibrium Expressions Involving Pressures Heterogeneous Equilibria Applications of the Equilibrium Constant Solving Equilibrium Problems Le Chatelier’s Principle.

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Chemical Equilibrium

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  1. Chemical Equilibrium AP Chem Unit 13

  2. Chemical Equilibrium • The Equilibrium Condition • The Equilibrium Constant • Equilibrium Expressions Involving Pressures • Heterogeneous Equilibria • Applications of the Equilibrium Constant • Solving Equilibrium Problems • Le Chatelier’s Principle

  3. Introduction To this point, we have assumed that reactions proceed to completion, that is, until one of the reactants runs out. • Most reactions stop short of completion. In fact, the system reaches chemical equilibrium, the state where the concentrations of all reactants and products remain constant with time.

  4. Introduction Any chemical reactions carried out in a closed vessel will reach equilibrium • Some reactions, the equilibrium position favors the products so that the reaction appears to go to completion. The equilibrium position is said to lie “far to the right”. This reflects the direction of the products.

  5. Introduction Some reactions only occur to a slight extent. • In this case, the equilibrium position is said to lie “far to the left”. This reflects the direction of the reactants.

  6. The Equilibrium Constant 13.1

  7. The Equilibrium Condition Equilibrium is not static but a highly dynamic situation. On a molecular level many molecules are moving back and forth between reactants and products. • No net change in concentration of reactants and products.

  8. The Equilibrium Condition H2O(g) + CO(g) H2(g) + CO2(g)

  9. The Equilibrium Condition H2O(g) + CO(g) H2(g) + CO2(g)

  10. The Equilibrium Condition H2O(g) + CO(g) H2(g) + CO2(g) The equilibrium position lies far to the right. This reaction favors the products. But the reactants never reach a concentration of zero.

  11. The Equilibrium Condition H2O(g) + CO(g)H2(g) + CO2(g) What would happen if H2O(g) was added to the system? First, the forward reaction would increase, then the reverse reaction would increase. A new equilibrium would occur.

  12. Characteristics of Chemical Equilibrium The equilibrium position is determined by many factors: • initial concentrations. • relative energies of the reactants and products. • relative degree of “organization” of the reactants and products.

  13. The Equilibrium Constant 13.2

  14. Law of Mass Action The Law of Mass Action is a general description of the equilibrium condition. jA + kBlC + mD • The square brackets indicate the concentrations of the the reactants and products at equilibrium. K is the equilibrium constant.

  15. Practice Problem #1 Write the equilibrium expression for the following reaction: 4NH3(g)+ 7O2(g) 4NO2(g) + 6H2O(g)

  16. The Equilibrium Constant The value of the equilibrium constant at a given temperature can be calculated if we know the equilibrium concentrations of the reaction components. • Equilibrium constants are typically given without units.

  17. Practice Problem #2 The following equilibrium concentrations were observed for the Haber process for the synthesis of ammonia at 127°C: [NH3] = 3.1 x 10-2mol/l [N2] = 8.5 x 10-1mol/l [H2] = 3.1 x 10-3mol/l a) Calculate the value of K at 127°C for this reaction. 3.8 x 104

  18. Practice Problem #2 The following equilibrium concentrations were observed for the Haber process for the synthesis of ammonia at 127°C: [NH3] = 3.1 x 10-2mol/l [N2] = 8.5 x 10-1mol/l [H2] = 3.1 x 10-3mol/l b) Calculate the value of the equilibrium constant at 127°C for the reaction: 2NH3(g) N2(g) + 3H2(g) 2.6 x 10-5 (the reverse order reaction gives the reciprocal of K)

  19. Practice Problem #2 The following equilibrium concentrations were observed for the Haber process for the synthesis of ammonia at 127°C: [NH3] = 3.1 x 10-2mol/l [N2] = 8.5 x 10-1mol/l [H2] = 3.1 x 10-3mol/l c) Calculate the value of the equilibrium constant at 127°C for the reaction: 1.9 x 102 (When the coefficients are ½ of the balanced equation, new K = K1/2)

  20. Equilibrium Expression Summary • The equilibrium expression for a reaction is the reciprocal of that for the reaction written in reverse. • When the balanced equation for a reaction is multiplied by a factor n, the equilibrium expression for the new reaction is the original expression raised to the nth power. Knew =Kon • K values are customarily written without units. • Law of mass action can describe reactions in the solution and gas phase.

  21. Equilibrium Expression Summary • The equilibrium expression and constant for a reaction is the same at a given temperature, regardless of the initial amounts of the reaction components. • equilibrium concentrations will not always be the same. • See Table 13.1 p600

  22. Equilibrium Expression A set of equilibrium concentrations is called an equilibrium position. • There is only one equilibrium constant for a particular system at a given temperature, but there is an infinite number of equilibrium positions.

  23. Practice Problem #3 These results were collected for two experiments involving the reaction at 600°C between gaseous sulfur dioxide and oxygen to form gaseous sulfur trioxide: Show that the equilibrium constant is the same in both experiments. 4.36 and 4.32, within experimental error.

  24. Equilibrium Expressions Involving Pressures 13.3

  25. Pressure Equilibria So far we have described equilibria involving gases in terms of concentrations. Equilibria involving gases also can be described with pressure. • C represents the molar concentration of the gas. • jA + kBlC + mD

  26. Practice Problem #4 The reaction for the formation of nitrosyl chloride: 2NO(g) + Cl2(g) 2NOCl(g) was studied at 25°C. The presurres at equilibrium were found to be: NOCl =1.2 atm, NO = 5.0 x 10-2atm, Cl2 = 3.0 x 10-1 atm. Calculate the value of Kp for this reaction at 25°C. 1.9 x 103

  27. Kc vs. Kp Kp = Kc(RT)Δn • Δn is the sum of the coefficients of the gaseous products minus the sum of the coefficients of the gaseous reactants • jA + kBlC + mD • Δn=(l + m) – (j + k) • more moles of gas = more pressure

  28. Practice Problem #5 Using the value of Kp obtained in Problem #4, calculate the value of K at 25°C for the reaction: 2NO(g) + Cl2(g) 2NOCl(g) Kp=1.9 x 103 K = 4.6 x 104

  29. Heterogeneous Equilibria 13.4

  30. Homogenous vs. Heterogeneous • Homogenous equlibria is where all the reactants are in the same phase. Typically gases • Heterogeneous equilibria involve more than one phase. • The position of a heterogeneous equilibrium does not depend on the amounts of pure solids or liquids present. • Concentrations of pure solids and liquids cannot change. • Concentrations of pure solids and liquids are not included in the equilibrium expression for the reaction

  31. Example • CaCO3(s)CaO(s) + CO2(g) • This simplification only occurs with pure solids or liquids and not solutions or gases.

  32. Practice Problem #6 Write the expressions for K and Kp for the following processes: a) Solid phosphorus pentachloride decomposes to liquid phosphorus trichloride and chloride gas. b) Deep blue solid copper (II) sulfate pentahydrate is heated to drive off water vapor to form white solid copper (II) sulfate.

  33. Practice Problem #6 a) Solid phosphorus pentachloride decomposes to liquid phosphorus trichloride and chloride gas. K = [Cl2] and Kp=PCl2

  34. Practice Problem #6 b) Deep blue solid copper (II) sulfate pentahydrate is heated to drive off water vapor to form white solid copper (II) sulfate. K = [H2O]5Kp= (PH2O)5

  35. Applications of the Equilibrium Constant 13.5

  36. Applications of the Equilibrium Constant Knowing the equilibrium constant for a reaction allows us to predict several important features of the reaction • The tendency for the reaction to occur (but not the speed). • Whether or not a given set of concentrations represents an equilibrium condition.

  37. Applications of the Equilibrium Constant If the reaction is not at equilibrium, we can determine which way the reaction is moving by taking the current law of mass action ratio and comparing it to the equilibrium constant. • The ratio of non-equilibrium concentrations gives us the reaction quotient, Q.

  38. Applications of the Equilibrium Constant To determine which direction a system will shift to reach equilibrium, we compare the values of Q and K. • Q=K. The system is at equlibrium; no shift will occur. • Q>K. The initial concentrations of product to initial reactants is too large. To reach equilibrium, the system must shift left, consuming products and forming reactants. • Q<K. The ratio of initial concentrations of products to initial concentrations of reactants is too small. The system must shift to the right to form more products.

  39. Applications of the Equilibrium Constant

  40. Practice Problem #7 For the synthesis of ammonia at 500°C, the equilibrium constant is 6.0 x 10-2. Predict the direction in which the system will shift to reach equilibrium in each of the following cases: • [NH3]=1.0x10-3M, [N2]=1.0x10-5M, [H2]=2.0x10-3M • [NH3]=2.00x10-4M, [N2]=1.50x10-5M, [H2]=3.54x10-1M • [NH3]=1.0x10-4M, [N2]=5.0M, [H2]=1.0x10-2M a) Q>K, shift left. b)Q=K, no shift. c)Q<K, shift right.

  41. Practice Problem #8 Dinitrogen tetroxide in its liquid state was used as one of the fuels on the lunar lander for the NASA Apollo missions. In the gas phase it decomposes to gaseous nitrogen dioxide. N2O4(g) 2NO2(g) Consider an experiment in which gaseous N2O4 was placed in a flask and allowed to reach equilibrium at a temperature where Kp=0.133. At equilibrium, the pressure of N2O4 was found to be 2.71atm. Calculate the equilibrium pressure of NO2(g). .600 atm

  42. The ICE Table When initial concentrations and equilibrium constants are known, but none of the equilibrium positions are known it is helpful to write an ICE table. • I:TheInitial concentrations of products and reactants • C:The Change in concentrations needed to reach equilibrium is summarized in terms of variables. • E: The Equilibrium values are summarized as a combination of initial and change needed.

  43. ICE Table Example: Consider the reaction: N2(g) + 3H2(g) 2NH3(g) K = 6.0 x 10-2 at 500°C. The initial concentration of N2 is 3.0M and H2 is 2.0M. What are the equilibrium positions of this reaction?

  44. ICE Table Example: N2(g) + 3H2(g) 2NH3(g), K = 6.0 x 10-2

  45. Practice Problem #9 At a certain temperature a 1.00 L flask initially contained 0.298 mol PCl3(g) and 8.70x10-3mol of PCl5(g). After the system had reached equilibrium, 2.00 x 10-3mol Cl2(g) was found in the flask. Gaseous PCl5 decomposes according to the reaction: PCl5(g) PCl3(g) + Cl2(g). Calculate the equilibrium concentrations of all species and the value of K. >

  46. Practice Problem #9 Initial: 0.298 mol PCl3(g). 8.70x10-3mol of PCl5(g)in 1.00L equil: 2.00 x 10-3mol Cl2(g) PCl5(g) PCl3(g) + Cl2(g) Equilibrium expression: ICE Table: k=8.96 x 10-2

  47. Practice Problem #10 Carbon monoxide reacts with steam to produce carbon dioxide and hydrogen. At 700 K the equilibrium constant is 5.10. Calculate the equilibrium concentrations of all species if 1.00 mol of each component is mixed in a a 1.00L flask. >

  48. Practice Problem #10 Reaction: • CO(g) + H2O(g) CO2(g) + H2(g), K= 5.10 Which way does the equilibrium need to go? • Q=1.00 Q<K, shift right ICE Table:

  49. Practice Problem #10 x = 0.387 mol/L [CO] & [H2O] = .613M, [CO2] & [H2] = 1.387M Double check K with expression.

  50. Practice Problem #11 Assume that the reaction for the formation of gaseous hydrogen fluoride from hydrogen and fluorine has an equilibrium constant of 1.15 x 102 at a certain temperature. In a particular experiment, 3.000 mol of each component was added to a 1.500 L flask. Calculate the equilibrium concentration of all species. >

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