aA + bB → cC + dD

1. aA + bB→cC + dD The above assumes the reaction goes to completion and that no reverse reaction takes place. This is rarely true. Once products are formed they can react with each other and return to the reactants. Q = [C]c [D]d [A]a [B]b Reactions reach a state where the amounts of reactants and the amounts of product no longer change. This state is called equilibrium. aA + bB cC + dD Writing Reaction Quotients (Q) or Equilibrium Constant (K) expressions K = Q when the reaction is already at equilibrium

2. N2O4(g) 2NO2(g) Write down specific expression for Q for this reaction • Did you write ….. • [N2O4] b) [NO2]2c) [NO2] d) none of these • [NO2]2 [N2O4] [N2O4]

3. CH4(g)+ H2O(g)↔ CO(g) + 3H2(g) Q = ?? a) [CO] • [H2]3 [CH4] • [H2O] b) [CH4] • [H2O] [CO] • [H2]3 c) [CO] • [H2]3 [CH4] Solids and pure liquids are not found in Q expression.

4. equilibrium N2O4(g) 2NO2(g) Q = [NO2]2 [N2O4] Start with NO2 Start with N2O4 Start with NO2 & N2O4 At equilibrium kf = kr The reaction direction depends on the value of Q. In this case the value of Q at the start is …. a) = K b) < K c) > K

5. N2O4(g) 2NO2(g) Q = [NO2]2 [N2O4] Finding Kc for a reaction …. 1. Put reactants in closed system. • Allow the reaction to proceed until equilibrium • is reached. This occurs when there is no change in • [ ]s of reactants or products. 3. Measure the [ ]s of all reactants and products. 4. Sub these into the equilibrium expression. Initial I 0.670 0 Equilibrium E 0.643 0.0547 What is Q? A) 0 b) 0.670 c) 1/0.670 What is K? A) 0 b) 0.085 c) 0.00465 d) 215

6. N2O4(g)↔2NO2(g) K = [NO2]2 = 0.00465 [N2O4] Do you have sufficient information to determine K for the reverse reaction? 2NO2(g) ↔ N2O4(g) a) yes b) no Krev = 1/K = [N2O4]= 1/0.00465 = 215 [NO2] 2

7. constant KC = [NO2]2 [N2O4]

8. Kc Kp In most cases N2O4(g) 2NO2(g) KC = [NO2]2 = 0.00465 [N2O4] KP = (PNO2)2 = ? PN2O4 M = n/V … P = n• RT = MRT V PNO2 = 0.0547 • 0.08206 • 298 = 1.34 atm Equilibrium E 0.643 M 0.0547 M PN2O4 = 0.643 • 0.08206 • 298 = 15.7 atm KP = 1.342 = 0.114 15.7

9. N2O4(g) 2NO2(g) K = [NO2]2 = 0.00464 [N2O4] Initial I 0.670 0 Change C Equilibrium E -x +2x 0.670 - x +2x KC = (2x)2 = 0.00464 solve for x (0.670 – x) x = -b ± (b2 – 4ac)½ 2a KC = 4x2 + 0.00464x – 0.00311 = 0 a b c x = -.00464 ± (.004642 + .0498)½ 8 x = 0.0273 [NO2]2= 2 • 0.0273 = 0.0546 [N2O4] = 0.670 – 0.0273 = 0.643

10. KP = 158 at 1000 K for … 2 NO2(g)↔ 2 NO(g) + O2(g) What is the equilibrium pressure of O2 if the PNO2 = 0.400 atm and PNO = 0.270 atm? KP = PNO2• PO2 PNO22 158 = 0.2702• PO2 0.4002 = 347 atm

11. CaCO3(s)CaO(s) + CO2(g) Heterogenous equilibrium applies to reactions in which reactants and products are in different phases. KC ≠ [CaO] • [CO2] [CaCO3] KC = [CaO] • [CO2] b) 1/[CO2] c) [CO2] [CaCO3] The concentration of solids and pure liquids are not included in the expression for the equilibrium constant. or …. KP = PCO2 KC = [CO2]

12. Kc = [C]c [D]d [A]a [B]bat equilibrium Q = [C]c [D]d [A]a [B]b Predicting reaction direction given known [ ]s. For aA + bB  cC + dD …… where the system is not at equilibrium …. If Q < Kc the reaction proceeds in the forward direction. If Q > Kc the reaction proceeds in the reverse direction. If Q = Kc the reaction is at equilibrium.

13. Finding Kc for a reaction …. CO(g) + 3H2(g) CH4(g) + H2O(g) I 0.100 M 0.300 M 0.0 M 0.0 M C + 0.00387 + 0.00387 - 0.00387 - 3 • 0.00387 E 0.00387 M 0.00387 M 0.288 M 0.096 M 1.00 mol of CO and 3.00 mol H2 are placed in a 10.0L container at 1200K. At equilibrium there is 0.0387 mol H2O. What is Kc and what are the equilibrium concentrations of each species? [H2] at equilibrium = a) 0. 300M b) 0.296M c) 0.00387M d) 0.288M KC = (0.00387)2 (0.096)(0.288)3 = 0.653 KC = [CH4][H2O] [CO] [H2]3

14. KC = 1.93 at 1000 K for … 2 NO2(g)↔ 2 NO(g) + O2(g) What is the equilibrium [O2]if [NO2]= 0.00487 M and [NO]= 0.00329 M? a) 2.46 b) 4.23 c) 0.00243 d) 0.045 KC = [NO]2• [O2] [NO2]2 1.93 = 0.003292• [O2] 0.004872 = 4.23 M

15. N2O4(g) 2NO2(g) K = [NO2]2 = 0.00464 [N2O4] Initial I 0.670 0 Change C Equilibrium E -x +2x 0.670 - x +2x KC = (2x)2 = 0.00464 solve for x (0.670 – x) x = -b ± (b2 – 4ac)½ 2a KC = 4x2 + 0.00464x – 0.00311 = 0 a b c x = -.00464 ± (.004642 + .0498)½ 8 x = 0.0273 [NO2]2= 2 • 0.0273 = 0.0546 [N2O4] = 0.670 – 0.0273 = 0.643

16. Kc = [NH3]2 [N2] [H2]3 Kc = (0.0100 – 2x)2 (0.0200 + x)(0.0600 + 3x)3 A 50.0 L vessel contains 1.00 mol N2, 3.00 mol H2, and 0.500 mol NH3. Kc = 0.500 at 400˚C. Will direction will the reaction proceed? N2(g) + 3H2(g) 2NH3(g) [N2] = 0.0200 M; [H2] = 0.0600 M; [NH3] = 0.0100 M Q = 0.012/{(0.02)(0.06)3} = 23.1 > 0.500 The reaction goes to left …. NH3will dissociate Set up the equation for determining the final [ ]s of all reagents Let x = the increase in the Molarity of N2(g) [N2] = 0.0200 + x M; [H2] = 0.0600 + 3x M; [NH3] = 0.0100 – 2x M This is a 4th order polynomial that we will not attempt to solve!!

17.  -b ± b2 – 4ac x = 2a ICE Charts = try 14.43 on your own Br2(g) ↔ 2Br(g) at 1280 ºC, Kc = 1.1 x 10-3 What are the equilibrium concentrations of Br and Br2 if you start with only 0.50 M Br2 in a flask? Let x = ? a) -change in [Br2] b) change in [Br] [Br2] (M) [Br] (M) Initial 0.50 0 Change - x +2x Equilibirum 0.50 – x 2x [Br2] (M) [Br] (M) Initial 0.50 0 Change Equilibirum = 0.024 M = 0.49 M 1.1 x 10-3 = (2x)2 = 4x2 0.50 - x 0.50 - x KC = [Br]2 [Br2] 0.0938 -.0011 5.5 x 10-4 - 1.1 x 10-3x = 4x2 4x2 + 1.1 x 10-3x - 5.5 x 10-4 = 0 8 x = 0.012 M

18.  -b ± b2 – 4ac x = 2a ICE Charts If you assign x differently Br2(g) ↔ 2Br(g) at 1280 ºC, Kc = 1.1 x 10-3 What are the equilibrium concentrations of Br and Br2 if you start with only 0.50 M Br2 in a flask? Let x = ? a) -change in [Br2] b) change in [Br] [Br2] (M) [Br] (M) Initial 0.50 0 Change - 0.5x +x Equilibirum 0.50 – 0.5x x [Br2] (M) [Br] (M) Initial 0.50 0 Change Equilibirum = 0.023 M = 0.49 M 1.1 x 10-3 = x2 0.50 – 0.5x KC = [Br]2 [Br2] 0.0469 -.00055 5.5 x 10-4 - 5.5 x 10-4x = x2 x2 + 5.5 x 10-4x - 5.5 x 10-4 = 0 2 x = 0.023 M

19.  -b ± b2 – 4ac x = 2a ICE Charts = try 14.43 on your own H2(g) + Br2(g) ↔ 2HBr(g) at 730 ºC, Kc = 2.18 x 106 [H2] (M) [Br2] (M) [Br] (M) Initial 0 0 0.267 Change +x +x -2x Equilibirum x x 0.267 – 2x = 0.267 M 1.81 x 10-4 M 2.18 x 106 = (0.267 – 2x)2 = 4x2 – 0.801x + 0.0713 x2 x2 KC = [HBr]2 [H2] [Br2] 788.5 -.801 2.18 x 106x2 + 0.801x - 0.0713 = 0 x = 1.81 x 10-4 M 4.36 x 106

20. LeChatelier’s Principle If a system at equilibrium is altered by applying a ‘stress’, the equilibrium will shift to reduce the stress. Stress can be ….. 1) Changes in [ ]s of select reagents. 2) Temperature changes (heat or cool). 3) Pressure changes. 2 NO2(g) N2O4(g) + heat (DH˚ = -57.2 kJ mol-1) Brown colorless What will happen to the equilibrium if the gas is cooled? a) darker color b) lighter color c) stays same What will happen to the equilibrium if Ptot↑? a) darker color b) lighter color c) stays same

21. N2O4(g) 2NO2(g) KC = 0.00464 = [NO2]2 [N2O4] 0.643 0.0547 Add 0.20M additional N2O4 to the flask. Which way will the reaction go? a) toward reactants b) toward products c) still at equilibrium What is Q just after additional N2O4added? a) 0.00464 b) 0.00572 c) 0.00355 Initial I 0.670 0 Equilibrium E What are new [ ]s of each reagent?

22.  -b ± b2 – 4ac x = 2a N2O4(g) 2NO2(g) KC = 0.00464 = [NO2]2 [N2O4] = 0.624 M 0.843 -x 0.0547 + 2x = 0.839 M Add 0.20M additional N2O4 to the flask. What are new [ ]s of each reagent? KC = 0.00464 = (0.0547 + 2x)2 0.843 - x 0.254 -.223 4x2 + 0.223x - 0.000919 = 0 Initial I 0.670 0 Equilibrium E 8 x = 0.00385 M Note that KC is still 0.00464

23. Which reaction occurs at a faster rate? A) A + B → C + D B) C + D → A + B Equilibrium occurs when the rate of the forward reaction equals the rate of the reverse reaction … in an elementary reaction ….. (partial orders = coefficients) rf = kf [A] [B] rr = kr [C] [D] rf= rr A + B ↔ C + D • kf[A] [B] = kr [C] [D] • kf /kr = [C] [D]/([A] [B])=KC The rate of a reaction depends upon …. 1) The frequency of collisions 2) The energy of each collision (what fraction of collisions will succeed?)