1 / 45

Ch. 15 EQUILIBRIUM & Le CHÂTELIER

Ch. 15 EQUILIBRIUM & Le CHÂTELIER. EQUILIBRIUM >Constant, Kc >Calculations, I.C.E. Tables RXN QUOTIENT, Q c >Compare K - Q Le CHÂTELIER’S PRINCIPLE. KINETICS RELATIONSHIP USE [ ] & P. EQUILIBRIUM *State of balance

avalon
Download Presentation

Ch. 15 EQUILIBRIUM & Le CHÂTELIER

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. Ch. 15 EQUILIBRIUM&Le CHÂTELIER EQUILIBRIUM >Constant, Kc >Calculations, I.C.E. Tables RXN QUOTIENT, Qc >Compare K - Q Le CHÂTELIER’S PRINCIPLE KINETICS RELATIONSHIP USE [ ] & P

  2. EQUILIBRIUM *State of balance *2 = & opp opposing forces occur @ same rate (still reacting) Chemistry *fwd & rev rxns @ same rate *no  in concentratons N2O4(g) 2 NO2 (g) colorless brown REVERSIBLE RXN rf = kf[N2O4] rr = kr[NO2]2

  3. @ EQUILIBRIUM set = & rearrange rf = kf[N2O4] rr = kr[NO2]2 kf[N2O4] = kr[NO2]2 [NO2]2/[N2O4] = kf/kr = a constant; equilibrium constant; Kc “unitless” HABER PROCESS 3 H2(g) + N2(g) 2 NH3 (g)

  4. LAW MASS ACTION aA + bB  cC + dD Equilibrium Expression Kc depends on………… not ……….. nature of rxn how (mechanism) value not depend on……… & …….. reactant product amounts so, depends only on .… & ….. rxn Temp

  5. Le Châtelier Equilibrium: balance between 2 opposing reactions How sensitive is this balance to changes in conditions? What can be done to change the equilibrium state? If pdts. be withdrawn continuously, then reacting system can be kept constantly off-balance More reactants used, more pdts. formed Most useful if 1 pdt. can 1. escape as gas 2. Condensed or frozen from gas phase as solid or liquid 3. Washed out of gas mixture by liquid spray which is especially soluble 4. Precipitated from gas or solution Forward Rxn.: forward direction; reaction favored reactants to pdts. Reverse Rxn.: reverse direction; reaction favored pdts. to reactants

  6. CaO (s) + 3 C (s) CaC2(s) + CO (g) TiCl4(g) + O2(g) TiO2(s) + 2 Cl2(g) Production of calcium carbide Remove CO: reaction tipped toward CaC2 formation Production of titanium dioxide TiO2 separates from gases as fine powdered solid thus, rxn. kept moving in forward direction

  7. CH3COOH + HOCH2CH3 CH3COOCH2CH3 + H20 N2(g) + 3 H2(g) 2 NH3(g) Synthesize Rxn. Remove H2O: forward rxn. favored Production of ammonia NH3 more soluble in water than H2 or N2 Wash out NH3 out of equilibrium mixture

  8. H2(g) + I2(g) 2 HI (g) + heat (givenoff) Temperature Equilibrium usually temp. dependent Increase temp. both forward/reverse rxns. speed up. Why? molecules move faster, increase molecular collisions Adding heat shifts rxn. to left Pressure Rxn shifts in direction of fewer gas molecules w/ P increase Looking at rxn above: 1. Which direction would rxn favor with an increase in P? 2. If rxn were at a temp at which iodine was a solid, which direction would rxn favor at an increase in P? total 2 moles gas reactants -------> total 2 moles gas pdts; no change 1 mole gas reactant -------> 2 moles gas pdts; left (reverse) favored

  9. Catalyst No effect, but can increase speed which equilibrium is reached Le Châtelier’s Principle If an external stress is applied to a system at chemical equilibrium, then the equilibrium pt. will change in such a way as to counteract (alleviate) the effects of that stress. The amounts of reactants & pdts will shift in such a manner as to minimize the stress

  10. Problem Set #1 1. Write Kc for: CH4(g) + H2S(g) CS2 (g) + H2(g) 2. Write Kc for: Al(s) + HCl(aq) AlCl3 (aq) + H2(g)

  11. Use Partial Pressure - atm 760 mmHg(torr) = 1 atm Kc -----> Kp HABER PROCESS 3 H2(g) + N2(g) 2 NH3 (g) Kc Kp convert using PV = nRT

  12. P = (n/V)RT Relate Kp - Kc Kp = Kc(RT)ngas ngas = ( moles gas pdts) - ( moles gas reacts)

  13. Problem Set #2 1. When does Kc = Kp? If ever!! 2. CH4(g) + H2S(g) CS2(g) + H2(g)Kc = 1.3*10-2 @ 475oC a) find ngas b) write Kp expression c) find Kp • when (sum moles gas pdts) = (sum moles gas reacts), then ngas = 0 • a) 1 CH4(g) + 2 H2S(g) 1 CS2(g) + 4 H2(g)ngas = 5 – 3 = 2 • b) Kp = (0.013)[(0.0821)(748)]2 • c) Kp = (0.013)(3771.2863) = 49.03

  14. EVALUATE K 1- Kc = 4.56*109 2- Kc = 4.56*10-9 3- Kp = 49 4- Kp = 0.49 Gives info of mixture @ equilibrium K = [pdt]/[react] K >> 1 lies rgt; pdts K = 1 50 - 50; = amts K << 1 lies left; reacts Evaluate List 1- rgt; pdts 2- left; reacts 3- slightly more pdts than reacts 4.  = amts

  15. Figure 15.07ab

  16. EQUILIBRIUM SYSTEMS Systems look at: Reversible RXN Temp P [ ]; 2 special rules

  17. 1: Driving Reversible RXN state: rev rxn @ equilib - sys open; react/pdt allowed to escape - no longer @ equilib due to escaping particle - rxn shift to side that contains escaping particle Used to drive rev rxn to pdts that are wanted Problem Set #2 HABER PROCESS 3 H2(g) + N2(g) 2 NH3 (g) 5. If [H2] is increased a) Equilib shift direction? ____ b) [N2] will incr/decr? _____ c) [NH3] will incr/decr? ____ 6. If [N2] is decreased a) Equilib shift direction? ____ b) [H2] will incr/decr? _____ c) [NH3] will incr/decr? ____ => Decr Incr <= Incr Decr

  18. 2: Temperature Changes state: Energy shown as term in rxn @ equilib - follows Le Châtelier same manner as w/ [ ] - Endo add E, shifts equilib away from E side - Exo remove E, shifts equilib toward E side E added by incr Temp E removed by cooling Energy E added by incr Temp E produced, Temp incr E removed by cooling E used, Temp decr PS#2, cont. N2(g) + O2(g) +90 kJ  2 NO(g) 7. If temp is incr a) Equilib shift direction? ____ b) [O2] will incr/decr? _____ c) [NO] will incr/decr? ____ 8. If [O2] is decreased a) Equilib shift direction? ____ b) [N2] will incr/decr? _____ c) Temp will incr/decr? ____  Decr Incr  Incr Incr

  19. 3: Pressure Changes state: affect sys w/ gases @ equilib - Incr P, shift to side w/ less total gas moles - Decr P, shift to side w/ more total gas moles PS#3 2 NO2(g) N2O4 (g) + E 9. If temp is incr a) Equilib shift direction? ____ b) [NO2] will incr/decr? _____ c) [N2O4] will incr/decr? ____ d) Temp will incr/decr? ____  Incr Decr Decr 10. If volume is decreased a) Equilib shift direction? ____ b) [NO2] will incr/decr? _____ c) [N2O4] will incr/decr? ____ d) Temp will incr/decr? ____  Decr Incr Incr

  20. 4: Concentration Changes 2 Cases - add/remove solid or liquid @ equilib no shift - liquid solvent; add/remove no shift I - Solids [consistent] by density add/remove, no  in solid’s [ ] 11. NaCl(s) NaCl(aq) add salt till soln saturated a) When is equilib reached? b) amt NaCl (s) incr/decr? _____ c) [NaCl(s)] incr/decr? _____ d) [NaCl(aq)] incr/decr? _____ Add more NaCl @ equilib Incr No  No

  21. Liquid Solvents Liq solvent list as term in eqn then +/- solvent no  equlib Subst dissolved in solvent, [liq solv] can . But, in practice nearly always negligible. Solvent at such higher concentration than reaction substs. PS#3, cont 12. Ca(NO3)2(s) + H2O(l) Ca+2(aq) + 2 NO3-1(aq) No   a) Equilib shift direction, add water? ____ b) Equilib shift direction, add nitrate? _____

  22. DIRECTION N2(g) + O2(g) 2 NO(g) Kc = 1*10-30 @ 25oC Write Kc and find for reverse Kc = [NO]2/[[N2][O2]] = 1*10-30 Kc = [[N2][O2]]/[NO]2 = 1/1*10-30 = 1*10+30 Evaluate What is favored in this rxn? What need to  to favor NO? N2 - O2 Incr Temp

  23. PS #4 13. Haber Process @ 300oC, Kp = 4.34*10-3 Find Kp reverse? 14. For each: a) CaCO3(s) CaO(s) + CO2(g)(concen in M & atm) Kc= Kp= b) CO2(g) + H2(g) CO(g) + H2O(l) (concen in M & atm) Kc= Kp= c) Fe(s) + H2O(g) Fe3O4(s) + H2(g) (all concen listed in “atm”) Kc= Kp= 1/(4.34*10-3 ) = 2.3*102 [CO2] [PCO2] [CO]/[[CO2][H2]][PCO]/[[PCO2][PH2]] 344 (PH2)4/(PH2O)4

  24. PS #1- revisit 2. Write Kc for: Al(s) + HCl(aq) AlCl3 (aq) + H2(g) Ignore pure “solids & liquids” Diff. Phases: heterogeneous equilibrium solids: same concen @ given temp; same # mols/L; no DV liquids: same applies therfore, we ignore pure solids & liquids only concerned w/ those that will D [ ] in rxn

  25. SUMMARY 1) Kc fwd is 1/Kc reverse 2) Kc “unitless” 3) Kc = [pdts]x/[reacts]y ignore pure solids - liquids 4) Kp = Kc(RT)ngas 5) Kc depends rxn & Temp 6) K >> 1 K <<<<1 direction - evaluate

  26. Calculate Kc 2 methods I. Know amts @ equilibrium [ ]/P @ spec Temp. ex.Haber process analyze equil @ 472oC mixture: 7.38 atm H2, 1870 torr N2, & NH3; Ptot = 10.00 atm Determine Kp convert: N2 atm; find atm NH3; bal eqn

  27. II. I.C.E. TABLE SET UP -- find equil quantities & K write balanced eqn, list given & unkn [ ] or P label conditions [initial, change, equlib] ex. A mixture is analyzed to find [A] = 2.000*10-3 & [B]= 4.000*10-3 and allowed to react. The reaction is A(g) + 2 B(g) C(g) Find [ ] @ equilib Kc = [C]/[[A][B]2] initial [A], [B], [C] need to find  [all]

  28. [A] + 2 [B] <------> [C] initial change equilibrium 2.000*10-3 4.000*10-3 0.000 -x -2x +x 0.002-x 0.004-2x 1.56*10-3 Can deduce from PDT amt  [C] Now do mole conversion [A] = [C] * mole ratio = (1.56*10-3)*(1A/1C) = -1.56*10-3 [A] [B] = (1.56*10-3)*(2B/1C) = -3.12*10-3 [B]

  29. [A] + 2 [B] <------> [C] initial change equilibrium 2.000*10-3 4.000*10-3 0.000 -1.56*10-3-3.12*10-3+1.56*10-3 4.4*10-4 8.8*10-41.56*10-3

  30. PS #4, cont #15. find quantity X from know equil [ ] & K 2 NO (g) <----> N2 (g) + O2 (g) Kc = 4.6*10-1 [.976] [.781] X #16. 0.500 mol ICl gas decomposes into two diatomic gases in a 5.00 L container. 1) construct a reaction table 2) concen @ equilb, Kc = 0.110 Determine initial [ ]: ICl = 0.500 mol/2.00 L = 0.100 M since no rxn started; Cl2 & I2 = 0 mol

  31. 2 ICl (g) <------> Cl2 (g) + I2 (g) initial change equilibrium 0.100 0 0 -2x +x +x 0.100-2x +x +x

  32. 0.332[0.100-2x] = x 0.0332 = 1.664x x = 0.02 2 ICl <------> Cl2 + I2 initial change equilibrium 0.100 0.00 -0.02 +0.02since Cl2 = I2 [0.08] [0.02] [0.02]

  33. RXN QUOTIENT, Qc K: equilibrium constant Q: rxn quotient only 1 value @ equilb @ spec Temp varies as rxn proceeds @ same spec Temp Qc = Kc @ equilb CH4 + Cl2 <-----> CH3Cl + HCl @ 1500 K Patm 0.13 0.035 0.24 0.47 Kp = 1.6*104 Find Qp, which direction? Qp < Kp ??

  34. COMPARE Q <----> K Q < K more pdts, fwd rxn Q = K no D, equilibrium Q > K more reactant, rev rxn now, values are “mol concen”; rxn in 250 mL flask CH4 + Cl2 <-----> CH3Cl + HCl @ 1500 K [mol] 0.13 0.035 0.24 0.47 Kc = 1.6*104 Find Qc, which direction? 1) 0.25 L 2) find M, mols/L 3) use values Qc eqn solve & compare

  35. Solution: as reference: Kref = a) reverse of original rxn by factor ½. Qc = (1/Kref)½ EX. At 500oC, Kc = 1.6*10-2 for the rxn: 2 H2S(g) <----> 2 H2(g) + S2(g) Calculate Kc for each: a) 0.5 S2(g) + H2(g) <----> H2S(g) b) 5 H2S(g) <----> 5 H2(g) + 5/2 S2(g) Kc = (1/1.6*10-2)½ = 7.9

  36. b) original rxn by factor 5/2. Qc = (1/Kref)5/2 Kc = (1.6*10-2)5/2 = 3.2*10-5 PS #5 #17. H2 + Cl2 <-----> 2 HCl Kc = 7.6*108 Find Kc: 0.5 H2 + 0.5 Cl2 <-----> HCl #18. Find Kc: 4/3 HCl <-----> 2/3 H2 + 2/3 Cl2

  37. PS #5

  38. EX. Mixture of 5.00 volumes of N2 & 1.00 volume of O2 reaches equlibrium @ 900 K & 5.00 atm: N2(g) + O2(g) <----> 2 NO(g) Kp = 6.70*10-10 What is if partial pressure of NO? Solution: construct I.C.E. table initial: 6 vol of gas @ 5 atm & 900 K. Vols are proportional to moles, so vol fraction = mole fraction PN2 + PO2 = 5.00 atm PN2 = XN2Ptot = (5.00/6.00) = 4.17 atm PO2 = XO2Ptot = (1.00/6.00) = 0.83 atm

  39. N2 (g) + O2 (g) <------> 2 NO (g) Initial 4.17 0.83 0 Change -X -X +2X Equilibrium4.17-X 0.83-X 2X Kp very small, assume [N2]eq = 4.17 - X = 4.17 & same O2 = 0.83 X = 2.41*10-5 NO: 2*(2.41*10-5) = 4.82*10-5 atm assumption: [(2.41*10-5)/4.17]*100 = 0.0006% assumption < 5%

  40. PS #5, cont #19. To obtain cleaner fuel from coal, a water-gas shift rxn is used. CO(g) + H2O(g) <-----> CO2(g) + H2(g) @ equilb: [CO] = [H2O] = [H2] = 0.10 M & [CO2] = 0.40 M. 0.60 mol H2 is added to the 2.0-L vessel & new equilbr reached. What are new equilbr concentrations? Solution: find Kc, find new initial [H2], construct I.C.E. table, find new [ ]s [H2] = 0.10 M + (0.60 mol/2.0-L) = 0.40 M CO(g) + H2O(g) <-----> CO2(g) + H2(g) Initial 0.10 0.10 0.40 0.40 Change +X +X -X -X Equilibrium0.10+X 0.10+X 0.40-X 0.40-X

  41. CO(g) + H2O(g) <-----> CO2(g) + H2(g) Initial 0.10 0.10 0.40 0.40 Change +X +X -X -X Equilibrium0.10+X 0.10+X 0.40-X 0.40-X 2.0(0.10+X) = 0.40-X X = 0.067 CO(g) + H2O(g) <-----> CO2(g) + H2(g) Equilibr 0.10+0.067 0.40-0.067 [CO]=[H2O] = 0.167 M [CO2]=[H2]= 0.333 M

  42. PS#6 #20. Phosgene (COCl2) that forms from CO & Cl2 at high temps. CO(g) + Cl2(g) <-----> COCl2(g) 0.350 mols of each reactant placed in 0.500-L flask @ 600 K. What are all [ ]s @ equilibrium? Kc = 4.95 Solution: find initial [CO & Cl2], construct I.C.E. table, find new [ ]s 0.350mol/0.5 L =0.700 CO(g) + Cl2(g) <-----> COCl2(g) Initial 0.700 0.700 0.0 Change -X -X +X Equilibrium0.700-X 0.700-X X

  43. 4.95X2 - 7.93X + 2.4255 = 0 Check solutions for viability CO(g) + Cl2(g) <-----> COCl2(g) Initial 0.700 0.700 0.0 Change -____ -____ +____ Equilib 0.700-___ 0.700-____ _____ 1.19 value not possible, since 0.700 - X result in “-” value Equilib 0.700-0.412 0.700-0.412 +0.412 [CO]=[Cl2] = 0.288 M 0.412 M = [COCl2]

More Related