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CHEMICAL EQUILIBRIUM Chapter 16

CHEMICAL EQUILIBRIUM Chapter 16. ALL BOLD NUMBERED PROBLEMS. Properties of an Equilibrium. Equilibrium systems are DYNAMIC (in constant motion) & REVERSIBLE. Cool. Heat. Blue to pink CoCl 4 2- (aq) + 6 H 2 O (l)  Co(H 2 O) 6 2+ (aq) . Cool. Pink to blue

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CHEMICAL EQUILIBRIUM Chapter 16

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  1. CHEMICAL EQUILIBRIUMChapter 16 ALL BOLD NUMBERED PROBLEMS

  2. Properties of an Equilibrium Equilibrium systems are • DYNAMIC (in constant motion) & REVERSIBLE

  3. Cool Heat Blue to pink CoCl42- (aq) + 6 H2O (l)  Co(H2O)62+ (aq) Cool Pink to blue Co(H2O)62+ (aq)  CoCl42- (aq) + 6 H2O (l) Heat

  4. + Fe(H2O)63+ + SCN- + H2O Fe(SCN)(H2O)52+ Chemical EquilibriumFe3+ + SCN-FeSCN2+ colorless red-orange

  5. Chemical EquilibriumFe3+ + SCN- FeSCN2+ • After a period of time, the concentrations of reactants and products are constant. • The forward and reverse reactions continue after equilibrium is attained.

  6. Examples of Chemical Equilibria Phase changes such as H2O(s) H2O(liq)

  7. Color Red Purple Violet Blue Blue-Green Green pH 2 4 6 8 10 12 +Na2CO3 H3O+(aq) + 2CO32-(aq) OH-(aq) + 2HCO3-(aq) +CO2 Examples of Chemical Equilibria Acid Base Equilibria

  8. Examples of Chemical Equilibria Formation of stalactites (ceiling) and stalagmites (floor) CaCO3(s) + H2O(liq) + CO2(g) Ca2+(aq) + 2 HCO3-(aq)

  9. Chemical Equilibria At a given T and P of CO2, [Ca2+] and [HCO3-] can be found from the EQUILIBRIUM CONSTANT. CaCO3(s) + H2O(liq) + CO2(g) Ca2+(aq) + 2 HCO3-(aq)

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

  11. Determining K 2 NOCl(g) 2 NO(g) + Cl2(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 up a table of concentrations: [NOCl] [NO] [Cl2] Initial 2.00 0 0 Change Equilibrium 0.66

  12. Determining K 2 NOCl(g) 2 NO(g) + Cl2(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 a table of concentrations: [NOCl] [NO] [Cl2] Initial 2.00 0 0 Change Equilibrium 0.66 +0.66 +0.33 -0.66 1.34 0.33

  13. 2 [ NO ] [ Cl ] 2 K = 2 [ NOCl ] 2 2 [ NO ] [ Cl ] (0.66) (0.33) 2 K = = 0.080 = 2 2 [ NOCl ] (1.34) Determining K 2 NOCl(g) 2 NO(g) + Cl2(g) [NOCl] [NO] [Cl2] Initial 2.00 0 0 Change -0.66 +0.66 +0.33 Equilibrium 1.34 0.66 0.33

  14. [SO ] 2 K = [O ] 2 Writing and Manipulating K Expressions Solids and liquids NEVER appear in equilibrium expressions. S(s) + O2(g) SO2(g)

  15. - + [NH ][OH ] 4 K = [NH ] 3 Writing and Manipulating K Expressions Solids and liquids NEVER appear in equilibrium expressions. NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)

  16. [SO ] 3 K = 3/2 [O ] 2 2 [SO ] 3 K = new 3 [O ] 2 2 [SO ] 2 3 K = (K ) = new old 3 [O ] 2 Writing and Manipulating K Expressions Changing coefficients S(s) + 3/2 O2(g) SO3(g) 2 S(s) + 3 O2(g) 2 SO3(g)

  17. [SO ] 2 K = [O ] 2 [O ] 2 K = new [SO ] 2 [O ] 1 2 K = = new [SO ] K 2 old Writing and Manipulating K Expressions Changing direction S(s) + O2(g) SO2(g) SO2(g) S(s) + O2(g)

  18. [SO ] 3 K = K • K = net 1 2 3/2 [O ] 2 Writing and Manipulating K Expressions Adding equations for reactions S(s) + O2(g) SO2(g) K1 = [SO2] / [O2] SO2(g) + 1/2 O2(g) SO3(g) K2 = [SO3] / [SO2][O2]1/2 NET EQUATION S(s) + 3/2 O2(g) SO3(g)

  19. Manipulating K Expressions • The equilibrium expression is tied to the equation from which it is written. • Changing the equation changes the expression and thus the numerical value of K. • Three general cases need to be considered. 1. If the equation is multiplied by a number, “a”, then the K is raised to the “a” power. 2. If the equation is reversed, then the new K is the reciprocal of the old K. 3. If two equations are added, the new K is the product of the two old K's. • Using these rules, new K's can be derived for the modified equations. SAMPLE QUESTIONS

  20. Writing and Manipulating K Expressions Concentration Units We have been writing K in terms of mol/L. These are designated by Kc. But with gases, P = (n/V)•RT = [ ] • RT P is proportional to concentration, so we can write K in terms of P. These are designated by Kp. Kc and Kp may or may not be the same. Kp = Kc(RT)Dn

  21. Writing and Manipulating K Expressions Practice Problems • Write the equation and the Kc expression for the formation of two moles of gaseous ammonia from the elements in the standard state. • Kc for this reaction at 25oC is 3.5x108. • Calculate Kp for this reaction. • Calculate Kc for the reaction forming the elements from one mole of ammonia gas.

  22. 2 [NH ] 8 3 K = = 3.5 x 10 c 3 [N ][H ] 2 2 The Meaning of K We can tell if a reaction is product-favored or reactant-favored. For N2(g) + 3 H2(g) 2 NH3(g) Concentration of products is much greater than that of reactants at equilibrium. The reaction is strongly product-favored.

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

  24. Comparing Q and K • The relative magnitudes of Q and K tell us which direction the reaction will proceed to reach equilibrium. • If Q<K, not at equilibrium and Reactants ------> Products. • If Q=K, the system is at equilibrium. • If Q>K, not at equilibrium and Products ------> Reactants.

  25. Using Q Can tell if a reaction is at equilibrium. If not, which way it moves to approach equilibrium.

  26. n-butane iso-butane H H H H H H H H—C—C—C—C—H H—C—C—C—H H H H H H H H H C H [iso] K = = 2.5 [n] Using Q If [iso] = 0.35 M and [n] = 0.15 M, are you at equilibrium? Which way does the reaction “shift” to approach equilibrium?

  27. product concentrations Q = reactant concentrations conc. of iso 0.35 Q = = = 2.3 conc. of n 0.15 Using Q In general, all reacting chemical systems are characterized by their REACTION QUOTIENT, Q. If Q = K, then system is at equilibrium. [iso]=0.35 M and [n]=0.15 M, are you at equilibrium? Q (2.3) < K (2.5) Reaction is NOT at equilibrium, so [Iso] must become ________ and [n] must ____________. larger decrease SAMPLE QUESTIONS

  28. 2 [HI] K = = 55.3 c [H ][I ] 2 2 Using K PROBLEM: Place 1.00 mol each of H2 and I2 in a 1.00 L flask. Calculate the equilibrium concentrations. H2(g) + I2(g) 2 HI(g)

  29. H2(g) + I2(g) 2 HI(g), Kc = 55.3 Step 1. Set up table to define EQUILIBRIUM concentrations. [H2] [I2] [HI] Initial Change Equilib where x is defined as amount of H2 and I2 consumed on approaching equilibrium. 1.00 1.00 0 -x -x +2x 1.00-x 1.00-x 2x

  30. 2 [2x] K = = 55.3 c [1.00 - x][1.00 - x] 2x 7 . 44 = 1.00 - x H2(g) + I2(g) 2 HI(g), Kc = 55.3 Step 2. Put equilibrium concentrations into Kc expression. Step 3. Solve Kc expression - take square root of both sides.

  31. 2x 7 . 44 = 1.00 - x H2(g) + I2(g) 2 HI(g), Kc = 55.3 Step 3. Solve Kc expression - take square root of both sides. x = 0.788 Therefore, at equilibrium [H2] = [I2] = 1.00 - x = 0.21 M [HI] = 2x = 1.58 M

  32. Sample Problem At a certain temperature 8.00 atm of H2S(g) comes to equilibrium with H2(g) and S2(g). The equilibrium pressure of the sulfur gas is 0.60 atm. Write the equation for the decomposition of the hydrogen sulfide, calculate the equilibrium pressure for each gas, and calculate KP. Solution

  33. Nitrogen Dioxide EquilibriumN2O4(g) 2 NO2(g)

  34. 2 [NO ] 2 K = = 0.0059 at 298 K c [N O ] 2 4 Nitrogen Dioxide EquilibriumN2O4(g) 2 NO2(g) If initial concentration of N2O4 is 0.50 M, what are the equilibrium concentrations? Step 1. Set up an equilibrium table [N2O4] [NO2] Initial 0.50 0 Change +2x Equilib 0.50 - x 2x -x

  35. 2 2 [NO ] (2x) 2 K = 0.0059 = = c [N O ] (0.50 - x) 2 4 Nitrogen Dioxide EquilibriumN2O4(g) 2 NO2(g) Step 2. Substitute into Kc expression and solve. Rearrange: 0.0059 (0.50 - x) = 4x2 0.0029 - 0.0059x = 4x2 4x2 + 0.0059x - 0.0029 = 0 This is a QUADRATIC EQUATION ax2 + bx + c = 0 a = 4 b = 0.0059 c = - 0.0029

  36. 2 -b b - 4ac ± x = 2a 2 -0.0059 ( 0 . 0059 ) - 4(4)(-0.0029) ± x = 2(4) Nitrogen Dioxide EquilibriumN2O4(g) 2 NO2(g) Solve the quadratic equation for x. ax2 + bx + c = 0 a = 4 b = 0.0059 c = - 0.0029 x = - 0.00074 ± 1/8(0.046)1/2 = - 0.00074 ± 0.027

  37. 2 -0.0059 ( 0 . 0059 ) - 4(4)(-0.0029) ± x = 2(4) Nitrogen Dioxide EquilibriumN2O4(g) 2 NO2(g) x = -0.00074 ± 1/8(0.046)1/2 = -0.00074 ± 0.027 x = 0.026 or -0.028 But a negative value is not reasonable. Conclusion [N2O4] = 0.50 - x = 0.47 M [NO2] = 2x = 0.052 M More Sample Problems

  38. 38 EQUILIBRIUM AND 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.”

  39. EQUILIBRIUM AND EXTERNAL EFFECTS Henri Le Chatelier 1850-1936 Studied mining engineering. Interested in glass and ceramics.

  40. Figure 16.6 50o C 0o C NO2 / N2O4is temperature dependent.

  41. EQUILIBRIUM AND EXTERNAL EFFECTS • Temperature change ---> change in K • Consider the fizz in a soft drink CO2(g) + H2O(liq) CO2(aq) + heat • Decrease T. What happens to equilibrium position? To value of K? • K = [CO2] / P (CO2) K increases as T goes down because [CO2] increases and P(CO2) decreases. • Increase T. Now what? • Equilibrium shifts left and K decreases.

  42. 2 [NO ] 2 K = c [N O ] 2 4 Temperature Effects on Equilibrium N2O4 (colorless) + heat 2 NO2 (brown) DHo = + 57.2 kJ Kc = 0.00077 at 273 K Kc = 0.0059 at 298 K

  43. EQUILIBRIUM AND EXTERNAL EFFECTS • Add catalyst ---> no change in K • A catalyst only affects the RATE of approach to equilibrium. Catalytic exhaust system

  44. NH3 ProductionFritz Haber, 1909 • N2(g) + 3 H2(g) 2 NH3(g) • K = 3.5 x 108 at 298 K • K = 0.16 at 723 K

  45. EQUILIBRIUM AND EXTERNAL EFFECTS Concentration changes no change in K only the position of equilibrium changes

  46. Le Chatelier’s Principle Adding a “reactant” to a chemical system.

  47. Le Chatelier’s Principle Removing a “reactant” from a chemical system.

  48. Le Chatelier’s Principle Adding a “product” to a chemical system.

  49. Le Chatelier’s Principle Removing a “product” from a chemical system.

  50. Figure 16.7 (a) 7 n-butane molecules (b) 7 iso-butane are added (c) What are the new equilibrium concentration?

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