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Chemical Equilibrium AP Chem Unit 13. Chemical Equilibrium  The Equilibrium Condition  The Equilibrium Constant  Equilibrium Expressions Involving.

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Presentation on theme: "Chemical Equilibrium AP Chem Unit 13. Chemical Equilibrium  The Equilibrium Condition  The Equilibrium Constant  Equilibrium Expressions Involving."— Presentation transcript:

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 H 2 O (g) + CO (g) H 2(g) + CO 2(g)

9 The Equilibrium Condition H 2 O (g) + CO (g) H 2(g) + CO 2(g)

10 The Equilibrium Condition H 2 O (g) + CO (g) H 2(g) + CO 2(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 H 2 O (g) + CO (g) H 2(g) + CO 2(g) What would happen if H 2 O (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 + kB lC + 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: 4NH 3(g) + 7O 2(g) 4NO 2(g) + 6H 2 O (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: [NH 3 ] = 3.1 x mol/l [N 2 ] = 8.5 x mol/l [H 2 ] = 3.1 x mol/l a) Calculate the value of K at 127°C for this reaction. 3.8 x 10 4

18 Practice Problem #2 The following equilibrium concentrations were observed for the Haber process for the synthesis of ammonia at 127°C: [NH 3 ] = 3.1 x mol/l [N 2 ] = 8.5 x mol/l [H 2 ] = 3.1 x mol/l b) Calculate the value of the equilibrium constant at 127°C for the reaction: 2NH 3(g) N 2(g) + 3H 2(g) 2.6 x (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: [NH 3 ] = 3.1 x mol/l [N 2 ] = 8.5 x mol/l [H 2 ] = 3.1 x mol/l c) Calculate the value of the equilibrium constant at 127°C for the reaction: 1.9 x 10 2 (When the coefficients are ½ of the balanced equation, new K = K 1/2 )

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. K new =K o n  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. InitialEquilibrium [SO 2 ] = 2.00 M1.50 M [O 2 ] = 1.50 M1.25 M [SO 3 ] = 3.00 M3.50 M InitialEquilibrium [SO 2 ] = M0.590 M [O 2 ] = 0 M M [SO 3 ] = M0.260 M 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 + kB lC + mD 

26 Practice Problem #4 The reaction for the formation of nitrosyl chloride: 2NO (g) + Cl 2(g) 2NOCl (g) was studied at 25°C. The presurres at equilibrium were found to be: NOCl =1.2 atm, NO = 5.0 x atm, Cl 2 = 3.0 x atm. Calculate the value of K p for this reaction at 25°C. 1.9 x 10 3

27 K c vs. K p K p = K c (RT) Δn  Δn is the sum of the coefficients of the gaseous products minus the sum of the coefficients of the gaseous reactants  jA + kB lC + mD  Δn=(l + m) – (j + k)  more moles of gas = more pressure

28 Practice Problem #5 Using the value of K p obtained in Problem #4, calculate the value of K at 25°C for the reaction: 2NO (g) + Cl 2(g) 2NOCl (g) K p =1.9 x 10 3 K = 4.6 x 10 4

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  CaCO 3(s) CaO (s) + CO 2(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 K p 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 = [Cl 2 ] and K p =P Cl2

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 = [H 2 O] 5 K p = (P H2O ) 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

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 Predict the direction in which the system will shift to reach equilibrium in each of the following cases: a) [NH 3 ]=1.0x10 -3 M, [N 2 ]=1.0x10 -5 M, [H 2 ]=2.0x10 -3 M b) [NH 3 ]=2.00x10 -4 M, [N 2 ]=1.50x10 -5 M, [H 2 ]=3.54x10 -1 M c) [NH 3 ]=1.0x10 -4 M, [N 2 ]=5.0M, [H 2 ]=1.0x10 -2 M a) Q>K, shift left. b)Q=K, no shift. c)Q

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. N 2 O 4(g) 2NO 2(g) Consider an experiment in which gaseous N 2 O 4 was placed in a flask and allowed to reach equilibrium at a temperature where K p = At equilibrium, the pressure of N 2 O 4 was found to be 2.71atm. Calculate the equilibrium pressure of NO 2(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 : The I nitial concentrations of products and reactants  C: The C hange in concentrations needed to reach equilibrium is summarized in terms of variables.  E: The E quilibrium values are summarized as a combination of initial and change needed.

43 ICE Table Example: Consider the reaction: N 2(g) + 3H 2(g) 2NH 3(g) K = 6.0 x at 500°C. The initial concentration of N 2 is 3.0M and H 2 is 2.0M. What are the equilibrium positions of this reaction? Initial (M)ChangeEquilibrium (M) N2N2 3.0-x3.0 - x H2H x2.0 – 3x NH x2x

44 ICE Table Example: N 2(g) + 3H 2(g) 2NH 3(g), K = 6.0 x Initial (M)ChangeEquilibrium (M) N2N2 3.0-x3.0 - x H2H x2.0 – 3x NH x2x

45 Practice Problem #9 At a certain temperature a 1.00 L flask initially contained mol PCl 3(g) and 8.70x10 -3 mol of PCl 5(g). After the system had reached equilibrium, 2.00 x mol Cl 2(g) was found in the flask. Gaseous PCl 5 decomposes according to the reaction: PCl 5(g) PCl 3(g) + Cl 2(g). Calculate the equilibrium concentrations of all species and the value of K. >

46 Practice Problem #9 Initial: mol PCl 3(g). 8.70x10 -3 mol of PCl 5(g) in 1.00L equil: 2.00 x mol Cl 2(g) PCl 5(g) PCl 3(g) + Cl 2(g) Equilibrium expression: ICE Table: k=8.96 x Initial (M)ChangeEquilibrium (M) PCl 5(g) x PCl 3(g) 8.70x x Cl 2(g) x x 10 -3

47 Practice Problem #10 Carbon monoxide reacts with steam to produce carbon dioxide and hydrogen. At 700 K the equilibrium constant is 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) + H 2 O (g) CO 2(g) + H 2(g), K= 5.10 Which way does the equilibrium need to go?  Q=1.00 Q

49 Practice Problem #10 x = mol/L [CO] & [H 2 O] =.613M, [CO 2 ] & [H 2 ] = 1.387M Double check K with expression. Initial (M)ChangeEquilibrium (M) CO1.0-x1.0 - x H2OH2O1.0-x1.0 - x CO x1.0 + x H2H2 1.0+x1.0 + x

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 10 2 at a certain temperature. In a particular experiment, mol of each component was added to a L flask. Calculate the equilibrium concentration of all species. >

51 Practice Problem #11 Reaction:  H 2(g) + F 2(g) 2HF (g), K=1.15 x 10 2 Which way does the equilibrium need to go?  Initial concentrations:  3.000mol/1.500L = 2.00 M  Q=1.00 Q

52 Practice Problem #11 x=1.528 [H 2 ] & [F 2 ]= M [HF] = M Check K with equilibrium values Initial (M)ChangeEquilibrium (M) H2H2 2.0-x2.0 - x F2F2 2.0-x2.0 - x HF2.0+2x x

53 Solving Equilibrium Problems 13.6

54 Solving Equilibrium Problems Strategy: 1. Write the balanced equation for the reaction. 2. Write the equilibrium expression using law of mass action. 3. List the initial concentrations. 4. Calculate Q, and determine the direction of the shift needed for equilibrium.

55 Solving Equilibrium Problems Strategy continued: 5. Define the change needed to reach equilibrium, and define the equilibrium concentrations by applying the change to the initial concentrations. 6. Substitute the equilibrium concentrations into the equilibrium expression, and solve for the unknown. 7. Check your calculated equilibrium concentrations by making sure the give the correct value of K.

56 Solving Equilibrium Problems Typical systems do not produce an expression that can be solved by taking the square root of both sides.  To solve some expressions, we will use the quadratic equation.

57 Equilibrium Example Suppose for a synthesis of hydrogen fluoride from hydrogen and fluorine, mol H 2 and mol F 2 are mixed in a L flask. Assume the equilibrium constant for the synthesis reaction at this temperature is 1.15x10 2. What are the equilibrium concentrations of each component. >

58 Equilibrium Example 1. Write the balanced equation for the reaction. H 2(g) + F 2(g) 2HF (g) 2. What is the equilibrium expression? 3. What are the initial concentrations? [H 2 ] = 3.00mol/3.00 L = 1.000M [F 2 ] = 6.00mol/3.00 L = 2.000M [HF]= 0

59 Equilibrium Example 4. What is Q? Q does not need to be calculated in this example. Since HF is not present initially, we can assume that the reaction will shift to the right to reach equilibrium. 5. What change is required to reach equilibrium? Initial (M)ChangeEquilibrium (M) H2H2 1.0-x1.0 - x F2F2 2.0-x2.0 - x HF0.0+2x2x

60 Equilibrium Example 6. What is the value of K? (Use ICE in expression) collect terms and set = 0 ax 2 + bx + c = 0 a=1.11x10 2, b=-3.45x10 2, c=2.30x10 2

61 Equilibrium Example 6. What is the value of K? a=1.11x10 2, b=-3.45x10 2, c=2.30x10 2 Substituting these values give two answers for x: x=2.14 mol/L and mol/L Both of these results are not valid; the changes in concentration must be checked for validity

62 Equilibrium Example 6. What is the value of K? x=2.14 mol/L or mol/L [H 2 ] = – x, [F 2 ] = – x, [HF] = 2x [H 2 ] = 3.2x10 -2 M, [F 2 ] = 1.032M, [HF] = 1.936M 7. Check concentrations by substituting them into the equilibrium expression.

63 Practice Problem #12 Assume that gaseous hydrogen iodide is synthesized from hydrogen gas and iodine vapor at a temperature where the equilibrium constant is 1.00x10 2. Suppose HI at 5.000x10 -1 atm, H 2 at 1.000x10 -2 atm, and I 2 at 5.000x10 -3 atm are mixed in a 5.000L flask. Calculate the equilibrium pressures of all species.

64 Practice Problem #12 1. Write the balanced equation 2. What is the equilibrium expression? 3. What are the initial pressures? 4. What is the value of Q?

65 Practice Problem #12 5. What is the change required? 6. What is the value of K p (& equilibrium pressures)? 7. Expression check. P HI =4.29x10 -1 atm, P H2 =4.55x10 -2 atm, P I2 =4.05x10 -2 atm

66 Small Equilibrium Constants Sometimes there are simplifications that can be made to the math of some equilibrium problems.  When reactions lie far to the left, the equilibrium constants can be very small.  Changes in initial concentrations can be negligible and partially disregarded.

67 Small K Example Gaseous NOCl decomposes to form the gases NO and Cl 2. At 35°C the equilibrium constant is 1.6x In an experiment in which 1.0 mol NOCl is placed in a 2.0 L flask, what are the equilibrium concentrations? 1. What is the equation? 2NOCL (aq) 2NO (g) + Cl 2(g) 2. What is the expression?

68 Small K Example 3. What are the initial concentrations? [NOCl]=0.50M, [NO]=0, [Cl 2 ] = 0 4. What is Q? Direction must lie to the right for equilibrium. 5. What is the change required? Initial (M)ChangeEquilibrium (M) NOCl0.50-2x x NO0.0+2x2x Cl xx

69 Small K Example 6. What is the value of K (& concentrations) x must represent a relatively small number. In order to simplify this expression, we can assume that: 0.50 – 2x ≅.50 Therefore we can simplify the expression: x=1.0x10 -2 Initial (M) ChangeEquilibrium (M) NOCl0.50-2x x NO0.0+2x2x Cl xx

70 Small K Example [NOCl] =.50 – 2x ≈ 0.50 M [NO] = 2.0 x M [Cl 2 ] = 1.0 x M 7. Check the K expression.

71 Le Chatelier’s Principle 13.7

72 Le Chatelier’s Principle Several factors can control the position of a chemical equilibrium.  Changes in Concentration  Temperature (removal or addition of energy)  Pressure

73 Le Chatelier’s Principle Le Chatelier’s principle states that if a change is imposed on a system at equilibrium, the position of the equilibrium will shift in a direction that tends to reduce that change.  It is important to realize that although changes to the reaction may alter the equilibrium positions, they do not alter the equilibrium constant.

74 Le Chatelier’s Principle Change in Concentration:  If a component (reactant or product) is added to a reaction system at equilibrium (at constant T and P or constant T and V), the equilibrium position will shift in the direction that lowers the concentration of that component. If a component is removed, the opposite effect occurs.

75 Practice Problem #13 Arsenic can be extracted from its ores by first reacting the ore with oxygen (called roasting) to form solid As 4 O 6, which is then reduced using carbon: As 4 O 6(s) + 6C (s) As 4(g) + 6CO (g) Predict the direction of the shift of the equilibrium position in response to each of the following changes in concentration. a) Addition of carbon monoxide. b) Addition or removal of carbon or tetrarsenic hexoxide. c) Removal of gaseous arsenic. left shift, no effect, right shift

76 Le Chatelier’s Principle Change in Pressure:  There are three ways to change the pressure of a reaction system involving gaseous components: 1. Add or remove a gaseous reactant or product. 2. Add an inert gas (one not involved in the reaction). 3. Change the volume of the container

77 Le Chatelier’s Principle Change in Pressure:  When an inert gas is added, there is no effect on the equilibrium position. The addition of an inert gas increases the total pressure but has no effect on the concentrations of the reactants or products.  The system will remains at the original equilibrium position

78 Le Chatelier’s Principle Change in Volume:  When the volume of the container holding a gaseous system is reduced, the system responds by reducing its own volume. This is done by decreasing the total number of gaseous molecules in the system.  A reaction will shift in order to reduce the number of gas molecules.

79 Practice Problem #14 Predict the shift in equilibrium position that will occur for each of the following processes when the volume is reduced. a. The preparation of liquid phosphorus trichloride by the reaction. P 4(s) + 6Cl 2(g) 4PCl 3(l) right shift

80 Practice Problem #14 Predict the shift in equilibrium position that will occur for each of the following processes when the volume is reduced. b. The preparation of gaseous phosphorus pentachloride according to the equation: PCl 3(g) + Cl 2(g) PCl 5(g) right shift

81 Practice Problem #14 Predict the shift in equilibrium position that will occur for each of the following processes when the volume is reduced. c. The reaction of phosphorus trichloride with ammonia: PCl 3(g) + 3NH 3(g) P(NH 2 ) 3(g) + 3HCl (g) no effect

82 Le Chatelier’s Principle Change in Temperature:  If energy is added or removed from a system in equilibrium, the system will shift according to the heat of the reaction.  Heat is a product in an exothermic reaction.  Heat is a reactant in an endothermic reaction.  The effect of temperature on equilibrium changes the value of K because K changes with temperature.

83 Practice Problem #15 For each of the following reactions, predict how the value of K changes as the temperature is increased. a. N 2(g) + O 2(g) 2NO (g) ΔH = 181 kJ b. 2SO 2(g) + O 2(g) 2SO 3(g) ΔH = -198 kJ a) shift right b) shift left

84 THE END 3 more units to go!!!


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