Chemical Equilibrium CHAPTER 15 Chemistry: The Molecular Nature of Matter, 6 th edition - PowerPoint PPT Presentation

slide1 n.
Download
Skip this Video
Loading SlideShow in 5 Seconds..
Chemical Equilibrium CHAPTER 15 Chemistry: The Molecular Nature of Matter, 6 th edition PowerPoint Presentation
Download Presentation
Chemical Equilibrium CHAPTER 15 Chemistry: The Molecular Nature of Matter, 6 th edition

play fullscreen
1 / 98
Chemical Equilibrium CHAPTER 15 Chemistry: The Molecular Nature of Matter, 6 th edition
360 Views
Download Presentation
amber
Download Presentation

Chemical Equilibrium CHAPTER 15 Chemistry: The Molecular Nature of Matter, 6 th edition

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

  1. Chemical Equilibrium CHAPTER 15 Chemistry: The Molecular Nature of Matter, 6th edition By Jesperson, Brady, & Hyslop

  2. CHAPTER 15 Chemical Equilibrium • Learning Objectives: • Reversible Reactions and Equilibrium • Writing Equilibrium Expressions and the Equilibrium Constant (K) • Reaction Quotient (Q) • KcvsKp • ICE Tables • Quadratic Formula vs Simplifying Assumptions • LeChatelier’sPrinciple Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  3. CHAPTER 15 Chemical Equilibrium Lecture Road Map: Dynamic Equilibrium Equilibrium Laws Equilibrium Constant Le Chatelier’s Principle Calculating Equilibrium Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  4. CHAPTER 15 Chemical Equilibrium Dynamic Equilibrium Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  5. Dynamic Eq Equilibrium • Chemical equilibrium exists when • Rates of forward and reverse reactions are equal • Reaction appears to stop • Concentration of reactants and products do not change over time • Remain constant • Both forward and reverse reaction never cease • Equilibrium signified by double arrows ( ) Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  6. Dynamic Eq Equilibrium N2O4 2 NO2 • Initially have only N2O4 • Only forward reaction • As N2O4 reacts NO2 forms • As NO2forms • Reverse reaction begins to occur • NO2 collide more frequently as concentration of NO2increases • Eventually, equilibrium is reached • Concentration of N2O4 does not change • Concentration of NO2 does not change Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  7. Dynamic Eq Equilibrium Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  8. Dynamic Eq Equilibrium N2O4 2NO2 Closed system • Equilibrium can be reached from either direction • Independent of whether it starts with “reactants” or “products” • Always have the same composition at equilibrium under same conditions Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  9. Dynamic Eq Equilibrium Reactants Equilibrium Products N2O42NO2 Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  10. Dynamic Eq Mass Action Expression • Simple relationship among [reactants] and [products] for any chemical system at equilibrium • Called the mass action expression • Derived from thermodynamics • Forward reaction: AB • Reverse reaction: AB • At equilibrium: AB Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  11. Dynamic Eq Reaction Quotient • Uses stoichiometric coefficients as exponent for each reactant • For reaction:aA + bBcC + dD Reaction quotient • Numerical value of mass action expression • Equals “Q ” at any time, and • Equals “K ” only when reaction is known to be at equilibrium Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  12. Ex. 1 H2(g) + I2(g)2HI(g) 440˚C

  13. Ex. 1 H2(g) + I2(g) 2HI(g)440 ˚C

  14. Mass Action Expression = same for all data sets at equilibrium Average = 49.5

  15. Group Problem Write mass action expressions for the following: • 2NO2(g) N2O4(g) • 2CO(g) + O2(g) 2CO2(g) Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  16. Group Problem Which of the following is the correct mass action expression for the reaction: Cu2+(aq) + 4NH3(aq) [Cu(NH3)42+](aq)? Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  17. CHAPTER 15 Chemical Equilibrium Equilibrium Laws Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  18. Equilibrium Equilibrium Laws • For reaction H2(g) + I2(g) 2HI(g) at 440 ˚C at equilibrium write the following equilibrium law • Equilibrium constant = Kc = constant at given T • Use Kc since usually working with concentrations in mol/L • For chemical equilibrium to exist in reaction mixture, reaction quotient Q must be equal to equilibrium constant, Kc Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  19. Equilibrium Predicting Equilibrium Laws For general chemical reaction: • dD + eEfF + gG • Where D, E, F, and G represent chemical formulas • d, e, f, and g are coefficients • Mass action expression is • Note: Exponents in mass action expression are stoichiometric coefficients in balanced equation. • Equilibrium lawis: Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  20. Equilibrium Predicting Equilibrium Laws • Only concentrations that satisfy this equation are equilibrium concentrations • Numerator • Multiply concentration of products raised to their stoichiometric coefficients • Denominator • Multiply concentration reactants raised to their stoichiometric coefficients is scientists’ convention Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  21. Equilibrium Example 3H2(g) + N2(g) 2NH3(g) Kc= 4.26 × 108 at 25 °C What is equilibrium law? Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  22. Equilibrium Operations Various operations can be performed on equilibrium expressions 1. When direction of equation is reversed, new equilibrium constant is reciprocal of original A + BC + D C +DA + B Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  23. Equilibrium Operations 1. When direction of equation is reversed, new equilibrium constant is reciprocal of original 3H2(g) + N2(g) 2 NH3(g) at 25˚C 2NH3(g) 3H2(g) + N2(g) at 25 ˚C Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  24. Equilibrium Operations 2. When coefficients in equation are multiplied by a factor, equilibrium constant is raised to a power equal to that factor. A + BC + D 3A + 3B 3C + 3D Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  25. Equilibrium Operations • When coefficients in equation are multiplied by factor, equilibrium constant is raised to power equal to that factor 3H2(g) + N2(g) 2NH3(g) at 25 ˚C Multiply by 3 9H2(g) + 3N2(g) 6NH3(g) Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  26. Equilibrium Operations 3. When chemical equilibria are added, their equilibrium constants are multiplied A + BC + D C + EF + G A + B + ED + F + G Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  27. Equilibrium Operations 3. When chemical equilibria are added, their equilibrium constants are multiplied Therefore Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  28. Group Problem For: N2(g)+ 3H2(g) 2NH3(g)Kc= 500 at a particular temperature. What would be Kc for following? • 2NH3(g) N2(g)+ 3H2(g) • 1/2N2(g)+ 3/2H2(g) NH3(g) 0.002 22.4 Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  29. CHAPTER 15 Chemical Equilibrium Equilibrium Constant Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  30. Equilibrium Constant Kc • Most often Kcis expressed in terms of a ratio of concentrations of products and reactants as shown on previous slides • Sometimes partial pressures, in atmospheres, may be used in place of concentrations Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  31. Equilibrium Kp • Based on reactions in which all substances are gaseous • Gas quantities are expressed in atmospheres in mass action expression • Use partial pressures for each gas in place of concentrations e.g. N2(g)+ 3H2(g)2NH3(g) Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  32. Equilibrium Relationship between Kp and Kc • Start with ideal gas law PV = nRT • Rearranging gives • Substituting P/RT for molar concentration into Kc results in pressure-based formula • ∆n = moles of gas in product – moles of gas in reactant Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  33. Group Problem Consider the reaction: 2NO2(g) N2O4(g) If Kp = 0.480 for the reaction at 25 ˚C, what is value of Kc at same temperature? n = nproducts – nreactants= 1 – 2 = –1 Kc = 11.7 Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  34. Group Problem Consider the reaction A(g) + 2B(g) 4C(g) If the Kc for the reaction is 0.99 at 25 ˚C, what would be the Kp? • 0.99 • 2.0 • 24 • 2400 • None of these Δn = (4 – 3) = 1 Kp = Kc(RT)Δn Kp= 0.99 × (0.082057 × 298.15)1 Kp = 24 Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  35. Equilibrium Homogeneous and Hetergeneous Homogeneous reaction/equilibrium • All reactants and products in same phase • Can mix freely Heterogeneous reaction/equilibrium • Reactants and products in different phases • Can’t mix freely • Solutions are expressed in M • Gases are expressed in M • Governed by Kc Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  36. Equilibrium Heterogeneous 2NaHCO3(s) Na2CO3(s) + H2O(g) + CO2(g) • Equilibrium Law = • Can write in simpler form • For any pure liquid or solid, ratio of moles to volume of substance (M) is constant • e.g. 1 mol NaHCO3 occupies 38.9 cm3 2 mol NaHCO3 occupies 77.8 cm3 Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  37. Equilibrium Heterogeneous 2NaHCO3(s) Na2CO3(s) + H2O(g) + CO2(g) • Ratio (n/V ) or M of NaHCO3 is constant (25.7 mol/L) regardless of sample size • Likewise can show that molar concentration of Na2CO3 solid is constant regardless of sample size • So concentrations of pure solids and liquids can be incorporated into equilibrium constant, Kc • Equilibrium law for heterogeneous system written without concentrations of pure solids or liquids Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  38. Equilibrium Heterogeneous Write equilibrium laws for the following: Ag+(aq) + Cl–(aq) AgCl(s) H3PO4(aq) + H2O H3O+(aq) + H2PO4–(aq) Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  39. Interpreting KC • Large K (K >>1) • Means product rich mixture • Reaction goes far toward completion e.g. 2SO2(g) + O2(g) 2SO3(g) Kc = 7.0  1025 at 25 °C

  40. Interpreting KC • Small K (K << 1) • Means reactant rich mixture • Only very small amounts of product formed e.g. H2(g) + Br2(g) 2HBr(g) Kc = 1.4  10–21 at 25 °C

  41. Interpreting KC • K 1 • Means product and reactant concentrations close to equal • Reaction goes only about halfway

  42. Size of K gives measure of how reaction proceeds • K >> 1 [products] >> [reactants] • K = 1 [products] = [reactants] • K << 1 [products] << [reactants]

  43. CHAPTER 15 Chemical Equilibrium Le Chatelier’s Principle Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  44. Le Chatelier Definition • Equilibrium positions • Combination of concentrations that allow Q = K • Infinite number of possible equilibrium positions • Le Châtelier’s principle • System at equilibrium (Q = K) when upset by disturbance (Q ≠ K) will shift to offset stress • System said to “shift to right” when forward reaction is dominant (Q < K) • System said to “shift to left” when reverse direction is dominant (Q > K) Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  45. Le Chatelier Q & K Relationships • Q = K reaction at equilibrium • Q < K reactants go to products • Too many reactants • Must convert some reactant to product to move reaction toward equilibrium • Q > K products go to reactants • Too many products • Must convert some product to reactant to move reaction toward equilibrium Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  46. Le Chatelier Change in Concentration Cu(H2O)42+(aq) + 4Cl–(aq) CuCl42–(aq) + 4H2O blueyellow • Equilibrium mixture is blue-green • Add excess Cl– (conc. HCl) • Equilibrium shifts to products • Makes more yellow CuCl42– • Solution becomes green Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  47. Le Chatelier Change in Concentration Cu(H2O)42+(aq) + 4Cl–(aq) CuCl42–(aq) + 4H2O blueyellow • Add Ag+ • Removes Cl–: Ag+(aq) + Cl–(aq) AgCl(s) • Equilibrium shifts to reactants • Makes more blue Cu(H2O)42+ • Solution becomes increasingly more blue • Add H2O? Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  48. Le Chatelier Change in Concentration: Example For the reaction 2SO2(g) + O2(g) 2SO3(g) Kc = 2.4 × 10–3 at 700 °C Which direction will the reaction move if 0.125 moles of O2 is added to an equilibrium mixture? • Towards the products • Towards the reactants • No change will occur Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  49. Le Chatelier Change in Concentration • When changing concentrations of reactants or products • Equilibrium shifts to remove reactants or products that have been added • Equilibrium shifts to replace reactants or products that have been removed Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  50. Le Chatelier Change in Pressure or Volume • Consider gaseous system at constant T and n 3H2(g) + N2(g) 2NH3(g) • If volume is reduced • Expect pressure to increase • To reduce pressure, look at each side of reaction • Which has less moles of gas • Reactants = 3 mol + 1 mol = 4 mol gas • Products = 2 mol gas • Reaction favors products (shifts to right) Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E