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Chapter 14 Chemical Equilibrium Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA Chemistry: A Molecular Approach, 2nd Ed. Nivaldo Tro.

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Presentation on theme: "Chapter 14 Chemical Equilibrium Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA Chemistry: A Molecular Approach, 2nd Ed. Nivaldo Tro."— Presentation transcript:

1 Chapter 14 Chemical Equilibrium Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA Chemistry: A Molecular Approach, 2nd Ed. Nivaldo Tro

2 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 2 Also written as 

3 Hemoglobin Hemoglobin (Hb) is a large protein found in red blood cells, that binds with O 2 Hb is an abreviation, not a chemical formula Hb carries O 2 through the bloodstream Hb + O 2  HbO 2 the  is used to describe a process that is in dynamic equilibrium 3

4 Hemoglobin Equilibrium System Hb + O 2  HbO 2 [Hb], [O 2 ], and [HbO 2 ] are all interdependent Their relative amounts at equilibrium are related to the equilibrium constant, K the larger the value of K, the more product HbO 2 is found at equilibrium Changing the concentration of any one component results in changing the other concentrations to restore equilibrium 4

5 O 2 Transport Hb + O2O2  The lungs have a high [O 2 ], shifting the equilibrium toward HbO 2 (by combining Hb and O 2 ) In the cells, low [O 2 ] causes the equilibrium to shift back to free Hb and O 2 (by breaking down HbO 2 ) HbO 2 O 2 in lungs Hb HbO 2 O 2 in cells 5

6 HbF Hb Fetal Hemoglobin, HbF HbF + O 2  HbFO 2 Hb Fetal has a larger equilibrium constant than Hb mother Hb Fetal is more efficient at binding O 2, so O 2 is transferred from the Hb mother to Hb Fetal in the placenta Hb + O2O2  HbO 2 O2O2 O2O2 HbF + O2O2  HbFO 2 O2O2 6

7 Reaction Dynamics When a reaction starts (primarily a forward rxn) reactants are consumed  [reactants] decrease products are made  [products] increases Over time the forward reaction rate decreases Eventually, products can reform reactants (reverse rxn becomes significant assuming the products are not allowed to escape) as [products] increases, the reverse reaction rate increases products  reactants Processes that proceed in both the forward and reverse direction are said to be reversible reactants  products 7

8 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. Equilibrium does not mean the concentrations are equal! Once equilibrium is established, the rate of Red molecules turning into Blue = the rate of Blue molecules turning into Red 8

9 Hypothetical Reaction 2 Red  Blue 9

10 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 10

11 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 11

12 H 2 (g) + I 2 (g)  2 HI(g) At t=0, only reactants exist, so only the forward reaction can take place [H 2 ] = 8, [I 2 ] = 8, [HI] = 0 [H 2 ] = 6, [I 2 ] = 6, [HI] = 4 12 At t=16, both reactants and products are present in the mixture, so both the forward reaction and reverse reaction can take place

13 H 2 (g) + I 2 (g)  2 HI(g) At t=32, there are more products than reactants in the mixture the forward reaction slows down as the reactants run out, and the reverse reaction accelerates [H 2 ] = 4, [I 2 ] = 4, [HI] = 8 At t=48, the amount of products and reactants in the mixture don’t change the forward and reverse reactions are proceeding at the same rate equilibrium has been reached 13

14 At equilibrium, the forward reaction rate is the same as the reverse reaction rate As the reaction proceeds, the [H 2 ] and [I 2 ] decrease, and the [HI] increases Because the reactant concentrations are decreasing, the forward reaction rate slows down And because the product concentration is increasing, the reverse reaction rate speeds up H 2 (g) + I 2 (g)  2 HI(g) Equilibrium established Because the [HI] at equilibrium is larger than the [H 2 ] or [I 2 ], we say the position of the equilibrium favors the products Once equilibrium is established, the concentrations no longer change 14

15 Equilibrium  Equal The rates of the forward and reverse reactions are equal at equilibrium That does NOT mean the concentrations of reactants and products are equal Some reactions reach equilibrium after most of the reactants are consumed (the position of equilibrium favors the products) Other reactions reach equilibrium when only a small percentage of the reactants are consumed (the position of equilibrium favors the reactants) 15

16 When Country A citizens feel overcrowded, some will emigrate to Country B An Analogy: Population Changes 16 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

17 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 relates K, the equilibrium constant, with the concentrations of products and reactants always written as products / reactants K is unitless the exponents represent the coefficients of the balanced chemical equation 17

18 Writing Equilibrium Constant, K eq, 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 18

19 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 19

20 A Large Equilibrium Constant 20

21 A Small Equilibrium Constant 21

22 Write equilibrium constant expressions, K, and predict the position of equilibrium for: 2 SO 2 (g) + O 2 (g)  2 SO 3 (g)K = 8 x 10 25 N 2 (g) + 2 O 2 (g)  2 NO 2 (g)K = 3 x 10 −17 favors products favors reactants 22

23 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 23

24 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 24

25 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 total reaction aA  cC the equilibrium constant expression is 25

26 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) + 1.5 H 2 (g) 26 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) + 1.5 H 2 (g)

27 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 27

28 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 28

29 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 The relationship between them is  n is the difference between the number of moles of reactants and moles of products 29 K p = K c when  n = 0

30 Deriving the Relationship between K p & K c 30 moles of A gas volume =

31 Deriving the Relationship Between K p & K c for aA(g) + bB(g)  cC(g) + dD(g) substituting 31 Since we now know that

32 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 10 12 @ 25 °C 32 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 10 12 KcKc Check: Solution: Conceptual Plan: Relationships: Given: Find: KpKp KcKc 2 NO(g) + O 2 (g)  2 NO 2 (g)  n = 2  3 = −1

33 Calculate K p or K c at 27 °C 2 SO 2 (g) + O 2 (g)  2 SO 3 (g)K c = 8 x 10 25 N 2 (g) + 2 O 2 (g)  2 NO 2 (g)K p = 3 x 10 −17 33

34 Heterogeneous Equilibria Pure solids or pure liquids that are not in solution can take part in a reaction, but their concentration doesn’t change their amount can change think of them as infinite reservoirs of material in contact with the solution where the reaction takes place because their concentration doesn’t change, they are not included in the equilibrium constant expression For the reaction aA(s) + bB(aq)  cC(l) + dD(aq) the equilibrium constant expression is 34

35 Heterogeneous Equilibria Given different amounts of solid carbon, that oxidize to carbon monoxide and carbon dioxide in two sealed containers: The amount of solid carbon is different, but the amounts of CO and CO 2 remain the same. Therefore the amount of solid carbon has no effect on the position of equilibrium. 35

36 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 36

37 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 at constant temperature the value of the equilibrium constant is independent of the initial amounts of reactants and products 37

38 Initial and Equilibrium Concentrations for H 2 (g) + I 2 (g)  2HI(g) @ 445 °C Con Initial centration Eq Con uilibri centr um ation Equilibrium Constant [H 2 ][I 2 ][HI][H 2 ][I 2 ][HI] 0.50 0.00.11 0.78 0.0 0.500.055 0.39 0.50 0.165 1.17 1.00.50.00.530.0330.934 38

39 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 39

40 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: 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. +0.50-¼(0.50)-½(0.50) 0.75 0.88 40 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

41 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 ] = 0.115 M and the equilibrium [C 2 H 2 ] = 0.035 M 41 +0.035

42 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 ] = 0.115 M and the equilibrium [C 2 H 2 ] = 0.035 M 42 +0.035 −2(0.035) +3(0.035) 0.045 0.105 2 CH 4 (g)  C 2 H 2 (g) + 3 H 2 (g)

43 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 43 -0.0028+0.0014 0.0014 2 NO 2 (g)  N 2 O 4 (g)

44 A 2 nd experiment gave the following data for the same 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 to show that they will be the same as in the first experiment. 44 +0.0310-0.0155 0.0310 2 NO 2 (g)  N 2 O 4 (g)

45 The Reaction Quotient If a reaction mixture is not at equilibrium; we can determine which direction the rxn will proceed by comparing the product/reactant concentration ratios at any given time 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: 45

46 The Reaction Quotient: Predicting the Direction of Change If Q > K, the rxn proceeds in the reverse direction the [products] will decrease and [reactants] will increase If a reaction mixture contains just products, Q = ∞, and the reaction will proceed in the reverse direction If Q < K, the rxn proceeds in the forward direction the [products] will increase and [reactants] will decrease If a reaction mixture contains just reactants, Q = 0, and the reaction will proceed in the forward direction If Q = K, the rxn is at equilibrium the [products] and [reactants] will not change 46

47 Q, K, and the Direction of Reaction 47

48 If Q = K, equilibrium; If Q K, reverse Example 14.7: For the reaction below, which direction will it proceed if P I2 = 0.114 atm, P Cl2 = 0.102 atm. & P ICl = 0.355 atm 48 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

49 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 find whether the reaction is at equilibrium, has not reached equilibrium, or has gone past the equilibrium point. 49

50 Example 14.8: If [COF 2 ] eq = 0.255 M and [CF 4 ] eq = 0.118 M, and K c = 2.00 @ 1000 °C, find the [CO 2 ] eq for the reaction given. 50  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 = 0.255 M, [CF 4 ] eq = 0.118 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 ]

51 A sample of PCl 5 (g) is placed in a 0.500L container and heated to 160°C. The PCl 5 is decomposed into PCl 3 (g) and Cl 2 (g). At equilibrium, 0.203 moles of PCl 3 and Cl 2 are formed. Determine the equilibrium concentration of PCl 5 if K c = 0.0635 PCl 5  PCl 3 + Cl 2 51 M

52 Finding Equilibrium Concentrations when the Equilibrium Constant and Initial Concentrations or Pressures are given 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 the coefficient from the chemical equation is the coefficient of x the change in x is + for materials being formed the change in x is  for materials being consumed 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 52

53 Example 14.11: For I 2 (g) + Cl 2 (g)  2 ICl(g), K p = 81.9 @ 25 °C. The initial partial pressures are all 0.100 atm, find the equilibrium concentrations 53 Q p (1) < K p (81.9), so the reaction is proceeding forward I 2 (g) + Cl 2 (g)  2 ICl(g),

54 54 +2x xx xx 0.100  x 0.100+2x Example 14.11: For I 2 (g) + Cl 2 (g)  2 ICl(g), K p = 81.9 @ 25 °C. The initial partial pressures are all 0.100 atm, find the equilibrium concentrations I 2 (g) + Cl 2 (g)  2 ICl(g),

55 55 +2x xx xx 0.100  x 0.100+2x Example 14.11: For I 2 (g) + Cl 2 (g)  2 ICl(g), K p = 81.9 @ 25 °C. The initial partial pressures are all 0.100 atm, find the equilibrium concentrations I 2 (g) + Cl 2 (g)  2 ICl(g),

56 56 +2x xx xx 0.100  x 0.100+2x 0.027 0.246 −0.0729+2(0.0729)−0.0729 Example 14.11: For I 2 (g) + Cl 2 (g)  2 ICl(g), K p = 81.9 @ 25 °C. The initial partial pressures are all 0.100 atm, find the equilibrium concentrations

57 −0.0729 57 0.027 0.246 −0.0729 2(0.0729) K p (calculated) = K p (given) within significant figures Example 14.11: For I 2 (g) + Cl 2 (g)  2 ICl(g), K p = 81.9 @ 25 °C. The initial partial pressures are all 0.100 atm, find the equilibrium concentrations I 2 (g) + Cl 2 (g)  2 ICl(g),

58 For some problems, you will need to use the quadratic formula to solve for x 58 MEMORIZE THIS EQUATION !!!! The solution is found by the quadratic formula:

59 Given I 2 (g)  2 I(g). If 1.00 mole of I 2 is placed into a 2.00 L flask and heated, find the equilibrium concentrations of [I 2 ] and [I]? K c = 3.76 x 10 −5 @1000K because [I] initial = 0, Q = [I] 2 /[ I 2 ] = 0/[ I 2 ] = 0 and the reaction must proceed forward 59 1.00 mol I 2 /2.00L=0.500M I 2 Let x = change in [I 2 ] Therefore change in [I] = 2x I 2 (g)  2 I(g)

60 60-stop Quadratic equation required Equation must be in the form 0=ax 2 +bx+c I 2 (g)  2 I(g)

61 x = 0.00216 Substitute the value for x into the original table 0.500  0.00216 = 0.498 [I 2 ] = 0.498 M 2(0.00216) = 0.00432 [I] = 0.00432 M  61 I 2 (g)  2 I(g)

62 Approximations to Simplify the Math When K equilibrium is very small, the position of equilibrium favors the reactants For relatively large [reactant] initial reactant concentration will not change significantly when it reaches equilibrium we can approximate the equilibrium concentration of reactant to be the same as its initial concentration  [A] equilibrium = [A] initial  ax  [A] initial  assumes the reaction is proceeding forward 62

63 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 63

64 Example14.13: For the reaction 2 H 2 S(g)  2 H 2 (g) + S 2 (g) @ 800 °C, K c = 1.67 x 10 −7. If a 0.500 L flask initially containing 1.25 x 10 −4 mol H 2 S is heated to 800 °C, find the equilibrium concentrations. 64 1.because no products initially, Q c = 0, and the reaction is proceeding forward 2.Kc is small 3.Can use the x is small approximation 2 H 2 S(g)  2 H 2 (g) + S 2 (g)

65 2 H 2 S(g)  2 H 2 (g) + S 2 (g) @ 800 °C, K c = 1.67 x 10 −7. 2.50 x 10 −4 M H 2 S, find the equilibrium concentrations. 65 +x+x +2x 2x2x 2.50x10 −4  2x 2x2xx Let x = [S 2 ], then [H 2 ]=2x and [H 2 S] = -2x 2 H 2 S(g)  2 H 2 (g) + S 2 (g)

66 66 +x+x +2x 2x2x 2.50E−4  2x 2x2x x 2.50E−4 K c = 2 H 2 S(g)  2 H 2 (g) + S 2 (g)

67 67 +x+x +2x 2x2x 2x2xx 2.50E–4 the approximation is not valid!! Need to use the quadratic formula to solve X = 1.38 x 10 -5 2 H 2 S(g)  2 H 2 (g) + S 2 (g)

68 68 +x+x +2x 2x2x 2.50E−4  2x 2x2xx x current = 1.38 x 10 −5 2 H 2 S(g)  2 H 2 (g) + S 2 (g)

69 69 +x+x +2x 2x2x 2.50E−4  2x 2x2xx x current = 1.27 x 10 −5 since x current = x new, approx. OK x new = 1.28 x 10 −5 2 H 2 S(g)  2 H 2 (g) + S 2 (g)

70 70 +x+x +2x 2x2x 2.50E−4  2x 2x2xx x current = 1.28 x 10 −5 2.24E−42.56E−51.28E−5 2 H 2 S(g)  2 H 2 (g) + S 2 (g)

71 71 +x+x +2x 2x2x 2.24E−4 2.56E−51.28E−5 K c (calculated) = K c (given) within significant figures 2 H 2 S(g)  2 H 2 (g) + S 2 (g)

72 Because [I] initial = 0, Q = 0 and the reaction must proceed forward. Assume x is small. 72 I 2 (g)  2 I(g), K c = 3.76 x 10 −5 @1000 K. Given 1.00 mole I 2 in a 2.00 L flask that is heated. Find the equilibrium concentrations of [I 2 ] and [I]? (use the simplifying assumption to solve for x) I 2 (g)  2 I(g)

73 the approximation is valid!! 73 I 2 (g)  2 I(g)

74 x = 0.00217 [I 2 ] = 0.500  0.00217 = 0.498 M [I] = 2(0.00217) = 0.00434 M  74 I 2 (g)  2 I(g)

75 because [NO 2 ] initial = 0, Q = 0 and the reaction must proceed forward 75 Given N 2 O 4 (g)  2 NO 2 (g) and K c = 1.07 x 10 −5. If [N 2 O 4 ] initial = 0.0125 M, find the equilibrium concentrations of [N 2 O 4 ] and [NO 2 ]? K is small, assume x is small

76 with x is small approximation the approximation is valid!! 76 Given N 2 O 4 (g)  2 NO 2 (g) and K c = 1.07 x 10 −5. If [N 2 O 4 ] initial = 0.0125 M, find the equilibrium concentrations of [N 2 O 4 ] and [NO 2 ]? without approximation x = 1.82 x 10 −4

77 x = 1.83 x 10 −4 [N 2 O 4 ]= 0.0125  1.83 x 10 −4 [N 2 O 4 ] = 0.0123 M [NO 2 ]=2(1.83 x 10 −4 ) [NO 2 ] = 3.66 x 10 −4 M  77 Given N 2 O 4 (g)  2 NO 2 (g) and K c = 1.07 x 10 −5. If [N 2 O 4 ] initial = 0.0125 M, find the equilibrium concentrations of [N 2 O 4 ] and [NO 2 ]?

78 Disturbing and Restoring Equilibrium Once a reaction is at equilibrium, the concentrations of all the reactants and products remain the same if conditions other than temperature 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 If the temperature is changed, K can change 78

79 Le Ch âtelier’s Principle Le Châtelier's Principle guides us for 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 79

80 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 80 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.

81 Disturbing Equilibrium: Adding or Removing Reactants If, after equilibrium is established a reactant included in the equilibrium constant expression is added (not a solid or a 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? 81

82 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 rxn The rxn proceeds right until equilibrium is restored At the new equilibrium position, you will have more products than before, less non-added reactants than before, and less added reactant than the sum of the original reactant and the added portion At the new equilibrium position, the concentrations of reactants and products will be such that the value of the equilibrium constant is the same 82

83 Disturbing Equilibrium: Adding or Removing Reactants If, after equilibrium is established a reactant included in the equilibrium constant expression is removed 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? 83

84 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 rxn. so the reaction is going faster in reverse The rxn proceeds left until equilibrium is restored At the new equilibrium position, you will have less 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 84

85 The Effect of Concentration Changes on Equilibrium Adding a reactant can be used to drive a reaction to completion e.g. loading up on a cheap reactant to insure complete rxn of an expensive reactant 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 85

86 The Effect of Concentration Changes on Equilibrium When NO 2 is added, some of it combines to make more N 2 O 4 86

87 The Effect of Concentration Changes on Equilibrium When N 2 O 4 is added, some of it decomposes to make more NO 2 87

88 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 88

89 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? 89

90 Disturbing Equilibrium: Reducing the Volume Decreasing the container volume increases concentration for all gases same # moles, but fewer liters, results in higher molarity does not effect solids, liquids, or solutions Decreasing 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, according to Le Châtelier’s Principle 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 90

91 The Effect of Volume Changes on Equilibrium 91 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

92 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 on the equilibrium even though heat is not matter and not written in a proper equation 92

93 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 aA + bB  cC + dD + Heat 93

94 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 Heat + aA + bB  cC + dD 94

95 The Effect of Temperature Changes on Equilibrium 95

96 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 96

97 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? 97 shift to SO 3 shift to SO 2 no effect 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


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