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Chemical Equilibrium Part 1

Chemical Equilibrium Part 1. Ch. 15 in Textbook. Fanpop.com. I. Static vs. Dynamic. Generally speaking, an equilibrium is a state of balance. A static equilibrium is one in which there is no motion. Wirednewyork.com. Tutorvista.com.

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Chemical Equilibrium Part 1

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  1. Chemical EquilibriumPart 1 Ch. 15 in Textbook Fanpop.com

  2. I. Static vs. Dynamic • Generally speaking, an equilibrium is a state of balance. • A static equilibrium is one in which there is no motion. Wirednewyork.com Tutorvista.com

  3. A dynamic equilibrium is one in which there is motion despite there being no net change. • A chemical equilibrium is an example of a dynamic equilibrium. Ene.on.gov.ca

  4. II. Chemical Equilibria • Suppose we have the gaseous reactants I2 and H2. They undergo a synthesis reaction to form HI: • H2 + I2→ 2HI • As HI accumulates, some molecules have enough energy to decompose to H2 and I2: • 2HI → H2 + I2 Coolchaser.com

  5. As this process continues, eventually the rate of the forward reaction equals the rate of the reverse reaction. Tutorvista.com Link

  6. The final equilibrium mixture will contain both reactants and products. Tutorvista.com

  7. Although the amounts are NOT necessarily equal, the amounts must eventually remain constant since the forward and reverse reactions are occurring simultaneously and at the same rate. Mmsphyschem.com

  8. Although there is no noticeable net change in reactants or products, the reaction is still proceeding in the forward and reverse directions, making chemical equilibrium a dynamic equilibrium. Freefoto.com HW: 15.2 Tutorial

  9. III. Kinetics Flashback!!!! • Given: A  B • Assume that the forward and reverse reactions are both elementary steps. • Rate (forward) = kf [A] • Rate (reverse) = kr [B] • At equilibrium: kf [A] = kr [B] Motivatedphotos.com

  10. Rearranging: [B]/[A] = kf/kr = constant • What is the meaning of this?! • At a given temp. the products and reactants will ALWAYS be in the same ratio at equilibrium, no matter the starting point… Blog.al.com HW: 15.3

  11. IV. The Law of Mass Action • This proportion can be represented by the equilibrium expression. • The Law of Mass Action states that the equilibrium expression depends on the equilibrium concentrations of reactants and products. Marcvallee.wordpress.com

  12. For the reaction: aA + bB  pP + qQ the equilibrium expression is written as: K = [P]p [Q]q [A]a [B]b where Kis the equilibrium constant. Oudaily.com

  13. Ex) What is the equilibrium expression for the synthesis of hydrogen iodide? Threadless.com HW: 15.6, 15.20

  14. In KINETICS, the rate law did NOT depend on the stoichiometry of the net equation, only on the rate-determining step. • In EQUILIBRIUM, the equilibrium expression DOES depend on the stoichiometry of the net equation. Yes, typo Efuse.com

  15. The Equilibrium Constant, K,…A) Dependence • …does not depend on the reaction mechanism. • …does not depend on the initial concentrations of reactants and products. • …depends on the equilibrium concentrations of reactants and products. • ..also depends on temperature! (more on that later…) • We generally omit the final units. Tampabayrun.com

  16. B) Magnitude • Since products are divided by reactants in the expression: • A larger K value (>>1) means that products dominate the final reaction mixture and we say “the equilibrium lies to the right.” • A smaller K value (<<1) means that reactants dominate the final reaction mixture and we say “the equilibrium lies to the left.” Balthaus.org HW: 15.10

  17. C) Kc • This is the constant when concentrations are expressed in molarity. • It is the most common version of K. Dummidumwit.wordpress.com

  18. D) Kp • This is the constant when concentrations (of gases) are expressed in terms of partial pressures (atm). • K = (PP)p (PQ)q (PA)a (PB)b Lankapeacewatch.com

  19. E) Relationship between Kc and Kp Note: My notes are WAY better…this PowerPoint slide doesn’t even have a stupid picture in it…or a picture of another chemistry PowerPoint slide…of another PowerPoint Slide…of another PowerPoint slide…ad infinitum. Files.chem.vt.edu

  20. E) Relationship between Kc and Kp (fo’ real dis time) • PV = nRT • P = (n/V) RT • For substance A: PA = [A] RT • Do this for all partial pressures in an equilibrium mixture and we get: Kp = Kc (RT)Δn where Δn is the moles of gaseous products - the moles of gaseous reactants. Link

  21. Ex) Given: 2SO3(g)  2SO2(g) + O2(g) • Write the Kp expression. • Calculate Kp if Kc = 4.08 x 10-3 at 1000 K. W3.org HW: 15.8 (a) & (b), 15.12

  22. F) Direction • Equilibrium can be reached from any direction (all reactants or all products or any mixture of both). • Kc (forward) = 1/Kc (reverse) Fotolia.com HW: 15.16

  23. Breakdancing Baby Break

  24. VI. Heterogeneous Equilibria • These involve one or more reactants or products present in a different phase. • Many equilibria do not involve dissolved species or gases. • The concentration of a pure solid or liquid is constant and therefore does not appear in the equilibrium expression. Erroraccessdenied.com

  25. Ex) 3Fe(s) + 4H2O(g)  Fe3O4(s) + 4H2(g) • Write the Kp expression. Pig Iron from Bbc.co.uk

  26. Thus, only the partial pressures of water and hydrogen affect the equilibrium expression. • Don’t forget though, the solids must be present in order for the equilibrium to be established. Open.salon.com HW: 15.18

  27. VII. Calculating K • Use an ICE chart: Initial, Change, and Equilibrium. • First, determine all initial concentrations. • Second, determine any available equilibrium concentrations. • Third, use the reaction stoichiometry to determine the change in concentration from the initial to equilibrium. Burnblog.burningman.com

  28. Fourth, determine all equilibrium concentrations. • Fifth, plug and chug into equilibrium expression and find K. Nitrocotton.com

  29. Ex) 2SO3 (g)  2SO2(g) + O2(g) • A vessel at 1000 K contains 6.09 x 10-3M SO3. At equilibrium, the SO3 concentration is 2.44 x 10-3 M. What is the value of Kc? Portlandsentinel.com HW: 15.24, 15.26

  30. VIII. The Reaction Quotient (Q) • Given the initial concentrations of reactants and products, we can determine the direction the reaction will proceed in. • Substitute concentrations into the reaction quotient expression, which is the same as K except NOT at equilibrium. Mommylounge.com

  31. Erie.gov • K uses equilibrium concentrations!!!!!! • Q uses initial concentrations!!!!!! K-Dubs, 6th grade

  32. If Q>K, then there is an excess of products which will react to form reactants to establish equilibrium (shift left). • If Q<K, then there is an excess of reactants which will react to form products to establish equilibrium (shift right). • If Q=K, equilibrium is established. Ajobi.net HW: 15.28

  33. IX. Calculating Equilibrium Concentrations • Ex 1) At 500 K the value of Kp is 0.497 for the following reaction: PCl5(g)  PCl3(g) + Cl2(g) If the partial pressure of PCl5 is 0.860 atm and the partial pressure of PCl3 is 0.350 atm at equilibrium, what is the partial pressure of Cl2 in the mixture? Blogs.nature.com

  34. Ex 2) Using the equilibrium from Ex. 1, what are the equilibrium partial pressures of all species if PCl5 has an initial pressure of 1.66 atm? Desperatelyseekingsuddenlysusan. wordpress.com HW: 15.34, 15.36, 15.38, 15.40

  35. TO BE CONTINUED… Intlxpatr.wordpress.com

  36. Equilibrium Part 2 Ch. 15 in Textbook Bethedream.com

  37. Le Chatelier’s PrincipleA) Stress • Once an equilibrium is established, the forward and reverse reactions will proceed at equal rates and the amounts of reactants and products will not change. • Equilibrium will continue indefinitely unless conditions are somehow changed to disrupt it. • A disturbance to the equilibrium is called a stress. Wolfescape.com

  38. B) The Principle • When you deal with stress in your own life (e.g. taking this class) you do something to rid yourself of the stress and re-establish your personal equilibrium. • Similarly, French industrial chemist Henri-Louis Le Chatelier stated that when a stress is applied to a chemical equilibrium, the equilibrium will shift its position to eliminate the disturbance. Biocrawler.com

  39. C) Changes in Concentration • Adding a reactant or a product to an equilibrium mixture causes the equilibrium to speed up in the direction that will remove the excess. • This shift is only temporary and a new equilibrium is established with new equilibrium concentrations (although no change in K). Chem.ufl.edu

  40. Similarly, removing a reactant or product from an equilibrium mixture causes the equilibrium to speed up in the direction that will make up for the deficit. Stevenwolffinearts.com

  41. How can we maximize our ammonia output in the Haber Process? N2(g) + 3H2(g) ↔ 2NH3(g) Zazzle.com Link

  42. D) Changes In Volume/Pressure • The pressure may be increased by reducing the container volume or the addition of a non-reacting gas. • The equilibrium will speed up in the direction that will reduce the pressure through the production of fewer moles of gas. En.wikibooks.org

  43. Similarly, the pressure may be decreased by increasing the container volume or the removal of a non-reacting gas. • The equilibrium will speed up in the direction that will increase the pressure through the production of greater moles of gas. • This shift is only temporary and a new equilibrium is established with new equilibrium concentrations (although no change in K). Iforgottocitethis.com

  44. How can we maximize our ammonia output in the Haber Process? N2(g) + 3H2(g) ↔ 2NH3(g) Cafepress.com Link

  45. E) Changes in Temperature • Increasing the temperature results in an excess of heat, causing the equilibrium to speed up in the direction that removes the heat (endothermic). C-p-p.co.uk

  46. Similarly, lowering the temperature results in a deficit of heat, causing the equilibrium to speed up in the direction that produces heat (exothermic). Tutorvista.com

  47. Unlike all other stresses, however, the temperature changes the value of K. • In other words, a completely different equilibrium mixture is created at the different temperature. • If a reaction shifts to the right, then K increases. • If a reaction shifts to the left, then K decreases. Justhavingfunaroundthe house.com

  48. How can we maximize our ammonia output in the Haber Process? N2(g) + 3H2(g) ↔ 2NH3(g) + heat Wow.com Link

  49. F) Adding Catalysts • A catalyst lowers the activation energy for both the forward and the reverse reactions. • A catalyst therefore speeds up the forward and reverse reactions equally. • A catalyst speeds up the rate at which equilibrium is achieved, but the equilibrium mixture and K remain exactly the same. En.wikivisual.com

  50. Haber found that iron mixed with metal oxides allowed the Haber Process to run at sufficiently low temperatures. Dmotiv8.blogspot.com HW: 15.44, 15.46

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