1. 1 Equilibrium Equilibrium pp Online HW notes for the next 2 weeks: Some problems use ‘bar” or “millibar” for pressure. They are just another unit. One problem might give you -[ ]s. It’s bad chem, but good math & the problem is “doable.” Do not round when doing online HW (even if adding or subtracting).

2. 2 13.2 Law of Mass Action Z For any solution or gaseous reaction Z jA + kB lC + mD Z K c = [C] l [D] m PRODUCTS power [A] j [B] k REACTANTS power Z K c is called the equilibrium constant. Z K c often is written as just K Z is how we indicate a reversible reaction Z Does not apply to solids or pure liquids!

3. 3 Playing with K Z If we write the reaction in reverse. lC + mD jA + kB Z Then the new equilibrium constant is K ’ = [A] j [B] k = 1/K [C] l [D] m

4. 4 Playing with K Z If we multiply the equation by a constant Z njA + nkB nlC + nmD Z Then the equilibrium constant is Z K’ = [A] nj [B] nk = ([A] j [B] k ) n = K n [C] nl [D] nm ( [C] l [D] m ) n

5. 5 General Equation  K p = K c (RT)  n (R =.08206 l atm/K mol & T in Kelvin)  Where  n = change in moles of gas. Z Watch for mixed equilibria (solids & liquids in reaction)  Do not count their moles in calculating  n. Z Be sure to convert C to K for temp.

6. 6 What Q tells us Z If Q < K : Not enough products, so adjusts by : Shift to right (to make more) Z If Q > K : Too many products, so adjusts by : Shift to left (to reduce products) Z If Q = K system at =m and no shift

7. 7 13.6 Solving Equilibrium Problems Z Given the starting concentrations and one equilibrium concentration. Z Use stoichiometry to figure out other concentrations and K. Z Learn to create a table of initial and final conditions. Z Use “RICE”

8. 8 Z Consider the following at 600ºC Z 2SO 2 (g) + O 2 (g) 2SO 3 (g) Z In a certain experiment 2.00 mol of SO 2, 1.50 mol of O 2 and 3.00 mol of SO 3 were placed in a 1.00 L flask. At equilibrium 3.50 mol SO 3 were found to be present. Calculate: Z The equilibrium concentrations of O 2 and SO 2, K c and K P (procedure follows...)

9. 9 Strategy Z Make RICE table to show initial, change and final molarity (covert moles to M by dividing by volume) Z Use stoich & mole ratios as needed. Z Plug into =m expression to get K c Z Use K p = K c (RT) ∆n Z Remember: ∆n is change in gaseous moles only!!!

10. 10 Solving 2SO 2 (g) + O 2 (g 2SO 3 (g) initial2.00 M 1.5 M 3.00 M change equil.3.50 M K c = [SO 3 ] 2 [SO 2 ] 2 [O 2 ]

11. 11 Solving 2SO 2 (g) + O 2 (g 2SO 3 (g) initial2.00 M 1.5 M 3.00 M change0.50 M 0.25 M0.50 M final3.50 M K c = [SO 3 ] 2 [SO 2 ] 2 [O 2 ]

12. 12 Solving 2SO 2 (g) + O 2 (g 2SO 3 (g) initial2.00 M 1.5 M 3.00 M change-0.50 M-0.25 M +0.50 M final1.50 M 1.25 M3.50 M K c = [SO 3 ] 2 = (3.50) 2 [SO 2 ] 2 [O 2 ] (1.50) 2 (1.25) K c = 4.36

13. 13 Solving 600º C 2SO 2 (g) + O 2 (g 2SO 3 (g) initial2.00 M 1.5 M 3.00 M change-0.50 M-0.25 M +0.50 M final1.50 M 1.25 M3.50 M K c = 4.36 K p = K c (RT) ∆n and ∆n = -1 K p = 4.36(0.08206 x 873) -1 = 6.09 x 10 -2 atm (or 3.80 torr).

14. 14 What if not given equilibrium concentration? Z Easy if you get a “perfect square.” Z 3.000 mol each of hydrogen, fluorine and hydrofluoric acid were added to a 1.5000 L flask. Calculate the equilibrium concentrations of all species. K eq = 1.15 x 10 2. Z Write balanced equation, set up RICE Z Steps on next slides.

15. 15 Example continued. Z 3.00 mol each of hydrogen, fluorine and hydrofluoric acid were added to a 1.50 L flask. Calculate the equilibrium concentrations of all species. K eq = 1.15 x 10 2. Z Write balanced equation, determine Q and direction of shift, set up RICE Z Steps on next slides.

16. 16 Example cont. where K = 1.15 x 10 2 Z 3.00 mol each of H 2, F 2, HF added to a 1.50 L flask. So, M = 2.00 for each. Z H 2 (g) + F 2 (g) 2HF(g) Z Q = (2.00) 2 = 1.00 < K eq (2.00)(2.00) Z So, shifts to right. So, Left side is losing stuff and right side is gaining.

17. 17 Example continued. H 2 (g) + F 2 (g) 2HF(g) initial2.00 M 2.00 M 2.00 M change -x -x + 2x final 2.00-x 2.00-x 2.00+2x K = 1.15 x 10 2 = (2.00 + 2x) 2 (2.00 - x)(2.00 - x) K = 1.15 x 10 2 = (2.00 + 2x) 2 (2.00 - x) 2 Perfect square, so take square root of both sides.

18. 18 Example continued. √K = 10.74 = (2.00 + 2x) (2.00 - x) x = 1.528 H 2 (g) + F 2 (g) 2HF(g) final 2.00-x 2.00-x 2.00+2x 2.00 - x = [H 2 ] eq & [F 2 ] eq = 0.472 M (2.00 + 2x) = [HF] eq = 5.056 M

19. 19 What if you don’t get a “perfect square”? pp Z We are then stuck with having to solve a quadratic expression unless we can make simplifying assumptions. Z To avoid a quadratic, we will take different approaches depending on the size of K c.

20. 20 General Rules to avoid quadratics pp Z When K c is large - find limiting reactant and set it as “x” in the “equilibrium” line of RICE. (Example follows). Z When K c is small - easiest is to put “x” in the “change” line or RICE (see example). Z If putting “x” in “change line then requires a quadratic or you get a negative [ ], put it in the “equilibrium line” (see example). Z When K c is “middle” then must do quadratic.

21. 21 General Rules to avoid quadratics pp Z Why don’t we just always put “x” in the “equilibrium” line since it always works? Z We could, but most textbooks don’t show this method because the vast majority of problems involve a small K c where putting it the “change” line works. Z Also, when we do buffer equilibria we will put “x” in the “change” line so we mostly want to use that method unless we can’t because of a quadratic or we get a -[ ].

22. 22 Large K c example pp Z H 2 (g) + I 2 (g) 2HI(g) Z K = 7.1 x 10 2 at 25ºC Z Calculate the equilibrium concentrations if a 5.00 L container initially contains 15.7 g of H 2 and 294 g I 2. Z What is Q?... (can do without a calculator) Z Q = 0 because initially no product. Z [ H 2 ] 0 = (15.7g/2.02)/5.00 L = 1.56 M Z [I 2 ] 0 = (294g/253.8)/5.00L = 0.232 M Z [HI] 0 = 0 Z Problem: Need to know limiting reactant.

23. 23 Z H 2 (g) + I 2 (g) 2HI(g) pp Z Q = 0 < K so more product formed. Z Assumption: Since K is large, reaction will go to completion (I.e., forward arrow only). Z But, it doesn’t quite go to completion. Z Use Stoich to find I 2 is LR, so it will be smallest at =m (since essentially used up. So, let it be x in “equilibrium” line. Z Set up RICE table in concentrations (use moles with stoich,but M in RICE)

24. 24 Z Choose X so it is small (so use I 2 since it was limiting reactant. Z For I 2 the change in X must be X - 0.232 M Z Final must = initial + change H 2 (g) + I 2 (g) 2HI(g) pp initial1.56 M0.232 M 0 M change Equilibrium X

25. 25 Z H 2 (g) + I 2 (g) 2HI(g) Z Using stoichiometry we can find: Z Change in H 2 = X-0.232 M Z Change in HI = - twice change in H 2 Z Change in HI = 0.464-2X H 2 (g) I 2 (g) 2HI(g) pp initial1.56 M0.232 M 0 M changeX-0.232 final X

26. 26 Z Now we can determine the final concentrations by adding. H 2 (g) I 2 (g) 2HI(g) pp initial 1.56 M0.232 M 0 M change X-0.232 X-0.2320.464-2X final X

27. 27 Z Now plug these values into the equilibrium expression Z K = (0.464-2X) 2 = 7.1 x 10 2 (1.328+X)(X) Z Now you can see why we made sure X was small (by using the LR). H 2 (g) I 2 (g) 2HI(g) pp initial 1.56 M0.232 M 0 M change X-0.232 X-0.232 0.464-2X final 1.328+X X 0.464-2X

28. 28 Why We Chose x to be Small pp Z K = (0.464-2X) 2 = 7.1 x 10 2 (1.328+X)(X) Z Since x is going to be small, we can ignore it in relation to 0.464 and 1.328 Z So we can rewrite the equation Z 7.1 x 10 2 = (0.464) 2 (1.328)(X) Z Makes the algebra easy

29. 29 Z When we solve for X we get 2.3 x 10 -4 Z So we can find the other concentrations Z I 2 = 2.8 x 10 -4 M Z H 2 = 1.328 M Z HI = 0.464 M Z Note: If had set “x” in change line we would have got (-) =m concentrations for H 2 & I 2. H 2 (g) I 2 (g) 2HI(g) pp initial 1.56 M0.232 M 0 M change X-0.232 X-0.232 0.464-2X final 1.328+X X 0.464-2X

30. 30 Checking the assumption pp Z Rule of thumb: If the value of X is less than 5% of all the other concentrations, assumption is valid. Z If not we would have had to use the quadratic equation Z More on this later. Z Our assumption was valid.

31. 31 H 2 (g) I 2 (g) 2HI(g) initial 1.56 M0.232 M 0 M change -x -x +2x final 1.56-x0.232-x 0+2x pp Why we couldn’t put x in “change” line pp Z (2x) 2 = 7.1 x 10 2 (flies on Brian) (1.56-x)(.232-x) Z x = 8.01, and plugging into “final” we get: Z -6.45-7.78 +16.02 (-) concentrations for H 2 & I 2 !!!

32. 32 Problems with small K pp K<.01

33. 33 Example of small K c pp Z 90% of the time you can x in “change” line with a small K c, Z For the reaction 2NOCl(g) 2NO(g) + Cl 2 (g) Z K = 1.6 x 10 -5 M at 35ºC Z In an experiment 1.00 mol NOCl, is placed in a 2.0 L flask. Z Find the =m concentrations

34. 34 Example of small K c pp Z 1 mol NOCl in 2 L, K = 1.6 x 10 -5 2NOCl(g) 2NO(g) + Cl 2 (g) 0.50 M00 -2x +2x+x 0.5 - 2x +2x +x Z K c = 1.6 x 10 -5 = (2x) 2 (x) (0.50 - 2x) 2 Z K c ≈ (2x) 2 (x) ≈ 4x 3 (0.50) 2 (0.50) 2 Z x = 1.0 x 10 -2

35. 35 Example of small K c pp Z 1 mol NOCl in 2 L, K = 1.6 x 10 -5 2NOCl(g) 2NO(g) + Cl 2 (g) 0.50 M00 -2x +2x+x 0.5 - 2x +2x +x Z Since x = 1.0 x 10 -2, =m [ ]s are 0.48 0.020.01

36. 36 If can’t solve with “x” in change line pp Z Set up table of initial, change, and final concentrations (RICE). Z Choose X to be small (find LR) and put in “final” line of RICE. Z For this case it will be a product because K c is small. Z The following problem must be done this way to avoid a quadratic.

37. 37 For example pp Z For the reaction 2NOCl 2NO + Cl 2 Z K= 1.6 x 10 -5 Z If 1.20 mol NOCl, 0.45 mol of NO and 0.87 mol Cl 2 are mixed in a 1 L container Z What are the equilibrium concentrations Z Q = [NO] 2 [Cl 2 ] = (0.45) 2 (0.87) = 0.15 M [NOCl] 2 (1.20) 2 Z Since Q > K, shift to left

38. 38 Z Choose X to be small Z NO will be LR Z Choose NO to be X 2NOCl 2NO+Cl 2 pp Initial 1.20 0.450.87 Change Final

39. 39 Z Figure out change in NO Z Change = final - initial Z change = X-0.45 2NOCl 2NOCl 2 pp Initial 1.20 0.450.87 Change Final X

40. 40 Z Now figure out the other changes Z Use stoichiometry Z Change in Cl 2 is 1/2 change in NO Z Change in NOCl is - change in NO 2NOCl 2NOCl 2 pp Initial 1.20 0.450.87 Change X-.45 Final X

41. 41 l Now we can determine final concentrations l Add 2NOCl 2NOCl 2 pp Initial 1.20 0.450.87 Change0.45-X X-.45 0.5X -.225 Final X

42. 42 l Now we can write equilibrium constant l K = (X) 2 (0.5X+0.645) (1.65-X) 2 l Now we can test our assumption X is small so ignore it in (+) and (-) 2NOCl 2NOCl 2 pp Initial 1.20 0.450.87 Change0.45-X X-.45 0.5X -.225 Final1.65-X X 0.5X + 0.645

43. 43 l K = (X) 2 (0.645) = 1.6 x 10 -5 (1.65) 2 l X= 8.2 x 10 -3 l Figure out final concentrations 2NOCl 2NOCl 2 pp Initial 1.20 0.450.87 Change0.45-X X-.45 0.5X -.225 Final1.65-X X 0.5X +0.645

44. 44 l X= 8.2 x 10 -3 l [NOCl] = 1.65 l [Cl 2 ] = 0.649 l Check assumptions (x < all other [ ]s) l.0082/.45 = 1.8 % OKAY!!! 2NOCl 2NO+Cl 2 2NOCl 2NO+Cl 2 pp Initial 1.20 0.450.87 Change0.45-X X-.45 0.5X -.225 Final1.65-X X 0.5X +0.645

45. 45 2NOCl 2NO+Cl 2 Initial 1.20 0.450.87 Change -2x +2x +x Final1.20-2X 0.45+2X 0.87+x Quadratic if put x in “change” pp Z (0.45+2x (0.87+x) 2 = 1.6 x 10 -5 (1.20-2x) 2 Z So, here we need “x” in the “final” line

46. 46 Super Hints pp Z Always look first for a perfect square!! Z Large K - put “x” in “final row Z Small K - 98% of time can put “x” in change row, but... Z If with small K you get (-) [ ]s or must use quadratic then put “x” in final row. Z Putting “x” in final row always works, but is difficult to use when we get to acid/base/buffer equilibrium. Z So for small K try “x” in change row first.