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13 Dec. 2010 Take Out Homework: Lab Notebook and Kinetics AP Questions. Objective: SWBAT Do now:

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Presentation on theme: "13 Dec. 2010 Take Out Homework: Lab Notebook and Kinetics AP Questions. Objective: SWBAT Do now:"— Presentation transcript:

1 13 Dec. 2010 Take Out Homework: Lab Notebook and Kinetics AP Questions. Objective: SWBAT Do now:

2 Agenda Do now Kinetics Problem Set Solutions
Introduction to Equilibrium Demo Homework: p. #

3 Chemical Equilibrium Chapter 14

4 Equilibrium is a state in which there are no observable changes as time goes by.
Chemical equilibrium is achieved when: the rates of the forward and reverse reactions are equal and the concentrations of the reactants and products remain constant However, there is a lot of activity at the molecular level! Reactants continue to form products, while products continue to yield reactants!

5 Physical vs. Chemical Equilibrium
Physical equilibrium H2O (l) H2O (g) NO2 Chemical equilibrium N2O4 (g) 2NO2 (g)

6 N2O4 (g) 2NO2 (g) equilibrium equilibrium equilibrium Start with NO2
Start with NO2 & N2O4

7 constant

8 N2O4 (g) NO2 (g) Equilibrium Constant K = [NO2]2 [N2O4] = 4.63 x 10-3 aA + bB cC + dD K = [C]c[D]d [A]a[B]b Law of Mass Action

9 Law of Mass Action For a reversible reaction at equilibrium and a constant temperature, a certain ratio of reactant and product concentration has a constant value, K (the equilibrium constant) concentrations may vary but as long as temperature stays the same and the reaction is at equilibrium, K will not change.

10 Equilibrium Will K = [C]c[D]d [A]a[B]b aA + bB cC + dD K >> 1
Lie to the right Favor products K << 1 Lie to the left Favor reactants

11 Homogenous equilibrium applies to reactions in which all reacting species are in the same phase.
N2O4 (g) NO2 (g) Kp = NO2 P 2 N2O4 P Kc = [NO2]2 [N2O4] In most cases Kc  Kp aA (g) + bB (g) cC (g) + dD (g) Kp = Kc(RT)Dn Dn = moles of gaseous products – moles of gaseous reactants = (c + d) – (a + b)

12 Homogeneous Equilibrium
CH3COOH (aq) + H2O (l) CH3COO- (aq) + H3O+ (aq) [CH3COO-][H3O+] [CH3COOH][H2O] Kc = [H2O] = constant [CH3COO-][H3O+] [CH3COOH] = Kc [H2O] Kc = General practice not to include units for the equilibrium constant.

13 Example 1 Write expressions for Kc, and KP if applicable, for the following reversible reactions at equilibrium: HF(aq) + H2O(l)  H3O+(aq) + F-(aq) 2NO(g) + O2(g)  2NO2(g) CH3COOH(aq) + C2H5OH(aq)  CH3COOC2H5(aq) + H2O(l) Hints: KP applies only to gaseous reactions; the concentration of the solvent (usually water) does not appear in the equilibrium constant expression.

14 Problem 1 Write Kc and KP for the decomposition of dinitrogen pentoxide: 2N2O5(g)  4NO2(g) + O2(g)

15 Example 2 The following equilibrium process has been studied at 230oC: 2NO(g) + O2(g)  2NO2(g) In one experiment, the concentrations of the reacting species at equilibrium are found to be [NO]= M, [O2]=0.127 M and [NO2]=15.5 M. Calculate the equilibrium constant (Kc) of the reaction at this temperature.

16 Problem 2 Carbonyl chloride (COCl2), also called phosgene, was used in WWI as a poisonous gas. The equilibrium concentrations for the reaction between carbon monoxide and molecular chlorine to form carbonyl chloride CO(g) + Cl2(g)  COCl2(g) at 74oC are [CO]=1.2x10-2 M, [Cl2]=0.054 M and [COCl2]=0.14 M. Calculate the equilibrium constant (Kc).

17 Example 3 At the equilibrium constant KP for the decomposition of phosphorus pentachloride to phosphorus trichloride and molecular chlorine PCl5(g)  PCl3(g) + Cl2(g) is found to be 1.05 at 250oC. If the equilibrium partial pressures of PCl5 and PCl3 are atm and atm, respectively, what is the equilibrium partial pressure of Cl2 at 250oC?

18 Problem 3 The equilibrium constant KP for the reaction 2NO2(g)  2NO(g) + O2(g) is 158 at 1000 K. Calculate PO2 if PNO2 = atm and PNO = atm.

19 Example 4 Methanol (CH3OH) is manufactured industrially by the reaction CO(g) + 2H2(g)  CH3OH(g) The equilibrium constant Kc for the reaction is 10.5 at 220oC. What is the value of KP at this temperature?

20 Problem 4 For the reaction N2(g) + 3H2(g)  2NH3(g) KP is 4.3x10-4 at 375oC. Calculate Kc for the reaction.

21 14 Dec. 2010 Objective: SWBAT write equilibrium constant expressions for reactions with heterogeneous equilibria, and calculate reaction quotients. Do now: The equilibrium constant KP for the reaction 2SO3(g) ⇄ 2SO2(g) + O2(g) is 1.8x10-5 at 350oC. What is KC for this reaction?

22 Agenda Do now Homework solutions Heterogeneous equilibrium constants
Reaction quotients p. 649 #23, 25, 27, 29, 31, 35, 37

23 Heterogenous equilibrium applies to reactions in which reactants and products are in different phases. CaCO3 (s) CaO (s) + CO2 (g) [CaO][CO2] [CaCO3] Kc = [CaCO3] = constant [CaO] = constant [CaCO3] [CaO] Kc x Kc = [CO2] = Kp = PCO 2 The concentration of solids and pure liquids are not included in the expression for the equilibrium constant.

24 does not depend on the amount of CaCO3 or CaO
CaCO3 (s) CaO (s) + CO2 (g) PCO 2 = Kp PCO 2 does not depend on the amount of CaCO3 or CaO

25 Example 5 Write the equilibrium constant expression Kc, and KP if applicable, for each of the following heterogeneous sytems: (NH4)2Se(s)  2NH3(g) + H2Se(g) AgCl(s)  Ag+(aq) + Cl-(aq) P4(s) + 6Cl2(g)  4PCl3(l)

26 Problem 5 Write equilibrium constant expressions for Kc and KP for the formation of nickel tetracarbonyl, which is used to separate nickel from other impurities: Ni(s) + 4CO(g)  Ni(CO)4(g)

27 Example 6 Consider the following heterogeneous equilibrium: CaCO3(s)  CaO(s) + CO2(g) At 800oC, the pressure of CO2 is atm. Calculate a) Kp b) Kc for the reaction at this temperature.

28 Problem 6 Consider the following equilibrium at 395 K: NH4HS(s)  NH3(g) + H2S(g) The partial pressure of each gas is atm. Calculate KP and Kc for the reaction.

29 For which of the following reactions is Kc equal to KP?
4NH3(g) + O2(g) ⇄ 4NO(g) + 6H2O(g) 2H2O2(aq) ⇄ 2H2O(l) + O2(g) PCl3(g) + 3NH3(g) ⇄ 3HCl(g) + P(NH2)3(g)

30 What if the product molecules of one reversible reaction are involved in a second reaction as the products? A + B C + D C + D E + F

31 ′ ′ ′ ′ ′′ Kc = [C][D] [A][B] Kc = [E][F] [C][D] A + B C + D Kc Kc ′′
C + D E + F [E][F] [A][B] Kc = A + B E + F Kc Kc = Kc ′′ x If a reaction can be expressed as the sum of two or more reactions, the equilibrium constant for the overall reaction is given by the product of the equilibrium constants of the individual reactions.

32 ′ N2O4 (g) 2NO2 (g) 2NO2 (g) N2O4 (g) = 4.63 x 10-3 K = [NO2]2 [N2O4]
= 216 When the equation for a reversible reaction is written in the opposite direction, the equilibrium constant becomes the reciprocal of the original equilibrium constant.

33 Writing Equilibrium Constant Expressions
The concentrations of the reacting species in the condensed phase are expressed in M. In the gaseous phase, the concentrations can be expressed in M or in atm. The concentrations of pure solids, pure liquids and solvents do not appear in the equilibrium constant expressions. The equilibrium constant is a dimensionless quantity. In quoting a value for the equilibrium constant, you must specify the balanced equation and the temperature. If a reaction can be expressed as a sum of two or more reactions, the equilibrium constant for the overall reaction is given by the product of the equilibrium constants of the individual reactions.

34 15 Dec. 2010 Objective: SWBAT calculate reaction quotient and calculate an equilibrium concentration. Do now: If the equilibrium constant for the reaction N2O4 (g) ⇄ 2NO2 (g) is 4.63 x 10-3 calculate the equilibrium constant for the reaction 2NO2(g) ⇄ N2O4(g)

35 Agenda Do now Homework solutions Reaction quotient
Equilibrium concentrations Homework: p. 650 #23, 30, 36, 39

36 The Relationship between Chemical Kinetics and Chemical Equilibrium
ratef = kf [A][B]2 A + 2B AB2 kf kr rater = kr [AB2] Equilibrium ratef = rater kf [A][B]2 = kr [AB2] kf kr [AB2] [A][B]2 = Kc =

37 What does the Equilibrium Constant tell us?
We can use the equilibrium constant to calculate unknown equilibrium concentrations of products or reactants (at a constant temperature) We can predict the direction in which a reaction will proceed to achieve equilibrium

38 What if the reaction is not yet at equilibrium?
Calculate reaction quotient (Qc) by substituting the initial concentrations into the equilibrium constant expression.

39 The reaction quotient (Qc) is calculated by substituting the initial concentrations of the reactants and products into the equilibrium constant (Kc) expression. IF Qc > Kc system proceeds from right to left to reach equilibrium Qc = Kc the system is at equilibrium Qc < Kc system proceeds from left to right to reach equilibrium

40 Example 7 At the start of the reaction, there are mol N2, 3.21x10-2 mol H2 and 6.42x10-4 mol NH3 in a 3.50 L reaction vessel at 375oC. If the equilibrium constant Kc for the reaction N2(g) + 3H2(g) ⇄ 2NH3(g) is 1.2 at this temperature, decide whether the system is at equilibrium. If it is not, predict which way the net reaction will proceed.

41 Problem 7 The equilibrium constant (Kc) for the formation of nitrosyl chloride, an orange-yellow compound, from nitric oxide and molecular chlorine 2NO(g) + Cl2(g) ⇄ 2NOCl(g) is 6.5x104 at 35oC. In a certain experiment, 2.0x10-2 mole of NO, 8.3x10-3 mole of Cl2, and 6.8 moles of NOCl are mixed in a 2.0 L flask. In which direction will the system proceed to reach equilibrium?

42 What do you do with the Kc and KP?
If we know the equilibrium constant for a particular reaction, we can calculate the concentrations in the equilibrium mixture from the initial concentrations!

43 Calculating Equilibrium Concentrations
Express the equilibrium concentrations of all species in terms of the initial concentrations and a single unknown x, which represents the change in concentration. Write the equilibrium constant expression in terms of the equilibrium concentrations. Knowing the value of the equilibrium constant, solve for x. Having solved for x, calculate the equilibrium concentrations of all species.

44 Example 8 At 1280oC the equilibrium constant (Kc) for the reaction
Is 1.1 x If the initial concentrations are [Br2] = M and [Br] = M, calculate the concentrations of these species at equilibrium. Br2 (g) Br (g)

45 ICE At 1280oC the equilibrium constant (Kc) for the reaction
Is 1.1 x If the initial concentrations are [Br2] = M and [Br] = M, calculate the concentrations of these species at equilibrium. Br2 (g) Br (g) Let x be the change in concentration of Br2 Br2 (g) Br (g) Initial (M) 0.063 0.012 ICE Change (M) -x +2x Equilibrium (M) x x [Br]2 [Br2] Kc = Kc = ( x)2 x = 1.1 x 10-3 Solve for x

46 Kc = ( x)2 x = 1.1 x 10-3 4x x = – x 4x x = 0 -b ± b2 – 4ac 2a x = ax2 + bx + c =0 x = x = Br2 (g) Br (g) Initial (M) Change (M) Equilibrium (M) 0.063 0.012 -x +2x x x At equilibrium, [Br] = x = M or M At equilibrium, [Br2] = – x = M

47 Example 9 A mixture of mol H2 and mol I2 was placed in a 1.00 L stainless steel flask at 430oC. The equilibrium constant KC for the reaction H2(g) + I2(g) ⇄ 2HI(g) is 54.3 at this temperature. Calculate the concentrations of H2, I2 and HI at equilibrium.

48 Example 10 For the reaction below, the initial concentrations of each species are H2 = M, I2 = M and HI = M at 430oC. The equilibrium constant KC for the reaction is still Calculate the concentrations of these species at equilibrium. H2(g) + I2(g) ⇄ 2HI(g)

49 Homework p. 650 #23, 30, 36, 39

50 Problem 8 Consider the reaction below: H2(g) + I2(g) ⇄ 2HI(g) with an equilibrium constant of Starting with a concentration of M for HI, calculate the concentrations of HI, H2 and I2 at equilibrium.

51 Problem 9 At 1280oC the equilibrium constant (Kc) for the reaction Br2(g) ⇄ 2Br(g) is 1.1x10-3. If the initial concentrations are [Br2] = 6.3x10-2 M and [Br] = 1.2x10-2 M, calculate the concentrations of these species in equilibrium.

52 3 January 2011 Take Out AP Problem Set: Equilibrium
Objective: SWBAT review equilibrium concentration calculations and describe factors that affect chemical equilibrium. Do now: What does a very large Kc indicate? What does it mean if Qc is much smaller than Kc?

53 Agenda Do now AP Problem Set Questions?
Review ICE boxes and equilibrium concentrations Factors that affect chemical equilibrium: Demo Notes Homework: p. 651 #45, 49, 51, 52, odds

54 Demo Solid copper metal reacts with nitric acid to produce dinitrogen tetraoxide gas, water vapor and a solution of copper (II) nitrate. Dinitrogen tetraoxide gas forms nitrogen dioxide gas.

55 Le Châtelier’s Principle
If an external stress is applied to a system at equilibrium, the system adjusts in such a way that the stress is partially offset as the system reaches a new equilibrium position. Changes in Concentration N2 (g) + 3H2 (g) NH3 (g) Add NH3 Equilibrium shifts left to offset stress

56 Le Châtelier’s Principle
Changes in Concentration continued Remove Add Add Remove aA + bB cC + dD Change Shifts the Equilibrium Increase concentration of product(s) left Decrease concentration of product(s) right Increase concentration of reactant(s) right Decrease concentration of reactant(s) left

57 Le Châtelier’s Principle
Changes in Volume and Pressure A (g) + B (g) C (g) Change Shifts the Equilibrium Increase pressure Side with fewest moles of gas Decrease pressure Side with most moles of gas Increase volume Side with most moles of gas Decrease volume Side with fewest moles of gas

58 Example 1 At 720oC, the equilibrium constant Kc for the reaction N2(g) + 3H2(g) ⇄ 2NH3(g) is 2.37x10-3. In a certain experiment, the equilibrium concentrations are [N2]=0.683 M, [H2]=8.80M and [NH3]=1.05 M. Suppose some NH3 is added to the mixture so that its concentration is increased to 3.65 M. a) Use Le Châtelier’s principle to predict the shift in direction of the net reaction to reach a new equilibrium. b) Confirm your prediction by calculation the reaction quotient Qc and comparing its value to Kc.

59 Problem 1 At 430oC, the equilibrium constant (KP) for the reaction 2NO(g) + O2(g) ⇄ 2NO2(g) is 1.5x105. In one experiment, the initial pressures of NO, O2 and NO2 are 2.1x10-3 atm, 1.1x10-2 atm and 0.14 atm respectively. Calculate QP and predict the direction that the net reaction will shift to reach equilibrium.

60 Le Châtelier’s Principle
Changes in Temperature (gases only!) Change Exothermic Rx Endothermic Rx Increase temperature K decreases K increases Decrease temperature K increases K decreases N2O4 (g) NO2 (g) colder hotter

61 Example 2 Consider the following equilibrium systems:
2PbS(s) + 3O2(g) ⇄ 2PbO(s) + 2SO2(g) PCl5(g) ⇄ PCl3(g) + Cl2(g) H2(g) + CO2(g) ⇄ H2O(g) + CO(g) Predict the direction of the net reaction in each case as a result of increasing the pressure (decreasing the volume) on the system at constant temperature.

62 Problem 2 Consider the equilibrium reaction involving nitrosyl chloride, nitric oxide and molecular chlorine: 2NOCl(g) ⇄ 2NO(g) + Cl2(g) Predict the direction of the net reaction as a result of decreasing the pressure (increasing the volume) on the system at a constant temperature.

63 5 January 2011 Objective: SWBAT describe how stress on a system shifts equilibrium. Do now: Complete this table: Stress to system (exothermic) Effect Increase in temperature Increase in pressure increase in concentration of reactant increase in volume

64 Agenda Do now Homework answers Finish Le Chatlier’s examples
Equilibrium problem set (work time tomorrow, too) Test Monday, 2nd half of class

65 Le Châtelier’s Principle
Adding a Catalyst does not change K does not shift the position of an equilibrium system system will reach equilibrium sooner Catalyst lowers Ea for both forward and reverse reactions. Catalyst does not change equilibrium constant or shift equilibrium.

66 Example 3 Consider the following equilibrium process between dinitrogen tetrafluoride and nitrogen difluoride: N2F4(g) ⇄ 2NF2(g) ∆Ho=38.5 kJ/mol Predict the changes in the equilibrium if the reacting mixture is heated at a constant volume. some N2F4 gas is removed from the reacting mixture at constant temperature and volume the pressure on the reacting mixture is decreased at constant temperature a catalyst is added to the reacting mixture.

67 Problem 3 Consider the equilibrium between molecular oxygen and ozone:
3O2(g) ⇄ 2O3(g) ΔHo=284 kJ/mol What would be the effect of increasing the pressure on the system by decreasing the volume? adding O2 to the system at constant volume? decreasing the temperature? adding a catalyst?

68 Life at High Altitudes and Hemoglobin Production
Chemistry In Action Life at High Altitudes and Hemoglobin Production Hb (aq) + O2 (aq) HbO2 (aq) Kc = [HbO2] [Hb][O2]

69 Chemistry In Action: The Haber Process
N2 (g) + 3H2 (g) NH3 (g) DH0 = kJ/mol

70 Le Châtelier’s Principle - Summary
Change Shift Equilibrium Change Equilibrium Constant Concentration yes no Pressure yes* no Volume yes* no Temperature yes yes Catalyst no no *Dependent on relative moles of gaseous reactants and products

71 Homework p. 651 #45, 49, 51, 52, odds

72 20 Dec. 2010 Objective: SWBAT determine the stoichiometric relationship between reactants and products in a chemical reaction. Do now: Grab your lab notebook and complete the pre-lab.

73 Changes to the lab Use HALF the volume of each solution called for in the lab. Complete reactions in a 50 mL graduated cylinder. Measure reagents in a 25 mL graduated cylinder.

74 Notes Pour or transfer pipette OUT of reagent bottles.
Never pour or transfer pipette INTO reagent bottles. Pour excess down the sink while running the water. Pour products into the waste beaker.

75 Homework Post lab calculations and conclusions due tomorrow.

76 Equilibrium Review Problem Set
p. 648 #8d, 9a, 16, 18, 21, 28, 31, 40, 42, 48, 54, 58, 65, 71, 84, 88, 98 In-class work time: today and Thursday. Due Monday Test on Equilibrium: Monday Next unit: Acid-Base Reactions


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