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CH. 17 EQUILIBRIUM

CH. 17 EQUILIBRIUM. System of: Gases, Liquids, Solids (will look @ all gas sys to start) Extent of a rxn, based on connect of reacts/pdts @ moment. K: equilibrium constant Q: rxn quotient. based on concent @ equilb varies w / rxn. Q c = K c @ equilb.

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CH. 17 EQUILIBRIUM

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  1. CH. 17 EQUILIBRIUM System of: Gases, Liquids, Solids (will look @ all gas sys to start) Extent of a rxn, based on connect of reacts/pdts @ moment K: equilibrium constant Q: rxn quotient based on concent @ equilb varies w/ rxn Qc = Kc @ equilb @ equilb: [react] & [pdt] not D w/ time

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

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

  4. N2O4g, colorless <--------> 2 NO2g, brown ratefwd = raterev k[react]meq = k[pdt]neq k[N2O4]eq = k[NO2]2eq kfwd[N2O4] = krev[NO2]2 k: equilb const ratio of [react]:[pdt] at oT

  5. Magnitude of “K” K <<<<<< sm (10-20) little pdt, mostly react @ equilb K >>>>>> lrg (10+18) little react, mostly pdt @ equilb sm < K < lrg (10-3-10+3) varying = amts of both reacts & pdts write equilb const, Q, for: 2 2 POCl3 <------> PCl3 + O2 Cu2S + O2 <-----> Cu2O + SO2 2 3 2 2 reverse rxn:

  6. Overall RXN 1. N2O5 <------> NO2 + NO3 2. NO2 + NO3 <------> NO2 + NO + O2 3. NO3 + NO <-------> 2 NO2 k1 = 4.8*10-10 k2 = 1.1*10-5 k3 = 3.2*102 N2O5 + NO3 <------> 3 NO2 + O2 K = (4.8*10-10)*(1.1*10-5)*(3.2*102) = 1.69*10-12 Qoverall = Q1*Q2*Q3 = Koverall = k1*k2*k3 =

  7. Q = K Q = K NO2 [ ] N2O4 time

  8. reverse Qfwd = 1/Qrev Kfwd = 1/Krev 2 SO2 + O2 <-----> 2 SO3 Kfwd = 261 Krev = 1/261 = Coefficient Values H2 + Cl2 <-----> 2 HCl Kc = 7.6*108 Find Kc: 0.5 H2 + 0.5 Cl2 <-----> HCl Find Kc: 4/3 HCl <-----> 2/3 H2 + 2/3 Cl2

  9. What did we see? Specific K value has meaning only to that specific balanced equation Diff. Phases: heterogeneous equilibrium SnO2 (s) + 2 H2(g) <-----> Sn (s) + 2 H2O (g) What about liquids & solids? 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 Qc, Kc Expression Summarytable 17.2, pg 685

  10. PRESSURE, KP ideal gas behavior; Pgas easier to measure than [gas] PV = nRT ===> P = (n/v)RT ===> (P/RT) = (n/v) & (n/v) = M P ∞ M since R * T constant Kc – Kp related but not “=“ 2 SO2(g) + O2(g) <-----> 2 SO3(g) Qc = ?

  11. [ ] : concen, M Qc = Qp*RT @ equilibrium If no D in n, then = 0 No “RT” term, so Kp = Kc(RT)0 Kp = (Kc)(RT)Dngas

  12. calculate Kp: PCl3(g) + Cl2(g) <-----> PCl5(g) Kc = 1.67 @ 501 K Dngas = #mols gas pdt - # mols gas react 1 - 2 = -1 Kp = (Kc)*(RT) = (1.67)*(0.0821*501)-1 = 0.0407 COMPARE Q <----> K Q < K favors pdts Q = K no D, equilibrium Q > K favors reactant CH4 + Cl2 <-----> CH3Cl + HCl @ 1500 K Patm 0.13 0.035 0.24 0.47 Kp = 1.6*104 [mol]pt 2 Find Qp, which direction? Qp < Kp favors PDT

  13. now, values are “mol concen”; rxn in 250 mL flask 1) 0.25 L 2) find M, mols/L 3) use values step #2 in eqn CH4 = 0.13/0.25 = 0.52 M What did we see? Qc = QpRT @ equilibrium Qfwd = Qn Qrev = (1/Q)n Kfwd = Kn Krev = (1/K)n Qc = [CH3Cl][HCl]/[1.2][Cl2] SOLVE EQUILIBRIUM 1. Given equil [ ] or P then find “K” 2. Given K, initial [ ] or P then find [equilb]

  14. find quantity from know equil [ ] & K 2 NO (g) <----> N2 (g) + O2 (g) Kc = 4.6*10-1 [.976] [.781] X TABLE SET UP -- find equil quantities & K write balanced eqn, list given & unkn [ ] or P label conditions [initial, change, equlib]

  15. 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/5.00 L = 0.100 M since no rxn started; Cl2 & I2 = 0 mol 2 ICl (g) <------> Cl2 (g) + I2 (g) initial use/make equilibrium 0.100 0 0 -2x +x +x 0.100-2x +x +x

  16. 0.332[0.100-2x] = x x = 0.02 2 ICl <------> Cl2 + I2 used/made equilibrium 0.100-0.04 0.0+0.02 since Cl2 = I2 [0.06] [0.02] [0.02]

  17. LE CHATELIER’S PRINCIPLE When a sys at equilb is stressed, the sys will react in the direction to alleviate that stress & return to a new equilb position Q1 = K1 -------------> Q = K -----------> Q1 = K1 again * equilb shifts, Kc same Effects [ ] add/remove either react/pdt DP; DT; add catalyst/inert gas

  18. CO2 (g) + H2O (g) H2CO3 (g) >99% <1% Kc <<<< Shift RGT: add react; remove pdt Shift LEFT: add pdt; remove react Shift???: DP, DV Remember: sys adjusts thru [ ] changes, but Qc @ equilb same value as in orginal sys @ same given T Des [ ] +CO2; shifts rxn to ----> ; [H2CO3/CO2] 1:99 Des to 0.5:99.5 forms more H2CO3 shifts ratio back to 1:99

  19. Des TEMP Rise T, adds heat, favors endo dir drop T, removes heat, favors exo dir 2 SO2 + O2 <----> 2 SO3 + heat incr T shifts dir that absorbs heat think heat as pdt, exo shifts to left Incr Kc for +DH decr Kc for -DH Des P 3 H2 (g) + N2 (g) <----> 2 NH3 (g) equilb w/ only unequal # mols of gas 4 mols react 2 mols pdt effect incr/decr P??? Rise P, favors fewer gas mols dir drop P, favors more gas mols dir

  20. Add Catalyst 3 H2 (g) + N2 (g) + catalyst <----> 2 NH3 (g) no effect on equilb, no DKc effects rate fwd & rate rev equilb reached quicker Add Inert Gas 2 SO2 + O2 + inert gas <----> 2 SO3 inert gas not participate in equilb P’s don’t D, no D equilb

  21. Consider the following reaction: HI(g) <----> H2(g) + I2(g)a) At 298 K, Kc = 1.26*10-3. Calculate Kpb) Calculate Kc for the formation of HIc) Calculate DHorxn for HI from DHof values Consider the following reaction: NH3(g) <----> N2(g) + H2(g)a) In 1.0-L container at 727oC, 1.30 mol N2 & 1.65 mol H2 are added. At equilibrium, 0.100 mol of NH3 is present. Calculate the equilibrium concentration of reactants & Kcb) A second process, same size container & temp, equilibrium concentrations are: NH3 [8.34*10-2 mol], N2 [1.50 mol] H2 [1.25 mol] Calculate Kc for: NH3(g) <----> 1/2 N2(g) + 3/2 H2(g) c) Calculate DHorxn for NH3 from DHof values

  22. a) Kp = Kc(RT)0 ----> Kp = Kc -----> Kc = 1.26*10-3 b) Krev = 1/Kfwd = 1/(1.26*10-3) = 794 c) Hrxn = Hf(H2(g)) + Hf(I2(g)) - 2Hf(HI(g)) = 1mol[(0 kJ/mol) + (0 kJ/mol)] - 2 mol(25.9 kJ/mol) = -51.8 kJ reaction: NH3(g) <----> N2 (g) + H2 (g) *vol = 1 L, so, Molarity = mols present

  23. b) Kc = = = 20.5 c) Hrxn = Hf(N2(g)) + 3Hf(H2(g)) - 2Hf(NH3(g)) = (0 kJ/mol) + 3(0 kJ/mol) - 2 mol(45.9 kJ/mol) = -91.8 kJ

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

  25. 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

  26. 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

  27. 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

  28. H2(g) + I2(g) 2 HI (g) + heat (givenoff) Temperature Equilibrium usually temp. dependent Increase temp. both forward/reverse rxns. speeded 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

  29. 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

  30. 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) Ignore pure “solids & liquids”

  31. 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

  32. PS #1, cont 3. When does Kc = Kp? If ever!! 4. 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 ( moles gas pdts) = ( 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

  33. 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

  34. 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

  35. 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? ____  Incr 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

  36. 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) 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

  37. 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 12. Ca(NO3)2(s) + H2O(l) Ca+2(aq) +2NO3-1(aq) +H2O(l) No   a) Equilib shift direction, add water? ____ b) Equilib shift direction, add calcium cmpd? _____

  38. 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 to  to favor NO? N2 - O2 Incr Temp

  39. 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

  40. 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 therefore, we ignore pure solids & liquids only concerned w/ those that will D [ ] in rxn Qc, Kc Expression Summarytable 17.2, pg 685

  41. 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

  42. 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

  43. 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. 1.56*10-3 M of [C] is formed at equilibrium. The reaction is A(g) + 2 B(g) C(g) Find [ ] @ equilib & Kc concentrations? initial? think? no pdts? What do I need!!! Kc = [C]/[[A][B]2] initial [A], [B], [C] need to find  [all]

  44. [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*10-3C)*(1A/1C) = -1.56*10-3 [A] [B] = (1.56*10-3C)*(2B/1C) = -3.12*10-3 [B]

  45. [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 4.40*10-4 8.80*10-41.56*10-3

  46. 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 ICE table 2) concen @ equilb, Kc = 0.110 Determine initial [ ]: ICl = 0.500 mol/5.00 L = 0.100 M since no rxn started; Cl2 & I2 = 0 mol

  47. 2 ICl (g) <------> Cl2 (g) + I2 (g) Initial Change Equilibrium 0.100 0 0 -2x +x +x 0.100-2x +x +x

  48. 0.332[0.100-2x] = x 0.0332 = 1.664x x = 0.02 2 ICl <------> Cl2 + I2 Initial Change Equilibrium 0.100 0.00 2(-0.02) +0.02since Cl2 = I2 [0.06] [0.02] [0.02]

  49. 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(g) + Cl2(g) <-----> CH3Cl(g) + HCl(aq) @ 1500 K Patm 0.13 0.035 0.24 0.47 Kp = 1.6*104 Find Qp, which direction? Qp < Kp ??

  50. COMPARE Q <----> K Q < K favors pdts, fwd rxn Q = K no D, equilibrium Q > K facors reactant, rev rxn now, values are “mol concen”; rxn in 250 mL flask CH4(g) + Cl2(g) <----> CH3Cl(g) + HCl(aq) @ 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

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