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Chapter 14 Chemical Equilibrium

Chemistry II. Chapter 14 Chemical Equilibrium. Speed of a chemical reaction is determined by kinetics How fast a reaction goes Extent of a chemical reaction is determined by thermodynamics How far a reaction goes. Reaction Dynamics. forward reaction: reactants  products

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Chapter 14 Chemical Equilibrium

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  1. Chemistry II Chapter 14Chemical Equilibrium

  2. Speed of a chemical reaction is determined by kinetics • How fast a reaction goes Extent of a chemical reaction is determined by thermodynamics • How far a reaction goes

  3. Reaction Dynamics • forward reaction: reactants  products • therefore the [reactant] decreases and the [product] increases • as [reactant] decreases, the forward reaction rate decreases • reverse reaction: products  reactants • assuming the products are not allowed to escape • as [product] increases, the reverse reaction rate increases • processes that proceed in both the forward and reverse direction are said to be reversible • reactants ⇌ products

  4. Rate Forward Rate Reverse Reaction Dynamics Initially, only the forward reaction takes place. As the forward reaction proceeds it makes products and uses reactants. Because the reactant concentration decreases, the forward reaction slows. As the products accumulate, the reverse reaction speeds up. Once equilibrium is established, the forward and reverse reactions proceed at the same rate, so the concentrations of all materials stay constant. Eventually, the reaction proceeds in the reverse direction as fast as it proceeds in the forward direction. At this time equilibrium is established. Rate Time

  5. Concentration  Equilibrium Established Time  H2(g) + I2(g) ⇌ 2 HI(g) Since the reactant concentrations are decreasing, the forward reaction rate slows down Since the [HI] at equilibrium is larger than the [H2] or [I2], we say the position of equilibrium favors products As the reaction proceeds, the [H2] and [I2] decrease and the [HI] increases At equilibrium, the forward reaction rate is the same as the reverse reaction rate Once equilibrium is established, the concentrations no longer change And since the product concentration is increasing, the reverse reaction rate speeds up

  6. Dynamic Equilibrium • some reactions reach equilibrium only after almost all the reactant molecules are consumed – equilibrium favors the products • other reactions reach equilibrium when only a small percentage of the reactant molecules are consumed – equilibrium favors the reactants

  7. An Analogy: Pwad(left) ⇌ Pwad(right) Rules: • Each student wads up two paper wads. • You must start and stop as the timekeeper says. • Throw only one paper wad at a time. • If a paper wad lands next to you, you must throw it back.

  8. Equal Number of Students on Each Side of the Classroom Pwad (left) Pwad (right)

  9. Most Students on the Left Side– 2 Students on the Right Side Pwad (left) ⇌ Pwad (right)

  10. Most Students on the Left Side– 2 Students on the Right Side Pwad (left) ⇌ Pwad (right)

  11. Common Misconceptions • Equilibrium means equal amounts of reactant and product. • No - A reaction can be at equilibrium and have more paper wads on the product side of the room • A reaction at equilibrium has stopped. • No - The paper wads keep flying in both directions even after equilibrium is achieved • Equilibrium can only be achieved by starting with reactants. • No - Equilibrium can also be achieved when starting with all of the paper wads on the product side of the room.

  12. The Equilibrium Constant • [reactants] and [products] are not equal at equilibrium, there is a relationship between them Law of Mass Action: aA + bB ⇌cC + dD, Keq (or Kc) is called the equilibrium constant • Units vary from reaction to reaction, unitless in this case

  13. The Equilibrium Constant so for the reaction: 2 N2O5⇌ 4 NO2 + O2 Units: mol3/L3

  14. The Equilibrium ConstantSignificance of Keq • Keq >> 1, when the reaction reaches equilibrium there will be more product than reactant molecules • the position of equilibrium favors products • Keq << 1, when the reaction reaches equilibrium there will be more reactant molecules than product molecules • the position of equilibrium favors reactants

  15. A Large Equilibrium Constant Remember: does not tell us about how fast, only how far

  16. A Small Equilibrium Constant

  17. Relationships between Kand Chemical Equations • reaction is written backwards, the equilibrium constant is inverted for the reaction aA + bB ⇌cC + dD the equilibrium constant expression is: for the reaction cC + dD ⇌aA + bB the equilibrium constant expression is:

  18. Relationships between Kand Chemical Equations • coefficients of an equation are multiplied by a factor, K is raised to that factor for the reaction aA + bB ⇌cC equilibrium constant expression is: for the reaction 2aA + 2bB ⇌ 2cC equilibrium constant expression is:

  19. Relationships between Kand Chemical Equations • when you add equations to get a new equation, Keq for the new equation is the product of the equilibrium constants of the old equations for the reactions (1) aA ⇌ bB and (2) bB ⇌cC the K expressions are: for the overall reaction aA ⇌cC the K expression is:

  20. K K’ Ex 14.2 – Compute the equilibrium constant at 25°C for the reaction NH3(g) ⇌0.5 N2(g) + 1.5 H2(g) Given: Find: for N2(g) + 3 H2(g) ⇌ 2 NH3(g), Keq = 3.7 x 108 at 25°C Keq for NH3(g) ⇌ 0.5N2(g) + 1.5H2(g), at 25°C Concept Plan: Relationships: Kbackward = 1/Kforward, Knew = Koldn Solution: N2(g) + 3 H2(g)  2 NH3(g) K1 = 3.7 x 108 2 NH3(g)⇌ N2(g) + 3 H2(g) NH3(g)⇌ 0.5 N2(g) + 1.5 H2(g)

  21. Equilibrium Constant in Terms of PressureReactions Involving Gases • the concentration of a gas in a mixture is proportional to its partial pressure aA(g) + bB(g) ⇌cC(g) + dD(g) or

  22. Kc and Kp • when calculating Kp, partial pressures are always in atm • the values of Kp and Kc are not necessarily the same • because of the difference in units • the relationship between them is: Δn = c + d – (a + b) or no. moles of reactants minus no. Moles products When n = 0, Kp = Kc

  23. Kp Kc Ex 14.3 – Find Kc for the reaction 2 NO(g) + O2(g) ⇌ 2 NO2(g), given Kp = 2.2 x 1012 @ 25°C Given: Find: Kp = 2.2 x 1012 atm-1 Kc Concept Plan: Relationships: Solution: 2 NO(g) + O2(g) ⇌ 2 NO2(g) Dn = 2  3 = -1 Check: K has units L mol-1 since there are more moles of reactant than product, Kc should be larger than Kp, and it is

  24. Heterogeneous Equilibria: Reactions Involving Solids and Liquids aA(s) + bB(aq) ⇌cC(l) + dD(aq) • Concentrations of solids and liquids doesn’t change • its amount can change, but the amount of it in solution doesn’t because it isn’t in solution • solids and liquids are not included in the Keq expression

  25. Heterogeneous Equilibria The amount of C is different, amounts of CO and CO2 remain the same. C has no effect on the position of equilibrium.

  26. Calculating Equilibrium Constants from Measured Equilibrium Concentrations • To find K measure the amounts of reactants and products in a mixture at equilibrium • may have different amounts of reactants and products, but the value of the equilibrium constant will always be the same • as long as the temperature is kept constant

  27. Initial and Equilibrium Concentrations forH2(g) + I2(g) ⇌ 2HI(g) @ 445°C

  28. Calculating Equilibrium Constants from Measured Equilibrium Concentrations • Stoichiometry can be used to determine the equilibrium [reactants] and [products] if you know initial concentrations and one equilibrium concentration e.g. 2A(aq) + B(aq) ⇌ 4C(aq) with initial concentrations [A] = 1.00 M, [B] = 1.00 M, and [C] = 0. You then measure the equilibrium concentration of C as [C] = 0.50 M. -½(0.50) -¼(0.50) +0.50 0.88 0.75

  29. Calculating Equilibrium Concentrations e.g. 2A(aq) + B(aq) ⇌ 4C(aq) -½(0.50) -¼(0.50) +0.50 0.88 0.75 Referred to as ICE table (initial, change, equilibrium) Keq = [C]4/[A]2[B] = 0.13

  30. Ex 14.6  Find the value of Kc for the reaction2 CH4(g) ⇌ C2H2(g) + 3 H2(g) at 1700°C if the initial [CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M +0.035

  31. Ex 14.6  Find the value of Kc for the reaction2 CH4(g) ⇌ C2H2(g) + 3 H2(g) at 1700°C if the initial [CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M +0.035 -2(0.035) +3(0.035) 0.105 0.045

  32. The Reaction Quotient for the gas phase reaction aA + bB ⇌cC + dD the reaction quotient is: • reaction mixture not at equilibrium; how can we determine which direction it will proceed? • the answer is to compare the current concentration ratios to the equilibrium constant reaction quotient, Q

  33. The Reaction Quotient:Predicting the Direction of Change • if Q > K, the reaction will proceed fastest in the reverse direction • Q must decrease, the [products] will decrease and [reactants] will increase • if Q < K, the reaction will proceed fastest in the forward direction • Q must increase, the [products] will increase and [reactants] will decrease • if Q = K, the reaction is at equilibrium • the [products] and [reactants] will not change • if a reaction mixture contains just reactants, Q = 0, and the reaction will proceed in the forward direction • if a reaction mixture contains just products, Q = ∞, and the reaction will proceed in the reverse direction

  34. Q, K, and the Direction of Reaction

  35. PI2, PCl2, PICl Q Ex 14.7 – For the reaction below, which direction will it proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm & PICl = 0.355 atm Given: Find: for I2(g) + Cl2(g) ⇌ 2 ICl(g), Kp = 81.9 direction reaction will proceed Concept Plan: Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse Solution: I2(g) + Cl2(g) ⇌ 2 ICl(g) Kp = 81.9 since Q (10.8) < K (81.9), the reaction will proceed to the right

  36. K, [COF2], [CF4] [CO2] Ex 14.8 If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and Kc = 2.00 @ 1000°C, find the [CO2]eq for the reaction given. Given: Find: Sort: You’re given the reaction and Kc. You’re also given the [X]eq of all but one of the chemicals 2 COF2 ⇌ CO2 + CF4 [COF2]eq = 0.255 M, [CF4]eq = 0.118 M [CO2]eq Concept Plan: Relationships: Strategize: You can calculate the missing concentration by using the equilibrium constant expression Solution: Solve: Solve the equilibrium constant expression for the unknown quantity by substituting in the given amounts Check: Check: Round to 1 sig fig and substitute back in Units & Magnitude OK

  37. Finding Equilibrium Concentrations When Given the Equilibrium Constant and Initial Concentrations or Pressures • first decide which direction the reaction will proceed • compare Q to K • define the changes of all materials in terms of x • use the coefficient from the chemical equation for the coefficient of x • the x change is + for materials on the side the reaction is proceeding toward • the x change is  for materials on the side the reaction is proceeding away from • solve for x • for 2nd order equations, take square roots of both sides or use the quadratic formula • may be able to simplify and approximate answer for very large or small equilibrium constants

  38. Ex 14.11  For the reaction I2(g) + Cl2(g) ⇌ 2 ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures are all 0.100 atm, find the equilibrium concentrations since Qp(1) < Kp(81.9), the reaction is proceeding forward

  39. Ex 14.11  For the reaction I2(g) + Cl2(g) ⇌ 2 ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures are all 0.100 atm, find the equilibrium concentrations x x +2x 0.100+2x 0.100x 0.100x

  40. Ex 14.11  For the reaction I2(g) + Cl2(g) ⇌ 2 ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures are all 0.100 atm, find the equilibrium concentrations x x +2x 0.100+2x 0.100x 0.100x

  41. Ex 14.11  For the reaction I2(g) + Cl2(g) ⇌ 2 ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures are all 0.100 atm, find the equilibrium concentrations x x +2x -0.0729 -0.0729 2(-0.0729) 0.100+2x 0.100x 0.100x 0.027 0.027 0.246

  42. Disturbing and Re-establishingEquilibrium • once a reaction is at equilibrium, the concentrations of all the reactants and products remain the same • however if the conditions are changed, the concentrations of all the chemicals will change until equilibrium is re-established • the new concentrations will be different, but the equilibrium constant will be the same • unless you change the temperature

  43. Le Châtelier’s Principle • Le Châtelier's Principle: if a system at equilibrium is disturbed, the position of equilibrium will shift to minimize the disturbance • disturbances all involve making the system open • Concentration, temperature, volume, pressure

  44. The result will be people moving from Country B into Country A faster than people moving from Country A into Country B. This will continue until a new equilibrium between the populations is established, however the new populations will have different numbers of people than the old ones. When an influx of population enters Country B from somewhere outside Country A, it disturbs the equilibrium established between Country A and Country B. When the populations of Country A and Country B are in equilibrium, the emigration rates between the two states are equal so the populations stay constant. An Analogy: Population Changes

  45. The Effect of Concentration Changes on Equilibrium • Adding reactant : aA + bB ⇌ cC + dD System shifts in a direction to minimize the disturbance • increases the amount of products until a new equilibrium is found • that has the same K

  46. The Effect of Concentration Changes on Equilibrium • Removing product: aA + bB ⇌ cC + dD System shifts in a direction to minimize the disturbance • will increase the amounts of products and decrease the amounts of reactants • you can use this to drive a reaction to completion!

  47. The Effect of Adding a Gas to a Gas Phase Reaction at Equilibrium • adding a gaseous reactant increases its partial pressure, causing the equilibrium to the right • increasing its partial pressure increases its concentration • does not increase the partial pressure of the other gases in the mixture • adding an inert gas to the mixture has no effect on the position of equilibrium • does not effect the partial pressures of the gases in the reaction Tro, Chemistry: A Molecular Approach

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