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CPT 4. CHEMICAL EQUILIBRIA

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CPT 4. CHEMICAL EQUILIBRIA

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    1. CPT 4. CHEMICAL EQUILIBRIA Objectives: * Describe and explain the quantitative relations between the species involved in a chemical reaction equilibrium. * Calculate and apply the principles associates with chemical equilibria

    2. Figure: 14-12-02P25a-f Title: Problems by Topic, 25 Caption: H2 and I2 are combined in a flask and allowed to react according to the following reaction: H2(g) + I2(g)= 2 HI(g)Examine the figures below (sequential in time) and answer the following questions: a. Which figure represents the point at which equilibrium is reached? b. How would the series of figures change in the presence of a catalyst? c. Would the final figure (vi) have different amounts of reactants and products in the presence of a catalyst?Figure: 14-12-02P25a-f Title: Problems by Topic, 25 Caption: H2 and I2 are combined in a flask and allowed to react according to the following reaction: H2(g) + I2(g)= 2 HI(g)Examine the figures below (sequential in time) and answer the following questions: a. Which figure represents the point at which equilibrium is reached? b. How would the series of figures change in the presence of a catalyst? c. Would the final figure (vi) have different amounts of reactants and products in the presence of a catalyst?

    3. CHEMICAL EQUILIBRIA Thought teaser Examine the graphic on the previous slide showing the following reaction: H2 + I2 -> 2HI At what point can you tell that the reaction has reached equilibrium? Hint: watch the HI molecules. At what point do you notice that the proportion of products in the mixture does not changer any more?

    4. 4.1. Introduction: Equilibrium and equilibrium constant DEMO: NO2 <-> N2O4 CD E:\Chapter_14\Present\Media\AADZZSE0.html E:\Chapter_14\Present\Media\EquilibriumConstant\EquilibriumConstant.html Equilibrium: state of a reaction process in steady state, when the concentrations of reactants and products no longer change. Law of mass action : regulates relation between species at equilibrium. General case: aA + bB <-> cC + dD K = [C]c[D]d / ([A]a[B]b)

    5. Law of mass action (Continued) K = [C]c[D]d / ([A]a[B]b) K = Equilibrium constant = mathematical form of the law of mass action If K > 1, products are more abundant than reactants, reaction favors products. If K < 1, reactants is more abundant than products, reaction favors reactants. Demo: Tro/Media 14.10 Conclusion K = constant for each reaction at a given temp.

    6. RQ2-23B: Previous Material Review The equilibrium constant for the following reaction: 2C2H6 + 7O2 -> 4CO2 + 6H2O is K = a. [C2H6]2[O2]7/ [CO2]4[H2O]6 b. [CO2]4[O2]7/ [H2O]6 [C2H6]2 c. [CO2]4[H2O]6 / [C2H6]2[O2]7

    7. (RQ2-23B2): Previous Materal Review If the value of K in the previous reaction is 3, the reaction favors: a, reactants; b, products; c, both If you find more C2H6 and O2 than CO2 and H2O after the reaction has reached equilibrium, then the value of K is: a, <1; b, >1; c, <0

    8. Law of mass action (Continued 2) K = [C]c[D]d / ([A]a[B]b) Equilibrium positions: specific concentrations of reactants and products. Positions may change: K stays constant

    9. Figure: 14-12-01P24a-c Title: Problems by Topic, 24 Caption: Ethene (C2H4) can be halogenated by the following reaction: C2H4(g) + X2(g) = C2H4X2(g) where X2 can be Cl2 (green), Br2 (brown), or I2(purple). Examine the three figures below representing equilibrium concentrations in this reaction at the same temperature for the three different halogens. Rank the equilibrium constants for these three reactions from largest to smallest.Figure: 14-12-01P24a-c Title: Problems by Topic, 24 Caption: Ethene (C2H4) can be halogenated by the following reaction: C2H4(g) + X2(g) = C2H4X2(g) where X2 can be Cl2 (green), Br2 (brown), or I2(purple). Examine the three figures below representing equilibrium concentrations in this reaction at the same temperature for the three different halogens. Rank the equilibrium constants for these three reactions from largest to smallest.

    10. Law of mass action (RQ2-24) The graphics on the previous slide show the reaction C2H4 + X2 <-> C2H4X2 at equilibrium. X is Cl, Br, or I. Rank the equilibrium constants of the reactions from largest to smallest. Explain. a. Green rxn K > Red rxn K > Purple rxn K. The amount of X2 in the rxn mixture decreases in the same order. b. Purple rxn K > Green rxn K > Red rxn K. The amount of X2 in the rxn mixture decreases in the same order. c. Green rxn K > Red rxn K > Purple rxn K. The amount of X2 in the rxn mixture increases in the same order.

    11. Law of mass action (Illustrations) Example 1: 4NH3 + 7O2 <-> 4NO2 + 6H2O K = [NO2]4[H2O]6 / [NH3]4[O2]7 Example 2: Info provided: * 2SO2 + O2 <-> 2SO3 * [SO2] = 3.77E-3 mol/L; [O2] = 4.30E-3 mol/L; [SO3] = 4.13E-3 mol/L Info requested: K = ? Extra exercise: #25, pg 656

    12. Equilibrium Constant Determination (RQ2-24a) Consider the following equilibrium constant: K = [CO2]3[H2O]4 / [C3H8][O2]5 What reaction does it correspond to? 3CO2 + 4H2O -> C3H8 + 5O2 C3H8 + 5O2 -> 3CO2 + 4H2O C3H8 + 4H2O -> 3CO2 + 5O2

    13. Equilibrium Constant Calculation (Illustrations 2) Example 3: Info provided: * H2 + CO2 <-> H2O + CO * Reaction container volume: 1.00 L * Amounts of CO and H2O at equilibrium: 0.11 mol * Amounts of H2 and CO2 at equilibrium: 0.087 mol Info requested: (a) Calculate K for the reaction

    14. Equilibrium Constant Calculation (Illustrations 2 Solution) K = [H2O][CO] / [H2][CO2] [H2O] = [CO] = 0.11 mol/L (how come?) [H2] = [CO2] = 0.087 mol/L K = [0.11 M]2 / [0.087 M]2 = ?

    15. Equilibrium Constant Calculation (Illustration 2 Continued) Problem continued Info provided (b): * Reaction container volume: 2.00 L * amounts of H2 and CO2 at beginning of reaction: 0.050 mol Info requested (b): amounts of CO and H2O at equilibrium. Extra example: #37, pg 657

    16. Equilibrium Constant Calculation (Illustration 2 Continued) Assume K = 1.60 K = [H2O][CO] / [H2][CO2] = [CO]2 / [CO2]2 If at equilibrium x mol/L of CO are produced, then the initial concentration of CO2 is reduced by an amount equal to x. [CO2] = [CO2]0 – x Then 1.60 = x2 / ([CO2]0 – x)2

    17. Equilibrium Constant Calculation (Illustration 2 Continued 2) [CO2]0 = 0.050 mol / 2.00 L = 0.025 M 1.60 = x2 /(0.025 – x)2 which leads to x / 0.025 – x = 1.26. Solve for x Ultimately x = 0.025*1.26 / (1 + 1.26) = ? = [CO] = [H2O] [CO2] = 0.025 M – [CO] = ?

    18. Reaction Positions: Reaction Quotient(Q) Applies to reactions outside equilibrium conditions Q = [P]n / [R]m; [P]n , [R]m = concentrations of products and reactants before or after equilibrium. Q enables to calculate the reaction orientation.

    19. Reaction Quotient (Continued) If Q = K, reaction system is at equilibrium If Q < K, there still are more reactants than at equilibrium. The system shifts toward consumption of more reactants and formation of products If Q > K there are more products than at equilibrium. The system shifts toward consumption of products and formation of reactants.

    20. Reaction Quotient (Illustration) Example: Info provided: * N2 + O2 <->2NO * K = 4.0E-4 * At 2000 K, [N2] = 0.50 M; [O2] = 0.25 M; [NO] = 4.2E-3 M. Info requested: Is the reaction at equilibrium? If not which way will the reaction shift to reach equilibrium?

    21. Reaction Quotient (Illustration Solution) Expression of Q : [NO]2/[N2][O2]. Plug in numbers. Q = ? Compare Q to K = 4.0E-4. What conclusion do you draw from the value of Q ? See RQ2-25 Extra Exercise: #45, pg 657

    22. Reaction Quotient (RQ2-25) Based on the value of Q, what do you conclude? a. The reaction shifts toward the formation of more products because Q < K b. The reaction shifts toward the formation of more reactants because Q > K c. The reaction shifts toward the formation of more products because Q < 1

    23. 4.2 Equilibrium expression for gases For gases: PV = nRT <-> P = nRT / V = CRT, Therefore C = P/RT. C = molarity concentration. Case study: N2 + 3H2 <-> 2NH3 Kc = K = C2NH3 / (CN2*C3H2) = concentration-based equilibrium constant Replace C by P/RT in expression of Kc. After regroupment and simplification of RT factors: Kc = (RT)2*P2NH3 / (PN2*P3H2). Move RT term to constant side. Kc becomes Kp Final result: Kp = P2NH3 / (PN2*P3H2)

    24. Equilibrium expression for gases (Continued) Relation between Kc and Kp: Case of N2 + 3H2 <-> 2NH3 Kp = Kc/(RT)2 = Kc(RT)-2. -2 = NH3 coefficient - sum of N2 and H2 coefficients General Rule: Kp = Kc(RT)Dn Dn = change in #mol from reactants to products = sum of coefficients of products – sum of coefficients of reactants

    25. RQ2-25b Consider the relation Kp = Kc(RT)Dn In what conditions is Kp > Kc ? Justify your answer When the number of moles of products is > the number of moles of reactants, because Dn is negative. When the number of moles of products is < the number of moles of reactants, because Dn is negative. When the number of moles of products is > the number of moles of reactants, because Dn is positive.

    26. Equilibrium expression for gases (RQ2-26) Consider the following reaction: 4NH3 (g)+ 5O2 (g) <-> 4NO (g) + 6H2O (g) Express Kc of the reaction in terms of Kp Kc = Kp/(RT)Dn = Kp/(RT)(4+6-4-5)= Kp/RT = Kp(RT)-1 Kc = Kp(RT)Dn = Kp (RT)(4+6-4-5)= KpRT Kc = Kp/(RT)Dn = Kp/(RT)(4+5-4-6)= Kp/RT-1 = Kp(RT)

    27. Equilibrium Constants for Gases (Illustration) Example: Decomposition of N2O4 to NO2 Info provided: N2O4 <-> 2NO2 * Pressure of N2O4 at equilibrium: 0.85 atm * Kp = 0.15 (at 25 deg C) Info requested: Total pressure of the gas mixture in the container How do you proceed to solve the problem? See RQ2-25C

    28. RQ2-25C How do you proceed to solve the problem? a. Set up the expression of the final answer to Ptotal = PNO2 + PN2O4. PN204 is known. Figure out PNO2 using Kp = P2NO2 / PN2O4 b. Set up the expression of the final answer to Ptotal = PNO2 - PN2O4. PN204 is known. Figure out PNO2 using Kp = P2NO2 / PN2O4 c. Set up the expression of the final answer to Ptotal = PNO2 + PN2O4. PN204 is known. Figure out PNO2 using Kp = PN2O4 / P2NO2

    29. Kp (#45, pg 793: Solution) Kp = P2NO2 / PN2O4 = 0.15 Plug in value of PN2O4 0.15 = P2NO2 / 0.85 P2NO2 = ? PNO2 = ? Total pressure = PNO2 + PN2O4 = ? Extra Exercise: #39, PG 657

    30. 4.3. Heterogeneous equilibria Context: when components of a reaction are in different physical states. CaCO3(s) <-> CaO(s) + CO2(g) K’ = [CaO][CO2]/[CaCO3] [CaO] and [CaCO3]: Constant, because #mol/V of solid does not change. Conclusion: incorporate [CaO] and [CaCO3] into K’. Same consideration holds for pure liquids. Result: K = [CO2] Rule: Concentrations of pure liquids and solids are not included in the expression of the equilibrium constants of reactions.

    31. Heterogeneous equilibria (RQ2-27) The concentrations of pure solids and liquids are considered constant in a chemical reaction. What about the amounts of solids and liquids? Explain a. They change. If the #mol increases, V decreases, which keeps #mol/V constant. b. They do not change. Otherwise the concentrations would change too. c. They change. However the #mol and V change proportionately, which keeps #mol/V constant.

    32. Heterogeneous equilibria (Illustration) Example: Decomposition of NH4HS Info provided: * NH4HS(s) <-> NH3(g) + H2S(g). * Kp = 0.11 at 25 deg C Info requested: Total Pressure in the rxn container Kp = P2NH3 = 0.11; PNH3 = ? Extra exercise: 53, pg 657

    33. RQ2-27B: Previous Material Review When you have a reaction equilibrium involving substances in different physical states, you are dealing with a: a, heteroclite; b, heterogeneous; c, composite equilibrium. When expressing the equilibrium constant of a heterogeneous equilibrium, don’t take into account: a, pure solids and gases; b, pure liquids and gases; c, solids and pure liquids.

    34. 4.4. Combination of Equilibrium constants Study Case: Reactions resulting from combination of other reactions with known K’s General case: A + B <-> D + E K = … Equation 1: A+ B <-> C; K1 = … Equation 2: C <-> D + E; K2 = … Eq 1 + Eq 2 = A + B + C <-> C + D + E; K = [C][D][E] / ([A][B][C]) = K1*K2

    35. Combination of Equilibrium constants (Continued) Additional equilibrium constant relations If for the equation A <-> B, K1 = [B] / [A], then: For the equation B <-> A, K2 = [A] / [B] = 1/K1 For 2A <-> 2B, K3 = [B]2 / [A]2 = (K1)2

    36. Combination of Equilibrium constants (RQ2-28) Consider the following reactions: a. 2A <-> B + 2C b. 3D <-> B + 2C c. 2A <-> 3D How should you process equations a and b to produce equation c?

    37. Combination of Equilibrium constants (RQ2-28 Answer) a. Reverse equation a and add it to equation b to get equation c. b. Reverse equation b and add it to equation a to get equation c. c. Simply add equation a to equation b to get equation c.

    38. Combination of Equilibrium constants (Illustration) Info provided: * Fe(s) + H2O(g) <-> FeO(s) + H2(g) * H2O(g) + CO(g) <-> H2(g) + CO2(g) K1 = 1.6 * FeO(s) + CO(g) <-> Fe(s) + CO2(g) K2 = 0.67 Info requested: Find K for the 1st reaction. Extra exercise: 27, 29, pg 656;

    39. Combination of Equilibrium constants (Solution) Expression of K: [H2]/[H2O] Expression of K1: [H2][CO2]/([H2O][CO]) Expression of K2: [CO2]/[CO] What is needed in order to transform K1 into K? See RQ2-28b

    40. RQ2-28b Given the expressions of K, K1 and K2 n the previous slide, how can you change K1 to obtain K? a. Cancel [H2] and [H2O] by multiplying K1 by 1/K2 b. Cancel [CO2] and [CO] by multiplying K1 by K2 c. Cancel [CO2] and [CO] by multiplying K1 by 1/K2

    41. Multiply the expressions of K1 and 1/K2. What happens to [CO2] and [CO]? K = ? Extra ex: 27, pg 656

    42. 4.5. Applications of the Equilibrium Constants Equilibrium constants can be used to determine the amounts of reactants and products in a reaction. Tools needed: * Chemical Equation * Changes in concentrations of reactants caused by formation of products

    43. a. Calculating equilibrium concentrations and pressures Procedure: * Write the chemical equation * Set up the Initial-Change-Equilibrium (ICE) quantitative relationships between changes in concentrations of reactants due to formation of products. * Use the ICE relations in to set the expression of the equilibrium constant. * Figure out the concentrations of the components.

    44. Calculating equilibrium concentrations (Illustration) Example: Reaction of H2 and I2 to make HI. Info available: * H2 + I2 <-> 2HI; K = 55.64 at 425 deg. C. * Amounts of reactants before the reaction starts: 1 mol of H2 and I2 * Volume of the reaction container: 0.500 L Info requested: concentrations if each component at equilibrium. Extra exercise: #51, pg 658

    45. Calcultating equilibrium concentrations (Solution) ICE RELATIONS Concentrations (#mol / V(L’s) Initial Change At Equilibrium H2 2 -x 2 – x I2 2 -x 2 – x HI 0 2x(*) 2x (*) According to the mole ratios as shown in the equation of the reaction

    46. Solution (Continued) K = (2x)^2 / (2-x)^2 ? (K)^1/2 = 2x / 2-x = (55.64)^1/2 = 7.46 This leads to 7.46(2-x) = 2x = 14.9 - 7.46x Which leads to x(2+7.46) = 14.9 Therefore x = ? Conclusion: at equilibrium: [HI] = 2x = ? [H2] = [I2] = 2 – x = ?

    47. RQ2-28C: Previous Material Review If you know K1 and K2 for reactions A <-> B and B<-> C respectively, all you need to do to find K for reaction A<-> C is to: a, add; b, multiply; c, divide K1 and K2 In order to find the equilibrium concentration of a reactant for which you know the initial concentration in a reaction: a, add; b, multiply; c, subtract the change in concentration from the initial concentration.

    48. Calcultating equilibrium concentrations (RQ2-29) Consider the reaction A <-> 3C If you start with [A] = 5 mol/L, and form product with a 3n mol/L concentration, what are [A] and [C] at equilibrium? a. [A] = (5 –n) mol/L and [C] = 3n mol/L because [A] increases and [C] decreases during the reaction b. [A] = (3n) mol/L and [C] = 5 - n mol/L because [A] decreases and [C] increases during the reaction c. [A] = (5 – n) mol/L and [C] = 3n mol/L because [A] decreases and [C] increases during the reaction

    49. b. Equilibria with small constants General equation: A + B<-> C. Simplifications can be done to avoid use of complex math processes. Condition: [A]0 > 100 * K or [A]0 / K > 100 Simplification: [A]0 almost = [A]eq because very little products are formed and therefore changes in initial concentrations of reactants are negligible.

    50. Small equilibrium constants (Illustration) Example: Reaction of N2 and O2 to form NO Info available: * N2 + O2 ? 2NO; K = 1E-5 at 1500K [N2]0 = 0.80 M [O2]0 = 0.20M Info requested: equilibrium concentrations of the components Extra exercise: #57, pg 658

    51. Small equilibrium constants (Solution) ICE RELATIONS between components Concentrations (mol/l) Initial Change Equilibrium N2 0.8 -x 0.8 - x O2 0.2 -x 0.2 – x NO 0 2x(*) 2x (*) According to the mole ratios as shown in the equation of the reaction.

    52. Small equilibrium constants (Solution, continued) K = [NO]2 / [N2]*[O2] = (2x)2 / {(0.8 - x)(0.2 - x)} Compare K to the initial concentrations of reactants: [N2]0 / K = 0.8 / 1E-5 = 0.8E5 >> 100 and [O2]0 / K = 0.2E5 >> 100. Therefore the simplifications are justified: very little products are formed * [N2] – x = 0.8 – x = approximately 0.8; * [O2] – x = 0.2 – x = approximately 0.2

    53. Small equilibrium constants (RQ2-30) Given the simplifications on the previous slide, what is the next step in the problem solving process? a. Simplify the expression which becomes K = (2x)2 / (x2) b. Simplify the expression of K which becomes K = (2x)2 / (0.8*0.2) c. Simplify the expression which becomes K = (2x)2 / (0.8*0.2 – 1.0x – x2)

    54. Small equilibrium constants (Solution, continued 2) Expression of K = (2x)^2 / (0.8 x 0.2) (K)^1/2 = 2x / (0.16)^1/2 = (1E-5)^1/2 = 3.16E-3 = 2x Result: x = ? Conclusion: at equilibrium: [NO] = 2x = ? [N2] = 0.8 – x = ? [O2] = 0.2 – x = ?

    55. 4.7. Disturbing a Chemical Equilibrium : LeChatelier’s Principle If a change is made to an equilibrium, the equilibrium shifts in the direction that consumes the change

    56. LeChatelier Principle Case 1: Changing the amounts of reactants / products. The equilibrium shifts in a direction that consumes the change. Example: N2(g)+ 3H2(g) ?? 2NH3(g) Adding N2 or H2 results in the consumption of the addition. The system consumes the reagents et forms products. The same effect result if a product (NH3) is removed instead of adding reactants. Extra example: #63, pg 658 (hint: heterogeneous)

    57. LeChatelier Principle (RQ2-31) Consider this reaction: NH3 + H2O <-> NH4+ + OH- What would be the effect of removing water from the reaction? Justify your answer The equilibrium would shift to the left. It would seek to replace the removed H2O by consuming NH4+ and OH- b. The equilibrium would shift to the right. It would seek to replace the removed H2O by consuming NH4+ and OH- c. The equilibrium would shift to the left. It would seek to replace the removed H2O by consuming NH3 and OH-

    58. LeChatelier Principle (Continued) Case 2: Changing the volume by changing pressure. * Decreasing the volume: equilibrium reacts by shifting toward the side that produces less amounts of substances (Production of NH3 in the example) Example: running the previous reaction in a container half the size of the container in the original reaction.

    59. LeChatelier Principle (Continued) Increasing the volume: equilibrium reacts by shifting toward the side that produces more amounts of substances Example: running the reaction in a container with a volume twice the size of the original volume. Extra example: #69, pg 659

    60. LeChatelier Principle (Continued) Case 3: Changing the temperature. The equilibrium shifts in a direction that consumes the change. Note: Contrary to previous cases, changing T leads to changing the value of K. Exothermic reactions: heat is produced by the reaction. Heat is treated as a product. Adding more heat pushes the system toward consumption of products & formation of reactants. K is decreased Extra example: #69, pg 659

    61. LeChatelier Principle (Continued) Endothermic reactions: heat is consumed by the reaction. Heat is treated as a reactant. Adding more heat pushes the system toward consumption of reactants & formation of products. The value of K is increased Extra example: #69, pg 659

    62. LeChatelier Principle (RQ2-32) Consider this reaction endothermic reaction: NH3 + H2O <-> NH4+ + OH- How would you use temperature to make it produce more products? a. Lower the temperature. It is like removing the heat as a reactant. It pushes the equilibrium to the right b. Raise the temperature. It is like adding the heat as a reactant. It pushes the equilibrium to the right c. Raise the temperature. It is like removing the heat as a reactant. It pushes the equilibrium to the right

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