CHEMICAL EQUILIBRIUM. Chapter 16. Pb 2+ (aq) + 2 Cl (aq) e PbCl 2 (s) PLAY MOVIE Brooks/Cole - Cengage

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1 1 CHEMICAL EQUILIBRIUM Chapter 16 PLAY MOVIE Pb 2+ (aq) + 2 Cl (aq) e PbCl 2 (s)

2 Properties of an Equilibrium Equilibrium systems are DYNAMIC (in constant motion) REVERSIBLE can be approached from either direction 2 Pink to blue Co(H 2 O) 6 Cl 2 f Co(H 2 O) 4 Cl H 2 O Blue to pink Co(H 2 O) 4 Cl H 2 O f Co(H 2 O) 6 Cl 2 PLAY MOVIE

3 3 Chemical Equilibrium Fe 3+ + SCN - e FeSCN 2+ + Fe(H 2 O) SCN - e Fe(SCN)(H 2 O) H 2 O

4 Chemical Equilibrium Fe 3+ + SCN - e FeSCN 2+ 4 After a period of time, the concentrations of reactants and products are constant. The forward and reverse reactions continue after equilibrium is attained. PLAY MOVIE PLAY MOVIE

5 5 Examples of Chemical Equilibria Phase changes such as H 2 O(liq) H 2 O(s) e PLAY MOVIE

6 6 Examples of Chemical Equilibria Formation of stalactites and stalagmites CaCO 3 (s) + H 2 O(liq) + CO 2 (g) e Ca 2+ (aq) + 2 HCO 3- (aq)

7 7 Chemical Equilibria CaCO 3 (s) + H 2 O(liq) + CO 2 (g) e Ca 2+ (aq) + 2 HCO 3- (aq) At a given T and P of CO 2, [Ca 2+ ] and [HCO 3- ] can be found from the EQUILIBRIUM CONSTANT.

8 See Active Figure 16.2 Reaction Quotient & Equilibrium Constant 8 Equilibrium achieved Product conc. increases and then becomes constant at equilibrium Reactant conc. declines and then becomes constant at equilibrium

9 Reaction Quotient & Equilibrium Constant 9 At any point in the reaction H 2 + I 2 e 2 HI

10 Reaction Quotient & Equilibrium Constant 10 Equilibrium achieved In the equilibrium region

11 The Reaction Quotient, Q In general, ALL reacting chemical systems are characterized by their REACTION QUOTIENT, Q. 11 a A + b B e c C + d D If Q = K, then system is at equilibrium.

12 THE EQUILIBRIUM CONSTANT For any type of chemical equilibrium of the type a A + b B e c C + d D the following is a CONSTANT (at a given T) 12 If K is known, then we can predict concs. of products or reactants.

13 13 Determining K 2 NOCl(g) e 2 NO(g) + Cl 2 (g) Place 2.00 mol of NOCl is a 1.00 L flask. At equilibrium you find 0.66 mol/l of NO. Calculate K. Solution Set of an ICE table of concentrations [NOCl] [NO] [Cl 2 ] Initial Change Equilibrium 0.66

14 14 Determining K 2 NOCl(g) e 2 NO(g) + Cl 2 (g) Place 2.00 mol of NOCl in a 1.00 L flask. At equilibrium you find 0.66 mol/l of NO. Calculate K. Solution Set of an ICE table of concentrations [NOCl] [NO] [Cl 2 ] Initial Change Equilibrium

15 15 Determining K 2 NOCl(g) e 2 NO(g) + Cl 2 (g) [NOCl] [NO] [Cl 2 ] Initial Change Equilibrium

16 16 Writing and Manipulating K Solids and liquids NEVER appear in equilibrium expressions. S(s) + O 2 (g) SO 2 (g) Expressions e

17 17 Writing and Manipulating K Expressions Solids and liquids NEVER appear in equilibrium expressions. NH 3 (aq) + H 2 O(liq) e NH 4+ (aq) + OH - (aq)

18 18 The Meaning of K 1. Can tell if a reaction is productfavored or reactant-favored. For N 2 (g) + 3 H 2 (g) e 2 NH 3 (g) Conc. of products is much greater than that of reactants at equilibrium. The reaction is strongly product-favored.

19 19 The Meaning of K For AgCl(s) e Ag + (aq) + Cl - (aq) K c = [Ag + ] [Cl - ] = 1.8 x 10-5 Conc. of products is much less than that of reactants at equilibrium. The reaction with small K is strongly reactant-favored. Ag + (aq) + Cl - (aq) e AgCl(s) is product-favored.

20 20 Product- or Reactant Favored Product-favored K > 1 Reactant-favored K < 1

21 21 The Meaning of K K comes from thermodynamics. (See Chapter 19) G < 0: reaction is product favored G > 0: reaction is reactant-favored If K > 1, then G is negative If K < 1, then G is positive

22 The Meaning of K Can tell if a reaction is at equilibrium. If not, which way it moves to approach equilibrium. PLAY MOVIE PLAY MOVIE

23 The Meaning of K 23 If [iso] = 0.35 M and [n] = 0.15 M, are you at equilibrium? If not, which way does the reaction shift to approach equilibrium? See Chemistry Now

24 The Meaning of K 24 All reacting chemical systems are characterized by their REACTION QUOTIENT, Q. If Q = K, then system is at equilibrium. Q (2.33) < K (2.5) Reaction is NOT at equilibrium, so [iso] must become and [n] must.

25 25 Typical Calculations PROBLEM: Place 1.00 mol each of H 2 and I 2 in a 1.00 L flask. Calc. equilibrium concentrations. H 2 (g) + I 2 (g) e 2 HI(g) Æ

26 26 H 2 (g) + I 2 (g) e 2 HI(g) Step 1. Set up ICE table to define EQUILIBRIUM concentrations. [H 2 ] [I 2 ] [HI] Initial Change Equilib

27 27 H 2 (g) + I 2 (g) e 2 HI(g) Step 1. Set up ICE table to define EQUILIBRIUM concentrations. [H 2 ] [I 2 ] [HI] Initial Change -x -x +2x Equilib 1.00-x 1.00-x 2x where x is defined as am t of H 2 and I 2 consumed on approaching equilibrium.

28 28 H 2 (g) + I 2 (g) e 2 HI(g) Step 2. Put equilibrium concentrations into K c expression.

29 29 H 2 (g) + I 2 (g) e 2 HI(g) Step 3. Solve K c expression - take square root of both sides. x = 0.79 Therefore, at equilibrium [H 2 ] = [I 2 ] = x = 0.21 M [HI] = 2x = 1.58 M

30 30 Nitrogen Dioxide Equilibrium N 2 O 4 (g) e 2 NO 2 (g) e

31 31 Nitrogen Dioxide Equilibrium N 2 O 4 (g) e 2 NO 2 (g) If initial concentration of N 2 O 4 is 0.50 M, what are the equilibrium concentrations? Step 1. Set up an ICE table [N 2 O 4 ] [NO 2 ] Initial Change Equilib

32 32 Nitrogen Dioxide Equilibrium N 2 O 4 (g) e 2 NO 2 (g) If initial concentration of N 2 O 4 is 0.50 M, what are the equilibrium concentrations? Step 1. Set up an ICE table [N 2 O 4 ] [NO 2 ] Initial Change -x +2x Equilib x 2x

33 Nitrogen Dioxide Equilibrium N 2 O 4 (g) e 2 NO 2 (g) Step 2. Substitute into K c expression and solve. 33 Rearrange: ( x) = 4x x = 4x 2 4x x = 0 This is a QUADRATIC EQUATION ax 2 + bx + c = 0 a = 4 b = c =

34 Nitrogen Dioxide Equilibrium N 2 O 4 (g) e 2 NO 2 (g) Solve the quadratic equation for x. ax 2 + bx + c = 0 a = 4 b = c = x = ± 1/8(0.046) 1/2 = ± 0.027

35 Nitrogen Dioxide Equilibrium 35 N 2 O 4 (g) e 2 NO 2 (g) x = ± 1/8(0.046) 1/2 = ± x = or But a negative value is not reasonable. Conclusion: x = M [N 2 O 4 ] = x = 0.47 M [NO 2 ] = 2x = M

36 Writing and Manipulating K Expressions Adding equations for reactions 36 S(s) + O 2 (g) e SO 2 (g) SO 2 (g) + 1/2 O 2 (g) e SO 3 (g) Net equation S(s) + 3/2 O 2 (g) e SO 3 (g)

37 Writing and Manipulating K Expressions 37 Changing coefficients S(s) + 3/2 O 2 (g) e SO 3 (g) 2 S(s) + 3 O 2 (g) e 2 SO 3 (g)

38 Changing direction Writing and Manipulating K Expressions S(s) + O 2 (g) e SO 2 (g) 38 SO 2 (g) e S(s) + O 2 (g)

39 Writing and Manipulating K Expressions 39 Concentration Units We have been writing K in terms of mol/l. These are designated by K c But with gases, P = (n/v) RT = conc RT P is proportional to concentration, so we can write K in terms of P. These are designated by K p. K c and K p may or may not be the same.

40 Writing and Manipulating K Expressions K using concentration and pressure units 40 K p = K c (RT) n where n = moles gas productsmoles gas reactants For S(s) + O 2 (g) e SO 2 (g) n = 0 and K p = K c For SO 2 (g) + 1/2 O 2 (g) e SO 3 (g) n = 1/2 and K p = K c (RT) 1/2 (R =.0821 L-atm/mol-K)

41 EQUILIBRIUM AND 41 EXTERNAL EFFECTS Temperature, catalysts, and changes in concentration affect equilibria. The outcome is governed by LE CHATELIER S PRINCIPLE...if a system at equilibrium is disturbed, the system tends to shift its equilibrium position to counter the effect of the disturbance.

42 Le Chatelier s Principle 42 Change T change in K therefore change in P or concentrations at equilibrium Use a catalyst: reaction comes more quickly to equilibrium. K not changed. Add or take away reactant or product: K does not change Reaction adjusts to new equilibrium position

43 EQUILIBRIUM AND EXTERNAL EFFECTS 43 Temperature change f change in K Consider the fizz in a soft drink CO 2 (aq) + HEAT e CO 2 (g) + H 2 O(liq) K = P (CO 2 ) / [CO 2 ] Increase T. What happens to equilibrium position? To value of K? K increases as T goes up because P(CO 2 ) increases and [CO 2 ] decreases. Decrease T. Now what? Equilibrium shifts left and K decreases.

44 44 Temperature Effects on Equilibrium N 2 O 4 (colorless) + heat e 2 NO 2 (brown) H o = kj K c (273 K) = K c (298 K) = PLAY MOVIE

45 45 Temperature Effects on See Figure 16.8

46 46 EQUILIBRIUM AND EXTERNAL EFFECTS Catalytic exhaust system Add catalyst f no change in K A catalyst only affects the RATE of approach to equilibrium.

47 EQUILIBRIUM AND EXTERNAL EFFECTS 47 Concentration changes no change in K only the equilibrium composition changes.

48 48 Le Chatelier s Principle PLAY MOVIE Adding a reactant to a chemical system.

49 49 Le Chatelier s Principle PLAY MOVIE Removing a reactant from a chemical system.

50 50 Le Chatelier s Principle PLAY MOVIE Adding a product to a chemical system.

51 51 Le Chatelier s Principle PLAY MOVIE Removing a product from a chemical system.

52 52 Butane- butane Isobutane isobutane

53 Butane e Isobutane 53 At equilibrium with [iso] = 1.25 M and [butane] = 0.50 M. K = 2.5. Add 1.50 M butane. When the system comes to equilibrium again, what are [iso] and [butane]?

54 Butane e Isobutane Assume you are at equilibrium with [iso] = 1.25 M and [butane] = 0.50 M. Now add 1.50 M butane. When the system comes to equilibrium again, what are [iso] and [butane]? K = 2.5 Solution Calculate Q immediately after adding more butane and compare with K. 54 Q is LESS THAN K. Therefore, the reaction will shift to the.

55 You are at equilibrium with [iso] = 1.25 M and [butane] = 0.50 M. Now add 1.50 M butane. Solution Q is less than K, so equilibrium shifts right away from butane and toward isobutane. Set up ICE table Initial Change Equilibrium Butane e Isobutane [butane] [isobutane] x + x x x 55

56 Butane e Isobutane You are at equilibrium with [iso] = 1.25 M and [butane] = 0.50 M. Now add 1.50 M butane. Solution x = x; simplifies to 3.75 = 3.5 x x = 1.07 M At the new equilibrium position, [butane] = 0.93 M and [isobutane] = 2.32 M. Equilibrium has shifted toward isobutane.

57 Le Chatelier s Principle 57 Change P (by changing volume) affects gaseous reactants and products no change in K as long as temp is constant change in P at equilibrium as reaction adjusts to new equilibrium position. Increase P by decreasing volume on gaseous system shift toward fewer gas molecules Decrease P by increasing volume on gaseous system shift toward greater # of gas molecules

58 58 Nitrogen Dioxide Equilibrium N 2 O 4 (g) e 2 NO 2 (g) e Increase P in the system by reducing the volume (at constant T). PLAY MOVIE

59 59 Nitrogen Dioxide Equilibrium N 2 O 4 (g) e 2 NO 2 (g) Increase P in the system by reducing the volume. In gaseous system the equilibrium will shift to the side with fewer molecules (in order to reduce the P). Therefore, reaction shifts LEFT and P of NO 2 decreases and P of N 2 O 4 increases.

60 60 Haber-Bosch Ammonia Synthesis Fritz Haber Nobel Prize, 1918 Carl Bosch Nobel Prize, 1931

61 61 Haber-Bosch Process for NH 3 N 2 (g) + 3 H 2 (g) e 2 NH 3 (g) + heat K = 3.5 x 10 8 at 298 K

62 62 Ammonia Synthesis To favor product: Run at high pressure Continuously feed reactants into the system Continuously remove product (ammonia) from the system To speed up the reaction: Use a catalyst Run at high temperature (even though does not favor product).

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