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Copyright  2011 Pearson Education, Inc. Chapter 14 Chemical Equilibrium Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA.

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Presentation on theme: "Copyright  2011 Pearson Education, Inc. Chapter 14 Chemical Equilibrium Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA."— Presentation transcript:

1 Copyright  2011 Pearson Education, Inc. Chapter 14 Chemical Equilibrium Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA

2 Copyright  2011 Pearson Education, Inc. Arrow Conventions Chemists commonly use two kinds of arrows in reactions to indicate the degree of completion of the reactions A single arrow indicates all the reactant molecules are converted to product molecules at the end A double arrow indicates the reaction stops when only some of the reactant molecules have been converted into products  in these notes 2 Tro: Chemistry: A Molecular Approach, 2/e

3 Copyright  2011 Pearson Education, Inc. 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 re-form 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 Tro: Chemistry: A Molecular Approach, 2/e

4 Copyright  2011 Pearson Education, Inc. Hypothetical Reaction 2 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 Tro: Chemistry: A Molecular Approach, 2/e

5 Copyright  2011 Pearson Education, Inc. Hypothetical Reaction 2 Red  Blue 5 Tro: Chemistry: A Molecular Approach, 2/e

6 Copyright  2011 Pearson Education, Inc. Eventually, the reaction proceeds in the reverse direction as fast as it proceeds in the forward direction. At this time equilibrium is established. Reaction Dynamics Time Rate Rate Forward Rate Reverse 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 6 Tro: Chemistry: A Molecular Approach, 2/e

7 Copyright  2011 Pearson Education, Inc. Dynamic Equilibrium As the forward reaction slows and the reverse reaction accelerates, eventually they reach the same rate Dynamic equilibrium is the condition wherein 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 Tro: Chemistry: A Molecular Approach, 2/e

8 Copyright  2011 Pearson Education, Inc. 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 8 Tro: Chemistry: A Molecular Approach, 2/e

9 Copyright  2011 Pearson Education, Inc. When Country A citizens feel overcrowded, some will emigrate to Country B An Analogy: Population Changes 9 However, after a time, 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 Tro: Chemistry: A Molecular Approach, 2/e

10 Copyright  2011 Pearson Education, Inc. 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 10 Tro: Chemistry: A Molecular Approach, 2/e

11 Copyright  2011 Pearson Education, Inc. 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 N 2 O 5(g)  4 NO 2(g) + O 2(g) the equilibrium constant expression is 11 Tro: Chemistry: A Molecular Approach, 2/e

12 Copyright  2011 Pearson Education, Inc. What Does the Value of K eq Imply? When the value of K eq >> 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 K eq << 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 12 Tro: Chemistry: A Molecular Approach, 2/e

13 Copyright  2011 Pearson Education, Inc. A Large Equilibrium Constant 13 Tro: Chemistry: A Molecular Approach, 2/e

14 Copyright  2011 Pearson Education, Inc. A Small Equilibrium Constant 14 Tro: Chemistry: A Molecular Approach, 2/e

15 Copyright  2011 Pearson Education, Inc. Practice – Write the equilibrium constant expressions, K, and predict the position of equilibrium for the following 2 SO 2 (g) + O 2 (g)  2 SO 3 (g)K = 8 x N 2 (g) + 2 O 2 (g)  2 NO 2 (g)K = 3 x 10 −17 favors products favors reactants 15 Tro: Chemistry: A Molecular Approach, 2/e

16 Copyright  2011 Pearson Education, Inc. Relationships between K and Chemical Equations When the reaction is written backward, 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 16 Tro: Chemistry: A Molecular Approach, 2/e

17 Copyright  2011 Pearson Education, Inc. Relationships between K and 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 17 Tro: Chemistry: A Molecular Approach, 2/e

18 Copyright  2011 Pearson Education, Inc. Relationships between K and 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 18 Tro: Chemistry: A Molecular Approach, 2/e

19 Copyright  2011 Pearson Education, Inc. K backward = 1/K forward, K new = K old n Example 14.2: Compute the equilibrium constant at 25 °C for the reaction NH 3 (g)  0.5 N 2 (g) H 2 (g) 19 for N 2 (g) + 3 H 2 (g)  2 NH 3 (g), K = 3.7 x 10 8 at 25 °C K for NH 3 (g)  0.5N 2 (g) + 1.5H 2 (g), at 25 °C Solution: Conceptual Plan: Relationships: Given: Find: N 2 (g) + 3 H 2 (g)  2 NH 3 (g)K 1 = 3.7 x 10 8 K’K 2 NH 3 (g)  N 2 (g) + 3 H 2 (g) NH 3 (g)  0.5 N 2 (g) H 2 (g) Tro: Chemistry: A Molecular Approach, 2/e

20 Copyright  2011 Pearson Education, Inc. 2 B  AK 3 = 1/K rxn 1 = 6.25 x 10 −5 Z  2 BK 4 = 1/K rxn 2 = 2 Z  AK 3  K 4 = 1.25 x 10 −4 3 Z  3 A (K 3  K 4 ) 3 = 2.0 x 10 −12 Practice – When the reaction A (aq)  2 B (aq) reaches equilibrium [A] = 1.0 x 10 −5 M and [B] = 4.0 x 10 −1 M. When the reaction 2 B (aq)  Z (aq) reaches equilibrium [B] = 4.0 x 10 −3 M and [Z] = 2.0 x 10 −6 M. Calculate the equilibrium constant for each of these reactions and the equilibrium constant for the reaction 3 Z (aq)  3 A (aq) K rxn 1 = 1.6 x 10 4 K rxn 2 = 5.0 x 10 −1 20 Tro: Chemistry: A Molecular Approach, 2/e

21 Copyright  2011 Pearson Education, Inc. 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 21 Tro: Chemistry: A Molecular Approach, 2/e

22 Copyright  2011 Pearson Education, Inc. K c and K p In calculating K p, the partial pressures are always in atm The values of K p and K c are not necessarily the same because of the difference in units K p = K c when  n = 0 The relationship between them is  n is the difference between the number of moles of reactants and moles of products 22 Tro: Chemistry: A Molecular Approach, 2/e

23 Copyright  2011 Pearson Education, Inc. Deriving the Relationship between K p and K c 23 Tro: Chemistry: A Molecular Approach, 2/e

24 Copyright  2011 Pearson Education, Inc. Deriving the Relationship Between K p and K c for aA(g) + bB(g)  cC(g) + dD(g) substituting 24 Tro: Chemistry: A Molecular Approach, 2/e

25 Copyright  2011 Pearson Education, Inc. Example 14.3: Find K c for the reaction 2 NO(g) + O 2 (g)  2 NO 2 (g), given K p = 2.2 x °C 25 K is a unitless number because there are more moles of reactant than product, K c should be larger than K p, and it is K p = 2.2 x KcKc Check: Solution: Conceptual Plan: Relationships: Given: Find: KpKp KcKc 2 NO(g) + O 2 (g)  2 NO 2 (g)  n = 2  3 = −1 Tro: Chemistry: A Molecular Approach, 2/e

26 Copyright  2011 Pearson Education, Inc. Practice – Calculate the value of K p or K c for each of the following at 27 °C 2 SO 2 (g) + O 2 (g)  2 SO 3 (g)K c = 8 x N 2 (g) + 2 O 2 (g)  2 NO 2 (g)K p = 3 x 10 −17 26 Tro: Chemistry: A Molecular Approach, 2/e

27 Copyright  2011 Pearson Education, Inc. 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 Tro: Chemistry: A Molecular Approach, 2/e

28 Copyright  2011 Pearson Education, Inc. Heterogeneous Equilibria The amount of C is different, but the amounts of CO and CO 2 remain the same. Therefore the amount of C has no effect on the position of equilibrium. 28 Tro: Chemistry: A Molecular Approach, 2/e

29 Copyright  2011 Pearson Education, Inc. HNO 2(aq) + H 2 O (l)  H 3 O + (aq) + NO 2 − (aq) K = 4.6 x 10 −4 Ca(NO 3 ) 2(aq) + H 2 SO 4(aq)  CaSO 4(s) + 2 HNO 3(aq) K = 1 x 10 4 Practice – Write the equilibrium constant expressions, K, and predict the position of equilibrium for the following 29 Tro: Chemistry: A Molecular Approach, 2/e

30 Copyright  2011 Pearson Education, Inc. 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 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 30 Tro: Chemistry: A Molecular Approach, 2/e

31 Copyright  2011 Pearson Education, Inc. Initial and Equilibrium Concentrations for H 2 (g) + I 2 (g)  445 °C Con Initial centration Eq Con uilibri centr um ation Equilibrium Constant [H 2 ][I 2 ][HI][H 2 ][I 2 ][HI] Tro: Chemistry: A Molecular Approach, 2/e

32 Copyright  2011 Pearson Education, Inc. 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 Use the change in the concentration of the material that you know to determine the change in the other chemicals in the reaction 32 Tro: Chemistry: A Molecular Approach, 2/e

33 Copyright  2011 Pearson Education, Inc. Because Product C increased by 0.50 mol/L, you must have used ½ that amount of Reactant A, according to the stoichiometry. Calculating Equilibrium Concentrations: An Example 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 ¼(0.50)-½(0.50) Because Product C increased by 0.50 mol/L, you must have used ¼ that amount of Reactant B, according to the stoichiometry. 2 A (aq) + B (aq)  4 C (aq) Product C increased by 0.50 mol/L Tro: Chemistry: A Molecular Approach, 2/e

34 Copyright  2011 Pearson Education, Inc. Example 14.6: Find the value of K c for the reaction 2 CH 4 (g)  C 2 H 2 (g) + 3 H 2 (g) at 1700 °C if the initial [CH 4 ] = M and the equilibrium [C 2 H 2 ] = M Tro: Chemistry: A Molecular Approach, 2/e

35 Copyright  2011 Pearson Education, Inc. Example 14.6: Find the value of K c for the reaction 2 CH 4 (g)  C 2 H 2 (g) + 3 H 2 (g) at 1700 °C if the initial [CH 4 ] = M and the equilibrium [C 2 H 2 ] = M −2(0.035) +3(0.035) Tro: Chemistry: A Molecular Approach, 2/e

36 Copyright  2011 Pearson Education, Inc. Practice –The following data were collected for the reaction 2 NO 2 (g)  N 2 O 4 (g) at 100 °C. Complete the table and determine values of K p and K c. 36 Tro: Chemistry: A Molecular Approach, 2/e

37 Copyright  2011 Pearson Education, Inc. Practice – A second experiment was done and the following data were collected for the reaction 2 NO 2 (g)  N 2 O 4 (g) at 100 °C. Complete the table and determine values of K p and K c. Compare them to the first experiment. 37 Tro: Chemistry: A Molecular Approach, 2/e

38 Copyright  2011 Pearson Education, Inc. The Reaction Quotient If a reaction mixture, containing both reactants and products, is not at equilibrium; how can we determine in 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 for the gas phase reaction aA + bB  cC + dD the reaction quotient is: 38 Tro: Chemistry: A Molecular Approach, 2/e

39 Copyright  2011 Pearson Education, Inc. 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 39 Tro: Chemistry: A Molecular Approach, 2/e

40 Copyright  2011 Pearson Education, Inc. Q, K, and the Direction of Reaction 40 Tro: Chemistry: A Molecular Approach, 2/e

41 Copyright  2011 Pearson Education, Inc. If Q = K, equilibrium; If Q K, reverse Example 14.7: For the reaction below, which direction will it proceed if P I2 = atm, P Cl2 = atm. & P ICl = atm 41 for I 2 (g) + Cl 2 (g)  2 ICl(g), K p = 81.9 direction reaction will proceed Solution: Conceptual Plan: Relationships: Given: Find: I 2 (g) + Cl 2 (g)  2 ICl(g)K p = 81.9 QP I2, P Cl2, P ICl because Q (10.8) < K (81.9), the reaction will proceed to the right Tro: Chemistry: A Molecular Approach, 2/e

42 Copyright  2011 Pearson Education, Inc. Practice – For the reaction CH 4 (g) + 2 H 2 S(g)  CS 2 (g) + 4 H 2 (g) K c = 3.59 at 900 °C. For each of the following measured concentrations determine whether the reaction is at equilibrium. If not at equilibrium, in which direction will the reaction proceed to get to equilibrium? 42 Tro: Chemistry: A Molecular Approach, 2/e

43 Copyright  2011 Pearson Education, Inc. Example 14.8: If [COF 2 ] eq = M and [CF 4 ] eq = M, and K c = 1000 °C, find the [CO 2 ] eq for the reaction given. 43  Check: Check: substitute approximations back in to see if the value agrees Solution: Solve: Solve the equilibrium constant expression for the unknown quantity by substituting in the given amounts Conceptual Plan: Relationships: Strategize: You can calculate the missing concentration by using the equilibrium constant expression 2 COF 2  CO 2 + CF 4 [COF 2 ] eq = M, [CF 4 ] eq = M [CO 2 ] eq Given: Find: Sort: You’re given the reaction and K c. You’re also given the [X] eq of all but one of the chemicals K, [COF 2 ], [CF 4 ][CO 2 ] Tro: Chemistry: A Molecular Approach, 2/e

44 Copyright  2011 Pearson Education, Inc. Practice – A sample of PCl 5 (g) is placed in a L container and heated to 160 °C. The PCl 5 is decomposed into PCl 3 (g) and Cl 2 (g). At equilibrium, moles of PCl 3 and Cl 2 are formed. Determine the equilibrium concentration of PCl 5 if K c = PCl 5  PCl 3 + Cl 2 44 Tro: Chemistry: A Molecular Approach, 2/e

45 Copyright  2011 Pearson Education, Inc. Finding Equilibrium Concentrations When Given the Equilibrium Constant and Initial Concentrations or Pressures Step 1:decide which direction the reaction will proceed compare Q to K Step 2:define the changes of all materials in terms of x use the coefficient from the chemical equation as 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 Step 3:solve for x for 2 nd 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 45 Tro: Chemistry: A Molecular Approach, 2/e

46 Copyright  2011 Pearson Education, Inc. Example 14.11: For the reaction I 2 (g) + Cl 2 (g)  2 25 °C, K p = If the initial partial pressures are all atm, find the equilibrium concentrations 46 Q p (1) < K p (81.9), so the reaction is proceeding forward Tro: Chemistry: A Molecular Approach, 2/e

47 Copyright  2011 Pearson Education, Inc. Example 14.11: For the reaction I 2 (g) + Cl 2 (g)  2 25 °C, K p = If the initial partial pressures are all atm, find the equilibrium concentrations 47 +2x xx xx  x x Tro: Chemistry: A Molecular Approach, 2/e

48 Copyright  2011 Pearson Education, Inc. Example 14.11: For the reaction I 2 (g) + Cl 2 (g)  2 25 °C, K p = If the initial partial pressures are all atm, find the equilibrium concentrations 48 +2x xx xx  x x Tro: Chemistry: A Molecular Approach, 2/e

49 Copyright  2011 Pearson Education, Inc. Example 14.11: For the reaction I 2 (g) + Cl 2 (g)  2 25°C, K p = If the initial partial pressures are all atm, find the equilibrium concentrations 49 +2x xx xx  x x − (0.0729) − Tro: Chemistry: A Molecular Approach, 2/e

50 Copyright  2011 Pearson Education, Inc. − Example 14.11: For the reaction I 2 (g) + Cl 2 (g)  2 25 °C, K p = If the initial partial pressures are all atm, find the equilibrium concentrations − (0.0729) K p (calculated) = K p (given) within significant figures Tro: Chemistry: A Molecular Approach, 2/e

51 Copyright  2011 Pearson Education, Inc. Practice – For the reaction I 2 (g)  2 I(g) the value of K c = 3.76 x 10 −5 at 1000 K. If 1.00 mole of I 2 is placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I 2 ] and [I]? (Hint: you will need to use the quadratic formula to solve for x) 51 Tro: Chemistry: A Molecular Approach, 2/e

52 Copyright  2011 Pearson Education, Inc. Practice – For the reaction I 2 (g)  2 I(g) the value of K c = 3.76 x 10 −5 at 1000 K. If 1.00 mole of I 2 is placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I 2 ] and [I]? because [I] initial = 0, Q = 0 and the reaction must proceed forward 52 Tro: Chemistry: A Molecular Approach, 2/e

53 Copyright  2011 Pearson Education, Inc. Practice – For the reaction I 2 (g)  2 I(g) the value of K c = 3.76 x 10 −5 at 1000 K. If 1.00 moles of I 2 is placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I 2 ] and [I]? 53 Tro: Chemistry: A Molecular Approach, 2/e

54 Copyright  2011 Pearson Education, Inc. Practice – For the reaction I 2 (g)  2 I(g) the value of K c = 3.76 x 10 −5 at 1000 K. If 1.00 moles of I 2 is placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I 2 ] and [I]? x =  = [I 2 ] = M 2( ) = [I] = M  54 Tro: Chemistry: A Molecular Approach, 2/e

55 Copyright  2011 Pearson Education, Inc. Approximations to Simplify the Math When the equilibrium constant is very small, the position of equilibrium favors the reactants For relatively large initial concentrations of reactants, the reactant concentration will not change significantly when it reaches equilibrium the [X] equilibrium = ([X] initial  ax)  [X] initial  we are approximating the equilibrium concentration of reactant to be the same as the initial concentration assuming the reaction is proceeding forward 55 Tro: Chemistry: A Molecular Approach, 2/e

56 Copyright  2011 Pearson Education, Inc. Checking the Approximation and Refining as Necessary We can check our approximation by comparing the approximate value of x to the initial concentration If the approximate value of x is less than 5% of the initial concentration, the approximation is valid 56 Tro: Chemistry: A Molecular Approach, 2/e

57 Copyright  2011 Pearson Education, Inc. Example14.13: For the reaction 2 H 2 S(g)  2 H 2 (g) + S °C, K c = 1.67 x 10 −7. If a L flask initially containing 1.25 x 10 −4 mol H 2 S is heated to 800 °C, find the equilibrium concentrations. 57 because no products initially, Q c = 0, and the reaction is proceeding forward Tro: Chemistry: A Molecular Approach, 2/e

58 Copyright  2011 Pearson Education, Inc. Example 14.13: For the reaction 2 H 2 S(g)  2 H 2 (g) + S °C, K c = 1.67 x 10 −7. If a L flask initially containing 1.25 x 10 −4 mol H 2 S is heated to 800 °C, find the equilibrium concentrations. 58 +x+x +2x 2x2x 2.50E−4  2x 2x2xx Tro: Chemistry: A Molecular Approach, 2/e

59 Copyright  2011 Pearson Education, Inc. Example 14.13: For the reaction 2 H 2 S(g)  2 H 2 (g) + S °C, K c = 1.67 x 10 −7. If a L flask initially containing 1.25 x 10 −4 mol H 2 S is heated to 800 °C, find the equilibrium concentrations. 59 +x+x +2x 2x2x 2.50E−4  2x 2x2x x 2.50E−4 Tro: Chemistry: A Molecular Approach, 2/e

60 Copyright  2011 Pearson Education, Inc. Example 14.13: For the reaction 2 H 2 S(g)  2 H 2 (g) + S °C, K c = 1.67 x 10 −7. If a L flask initially containing 1.25 x 10 −4 mol H 2 S is heated to 800 °C, find the equilibrium concentrations. 60 +x+x +2x 2x2x 2x2xx 2.50E–4 the approximation is not valid!! Tro: Chemistry: A Molecular Approach, 2/e

61 Copyright  2011 Pearson Education, Inc. Example 14.13: For the reaction 2 H 2 S(g)  2 H 2 (g) + S °C, K c = 1.67 x 10 −7. If a L flask initially containing 1.25 x 10 −4 mol H 2 S is heated to 800 °C, find the equilibrium concentrations. 61 +x+x +2x 2x2x 2.50E−4  2x 2x2xx x current = 1.38 x 10 −5 Tro: Chemistry: A Molecular Approach, 2/e

62 Copyright  2011 Pearson Education, Inc. Example 14.13: For the reaction 2 H 2 S(g)  2 H 2 (g) + S °C, K c = 1.67 x 10 −7. If a L flask initially containing 1.25 x 10 −4 mol H 2 S is heated to 800 °C, find the equilibrium concentrations. 62 +x+x +2x 2x2x 2.50E−4  2x 2x2xx x current = 1.27 x 10 −5 since x current = x new, approx. OK Tro: Chemistry: A Molecular Approach, 2/e

63 Copyright  2011 Pearson Education, Inc. Example 14.13: For the reaction 2 H 2 S(g)  2 H 2 (g) + S °C, K c = 1.67 x 10 −7. If a L flask initially containing 1.25 x 10 −4 mol H 2 S is heated to 800 °C, find the equilibrium concentrations. 63 +x+x +2x 2x2x 2.50E−4  2x 2x2xx x current = 1.28 x 10 −5 2.24E−42.56E−51.28E−5 Tro: Chemistry: A Molecular Approach, 2/e

64 Copyright  2011 Pearson Education, Inc. Example 14.13: For the reaction 2 H 2 S(g)  2 H 2 (g) + S °C, K c = 1.67 x 10 −7. If a L flask initially containing 1.25 x 10 −4 mol H 2 S is heated to 800 °C, find the equilibrium concentrations. 64 +x+x +2x 2x2x 2.24E−4 2.56E−51.28E−5 K c (calculated) = K c (given) within significant figures Tro: Chemistry: A Molecular Approach, 2/e

65 Copyright  2011 Pearson Education, Inc. Practice – For the reaction I 2 (g)  2 I(g) the value of K c = 3.76 x 10 −5 at 1000 K. If 1.00 mole of I 2 is placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I 2 ] and [I]? (use the simplifying assumption to solve for x) 65 Tro: Chemistry: A Molecular Approach, 2/e

66 Copyright  2011 Pearson Education, Inc. Practice – For the reaction I 2 (g)  2 I(g) the value of K c = 3.76 x 10 −5 at 1000 K. If 1.00 mole of I 2 is placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I 2 ] and [I]? because [I] initial = 0, Q = 0 and the reaction must proceed forward 66 Tro: Chemistry: A Molecular Approach, 2/e

67 Copyright  2011 Pearson Education, Inc. Practice – For the reaction I 2 (g)  2 I(g) the value of K c = 3.76 x 10 −5 at 1000 K. If 1.00 mole of I 2 is placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I 2 ] and [I]? the approximation is valid!! 67 Tro: Chemistry: A Molecular Approach, 2/e

68 Copyright  2011 Pearson Education, Inc. Practice – For the reaction I 2 (g)  2 I(g) the value of K c = 3.76 x 10 −5 at 1000 K. If 1.00 mole of I 2 is placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I 2 ] and [I]? x =  = [I 2 ] = M 2( ) = [I] = M  68 Tro: Chemistry: A Molecular Approach, 2/e

69 Copyright  2011 Pearson Education, Inc. Practice – For the reaction N 2 O 4 (g)  2 NO 2 (g) the value of K c = 1.07 x 10 −5. If the initial concentration of N 2 O 4 is M, what will be the equilibrium concentrations of [N 2 O 4 ] and [NO 2 ]? 69 Tro: Chemistry: A Molecular Approach, 2/e

70 Copyright  2011 Pearson Education, Inc. Practice – For the reaction N 2 O 4 (g)  2 NO 2 (g) the value of K c = 1.07 x 10 −5. If the initial concentration of N 2 O 4 is M, what will be the equilibrium concentrations of [N 2 O 4 ] and [NO 2 ]? because [NO 2 ] initial = 0, Q = 0 and the reaction must proceed forward 70 Tro: Chemistry: A Molecular Approach, 2/e

71 Copyright  2011 Pearson Education, Inc. Practice – For the reaction N 2 O 4 (g)  2 NO 2 (g) the value of K c = 1.07 x 10 −5. If the initial concentration of N 2 O 4 is M, what will be the equilibrium concentrations of [N 2 O 4 ] and [NO 2 ]? without approximationwith approximation the approximation is valid!! 71 x = 1.82 x 10 −4 Tro: Chemistry: A Molecular Approach, 2/e

72 Copyright  2011 Pearson Education, Inc. Practice – For the reaction N 2 O 4 (g)  2 NO 2 (g) the value of K c = 1.07 x 10 −5. If the initial concentration of N 2 O 4 is M, what will be the equilibrium concentrations of [N 2 O 4 ] and [NO 2 ]? x = 1.83 x 10 −4 [N 2 O 4 ]=  1.83 x 10 −4 [N 2 O 4 ] = M [NO 2 ]=2(1.83 x 10 −4 ) [NO 2 ] = 3.66 x 10 −4 M  72 Tro: Chemistry: A Molecular Approach, 2/e

73 Copyright  2011 Pearson Education, Inc. Disturbing and Restoring Equilibrium 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 restored The new concentrations will be different, but the equilibrium constant will be the same unless you change the temperature 73 Tro: Chemistry: A Molecular Approach, 2/e

74 Copyright  2011 Pearson Education, Inc. Le Ch âtelier’s Principle Le Châtelier's Principle guides us in predicting the effect various changes in conditions have on the position of equilibrium It says that if a system at equilibrium is disturbed, the position of equilibrium will shift to minimize the disturbance disturbances all involve making the system open 74 Tro: Chemistry: A Molecular Approach, 2/e

75 Copyright  2011 Pearson Education, Inc. 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; the new populations will have different numbers of people than the old ones. An Analogy: Population Changes 75 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 countries are equal so the populations stay constant. Tro: Chemistry: A Molecular Approach, 2/e

76 Copyright  2011 Pearson Education, Inc. Disturbing Equilibrium: Adding or Removing Reactants After equilibrium is established, a reactant is added as long as the added reactant is included in the equilibrium constant expression  i.e., not a solid or liquid How will this affect the rate of the forward reaction? How will it affect the rate of the reverse reaction? How will it affect the value of K? 76 Tro: Chemistry: A Molecular Approach, 2/e

77 Copyright  2011 Pearson Education, Inc. Disturbing Equilibrium: Adding Reactants Adding a reactant initially increases the rate of the forward reaction, but has no initial effect on the rate of the reverse reaction The reaction proceeds to the right until equilibrium is restored At the new equilibrium position, you will have more of the products than before, less of the non- added reactants than before, and less of the added reactant but not as little of the added reactant as you had before the addition At the new equilibrium position, the concentrations of reactants and products will be such that the value of the equilibrium constant is the same 77 Tro: Chemistry: A Molecular Approach, 2/e

78 Copyright  2011 Pearson Education, Inc. Disturbing Equilibrium: Adding or Removing Reactants After equilibrium is established, a reactant is removed as long as the added reactant is included in the equilibrium constant expression  i.e., not a solid or liquid How will this affect the rate of the forward reaction? How will it affect the rate of the reverse reaction? How will it affect the value of K? 78 Tro: Chemistry: A Molecular Approach, 2/e

79 Copyright  2011 Pearson Education, Inc. Disturbing Equilibrium: Removing Reactants Removing a reactant initially decreases the rate of the forward reaction, but has no initial effect on the rate of the reverse reaction. so the reaction is going faster in reverse The reaction proceeds to the left until equilibrium is restored At the new equilibrium position, you will have less of the products than before, more of the non- removed reactants than before, and more of the removed reactant but not as much of the removed reactant as you had before the removal At the new equilibrium position, the concentrations of reactants and products will be such that the value of the equilibrium constant is the same 79 Tro: Chemistry: A Molecular Approach, 2/e

80 Copyright  2011 Pearson Education, Inc. The Effect of Concentration Changes on Equilibrium Adding a reactant will decrease the amounts of the other reactants and increase the amount of the products until a new position of equilibrium is found that has the same K Removing a product will increase the amounts of the other products and decrease the amounts of the reactants. you can use this to drive a reaction to completion! Equilibrium shifts away from side with added chemicals or toward side with removed chemicals Remember, adding more of a solid or liquid does not change its concentration – and therefore has no effect on the equilibrium 80 Tro: Chemistry: A Molecular Approach, 2/e

81 Copyright  2011 Pearson Education, Inc. The Effect of Concentration Changes on Equilibrium When NO 2 is added, some of it combines to make more N 2 O 4 81 Tro: Chemistry: A Molecular Approach, 2/e

82 Copyright  2011 Pearson Education, Inc. The Effect of Concentration Changes on Equilibrium When N 2 O 4 is added, some of it decomposes to make more NO 2 82 Tro: Chemistry: A Molecular Approach, 2/e

83 Copyright  2011 Pearson Education, Inc. 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 shift 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 affect the partial pressures of the gases in the reaction 83 Tro: Chemistry: A Molecular Approach, 2/e

84 Copyright  2011 Pearson Education, Inc. Effect of Volume Change on Equilibrium Decreasing the volume of the container increases the concentration of all the gases in the container increases their partial pressures It does not change the concentrations of solutions!! If their partial pressures increase, then the total pressure in the container will increase According to Le Châtelier’s Principle, the equilibrium should shift to remove that pressure The way the system reduces the pressure is to reduce the number of gas molecules in the container When the volume decreases, the equilibrium shifts to the side with fewer gas molecules 84 Tro: Chemistry: A Molecular Approach, 2/e

85 Copyright  2011 Pearson Education, Inc. Disturbing Equilibrium: Changing the Volume After equilibrium is established, the container volume is decreased How will it affect the concentration of solids, liquid, solutions, and gases? How will this affect the total pressure of solids, liquid, and gases? How will it affect the value of K? 85 Tro: Chemistry: A Molecular Approach, 2/e

86 Copyright  2011 Pearson Education, Inc. Disturbing Equilibrium: Reducing the Volume Decreasing the container volume increases the concentration of all gases, but not solids, liquid, or solutions same number of moles, but different number of liters, resulting in a different molarity Decreasing the container volume will increase the total pressure Boyle’s Law if the total pressure increases, the partial pressures of all the gases will increase – Dalton’s Law of Partial Pressures Because the total pressure increases, the position of equilibrium will shift to decrease the pressure by removing gas molecules shift toward the side with fewer gas molecules At the new equilibrium position, the partial pressures of gaseous reactants and products will be such that the value of the equilibrium constant is the same 86 Tro: Chemistry: A Molecular Approach, 2/e

87 Copyright  2011 Pearson Education, Inc. The Effect of Volume Changes on Equilibrium 87 Because there are more gas molecules on the reactants side of the reaction, when the pressure is increased the position of equilibrium shifts toward the side with fewer molecules to decrease the pressure When the pressure is decreased by increasing the volume, the position of equilibrium shifts toward the side with the greater number of molecules – the reactant side Tro: Chemistry: A Molecular Approach, 2/e

88 Copyright  2011 Pearson Education, Inc. The Effect of Temperature Changes on Equilibrium Position Exothermic reactions release energy and endothermic reactions absorb energy If we write Heat as a product in an exothermic reaction or as a reactant in an endothermic reaction, it will help us use Le Châtelier’s Principle to predict the effect of temperature changes even though heat is not matter and not written in a proper equation 88 Tro: Chemistry: A Molecular Approach, 2/e

89 Copyright  2011 Pearson Education, Inc. The Effect of Temperature Changes on Equilibrium for Exothermic Reactions For an exothermic reaction, heat is a product Increasing the temperature is like adding heat According to Le Châtelier’s Principle, the equilibrium will shift away from the added heat Adding heat to an exothermic reaction will decrease the concentrations of products and increase the concentrations of reactants Adding heat to an exothermic reaction will decrease the value of K How will decreasing the temperature affect the system? aA + bB  cC + dD + Heat 89 Tro: Chemistry: A Molecular Approach, 2/e

90 Copyright  2011 Pearson Education, Inc. The Effect of Temperature Changes on Equilibrium for Endothermic Reactions For an endothermic reaction, heat is a reactant Increasing the temperature is like adding heat According to Le Châtelier’s Principle, the equilibrium will shift away from the added heat Adding heat to an endothermic reaction will decrease the concentrations of reactants and increase the concentrations of products Adding heat to an endothermic reaction will increase the value of K How will decreasing the temperature affect the system? Heat + aA + bB  cC + dD 90 Tro: Chemistry: A Molecular Approach, 2/e

91 Copyright  2011 Pearson Education, Inc. The Effect of Temperature Changes on Equilibrium 91 Tro: Chemistry: A Molecular Approach, 2/e

92 Copyright  2011 Pearson Education, Inc. Not Changing the Position of Equilibrium: The Effect of Catalysts Catalysts provide an alternative, more efficient mechanism Catalysts work for both forward and reverse reactions Catalysts affect the rate of the forward and reverse reactions by the same factor Therefore, catalysts do not affect the position of equilibrium 92 Tro: Chemistry: A Molecular Approach, 2/e

93 Copyright  2011 Pearson Education, Inc. Practice – Le Châtelier’s Principle The reaction 2 SO 2 (g) + O 2 (g)  2 SO 3 (g) with  H° = −198 kJ is at equilibrium. How will each of the following changes affect the equilibrium concentrations of each gas once equilibrium is restored? adding more O 2 to the container condensing and removing SO 3 compressing the gases cooling the container doubling the volume of the container warming the mixture adding the inert gas helium to the container adding a catalyst to the mixture 93 Tro: Chemistry: A Molecular Approach, 2/e

94 Copyright  2011 Pearson Education, Inc. Practice – Le Châtelier’s Principle The reaction 2 SO 2 (g) + O 2 (g)  2 SO 3 (g) with  H° = −198 kJ is at equilibrium. How will each of the following changes affect the equilibrium concentrations of each gas once equilibrium is restored? adding more O 2 to the containershift to SO 3 condensing and removing SO 3 shift to SO 3 compressing the gasesshift to SO 3 cooling the containershift to SO 3 doubling the volume of the containershift to SO 2 warming the mixtureshift to SO 2 adding helium to the containerno effect adding a catalyst to the mixtureno effect 94Tro: Chemistry: A Molecular Approach, 2/e


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