1 / 34

LeChâtelier’s Principle

LeChâtelier’s Principle. more equilibrium concepts. LeChâtelier’s Principle. states that when a stress is applied to a system at equilibrium, the system will respond in a manner that attempts to bring the system back to equilibrium…stresses are: Concentration Pressure Temperature.

fuller
Download Presentation

LeChâtelier’s Principle

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. LeChâtelier’s Principle more equilibrium concepts

  2. LeChâtelier’s Principle • states that when a stress is applied to a system at equilibrium, the system will respond in a manner that attempts to bring the system back to equilibrium…stresses are: • Concentration • Pressure • Temperature

  3. Concentration • If you add more reactant, then… the rxn will shift to the right to use the excess reactant. • If you add more product, then… the rxn will shift to the left to use the excess product. • If you remove reactant, then… • If you remove product, then…

  4. Pressure • 3 means of affecting pressure of gaseous systems • Concentration • Volume of container • Addition of an inert gas

  5. Pressure—concentration’s effect • Same concept as concentration change

  6. Pressure—volume’s effect • If you decrease the volume of the container, then the pressure will increase and the system will adjust by… shifting to the side with the fewer number of moles in an effort to bring the pressure back down to what it was at equilibrium

  7. Pressure—volume’s effect • If you increase the volume of the container, then the pressure will decrease and the system will adjust by… shifting to the side with the greater number of moles in an effort to bring the pressure back up to what it was at equilibrium

  8. Pressure—volume’s effect • If you increase or decrease the volume of the container for a reaction who has equal number of moles on reactant and product side, then… the system cannot make the appropriate changes and the rxn will not be able to reach equlibrium

  9. Pressure—inert gas’s effect • If you add an inert gas to a system at equilibrium, then… there will be no shift because the inert gas only increases the total pressure of the system (think Dalton’s Law) and has no effect on the partial pressures of the reactants or products

  10. Temperature • effect depends on whether the rxn is endothermic or exothermic • endothermic—absorbs heat energy, thus the energy appears as a reactant or the change in enthalpy is positive • exothermic—releases heat energy, thus the energy appears as a product or the change in enthalpy is negative

  11. Temperature—Endothermic Rxn • If you add more heat, then… the rxn will shift to the right to use the excess heat. (think of the heat as a reactant) • If you remove heat, then… the rxn will shift to the left to replenish the missing heat.

  12. Temperature—Exothermic Rxn • If you add more heat, then… the rxn will shift to the left to use the excess heat. (think of the heat as a product) • If you remove heat, then… the rxn will shift to the right to replenish the missing heat.

  13. Le Châtelier Practice • Consider the reaction: • N2(g) + 3H2(g)  2NH3(g) + 33.3kJ • Explain the direction of shift it will • experience when the following stresses are • applied: • The system is heated. • Its container is squeezed. • Ammonia is added to the system. • Neon is added to the system.

  14. Le Châtelier Practice • Consider the reaction: • 2SO3(g) + 33.3kJ 2SO2(g) + O2(g) • Explain the direction of shift it will • experience when the following stresses are • applied: • The system is cooled. • Its container expands. • Sulfur trioxide is added to the system. • Argon is added to the system.

  15. Solubility Product Constant applying equilibrium stuff to salts

  16. Ksp • Unlike NaCl, not all salts dissolve completely in aqueous solutions. • The level to which a salt dissolves is expressed as its solubility product constant and is represented by Ksp (similar to K) • Remember that only gaseous and aqueous thingies can be represented in K expressions.

  17. Ksp • So, in a salt dissociation rxn, only the products are aqueous. • The reactant (or salt) is solid. • Consider the salt, silver hydroxide. Write its formula. AgOH

  18. Ksp • Now, write its dissociation into its ions… AgOH (s)  Ag1+(aq) + OH1-(aq) • Make sure you pay attention to • the phases. • Now write the Ksp expression.

  19. Ksp Ksp= [Ag1+][OH1-] • Note that the AgOH(s) is absent because it is a solid. • Also note that the power of the Ag1+ and OH1- are both one because the dissociation only yielded one of each ion.

  20. Ksp • This expression allows you to work three types of problems: • determine the concentrations of ions that have dissolved • determine the solubility of the salt (it will always equal x in your equilibrium chart) • determine the Ksp of a salt

  21. Ksp Practice Problem #1 Determine the concentration of each ion if the Ksp of calcium fluoride is 4.0 x 10-11.

  22. Ksp Practice Problem #1 Step 1—write the dissociation of salt. CaF2(s)  Ca2+(aq) + 2F1-(aq) Recognize that the mol:mol for the ions is 1:2. This is very important info that you will use in the problem, so make certain that the formula for the salt is correct!

  23. Ksp Practice Problem #1 Step 2—make a dissociation chart like you did for the other equilibrium rxns. Remember…only (g) & (aq)!! [CaF2] [Ca2+] [F1-] i -- 0 0 Δ -- +x +2x eq -- x 2x

  24. Ksp Practice Problem #1 Step 3—write your Kspexpression and plug in your stuff Ksp= [Ca2+][F1-]2 Note that the mol:mol is reflected in the power as well as in your eq line of your chart.

  25. Ksp Practice Problem #1 Step 3—continued Ksp= [Ca2+][F1-]2 4.0 x 10-11= [x][2x]2 4.0 x 10-11= 4x3 1.0 x 10-11= x3 2.15 x 10-4= x

  26. Ksp Practice Problem #1 Step 4—use your value for x and the eq line of your chart to figure out the concentrations of the ions. [Ca2+]= 2.15 x 10-4 M [F1-]= 2(2.15 x 10-4)= 4.30 x 10-4 M

  27. Ksp Practice Problem #2 Determine the solubility of calcium fluoride from your work in Problem #1.

  28. Ksp Practice Problem #2 Step 1—You’ve already done all of the work for this problem. Since the mol:mol:mol for the salt to its ions is 1:1:2, the solubility of the salt in this case is equal to the concentration of the calcium ion. Just remember that the solubility of the salt will always equal x from your eq line of your chart

  29. Ksp Practice Problem #2 Step 1—Thus, the answer to this problem is 2.15 x 10-4 M for the solubility of the calcium fluoride.

  30. Ksp Practice Problem #3 Determine the value of the solubility product constant of bismuth sulfide which has a solubility of 1.0 x 10-15 M.

  31. Ksp Practice Problem #3 Step 1—write the dissociation of salt and the Ksp expression. Bi2S3(s)  2Bi3+(aq) + 3S2-(aq) Ksp= [Bi3+]2[S2-]3 Ksp= [2x]2[3x]3 consolidated Ksp= 108x5

  32. Ksp Practice Problem #3 Step 2—Since your know the solubility (or x) of the salt, you can deduce the concentrations of the ions from the mol:mol of the ions. Therefore, [Bi3+] will be two times the solubility, and [S2-] will be three times the solubility.

  33. Ksp Practice Problem #3 Step 3—Plug your solubility into the expression as x and solve for Ksp. Ksp=[2(1.0 x 10-15)]2[3(1.0 x 10-15)]3 Or since, Ksp= 108x5… 108(1.0 x 10-15)5

  34. Ksp Practice Problem #3 Thus, the Ksp should be 1.08 x 10-73 M

More Related