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pp. 549 – 559

Ch. 18 – Reversible Reactions & Equilibrium. pp. 549 – 559 . A. Reversible Reactions. Reactions can be reversible Conversion of reactants to products and products to reactants happens simultaneously Forward reaction: 2SO 2 ( g ) + O 2 ( g ) → 2SO 3 ( g )

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pp. 549 – 559

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  1. Ch. 18 – Reversible Reactions & Equilibrium pp. 549 – 559

  2. A. Reversible Reactions • Reactions can be reversible • Conversion of reactants to products and products to reactants happens simultaneously Forward reaction: 2SO2(g) + O2(g) → 2SO3(g) Reverse reaction: 2SO2(g) + O2(g) ← 2SO3(g)

  3. A. Reversible Reactions • Chemical equilibrium – when the rates of the forward and reverse reaction are equal • Physical Equilibrium: vapor-liquid phase or solution equilibrium • No net change occurs in the actual amounts of the components of the system (reaction will never complete) • dynamic state • Concentration at equilibrium is temperature dependent

  4. A. Reversible Reactions • Equilibrium position – made up of relative concentration of reactants and products at equilibrium • A B • Different sized arrows indicate favored direction of a reaction • Catalyst speeds up forward and reverse reactions equally • decreases time it takes to reach equilibrium

  5. B. Equilibrium Constant • Law of Mass Action – expression of equilibrium condition using equilibrium constant, Keq jA + kB ↔lC + mD • Brackets indicate concentrations at equilibrium (*pure solids and liquids do not affect equilibrium values!*)

  6. B. Equilibrium Constant • Example 4NH3(aq) + 7O2(g) ↔ 4NO2(aq) + 6H2O(g) • Arrows indicate coefficients

  7. B. Equilibrium Constant • Write the balanced equation and the equilibrium expression for the production of ammonia N2 (g) + 3H2 (g)↔ 2NH3 (g)

  8. C. Evaluating Equilibrium Constant N2 (g) + 3H2 (g)↔ 2NH3 (g) • Keq > 1 • Products favored at equilibrium • Keq < 1 • Reactants favored at equilibrium • Keq = 1 • Large amount of both products and reactants at equilibrium

  9. Example • Solve for Keq in N2 (g) + 3H2 (g)↔ 2NH3 (g) if [N2] = .625, [H2] = 4.0 and [NH3] = 2.0 • Plug into Keq expression [2.0]2 Keq= .10 Keq= [.625] [4.0]3 • Are the reactants or products favored in this reaction? • Reactants

  10. B. Equilibrium Constant • Write the balanced equation and the equilibrium expression for: Zn (s) + 2 Ag+ (aq)↔ Zn2+(aq) + 2 Ag (s)

  11. Example • Solve for Keq in Zn (s) + 2 Ag+ (aq)↔ Zn2+(aq) + 2 Ag (s) • if [Zn] = 109 M, [Ag+] = 2.0 and [Zn2+] = 24M • Plug into Keq expression [24] Keq= 6.0 Keq= [2.0]2 • Are the reactants or products favored in this reaction? • Products

  12. Reaction Equilibrium • Concentrations in a problem, may not always be at equilibrium! • If you know initial molarities of reactants you can calculate the ending molarites of all substances if we know the Keq value. • Can also calculate the Keq if you know the initial molarities and one end molarity. • To help us with this we make a RICE chart! • R = Reaction • I = Initial • C = Change • E = Equilibrium

  13. Calculating Equilibrium Constant • Hydrogen & Iodine react to form hydrogen iodide. There are initially 2.00 moles of both hydrogen and iodine present in a 1.0 L flask. After the reaction has reached equilibrium, there are 1.20 moles of hydrogen remaining. Calculate the Keq for the reaction. • 1. First write out the reaction: • H2 + I2 2 HI

  14. Calculating Equilibrium Constant • 2. Then fill in the amounts given in molarity. • 2 moles H2 / 1 L = 2.0 M H2 • 2 moles I2 / 1 L = 2.0 M I2 • 1.2 moles H2 / 1 L = 1.2 M H2 • For now… the initial amount of products is always zero

  15. Calculating Equilibrium constant • 3. Figure out the change (subtraction) • You are given 1 ending amount,1.20 moles of H2 • Change: Initial – Equilibrium = .80 moles of H2 • 4. Use the coefficients of the equation to figure out the rest of the ‘change’ row. (mole ratio)

  16. Calculating Equilibrium constant • 5. Find the Equilibrium concentration by subtracting on the reactants side • But adding on the products side • 6. Substitute values into the Keq expression

  17. Example 2: • When 2.50 moles of nitrogen and 3.40 moles of hydrogen are mixed in a .500 liter container, they react to form ammonia. If the concentration of nitrogen is found to be 3.50 M when the reaction reaches equilibrium, what is the value of the equilibrium constant? • 1. First write out the reaction: • N2 + 3 H2 2 NH3

  18. Calculating Equilibrium Constant • 2. Then fill in the amounts given in molarity. • 2.5 moles N2 / 0.500 L = 5.00 M N2 • 3.40 moles H2 / 0.500 L = 6.80 M H2 • Initial amount of products is zero • 3. Calculate the Change • 4. Use mole ratio for rest of the Change row

  19. Calculating Equilibrium Constant • 5. Find the Equilibrium concentration by subtracting on the reactants side • But adding on the products side • 6. Substitue values in Keq expression

  20. Solubility Equilibrium • Special constant for the dissociation of solids that dissolve in water. • Ionic solids dissociate into their cations & anions • NaCl(s) Na+1 (aq) + Cl-1 (aq) • Ions formed carry an electrical current (electrolytes) • Solubility • The ratio of the maximum amount of solute to the volume of solvent in which this solute can dissolve at a specific Temperature H2O

  21. Solubility • All ionic solids are soluble up to a point… so they may not be completely soluble • Soluble: if concentration of solution is at least 0.1 M @ room temperature • Slightly Soluble: if concentration of solution is between 0.0001 M and 0.1 M @ room temp. • Insoluble: if concentration of solution is less than .0001 M @ room temperature

  22. Solubility Product Constant (Ksp) • This is the product of the equilibrium concentrations of ions in a saturated solution of a salt • *There is no denominator in solubility product equilibrium constant. • Ex: AgCl (s) Ag+ (aq) + Cl- (aq) • Ksp = [Ag+][Cl-] H2O

  23. Solubility Product Equilibrium • Example: The solubility of AgCl in pure water is 1.3 x10-5 M . Calculate the Ksp using the expression above. • The mole ratio of AgCl to both Ag+ and Cl-1 is 1:1, the solubility of each ions is equal to the solubility of AgCl. • Ksp = (1.3x10-5 ) (1.3x10-5 ) • = 1.69 x10-10 (.000000000169) • Is AgCl considered soluble, insoluble or slightly soluble? • insoluble

  24. Example • Calculate the solubility of CaF2 in grams/L if Ksp = 4.0 x 10-8 . • 1. Write Balanced Reaction: • CaF2 (s) Ca2+ (aq) + 2 F-1 (aq) • 2. Write the solubility expression: • Ksp = [Ca2+][F-1]2 • 4.0x10-8 = [Ca2+][F-1]2 • There are 2 unknowns so we have to write in terms using the mole ratios. For every one mole of Ca2+ there are 2 moles of F-1 • Assign a variable x for solubility of Ca2+ H2O

  25. Example • [Ca2+] = x and [F-] = 2x • Substitute the values into the equilibrium expression & solve for x • 4.0x10-8 = [x] [2x]2 • 4.0x10-8 = [x] [4x2] • 4.0x10-8 = 4x3 • 1.0x10-8 = x3 • √ 1.0x10-8 = √ x3 • X = 2.2 x 10-3 M 4 4 3 3

  26. Example • X = 2.2 x 10-3 M • We have assigned x as the solubility of the Ca2+ ion which is equal to the solubility of the salt, CaF2. • Now convert from mol/L to g/L 78.1 g CaF2 1 mole CaF2 2.2x10-3 mol CaF2 L = 0.017 g/L of CaF2

  27. E. Factors Affecting Equilibrium • Lechâtelier’s Principle • If a stress is applied to a system in dynamic equilibrium, the system will change in a way to relieve the stress • Reaction shifts towards products or reactants in order to relieve stress • Stresses that upset equilibrium include changes in: • Concentration • Temperature • Pressure

  28. E. Factors Affecting Equilibrium • Concentration •  in reactants causes forward rate to increase, then increase in reverse rate and equilibrium is re-established, but is shifted in direction of products H2CO3 (aq) CO2 (aq) + H2O (l) <1% add H2CO3 >99% •  in products has reverse effect •  in concentration shifts equilibrium toward that substance

  29. E. Factors Affecting Equilibrium • Temperature •  shifts away from side that contains the heat • Think of heat as a reactant or product, same as before with addition or removal Remove heat (cool) 2SO2 (g) + O2 (g) 2SO3 (l) + heat ‘ Add heat

  30. Temperature • Exothermic Reactions: give off heat • Heat in kJ or Joules is a product • Endothermic Reactions: take in heat • Heat in kJ or Joules is a reactant • Equilibrium is temperature dependent, Keq can only be affect overall by this change T!

  31. E. Factors Affecting Equilibrium • Pressure • Only affects gaseous solution with unequal number of moles of reactants and products • Increase shifts towards side w/ least amount of gas • Decrease shifts towards side w/ most amount of gas Decrease pressure N2 (g) + 3H2 (g) 2NH3 (g) Increase pressure

  32. Example • N2 (g) + 3 H2 (g)  2NH3 (g) + 92 kJ • What direction is the shift if you reduce the pressure? • How would cooling the system shift the reaction? • How would adding N2(g) shift the system? • What direction is the shift if you remove NH3 ? • left • right • right • right

  33. I. Rates of Reaction Reaction Energy and Reaction Kinetics

  34. Collision Theory • Reaction rate depends on the collisions between reacting particles. • The particles collide and make new substances. • Successful collisions occur if the particles... • collide with each other • have the correct orientation • have enough kinetic energy to break bonds

  35. Collision Theory Particle Orientation Required Orientation Unsuccessful Collisions Successful Collision

  36. Activation Energy Activation Energy Activation Energy (Ea) • minimum energy required for a reaction to occur • Activated Complex: the transitional structure in a collision that exists while old bonds are breaking and new bonds are being formed. • A. Collision Theory • Activation Energy (Ea) • minimum energy required for a reaction to occur

  37. Ea Activation Energy Cont… Activation Energy: - depends on reactants - is always positive - low Ea = fast reaction rate - takes less energy for the reaction to start. • Reaction Rate: the change in concentration of reactants per unit time as a reaction proceeds.

  38. Factors Affecting Rxn Rate 1. Nature of Reactants - substances vary greatly in their tendencies to react. - bonds are broken and other bonds are formed in reactions. - the rate of reaction depends on the particular reactants and the bonds involved. 2. Surface Area • high SA = fast rxn rate • more opportunities for collisions • Increase surface area by… • using smaller particles – if we make the pieces of the reactants smaller, we increase the number of particles on the surface which can react. • dissolving in water – gases & dissolved particles can mix & collide freely. Reactions happen rapidly.

  39. Factors Affecting Rxn Rate Cont. 3. Concentration - high conc = fast rxn rate - more opportunities for collisions because there are more particles in the same volume that can react. There are less red particles in the same volume so there is less chance of a collision There are more red particles in the same volume so there is more chance of a collision so the reaction goes faster

  40. Factors Affecting Rxn Rate Cont. 4. Temperature - high temp. = fast rxn. rate - high KE - when we increase the temperature, we give the particles energy - this makes the particles move faster - so there are more opportunities for collision - it is easier to reach activation energy

  41. Factors Affecting Rxn Rate Cont. 5. Catalyst • substance that increases rxn rate without being consumed in the rxn • lowers the activation energy

  42. II. Energy Diagrams Ch. 17 – Reaction Energy and Reaction Kinetics

  43. Terms and Symbols • Reactants: the chemicals you start with in a reaction. • Products: the chemicals formed during the reaction. • Heat of reaction (DE or DHrxn): the difference in energy between reagents and products (units are Joules). • Transition State: the highest point on the energy diagram, representing the point at which the reaction is half-completed. • Activation Energy (Ea): the amount of energy required for the reaction to take place. The higher the Ea, the slower the reaction (units are Joules).

  44. Energy Diagrams • Show relationship between time and energy during the course of a chemical reaction. Activated Complex (transition state) - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Ea Ea’ Energy (E) in kJ/mol Reactants - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - DE Products - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Forward Rxn (exothermic) Reverse Rxn (endothermic) Course of Reaction (time)

  45. Endothermic Reaction • A reaction in which heat is absorbed • Products have higher potential energy than the reactants. • The pink curve shows the uncatalyzed reaction. The blue curve shows what happens when a catalyst is present. • The energies and amounts of the products and reactants stays the same, and the DE stays the same. The catalyst just allows the reaction to reach equilibrium faster. Energy Course of Reaction

  46. Exothermic Reactions • Reactions in which heat is released. • Products have lower potential energy than the reactants. • The blue curve shows the uncatalyzed reaction. The red curve shows what happens when a catalyst is present. • Again, nothing changes but the amount of time it takes for the reaction to reach equilibrium. • Exothermic rxns are referred to as “spontaneous” because they can proceed to products without outside intervention. Energy Course of Reaction

  47. Formulas * DEforward = Eproducts – Ereactants * DEreverse = Ereactants – E products * Ea = energy of activated complex – energy of reactants * Ea’ = energy of activated complex – energy of products

  48. Practice #1 • For the energy diagram provided, label the reactants, products, DE, Ea, and Ea’. Also, determine the values of DE for the forward and reverse reactions, and the values of Ea and Ea’. 80 60 40 20 0 -20 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Ea’ - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Energy (kJ/mol) products Ea DE - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - reactants Forward Reverse

  49. Practice #1 Continued *DEforward = Eproducts – Ereactants = 55 kJ/mol – (-20 kJ/mol) = 75 kJ/mol *DEreverse = Ereactants – Eproducts = -20 kJ/mol – 55 kJ/mol = -75 kJ/mol 80 60 40 20 0 -20 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Ea’ - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Energy (kJ/mol) products Ea DE - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - reactants Forward Reverse

  50. Practice #1 Continued * Ea = energy of activated complex – energy of reactants = 80 kJ/mol – (-20 kJ/mol) = 100 kJ/mol * Ea’ = energy of activated complex – energy of products = 80 kJ/mol – 55 kJ/mol = 25 kJ/mol 80 60 40 20 0 -20 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Ea’ - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Energy (kJ/mol) products Ea DE - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - reactants Forward Reverse

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