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Equilibrium

Equilibrium. {15.1-15.4 Introduction to Equilibrium. What is Chemical Equilibrium?. The dynamic state of a reaction in which the rate of the forward reaction is equal to the rate of the reverse reaction

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Equilibrium

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  1. Equilibrium {15.1-15.4 Introduction to Equilibrium

  2. What is Chemical Equilibrium? • The dynamic state of a reaction in which the rate of the forward reaction is equal to the rate of the reverse reaction • Once an equilibrium is established, the [reactants] and [products] remains constant, but not necessarily equal to each other.

  3. What is Chemical Equilibrium? • "dynamic" means constantly changing • At equilibrium, the reaction is still going forward and backward, but the concentrations of reactants and products stops changing.

  4. Describing Chemical Equilibrim N2O42NO2 Rate forward = Rate reverse

  5. Describing Chemical Equilibrim N2O42NO2 constant [reactants] and [products]

  6. Describing Chemical Equilibrim Since, in a system at equilibrium, both the forward and reverse reactions are being carried out, we write its equation with a double arrow: N2O4⇄ 2NO2

  7. [C]c[D]d [A]a[B]b Kc = The Equilibrium Constant Expression • For the reaction aA + bB⇄cC + dD • the equilibrium constant, • Kcis independent of concentrations • Kcdepends ONLY on temperature

  8. [H2][I2] [HI]2 kf kr = Kc Where does this come from? 2HI(g) ⇄ H2(g) + I2(g) Rate forward = kf[HI]2 and Rate reverse = kr[H2][I2] At equilibrium, Rate forward = Rate reverse so…. kf[HI]2 = kr[H2][I2] =

  9. The Equilibrium Constant, Kc • Know how Kc changes as the chemical equation is manipulated! • For the forward reaction • 2NO + O2⇄ 2NO2 , Kc= • = 4.67 x 1013 • For the reverse reaction • 2NO2⇄ 2NO+ O2 , Kc’ = • = = 2.14 x 10-14

  10. The Equilibrium Constant, Kc For the reaction with equilibrium constant, Kc …the reverse reaction has the equilibrium constant

  11. The Equilibrium Constant, Kc • 2. For 2NO + O2⇄ 2NO2, Kc= • = 4.67 x 1013 • If the coefficients are multiplied by ½ … • NO + ½ O2⇄ NO2, Kc'= • = = Kc’1/2 =

  12. The Equilibrium Constant, Kc If you multiply the coefficients by some number “n” …you must raise Kc to the power of “n” EX: If you multiply equation by 3, Kc’=Kc3 If you multiply equation by ½, Kc’ = Kc1/2

  13. The Equilibrium Constant, Kc • 3. For N2O + 3/2 O2⇄ 2NO2 , a possible mechanism is… • N2O + ½ O2⇄2NOKc1= 1.7 x 10-13 • 2NO + O2⇄2NO2 Kc2= 4.67 x 1013 • ____________________ • N2O + 3/2 O2⇄2NO2 Kc3= Kc1 x Kc2

  14. The Equilibrium Constant, Kc When equations are added together, to get a net overall equation, …the equilibrium constants must be multiplied to get the overall equilibrium constant

  15. (PC)c(PD)d (PA)a(PB)b Kp= The Equilibrium Constant, Kc • 4. For gas phase equilibrium, sometimes we measure partial pressures instead of molarities, then we get a partial pressure equilibrium constant, Kp. • aA + bB⇄cC + dD

  16. How are Kc and Kp related? • Consider 2NO + O2⇄2NO2 • Kc = and Kp =

  17. How are Kc and Kp related?

  18. How are Kc and Kp related? • Where the exponent is equal to the • change in number of moles of gas in the forward reaction. • Here, 3 molof gas  2 molof gas so Dn = (-1)

  19. How are Kc and Kp related? So… R = 0.0821 Latm/molK (if pressure is in atm) T is in Kelvins

  20. Heterogeneous Equilibria • So far we have only considered homogeneous phase reactions in which ALL reactants and products were gases. • In heterogeneous reactions, reactants and products are NOT all in the same phase.

  21. Heterogeneous Equilibria • Pure liquids (l) and solids (s) are OMITTED from equilibrium expressions because… • their concentrations do not change during reactions. Their amount may change, but their concentration does not.

  22. Heterogeneous Equilibria • EXAMPLE: • CaCO3(s) ⇄ CaO(s) + CO2(g) • • CaCO3(s) is 100% CaCO3 regardless of amount • • CaO(s) is 100% CaO regardless of amount • • CO2(g) is pure CO2, but its concentration and pressure increase as more CO2 is produced in a • sealed container • So...CO2 goes in the equilibrium expression, but nothing else!!! • Kc = [CO2] and Kp = PCO2

  23. What does Kc mean anyway? • Large value of Kc or Kp means reaction goes to completion. • If K >>> 1 (large K): Equilibrium lies to the right; toward the product side. • Essentially no equilibrium because reaction is not reversible • “large” means double digit on power of ten (Kc = 2.5 x 1012)

  24. What does Kc mean anyway? • Small value of Kc or Kp means forward reaction does NOT go to any significant extent. • If K <<< 1 (small K): Equilibrium lies to the left; toward the reactant side. • No equilibrium because reaction doesn’t go forward • “small” means double digit negative power of ten (Kc = 2.5 x 10-10)

  25. Equilibrium vs. Rate Constant • Equilibrium constantspredict HOW MUCH/LITTLE product will be made • Rate law constants predict HOW FAST it reaches completion or equilibrium

  26. PRACTICE 15.1. Writing Equilibrium-Constant ExpressionsWrite the equilibrium-constant expression Kc for: H2(g) + I2(g) ⇄ 2HI(g) (b) Cd2+(aq) + 4Br-(aq) ⇄ CdBr42-(aq) Answer: (a) (b)

  27. PRACTICE 15.2. Converting between Kc and KpFor the equilibrium, 2SO3(g) ⇄ 2SO2(g) + O2(g) Kc is 4.08 x 10-3 at 1000K. Calculate the value of Kp. Answer: Kp = (4.08 x 10-3)(0.0821Latm/molK* 1000K)(3-2) Kp = 0.335

  28. PRACTICE 15.3. Interpreting the magnitude of an Equilibrium Constant The following diagrams represent three systems at equilibrium, all in the same-size containers. Without doing any calculations, rank the systems in order of increasing Kc. Answer: The more product present at equilibrium, relative to reactant, the larger the equilibrium constant. (ii) < (i) < (iii), from smallest (most reactant) to largest (most products)

  29. PRACTICE 15.3. Interpreting the magnitude of an Equilibrium Constant (b) If the volume of the containers is 1.0 L and each sphere represents 0.10 mol, calculate Kcfor each system. Answer: In (i) we have 0.60 mol/L product and 0.40 mol/L reactant, Kc = 0.600/0.40 = 1.5 In (ii) we have 0.10 mol/L product and 0.90 mol/L reactant, Kc= 0.10/0.90 = 0.11 In (iii) we have 0.80 mol/L product and 0.20 mol/L reactant, Kc= 0.80/0.20 = 4.0 These calculations verify the order in (a).

  30. PRACTICE 15.3. Interpreting the magnitude of an Equilibrium Constant (c) For the reaction H2(g) + I2(g) ⇄ 2HI(g), Kp = 794 at 298 K and Kp= 55 at 700 K. Is the formation of HI favored more at the higher or lower temperature? Answer: At the lower temperature because Kpis larger at the lower temperature and reactants are favored for larger K values.

  31. PRACTICE 15.4. Evaluating an Equilibrium Constant When an Equation is Reversed For the reaction N2(g) + 3H2(g) ⇄ 2NH3(g), Kp = 4.34 x 10-3at 300oC. What is the value of Kp for the reverse reaction? Answer: Kp’ = (4.34 x 10-3)-1 = 2.30 x 102

  32. PRACTICE 15.5. Combining Equilibrium Expressions (a) Given the reactions HF(aq) ⇄ H+(aq) + F-(aq), Kc= 6.8 x 10-4 H2C2O4(aq) ⇄ 2H+(aq) + C2O42-(aq), Kc = 3.8 x 10-6 determine the value of Kc for the reaction 2HF(aq)+ C2O42-(aq)⇄ 2 F-(aq) + H2C2O4(aq)

  33. PRACTICE 15.5. Combining Equilibrium Expressions Answer:

  34. PRACTICE 15.5. Combining Equilibrium Expressions (b) Given that, at 700 K, Kp = 54.0 for the reaction H2(g) + I2(g) ⇄ 2HI(g) and Kp = 1.04  104 for the reaction N2(g) +3H2(g) ⇄ 2NH3(g) determine the value of Kpfor the reaction 2NH3(g)+ 3I2(g) ⇄ 6HI(g) + N2(g) at 700K. Answer:

  35. PRACTICE 15.6. Writing Equilibrium-Constant Expressions for Heterogeneous Reactions Write the equilibrium-constant expression, K, for (a) Kc for (b) Kp for (a) (b) Answer:

  36. PRACTICE 15.7. Analyzing a Heterogeneous Reactions Each of these mixtures was placed in a closed container and allowed to stand: CaCO3(s) CaO(s) and CO2(g) at a pressure greater than the value of Kp CaCO3(s) and CO2(g) at a pressure greater than the value of Kp CaCO3(s) and CaO(s) (a)Determine whether or not each mixture can attain the equilibrium Answer: For equilibrium to be achieved, it must be possible for both the forward process and the reverse process to occur. For the forward process to occur, there must be some calcium carbonate present. For the reverse process to occur, there must be both calcium oxide and carbon dioxide. In both cases, either the necessary compounds may be present initially or they may be formed by reaction of the other species.

  37. PRACTICE 15.7. Analyzing a Heterogeneous Reactions Each of these mixtures was placed in a closed container and allowed to stand: CaCO3(s) CaO(s) and CO2(g) at a pressure greater than the value of Kp CaCO3(s) and CO2(g) at a pressure greater than the value of Kp CaCO3(s) and CaO(s) (a)Determine whether or not each mixture can attain the equilibrium Answer: Equilibrium can be reached in all cases except (c) as long as sufficient quantities of solids are present.

  38. PRACTICE 15.7. Analyzing a Heterogeneous Reactions Each of these mixtures was placed in a closed container and allowed to stand: CaCO3(s) CaO(s) and CO2(g) at a pressure greater than the value of Kp CaCO3(s) and CO2(g) at a pressure greater than the value of Kp CaCO3(s) and CaO(s) (a)Determine whether or not each mixture can attain the equilibrium Answer: (a) CaCO3 simply decomposes, forming CaO(s) and CO2(g) until the equilibrium pressure of CO2 is attained. There must be enough CaCO3, however, to allow the CO2 pressure to reach equilibrium.

  39. PRACTICE 15.7. Analyzing a Heterogeneous Reaction Each of these mixtures was placed in a closed container and allowed to stand: CaCO3(s) CaO(s) and CO2(g) at a pressure greater than the value of Kp CaCO3(s) and CO2(g) at a pressure greater than the value of Kp CaCO3(s) and CaO(s) (a)Determine whether or not each mixture can attain the equilibrium Answer: (b) CO2 continues to combine with CaO until the partial pressure of the CO2 decreases to the equilibrium value.

  40. PRACTICE 15.7. Analyzing a Heterogeneous Reactions Each of these mixtures was placed in a closed container and allowed to stand: CaCO3(s) CaO(s) and CO2(g) at a pressure greater than the value of Kp CaCO3(s) and CO2(g) at a pressure greater than the value of Kp CaCO3(s) and CaO(s) (a)Determine whether or not each mixture can attain the equilibrium Answer: (c) There is no CaO present, so equilibrium cannot be attained because there is no way the CO2 pressure can decrease to its equilibrium value (which would require some of the CO2 to react with CaO).

  41. PRACTICE 15.7. Analyzing a Heterogeneous Reactions Each of these mixtures was placed in a closed container and allowed to stand: CaCO3(s) CaO(s) and CO2(g) at a pressure greater than the value of Kp CaCO3(s) and CO2(g) at a pressure greater than the value of Kp CaCO3(s) and CaO(s) (a)Determine whether or not each mixture can attain the equilibrium Answer: (d) The situation is essentially the same as in (a): CaCO3 decomposes until equilibrium is attained. The presence of CaO initially makes no difference.

  42. PRACTICE 15.7. Analyzing a Heterogeneous Reactions When added to Fe3O4(s) in a closed container, which one of the following substances — H2(g), H2O(g), O2(g) — allows equilibrium to be established in the reaction 3Fe(s) + 4H2O(g) ⇄ Fe3O4(s) + 4 H2(g)? Answer: H2(g)

  43. Equilibrium {15.5-15.6 – Equilibrium Problem Solving

  44. Reaction Quotient, Q • Q gives the same ratio the equilibrium expression gives, but for a system that is not at equilibrium. • To calculate Q, one substitutes the initial concentrations on reactants and products into the equilibrium expression. • Q tells us if the reaction is at equilibrium yet or how the reaction needs to "shift" to reach equilibrium.

  45. Reaction Quotient, Q EXAMPLE: N2+ O2⇄ 2NO At equilibrium: Not at equilibrium: Equilibrium concentrations Equilibrium constant Non-quilibrium concentrations (any given concentrations) Reaction quotient

  46. Reaction Quotient, Q • If Q = K… • …rxn is at equilibrium. • If Q < K… • …Q needs to get bigger to reach equilibrium, so rxn shifts RIGHT (RP) • If Q > K… • …Q needs to get smaller to reach equilibrium, so rxn shifts LEFT (RP)

  47. PRACTICE 15.8. Predicting the Direction of Approach to Equilibrium At 1000 K the value of Kpfor the reaction 2SO3(g) ⇄ 2SO2(g) + O2(g) is 0.338. Predict the direction in which the reaction proceeds toward equilibrium if the initial partial pressures are PSO3 = 0.16 atm; PSO2 = 0.41 atm; PO2 = 2.5 atm. Answer: Qp= 16; Qp> Kp, and so the reaction will proceed from right to left, forming more SO3.

  48. Common Types of Equilibrium Problems • Find Kc or Kp from experimental data • LAB: The Determination of Keq for FeSCN2+ • 2. Find equilibrium concentrations from Kc or Kp and initial concentrations • Use ICE BOX method.

  49. Example 1 KNOWN: Concentrations  UNKNOWN: Kc In a 10.0 L vessel at 1000K, 0.250 mol SO2 and 0.200 mol O2 react to form 0.162 mol SO3 at equilibrium. What is Kc at 1000K?

  50. Example 2 KNOWN: Kc UNKNOWN: Equilibrium concentrations If we start with 0.500 mol H2 and 0.500 mol I2 in a 5.25 L vessel at 698K, how many moles of each substance are present at equilibrium? Kc = 54.3 at 698K

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