Collision Theory. Collision theory: 1. atoms, ions, and molecules must collide in order to react. Only a small number of collisions produce reactions

Transcription

1 UNIT 16: Chemical Equilibrium collision theory activation energy activated complex reaction rate reversible reaction chemical equilibrium law of chemical equilibrium equilibrium constant homogeneous equilibrium heterogeneous equilibrium LeChˆatlier s Principle solubility product constant common ion common ion affect

2 Collision Theory Collision theory: 1. atoms, ions, and molecules must collide in order to react. Only a small number of collisions produce reactions

3 2. Reacting substances must collide in the correct orientation. CO + NO 2 CO 2 + NO

4 3. Reacting substances must collide with sufficient energy to form an activated complex. Activated complex (transition state) temporary, unstable arrangement of atoms in which old bonds are breaking and new bonds are forming.

5 3. Reacting substances must collide with sufficient energy to form an activated complex. Activated complex (transition state) temporary, unstable arrangement of atoms in which old bonds are breaking and new bonds are forming.

6 Activation energy (E a ) the minimum amount of energy needed to form the activated complex and lead to a reaction Collision > E a Collision < E a

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9 Exothermic Reaction

10 Endothermic Reaction

11 Factors Affecting Reaction Nature of reactants Some things are more reactive than others (activity series) Concentration of reactants Temperature Catalysts

12 Lowers activation energy required for a reaction to take place

13 Lowers activation energy required for a reaction to take place

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16 Chemical Equilibrium Equilibrium the exact balancing of two processes, one of which is the exact opposite of the other

17 Chemical Equilibrium Equilibrium the exact balancing of two processes, one of which is the exact opposite of the other

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20 Reversible Reactions and Chemical Equilibrium Not all reactions go to completion Reversible reaction a reaction that occurs in both the forward and reverse direction Forward: Reverse: Both occurring:

21 Chemical equilibrium a dynamic state in which the forward and reverse reactions balance each other out because they take place at equal rates Rate forward reaction = Rate reverse reaction Concentration of all reactants and products remains constant

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23 Equal numbers of moles of H 2 O and CO are mixed in a closed container. The reaction begins to occur, and some products (H 2 and CO 2 ) are formed. The reaction continues as time passes and more reactants are changed to products. Although time continues to pass, the numbers of reactant and product molecules are the same as in (c). No further changes are seen as time continues to pass. The system has reached equilibrium.

24 Why does equilibrium occur? As concentration of reactants decreases, forward reaction slows down As concentration of products increases, reverse reaction speeds up

25 v=jsoawkguu6a

26 Equilibrium Expressions Law of chemical equilibrium at a given temperature a chemical system reaches a state in which the ratio of reactant and product concentration has a constant value

27 The Equilibrium Constant

28 The Equilibrium Constant Law of chemical equilibrium

29 The Equilibrium Constant Law of chemical equilibrium For a reaction of the type

30 The Equilibrium Constant Law of chemical equilibrium For a reaction of the type aa + bb cc + dd

31 The Equilibrium Constant Law of chemical equilibrium For a reaction of the type aa + bb cc + dd Equilibrium expression

32 The Equilibrium Constant Law of chemical equilibrium For a reaction of the type aa + bb cc + dd Equilibrium expression

33 The Equilibrium Constant Law of chemical equilibrium For a reaction of the type aa + bb cc + dd Equilibrium expression Each set of equilibrium concentrations is called an equilibrium position.

34 Interpreting K eq K eq > 1: K eq < 1:

35 Homogeneous Equilibria All reactants and products are in the same physical state Write the equilibrium expression for the following: H 2 (g) + F 2 (g) 2HF(g) N 2 (g) + 3H 2 (g) 2NH 3 (g)

36 N 2 O 4 (g) 2NO 2 (g) CO(g) + 3H 2 (g) CH 4 (g) + H 2 0(g)

37 Suppose that for the reaction: 2SO 2 (g) + O 2 (g) 2SO 3 (g) it is determined that at a particular temperature the equilibrium concentrations are: [SO 2 ] = 1.50M, [O 2 ] = 1.25M, [SO 3 ] = 3.50M. Calculate the value of K for this reaction.

38 Heterogeneous Equilibria Reactants and products are present in more than one state K eq does not depend on amounts of solids or liquids because their concentration cannot change.

39 Heterogeneous Equilibria Reactants and products are present in more than one state K eq does not depend on amounts of solids or liquids because their concentration cannot change.

40 So Equilibrium Expressions only include materials whose concentrations can change, namely GASES and AQUEOUS substances. 26

41 Write the equilibrium expression for the following: 2NaHCO 3 (s) Na 2 CO3(s) + CO 2 (g) + H 2 O(g) 2H 2 O(l) 2H 2 (g) + O 2 (g)

42 Equilibrium Constants For a reaction at a constant temperature K eq will always be the same regardless of initial concentration of reactants and products N 2 + H 2 2NH 3

43 Equilibrium Characteristics Reaction must take place in a closed system Temperature must remain constant All reactants and products are in constant dynamic motion

44 Factors Affecting Chemical Equilibrium When changes are made to a system at equilibrium the system shifts to a new equilibrium position

45 Le Chatelier s principle: if a stress is applied to a system at equilibrium, the system shifts in the direction that relieves that stress.

46 Applying Le Chatelier s Change in Concentration N 2 (g) + 3H 2 (g) 2NH 3 (g)

47 Applying Le Chatelier s Change in Concentration N 2 (g) + 3H 2 (g) 2NH 3 (g)

48 Concentration Changes 1. Add reactant/product - shift to consume it When a reactant or product is added to a system at equilibrium, the system shifts away from the added component 2. Remove product - shift R to replace it 3. Remove reactant - shift L to replace it If a reactant or product is removed from a system at equilibrium, the system shifts toward the removed component

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50 For the reaction: As 4 O 6 (s) + 6C(s) As 4 (g) + 6CO(g) Predict the direction of the shift in the equilibrium position in response to each of the following: Addition of carbon monoxide Addition of As 4 O 6 (s) removal of C(s) Removal of As 4 (g)

51 Pressure Changes - affects gases only Compare the number of moles of gases on each side of the equation. If UNEQUAL, affected by pressure changes Increase P shift to side with fewer particles (decrease pressure) Decrease P shift to side with more particles (increase pressure)

52 Changes in volume and pressure (gas only) Remember: for gasses there is an inverse relationship between volume and pressure

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54 The system is initially at equilibrium. The piston is pushed in, decreasing the volume and increasing the pressure. The system shifts in the direction that consumes CO 2 molecules, lowering the pressure again.

55 Decreasing the volume system shifts in direction that gives fewest number of gas molecules

56 Decreasing the volume system shifts in direction that gives fewest number of gas molecules

57 Increasing the volume system shifts in the direction to increase the number of gas molecules (increase the pressure)

58 Predict the direction of the equilibrium shift for each of the following when the volume of the container is decreased: CO(g) + 2H 2 (g) CH 3 OH(g) H 2 (g) + F 2 (g) 2HF(g)

59 Change in Temperature: Exothermic reaction heat is a product - add heat - shifts L - remove heat - shifts R Adding energy shifts equilibrium away from products Endothermic reaction heat is a reactant - add heat - shifts R - remove heat - shifts L Adding energy shifts equilibrium away from reactants Predict the direction of the shift in equilibrium position the same way as if a product or reactant were added or removed

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61 Decide whether higher or lower temperatures will produce more CH 3 CHO in the reaction: C 2 H 2 (g) + H 2 O(g) CH 3 CHO ΔH = -151 kj

62 For the exothermic reaction: 2SO 2 (g) + O 2 (g) 2SO 3 (g) Predict the equilibrium shift caused by each of the following changes: SO 2 is added SO 3 is removed Volume is decreased Temperature is decreased

63 Catalysts speeds up a reaction equally in both directions therefore does not affect equilibrium

64 Using Equilibrium Constants Equilibrium constant expressions can be used to calculate concentrations and solubilities

65 Calculating Equilibrium Concentrations At a temperature of 1405 K, hydrogen sulfide decomposes to form hydrogen and diatomic sulfur according to the reaction: 2H 2 S(g) 2H 2 (g) + S 2 (g). The equilibrium constant for the reaction is 2.27 x What is the concentration of hydrogen gas if [S 2 ] = M and [H 2 S] = 0.184M

66 2H 2 S(g) 2H 2 (g) + S 2 (g). The equilibrium constant for the reaction is 2.27 x What is the concentration of hydrogen gas if [S 2 ] = M and [H 2 S] = 0.184M Keq = [H2] 2 x [S2] [H2S] 2 [H2] 2 = Keq x [H2S] 2 = [S2] [H2] = 2.27 x 10-3 x (0.184)

67 PCl 5 PCl 3 + Cl 2. In a certain experiment, at a temperature where K = 8.96 x 10-2, the equilibrium concentrations of PCl 5 = 6.70 x 10-3 M and PCl 3 = M. What was the equilibrium concentration of Cl 2?

68 The Solubility Product Upon dissolving, all ionic compounds dissociate into ions NaCl(s) BaSO 4 (s) Mg(OH) 2 (s)

69 Solubility Equilibria CaF 2 (s) Ca 2+ (aq) + 2F - (aq) At equilibrium, solution is saturated K sp = [Ca 2+ ][F - ] 2 K sp = solubility product constant

70 Write the balanced equation describing the reaction for dissolving each of the following in water, also write the K sp expression for each solid: PbCl 2 (s) Ag 2 CrO 4 (s)

71 Small K sp = not very soluble compound

72 The K sp value for solid AgI is 8.5 x Calculate the solubility of AgI in water.

73 The K sp value for solid CuCO 3 is 2.5 x Calculate the solubility of AgI in water.

74 The solubility of CuBr is 2.0 x 10-4 mol/l. Calculate the K sp for copper (I) bromide.