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This is not the final version of this PowerPoint for chapter 16.

This is not the final version of this PowerPoint for chapter 16. Remember to check back and see when the final version has been posted. This first slide will indicate when the final version has been posted. Chemical Equilibrium Chapter 16.

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This is not the final version of this PowerPoint for chapter 16.

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  1. This is not the final version of this PowerPoint for chapter 16. Remember to check back and see when the final version has been posted. This first slide will indicate when the final version has been posted.

  2. Chemical Equilibrium Chapter 16 credit given to Larry Emme Chemeketa Community College

  3. Introduction A. A reaction at “equilibrium” is able to react in both directions: A + B <=> C + D B. At equilibrium, the rates of the forward and reverse reactions are equal, and the concentrations of the reactants and products are constant.

  4. D. The equilibrium constant (K) is given a subscript to indicate the type of reaction: Ka = weak acid ionization Kb = weak base ionization Kc = concentrations of non acid/base Kp = pressures of non acid/base Ksp = “insoluble” solids that “dissolve” Keq = generic “equilibrium” constant C. The equilibrium constant (K) is a value representing the unchanging concentrations of the reactants and the products in a chemical reaction at equilibrium.

  5. aA + bB cC + dD → → II. Equilibrium Constants A. For the general reaction: at a given temperature …. ONLY (aq) and (g) ARE ALLOWED IN EXPRESSION !!!

  6. 3H2(g) + N2 (g) 2NH3 (g) → → Ex: Write the equilibrium expression for the reaction:

  7. 4NH3 (g) + 3O2 (g) 2N2(g)+ 6H2O(g) → → Ex: Write the equlibrium expressionfor the reaction:

  8. Ex: Write the equlibrium expression for the reaction: 2KClO3(s)<=>2KCl (s) +3O2 (g) Keq = [O2]3

  9. B. The magnitude of an equilibrium constant indicates the extent to which the forward and reverse reactions take place. H2(g) + I2(g) 2HI(g) At equilibrium more product than reactant exists. At equilibrium more reactant than product exists. COCl2(g) CO(g) + Cl2(g) → → → →

  10. C. When the molar concentrations of all species in an equilibrium reaction are known, the Keq can be calculated by substituting the concentrations into the equilibrium constant expression.

  11. Calculate Keq for the following reaction if the equilibrium concentrations of PCl5 = 0.030 mol/L, PCl3 = 0.97 mol/L and Cl2 = 0.97 mol/L at 300oC. PCl5(g )PCI3(g) + Cl2(g) → →

  12. Calculate Kp for the following reaction if the equilibrium pressures of PCl5 = 1.46 atm, PCl3 = 2.50 atm and Cl2 = 2.50 atm at 300oC. PCl5(g )PCI3(g) + Cl2(g) → → Kp = 4.28

  13. III. Calculating Equilibrium Concentrations If K is known, then concentrations of all species at equilibrium can be calculated The procedure requires constructing an ICE chart: I = initial C = change (pay attention to coefficients!) E = equilibrium C. Substitute the E values into the equilibrium expression and solve for x

  14. The equilibrium constant for the reaction below is 54.8 at 425 oC. If .500 moles each of H2(g) and I2(g) are introduced into a 2.00 L container, what will be the concentrations at equilibrium? I C E H2(g) + I2(g) <=> 2 HI (g) 0.250 0.250 -x - x + 2x (.250-x) (2x) (.250-x) Solve for x and substitute back in = 54.8

  15. The equilibrium constant for the reaction below is 54.8 at 425 oC. If .500 moles of H2(g) and 0.750 mole I2(g) are introduced into a 1.50 L container, what will be the concentrations of all species at equilibrium? I C E H2(g) + I2(g) <=> 2 HI (g)

  16. A mixture of hydrogen and nitrogen in a reaction vessel is allowed to attain equilibrium at 472 oC. The equilibrium mixture of gases ws analyzed and found to contain 7.38 atm H2, 2.46 atm N2, and 0.166 atm NH3. From these data, calculate the Kp.? 3 H2(g) + N2(g) <=> 2 NH3(g)

  17. Enough ammonia is dissolved in 5.00 liters of water to produce a solution that is 0.0124 M in ammonia. The solution is then allowed to come to equilibrium. Analysis of the equilibrium mixture shows that the concentration of OH- is 4.64 x 10-6 M. Calculate Kc for the reaction. NH3(aq) + H2O (l) <=> NH4+(aq)+OH-(aq)

  18. Sulfur trioxide decomposes at high temperatures in a sealed container. Initially the vessel is filled with SO3 at a partial pressure of 0.500 atm. At equilibrium the SO3 partial pressure is 0.200 atm. Calculate Kp. 2 SO3(g) <=> 2 SO2(g)+O2(g)

  19. Ionization Constants

  20. In addition to Kw, several other ionization constants are used.

  21. Ka

  22. HC2H3O2(aq) H+ + C2H3O2 → → When acetic acid ionizes in water, the following equilibrium is established: Ka is the ionization constant for this equilibrium. Ka is called the acid ionization constant. Since the concentration of water is large and does not change appreciably, it is omitted from Ka.

  23. At 25oC, a 0.100 M solution of HC2H3O2 is 1.34% ionized and has an [H+] of 1.34 x 10-3 mol/L. Calculate Ka for acetic acid. HC2H3O2(aq) H+ + C2H3O2 Because each molecule of HC2H3O2 that ionizes yields one H+ and one C2H3O2 , the concentrations of the two ions are equal. → → The moles of unionized acetic acid per liter are 0.100 mol/L – 0.00134 mol/L = 0.099 mol/L

  24. Substitute these concentrations into the equilibrium expression and solve for Ka. [HC2H3O2] = 0.099 mol/L

  25. What is the [H+] in a 0.50 M HC2H3O2 solution? The ionization constant, Ka, for HC2H3O2 is 1.8 x 10-5. HC2H3O2(aq) H+ + C2H3O2 Because each molecule of HC2H3O2 that ionizes yields one H+ and one C2H3O2 , the concentrations of the two ions are equal. → Let Y = [H+] = → The equilibrium expression and Ka for HC2H3O2 are [HC2H3O2] at equilibrium is 0.50 – Y.

  26. Y = [H+] = Substitute these values into Ka for HC2H3O2. HC2H3O2 = 0.50 - Y Solve for Y2. Assume Y is small compared to 0.50 -Y. Then 0.50 – Y  0.50

  27. Take the square root of both sides of the equation. Making no approximation and using the quadratic equation the answer is 2.99 x 10-3 mol/L, showing that it was justified to assume Y was small compared to 0.5.

  28. Calculate the percent ionization in a 0.50 M HC2H3O2 solution. HA H+ + A- → → The ionization of a weak acid is given by The percent ionization is given by

  29. Calculate the percent ionization in a 0.50 M HC2H3O2 solution. HC2H3O2(aq) H+ +C2H3O2 → → [H+] was previously calculated as 3.0 x 10-3 mol/L The ionization of acetic acid is given by The percent ionization of acetic acid is given by

  30. Ionization Constants (Ka) of Weak Acids at 25oC

  31. Ionic Equilibria Problems

  32. Use Ka as a measure of acid strength. Let 1 = Hi and 5 = Lo

  33. Use Ka as a measure of acid strength. Let 1 = Hi and 5 = Lo

  34. → → Find the pH in 0.1 M HCN if Ka = 7.2  10–10 Step 1: ionize the acid Step 2: write the Ka expression

  35. Step 3: let x = [H+] = [CN–] [HCN] = 0.1 M – x≈ 0.1 M Step 4: substitute into Ka expression Step 5: solve for x:

  36. Step 6: solve for pH

  37. → → Find [H+] in 0.1 M H2CO3 if Ka = 4.0  10–17 Step 1: ionize the acid Step 2: write the Ka expression

  38. Step 3: let x = [H+] , [CO3–2] = ½ x [H2CO3] = 0.1 M – ½x≈ 0.1 M Step 4: substitute into Ka expression Step 5: solve for x:

  39. → → Find the pH of a 0.001 M solution of bombastic hydroxide (BmOH) if Kb = 1.6  10–10 Step 1: ionize the base Step 2: write the Kb expression

  40. Step 3: let x = [Bm+] = [OH–] [BmOH] = 0.001 M – x≈ 0.001 M Step 4: substitute into Kb expression Step 5: solve for x:

  41. Step 6: solve for pOH Step 7: solve for pH

  42. → → Find the Ka for a 0.01 M HCN solution if the pH = 6.3 Step 1: ionize the acid Step 2: write the Ka expression

  43. Step 3: use pH to find [H+] [H+] = 10–pH = 10–6.3 = 5.0  10–7 [H+] = [CN–], [HCN] = 0.01- 5.0  10–7≈ 0.01 Step 4: substitute into Ka expression

  44. Solubility Product Constant

  45. The solubility product constant, Ksp, is the equilibrium constant of a slightly (sparingly) soluble salt.

  46. AgCl(s) Ag+(aq) + Cl-(aq) The product of Keq and [AgCl(s) is a constant. → → Silver chloride is in equilibrium with its ions in aqueous solution. The equilibrium constant is Rearrange The amount of solid AgCl does not affect the equilibrium. The concentration of solid AgCl is a constant.

  47. Ksp for Sparingly Soluble Solids To find Ksp: 1. ionize the salt 2. write the Ksp expression 3. determine the M’s of the ions 4. substitute into the Ksp expression

  48. AgCl(s) Ag+(aq) + Cl-(aq) → → The solubility of AgCl in water is 1.3 x 10-5 mol/L. Because each formula unit of AgCl that dissolves yields one Ag+ and one Cl-, the concentrations of the two ions are equal. [Ag+] = [Cl-] = 1.3 x 10-5 mol/L Ksp = [Ag+][Cl-] The Ksp has no denominator. = (1.3 x 10-5)(1.3 x 10-5) = 1.7 x 10-10

  49. Molar Solubility for Sparingly Soluble Solids 1. ionize the salt 2. write the Ksp expression 3. let y equal the amount of salt that dissolves 4. find concentrations of ions in terms of y 5. substitute in the Ksp expression 6. solve for y

  50. The Kspvalue forlead sulfate is 1.3 x 10-8. Calculate the solubility of PbSO4 in moles per liter. → → Because each formula unit of PbSO4 that dissolves yields one Pb2+ and one , the concentrations of the two ions are equal. The equilibrium equation of PbSO4 is The Ksp of PbSO4 is

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