1 / 24

15.1

15.1. A  B K=[B]=Products [A] Reactants K c - concentration K eq - equilibrium K a - acid K b - base K sp - solubility product K p - pressure K i - ionization K w - water. ibrium. 15.1. Haber Process N 2(g) + 3H 2(g) ↔ 2 NH 3(g) K = [NH 3 ] 2

bhall
Download Presentation

15.1

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. 15.1 • A  B K=[B]=Products [A] Reactants • Kc- concentration • Keq- equilibrium • Ka- acid • Kb- base • Ksp- solubility product • Kp- pressure • Ki- ionization • Kw- water ibrium

  2. 15.1 • Haber Process N2(g) + 3H2(g)↔ 2 NH3(g) K = [NH3]2 [N2] [H2]3

  3. 15.1 Equilibrium Constant Fig 15.6 • The equilibrium cond. can be reached from either direction [conc] H2 NH3 N2 time time

  4. 15.2 • Law of Mass- Action aA + bB pP + qQ Kc= [P]p [Q]q [A]a [B]b Called equilibrium expressions

  5. 15.2 • Kp = Kc(RT) n n= moles gas product – moles gas reactant R= .0821 atm-l mol-K • K > 1 product is favored • K < 1 reactant is favored • K = forward reaction • 1/K= backwards reaction

  6. 15.3 • Heterogeneous Equilibriums CaCO3(s)  CaO(s) + CO2 (g) K= [CO2] • The density of a pure liquid or solid is a CONSTANT

  7. 15.5 Calc. Equilibrium Constants H2 + I2 2 HI I.001 .0020 C -.935x10-3-.935x10-3 +1.87x10-3 E .065x10-3 1.065x01-31.87x10-3M Kc= [HI]2 = (1.87x10-3)2 =51 [H2][I2] (.065x10-3)(1.065x10-3)

  8. Example; what if we start with….. • N2 + 3H2 2 NH3 1.0 M 2.0M 2.0M Q= [NH3]2 = (2)2= .500 [N2][H2]3 (1)(2)3 Keq = .105 Q=.500 Rxn goes toward reactant

  9. 15.3 • If K > Q goes toward product • If K = Q at equilibrium • If K < Q goes toward reactant Q= reactant quotient

  10. 15.4 (psssst, he was FRENCH) • LeChatelier’s Principle If a system at equilibrium is disturbed by a change in temperature, or concentration of one compound the system will shift to counteract the effect of the disturbance.

  11. N2 + 3H2 2NH3 • What happens when we add hydrogen? • Reaction shifts to the right.

  12. clear brown N2O4  2NO2

  13. http://teachertube.com/viewVideo.php?video_id=100858

  14. 15.4 • Sample 15.12 N2O4 (g) 2 NO2 (g) H= +58.0KJ • Add N2O4 • Remove NO2 • Increase total pressure by adding N2 or any noble gas • Increase volume (Decrease pressure) • Decrease temp • Heat on left- shift to side with more heat Shift to right Shift to right No effect Shift to right (Shift to side with more moles of gas) Shift to left

  15. 15.4 CaCO3(s) CaO (s) + CO2 (g) + heat • Add CO2 • Remove some CaCO3 • Remove all CaCO3 • Increase pressure • Decrease temperature Shift to left No effect Shift to left Shift to left Shift to right

  16. A 1-liter container initially holds .015 mol of H2 and .02 mol of I2 at 721K. What are the concentrations of H2, I2, and HI at equilibrium H2 + I2 2HI I .015M .020M 0M(Start conc.) C -x -x +2x (change) E (.015M-x) (.020M-x) 2x Got it? “Now what?” you ask? Well……… Kc = 50.5

  17. Put it in the expression silly! Kc= [HI]2 = 50.5 = (2x)2 [H2] [I2] (.015-x) (.02-x) Simplify!!  50.5= 4x2 (x2- .035x+ .0003) 4x2= 50.5x2 - 1.77x + .0152 0= 46.5x2 – 1.77x + .0152 0= x2 – 0.0381x + .000327 “Now what?” you ask again? Well you have two options!….

  18. Option 1: Quadratic equation! (and you thought you would never use it again….. tsk tsk) x= -b + b2 – 4ac go ahead, do all that looooong math! 2a You’ll get… x= .025 and x= .0131 [H2]= .015 – x= 0.002 M [I2]= .020 – x= 0.007 M [HI]= 2x = 0.026 M Congrats you’re done!

  19. Option #2 • Get a program called “quad” for your calculator. • Plug in a = 1 b = -0.0381 c = 0.000327 And you get your two roots.

  20. More Chapter 15 notes • We have done kinetics- rates (I am so sorry) • Along the rxn.pathway equilibrium is one point : The point in a reaction at which rate of forward rxn. Equals the rate of reverse rxn. • Equilibrium does not mean the quantities of reactants and products are equal • Measuring equilibrium- how recognized • Concentration- but we don’t have a molar meter  • Can measure: • color  absorption • pressure  gasses • mass • pH

  21. Requirements- • Closed System (HEAT IT!! YOU KNOW YOU WANT TO) • Water boiling on a stove will never hit equilibrium • Constant Temperature • If temperature changes the position of equilibrium changes. • Steady States • Something that stays the same without being in equilibrium • Ex. School population- even though the number in each class may be the same, this is steady state not equilibrium. The seniors do not come back as freshman. • New in – New out • Ex. Level of Lake Ontario. Flow system. Not equilibrium. New water flows in, new water flows out. New mass in, new mass out.

  22. The Great Basketball of Equilibrium Examples of Equilibrium(because you really want to know) • EQUILIBRIUM • Basketball game: Number of people playing and number of people on the bench is always the same • # playing = #on bench • If a new player comes in, one on floor must come out • YES- this IS equilibrium • EQUILIBRIUM • A capped bottle of water. • Rate of evaporation equals rateof condensation, • No new mass in, no old mass out

More Related