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Chemical Equilibriumq

Chemical Equilibriumq. Reaction Dynamics. when a reaction starts, the reactants are consumed and products are made forward reaction = reactants  products therefore the reactant concentrations decrease and the product concentrations increase

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Chemical Equilibriumq

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  1. ChemicalEquilibriumq

  2. Reaction Dynamics • when a reaction starts, the reactants are consumed and products are made • forward reaction = reactants  products • therefore the reactant concentrations decrease and the product concentrations increase • as reactant concentration decreases, the forward reaction rate decreases • eventually, the products can react to reform some of the reactants • reverse reaction = products  reactants • assuming the products are not allowed to escape • as product concentration increases, the reverse reaction rate increases • processes that proceed in both the forward and reverse direction are said to be reversible • reactants products

  3. Hypothetical Reaction2 RedBlue The reaction slows over time, But the Red molecules never run out! At some time between 100 and 110 sec, the concentrations of both the Red and the Blue molecules no longer change – equilibrium has been established. Notice that equilibrium does not mean that the concentrations are equal! Once equilibrium is established, the rate of Red molecules turning into Blue is the same as the rate of Blue molecules turning into Red

  4. Hypothetical Reaction2 RedBlue

  5. 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

  6. Dynamic Equilibrium • as the forward reaction slows and the reverse reaction accelerates, eventually they reach the same rate • dynamic equilibrium is the condition where the rates of the forward and reverse reactions are equal • once the reaction reaches equilibrium, the concentrations of all the chemicals remain constant • because the chemicals are being consumed and made at the same rate

  7. H2(g) + I2(g)  2 HI(g) at time 0, there are only reactants in the mixture, so only the forward reaction can take place [H2] = 8, [I2] = 8, [HI] = 0 at time 16, there are both reactants and products in the mixture, so both the forward reaction and reverse reaction can take place [H2] = 6, [I2] = 6, [HI] = 4

  8. H2(g) + I2(g)  2 HI(g) at time 32, there are now more products than reactants in the mixture − the forward reaction has slowed down as the reactants run out, and the reverse reaction accelerated [H2] = 4, [I2] = 4, [HI] = 8 at time 48, the amounts of products and reactants in the mixture haven’t changed – the forward and reverse reactions are proceeding at the same rate – it has reached equilibrium

  9. 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

  10. Equilibrium  Equal • the rates of the forward and reverse reactions are equal at equilibrium • but that does not mean the concentrations of reactants and products are equal • some reactions reach equilibrium only after almost all the reactant molecules are consumed – we say the position of equilibrium favors the products • other reactions reach equilibrium when only a small percentage of the reactant molecules are consumed – we say the position of equilibrium favors the reactants

  11. When Country A citizens feel overcrowded, some will emigrate to Country B. However, as time passes, emigration will occur in both directions at the same rate, leading to populations in Country A and Country B that are constant, though not necessarily equal An Analogy: Population Changes

  12. Equilibrium Constant • even though the concentrations of reactants and products are not equal at equilibrium, there is a relationship between them • the relationship between the chemical equation and the concentrations of reactants and products is called the Law of Mass Action • for the general equation aA + bB cC + dD, the Law of Mass Action gives the relationship below • the lowercase letters represent the coefficients of the balanced chemical equation • always products over reactants • K is called the equilibrium constant • unitless

  13. Writing Equilibrium Constant Expressions • for the reaction aA(aq) + bB(aq)cC(aq) + dD(aq) the equilibrium constant expression is • so for the reaction • 2 N2O5 4 NO2 + O2 the • equilibrium constant • expression is:

  14. What Does the Value of Keq Imply? • when the value of Keq >> 1, we know that when the reaction reaches equilibrium there will be many more product molecules present than reactant molecules • the position of equilibrium favors products • when the value of Keq << 1, we know that when the reaction reaches equilibrium there will be many more reactant molecules present than product molecules • the position of equilibrium favors reactants

  15. A Large Equilibrium Constant

  16. A Small Equilibrium Constant

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

  19. Relationships between Kand Chemical Equations • when you add equations to get a new equation, the equilibrium constant of 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 equilibrium constant expressions are: for the reaction aA cC the equilibrium constant expression is:

  20. K K’ 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), K = 3.7 x 108 at 25°C K 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 Constants for Reactions Involving Gases • the concentration of a gas in a mixture is proportional to its partial pressure • therefore, the equilibrium constant can be expressed as the ratio of the partial pressures of the gases • for aA(g) + bB(g) cC(g) + dD(g) the equilibrium constant expressions are or

  22. Kc and Kp • in calculating Kp, the partial pressures are always in atm • the values of Kp and Kc are not necessarily the same • because of the difference in units • Kp = Kc when Dn = 0 • the relationship between them is: Dn is the difference between the number of moles of reactants and moles of products

  23. Deriving the Relationshipbetween Kp and Kc Konsentrasi nitrogen 75% = misal terdapat 100 mol gas yang ada dalam udara, 75 mol merupakan gas nitrogen dalam udara, 25 mol Merupakan gas h dan o % volum merupakan Banyaknya volume gas dalam volum gas total

  24. Deriving the RelationshipBetween Kp and Kc for aA(g) + bB(g) cC(g) + dD(g) substituting

  25. Kp Kc 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 Kc Concept Plan: Relationships: Solution: 2 NO(g) + O2(g)  2 NO2(g) Dn = 2  3 = -1 Check: K is a unitless number since there are more moles of reactant than product, Kc should be larger than Kp, and it is

  26. Heterogeneous Equilibria • pure solids and pure liquids are materials whose concentration doesn’t change during the course of a reaction • its amount can change, but the amount of it in solution doesn’t • because it isn’t in solution • because their concentration doesn’t change, solids and liquids are not included in the equilibrium constant expression • for the reaction aA(s) + bB(aq) cC(l) + dD(aq) the equilibrium constant expression is:

  27. Heterogeneous Equilibria The amount of C is different, but the amounts of CO and CO2 remains the same. Therefore the amount of C has no effect on the position of equilibrium.

  28. Calculating Equilibrium Constants from Measured Equilibrium Concentrations • the most direct way of finding the equilibrium constant is to measure the amounts of reactants and products in a mixture at equilibrium • actually, you only need to measure one amount – then use stoichiometry to calculate the other amounts • the equilibrium mixture 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 • the value of the equilibrium constant is independent of the initial amounts of reactants and products

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

  30. Calculating Equilibrium Concentrations • Stoichiometry can be used to determine the equilibrium concentrations of all reactants and products if you know initial concentrations and one equilibrium concentration • suppose you have a reaction 2 A(aq) + B(aq) 4 C(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

  31. 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

  32. 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

  33. The following data were collected for the reaction 2 NO2(g)  N2O4(g) at 100°C. Complete the table and determine values of Kp and Kc for each experiment.

  34. The following data were collected for the reaction 2 NO2(g)  N2O4(g) at 100°C. Complete the table and determine values of Kp and Kc for each experiment.

  35. The Reaction Quotient for the gas phase reaction aA + bB cC + dD the reaction quotient is: • if a reaction mixture, containing both reactants and products, is 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 • the concentration ratio of the products (raised to the power of their coefficients) to the reactants (raised to the power of their coefficients) is called the reaction quotient, Q

  36. The Reaction Quotient:Predicting the Direction of Change • if Q > K, the reaction will proceed fastest in the reverse direction • the [products] will decrease and [reactants] will increase • if Q < K, the reaction will proceed fastest in the forward direction • 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

  37. Q, K, and the Direction of Reaction

  38. PI2, PCl2, PICl Q 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

  39. K, [COF2], [CF4] [CO2] 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

  40. A sample of PCl5(g) is placed in a 0.500 L container and heated to 160°C. The PCl5 is decomposed into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3 and Cl2 are formed. Determine the equilibrium concentration of PCl5 if Kc = 0.0635

  41. A sample of PCl5(g) is placed in a 0.500 L container and heated to 160°C. The PCl5 is decomposed into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3 and Cl2 are formed. Determine the equilibrium concentration of PCl5 if Kc = 0.0635 PCl5 PCl3 + Cl2

  42. 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

  43. 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

  44. 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

  45. 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

  46. 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

  47. 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 -0.0729 -0.0729 2(0.0729) 0.027 0.027 0.246 Kp(calculated) = Kp(given) within significant figures

  48. For the reaction I2(g)  2 I(g) the value of Kc = 3.76 x 10-5 at 1000 K. If 1.00 moles of I2 is placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I2] and [I]?(Hint: you will need to use the quadratic formula to solve for x)

  49. For the reaction I2(g)  2 I(g) the value of Kc = 3.76 x 10-5 at 1000 K. If 1.00 moles of I2 is placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I2] and [I]? since [I]initial = 0, Q = 0 and the reaction must proceed forward

  50. For the reaction I2(g)  2 I(g) the value of Kc = 3.76 x 10-5 at 1000 K. If 1.00 moles of I2 is placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I2] and [I]?

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