1. 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

2. EQUILIBRIUM *State of balance *2 = & opp opposing forces occur @ same rate (still reacting) Chemistry *fwd & rev rxns @ same rate *no  in concentratons N 2 O 4 (g)  2 NO 2 (g) colorless brown REVERSIBLE RXN r f = k f [N 2 O 4 ] r r = k r [NO 2 ] 2

4. “unitless” @ EQUILIBRIUM set = & rearrange r f = k f [N 2 O 4 ] r r = k r [NO 2 ] 2 k f [N 2 O 4 ] = k r [NO 2 ] 2 [NO 2 ] 2 / [N 2 O 4 ] = k f / k r = a constant; equilibrium constant; K c HABER PROCESS 3 H 2 (g) + N 2 (g)  2 NH 3 (g)

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

7. 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 be withdrawn continuously, then reacting system can be kept constantly off-balance More reactants used, more pdts.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

8. CaO (s) + 3 C (s) CaC 2 (s) + CO (g) Remove CO:reaction tipped toward CaC 2 formation TiCl 4 (g) + O 2 (g) TiO 2 (s) + 2 Cl 2 (g) Production of calcium carbide Production of titanium dioxide TiO 2 separates from gases as fine powdered solid thus, rxn. kept moving in forward direction

9. Synthesize Rxn. CH 3 COOH + HOCH 2 CH 3 CH 3 COOCH 2 CH 3 + H 2 0 Remove H 2 O: forward rxn. favored Production of ammonia N 2 (g) + 3 H 2 (g) 2 NH 3 (g) NH 3 more soluble in water than H 2 or N 2 Wash out NH 3 out of equilibrium mixture

10. Equilibrium usually temp. dependent Increase temp. both forward/reverse rxns. speed up. Why? molecules move faster, increase molecular collisions Pressure H 2 (g) + I 2 (g) 2 HI (g) + heat (given off) Adding heat shifts rxn. to left Temperature 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

11. Catalyst No effect, but can increase speed which equilibrium is reached 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. Le Châtelier’s Principle The amounts of reactants & pdts will shift in such a manner as to minimize the stress

12. Problem Set #1 1. Write Kc for: CH 4 (g) + H 2 S (g)  CS 2 (g) + H 2 (g) 2. Write Kc for: Al (s) + HCl (aq)  AlCl 3 (aq) + H 2 (g) Ignore pure “solids & liquids”

13. Use Partial Pressure - atm 760 mmHg(torr) = 1 atm K c -----> K p HABER PROCESS 3 H 2 (g) + N 2 (g)  2 NH 3 (g) K c  K p convert using PV = nRT

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

15. Problem Set #2 1. When does K c = K p ? If ever!! 2. CH 4 (g) + H 2 S (g)  CS 2 (g) + H 2 (g) K c = 1.3*10 -2 @ 475 o C a) find  n gas b) write K p expression c) find K p when (  moles gas pdts) = (  moles gas reacts), then  n gas = 0 a) 1 CH 4 (g) + 2 H 2 S (g)  1 CS 2 (g) + 4 H 2 (g)  n gas = 5 - 3 = 2 b) Kp = (0.013)[(0.0821)(748)] 2 c) Kp = (0.013)(3771.2863) = 49.03

16. EVALUATE K 1- K c = 4.56*10 9 2- K c = 4.56*10 -9 3- K p = 49 4- K p = 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

17. Figure 15.07ab

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

19. 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 H 2 (g) + N 2 (g)  2 NH 3 (g) 5. If [H 2 ] is increased a) Equilib shift direction? ____ b) [N 2 ] will incr / decr? _____ c) [NH 3 ] will incr / decr? ____ 6. If [N 2 ] is decreased a) Equilib shift direction? ____ b) [H 2 ] will incr / decr? _____ c) [NH 3 ] will incr / decr? ____  Decr Incr  Incr Decr

20. 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 PS#2, cont. N 2 (g) + O 2 (g) +90 kJ  2 NO (g) 7. If temp is incr a) Equilib shift direction? ____ b) [O 2 ] will incr / decr? _____ c) [NO] will incr / decr? ____ 8. If [O 2 ] is decreased a) Equilib shift direction? ____ b) [N 2 ] will incr / decr? _____ c) Temp will incr / decr? ____ Energy E added by incr Temp E produced, Temp incr E removed by cooling E used, Temp decr  Decr Incr  Incr

21. 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 NO 2 (g)  N 2 O 4 (g) + E 9. If temp is incr a) Equilib shift direction? ____ b) [NO 2 ] will incr / decr? _____ c) [N 2 O 4 ] will incr / decr? ____ 10. If volume is decreased a) Equilib shift direction? ____ b) [NO 2 ] will incr / decr? _____ c) [N 2 O 4 ] will incr / decr? ____ d) Temp will incr / decr? ____  Incr Decr  Decr Incr

22. 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 

23. 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(NO 3 ) 2 (s) + H 2 O (l)  Ca +2 (aq) + 2 NO 3 -1 (aq) a) Equilib shift direction, add water? ____ b) Equilib shift direction, add calcium compound? _____ No  

24. DIRECTION N 2 (g) + O 2 (g)  2 NO (g) K c = 1*10 -30 @ 25 o C Write K c and find for reverse K c = [NO] 2 / [ [N 2 ][O 2 ] ] = 1*10 -30 K c = [ [N 2 ][O 2 ] ] / [NO] 2 = 1 / 1*10 -30 = 1*10 +30 Evaluate What is favored in this rxn? What need to  to favor NO? N 2 - O 2 Incr Temp

25. PS #4 13. Haber Process @ 300 o C, Kp = 4.34*10 -3 Find K p reverse? 14. For each: a) CaCO 3 (s)  CaO (s) + CO 2 (g) (concen in M & atm) K c = K p = b) CO 2 (g) + H 2 (g)  CO (g) + H 2 O (l) K c = K p = c) Fe (s) + H 2 O (g)  Fe 3 O 4 (s) + H 2 (g) (all concen listed in “atm”) K c = K p = 1 / (4.34*10 -3 ) = 2.3*10 2 [CO 2 ] [P CO2 ] [CO]/ [ [CO 2 ][H 2 ] ] [P CO ]/ [ [P CO2 ][P H2 ] ] 3 4 4 (P H2 ) 4 /(P H2O ) 4

26. PS #1- revisit 2. Write Kc for: Al (s) + HCl (aq)  AlCl 3 (aq) + H 2 (g) Ignore pure “solids & liquids” Diff. Phases: heterogeneous equilibrium solids: same concen @ given temp; same # mols / L; no  V liquids: same applies therfore, we ignore pure solids & liquids only concerned w/ those that will  [ ] in rxn Qc, Kc Expression Summary fig 15.9, pg 645

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

28. Calculate K c 2 methods I. Know amts @ equilibrium [ ] / P @ spec Temp. ex. Haber process analyze equil @ 472 o C mixture: 7.38 atm H 2, 1870 torr N 2, & NH 3 ; P tot = 10.00 atm Determine K p convert: N 2 atm; find atm NH 3 ; bal eqn

29. 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 K c = [C] / [ [A][B] 2 ] initial [A], [B], [C] need to find  [all]

30. [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 +x 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]

31. [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 -4 1.56*10 -3

32. PS #4, cont #15. find quantity X from know equil [ ] & K 2 NO (g) N 2 (g) + O 2 (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, K c = 0.110 Determine initial [ ]: ICl = 0.500 mol / 2.00 L = 0.100 M since no rxn started; Cl 2 & I 2 = 0 mol

33. 2 I Cl (g) Cl 2 (g) + I 2 (g) initial change equilibrium 0.100 0 0 -2x +x +x 0.100-2x +x +x

34. 2 ICl Cl 2 + I 2 initial change equilibrium 0.0332 = 1.664x x = 0.02 0.332[0.100-2x] = x 0.100 0.00 -2(0.02) +0.02 since Cl 2 = I 2 [0.04] [0.02] [0.02]

35. 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 CH 4 + Cl 2 CH 3 Cl + HCl @ 1500 K P atm 0.13 0.035 0.24 0.47 K p = 1.6*10 4 Find Q p, which direction? Q p < Kp ? ?

36. COMPARE Q K Q < K more pdts, fwd rxn Q = K no , equilibrium Q > K more reactant, rev rxn now, values are “mol concen”; rxn in 250 mL flask CH 4 + Cl 2 CH 3 Cl + HCl @ 1500 K [mol] 0.13 0.035 0.24 0.47 Kc = 1.6*10 4 Find Qc, which direction? 1) 0.25 L 2) find M, mols/L 3) use values Qc eqn solve & compare

37. EX. At 500 o C, Kc = 1.6*10 -2 for the rxn: 2 H 2 S (g) 2 H 2 (g) + S 2 (g) Calculate Kc for each: a) 0.5 S 2 (g) + H 2 (g) H 2 S (g) b) 5 H 2 S (g) 5 H 2 (g) + 5 / 2 S 2 (g) Solution : as reference: Kref = a) reverse of original rxn by factor ½. Qc = (1 / Kref) ½  Kc = (1 / 1.6*10 -2 ) ½ = 7.9

38. 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. H 2 + Cl 2 2 HCl Kc = 7.6*10 8 Find Kc: 0.5 H 2 + 0.5 Cl 2 HCl #18. Find Kc: 4 / 3 HCl 2 / 3 H 2 + 2 / 3 Cl 2

39. PS #5

40. EX. Mixture of 5.00 volumes of N 2 & 1.00 volume of O 2 reaches equlibrium @ 900 K & 5.00 atm: N 2 (g) + O 2 (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 P N2 + P O2 = 5.00 atm P N2 = X N2 P tot = (5.00 / 6.00) = 4.17 atm P O2 = = (1.00 / 6.00) = 0.83 atm

41. N 2 (g) + O 2 (g) 2 NO (g) Initial 4.17 0.83 0 Change -X -X +2X Equilibrium 4.17-X 0.83-X 2X Kp very small, assume [N 2 ]eq = 4.17 - X = 4.17 & same O 2 = 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%

42. PS #5, cont #19. To obtain cleaner fuel from coal, a water-gas shift rxn is used. CO (g) + H 2 O (g) CO 2 (g) + H 2 (g) @ equilb: [CO] = [H 2 O] = [H 2 ] = 0.10 M & [CO 2 ] = 0.40 M. 0.60 mol H 2 is added to the 2.0-L vessel & new equilbr reached. What are new equilbr concentrations? Solution: find Kc, find new initial [H 2 ], construct I.C.E. table, find new [ ]s CO (g) + H 2 O (g) CO 2 (g) + H 2 (g) Initial 0.10 0.10 0.40 0.40 Change +X +X -X -X Equilibrium 0.10+X 0.10+X 0.40-X 0.40-X [H 2 ] = 0.10 M + (0.60 mol / 2.0-L) = 0.40 M

43. CO (g) + H 2 O (g) CO 2 (g) + H 2 (g) Initial 0.10 0.10 0.40 0.40 Change +X +X -X -X Equilibrium 0.10+X 0.10+X 0.40-X 0.40-X 2.0(0.10+X) = 0.40-X X = 0.067 CO (g) + H 2 O (g) CO 2 (g) + H 2 (g) Equilibr 0.10+0.067 0.40-0.067 [CO]=[H 2 O] = 0.167 M [CO 2 ]=[H 2 ]= 0.333 M

44. PS#6 #20. Phosgene (COCl 2 ) that forms from CO & Cl 2 at high temps. CO (g) + Cl 2 (g) COCl 2 (g) 0.350 mols of each reactant placed in 0.500-L flask @ 600 K. What are all [ ]s @ equilibrium? Kc = 4.95 CO (g) + Cl 2 (g) COCl 2 (g) Initial 0.700 0.700 0.0 Change -X -X +X Equilibrium 0.700-X 0.700-X X Solution: find initial [CO & Cl 2 ], construct I.C.E. table, find new [ ]s 0.350mol/0.5 L =0.700

45. CO (g) + Cl 2 (g) COCl 2 (g) Initial 0.700 0.700 0.0 Change -____ -____ +____ Equilib 0.700-___ 0.700-____ _____ 4.95X 2 - 7.93X + 2.4255 = 0 Check solutions for viability 1.19 value not possible, since 0.700 - X result in “-” value Equilib 0.700-0.412 0.700-0.412 +0.412 [CO]=[Cl 2 ] = _______ M _______ M = [COCl 2 ]