1. CH. 15 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 @ equilb: [react] & [pdt] not  w / time

2. N 2 O 4 g, colorless 2 NO 2 g, brown rate fwd = rate rev k[react] m eq = k[pdt] n eq k[N 2 O 4 ] eq = k[NO 2 ] 2 eq k fwd [N 2 O 4 ] = k rev [NO 2 ] 2 k: equilb const ratio of [react]:[pdt] at o T

3. 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, K, for: POCl 3 PCl 3 + O 2 Cu 2 S + O 2 Cu 2 O + SO 2 2 2 2 3 2 2 reverse rxn:

4. Overall RXN 1. N 2 O 5 NO 2 + NO 3 2. NO 2 + NO 3 NO 2 + NO + O 2 3. NO 3 + NO 2 NO 2 N 2 O 5 + NO 3 3 NO 2 + O 2 k 1 = 4.8*10 -10 k 2 = 1.1*10 -5 k 3 = 3.2*10 2 Q overall = Q 1 *Q 2 *Q 3 = K overall = k 1 *k 2 *k 3 = K = (4.8*10 -10 )*(1.1*10 -5 )*(3.2*10 2 ) = 1.69*10 -12

5. NO 2 N2O4N2O4 [ ] time Q = K

6. reverse Q fwd = 1 / Q rev K fwd = 1 / K rev 2 SO 2 + O 2 2 SO 3 K fwd = 261 K rev = 1 / 261 = Coefficient Values H 2 + Cl 2 2 HCl Kc = 7.6*10 8 Find Kc: 0.5 H 2 + 0.5 Cl 2 HCl Find Kc: 4/3 HCl 2/3 H 2 + 2/3 Cl 2

7. What did we see? Specific K value has meaning only to that specific balanced equation Diff. Phases: heterogeneous equilibrium SnO 2 (s) + 2 H 2 (g) Sn (s) + 2 H 2 O (g) 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 What about liquids & solids?

8. PRESSURE, K P ideal gas behavior; P gas 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 SO 2 (g) + O 2 (g) 2 SO 3 (g) Qc = ?

9. [ ] : concen, M Q c = Q p *RT @ equilibrium K c = K p /RT K p = (K c )(RT) If no  in n, then = 0 No “RT” term, so K p = K c K p = (K c )(RT)  ngas

10. calculate K p : PCl 3 (g) + Cl 2 (g) PCl 5 (g) Kc = 1.67 @ 501 K K p = (K c )(RT)= (1.67)(0.0821*501) -1 = 0.0406 COMPARE Q K Q < K more pdts Q = K no , equilibrium Q > K more reactant 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 more PDT  n gas = #mols gas pdt - # mols gas react 1 - 2 = -1

11. SOLVE EQUILIBRIUM 1. Given equil [ ] or P then find “K” 2. Given K, initial [ ] or P then find [equilb] What did we see? Q c = Q p RT @ equilibrium Q fwd = Q n Q rev = (1 / Q) n K fwd = K n K rev = (1 / K) n

12. find quantity from know equil [ ] & K 2 NO (g) N 2 (g) + O 2 (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]

13. 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 2 I Cl (g) Cl 2 (g) + I 2 (g) initial use/make equilibrium Determine initial [ ]: ICl = 0.500 mol / 5.00 L = 0.100 M since no rxn started; Cl 2 & I 2 = 0 mol 0.100 0 0 -2x +x +x 0.100-2x +x +x

14. 2 ICl Cl 2 + I 2 used/made equilibrium 0.332[0.100-2x] = 0.332 x = 0.02 0.100-0.04 0.0+0.02 since Cl 2 = I 2 [0.06] [0.02] [0.02]

15. 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 Q 1 = K 1 -------------> Q = K -----------> Q 1 = K 1 again * equilb shifts, Kc same Effects [ ] add/remove either react/pdt  P;  T; add catalyst / inert gas

16. Shift RGT : add react; remove pdt Shift LEFT : add pdt; remove react Shift???:  P,  V Remember: sys adjusts thru [ ] changes, but Qc @ equilb same value as in orginal sys @ same given T  es [ ] + CO 2 ; shifts rxn to ----> ; [H 2 CO 3 / CO 2 ] 1 : 99  es to 0.5 : 99.5 forms more H 2 CO 3 shifts ratio back to 1 : 99 CO 2 (g) + H 2 O (g) H 2 CO 3 (g) >99% <1% Kc <<<<

17.  es TEMP 2 SO 2 + O 2 2 SO 3 + heat incr T shifts dir that absorbs heat think heat as pdt, exo shifts to left Rise T, adds heat, favors endo dir drop T, removes heat, favors exo dir Incr Kc for +  H decr Kc for -  H  es P 3 H 2 (g) + N 2 (g) 2 NH 3 (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

18. Add Catalyst 3 H 2 (g) + N 2 (g) + catalyst 2 NH 3 (g) no effect on equilb, no  Kc effects rate fwd & rate rev equilb reached quicker Add Inert Gas 2 SO 2 + O 2 + inert gas 2 SO 3 inert gas not participate in equilb P ’ s don ’ t , no  equilb