Upcoming SlideShare
×

# Ch13 z5e equilibrium

782 views

Published on

0 Likes
Statistics
Notes
• Full Name
Comment goes here.

Are you sure you want to Yes No
• Be the first to comment

• Be the first to like this

Views
Total views
782
On SlideShare
0
From Embeds
0
Number of Embeds
1
Actions
Shares
0
6
0
Likes
0
Embeds 0
No embeds

No notes for slide
• Ch 13 Section 13.1 Z5e
• Z5e 13.1 p611 Analogy: Flow of cars across a bridge connecting two island cities
• Z5e 13.2 p615 Law of Mass Action does NOT apply to solids or pure liquids.
• Z5e 650 #21
• Z5e p617
• Z5e 618 Table 13.1
• Rf Z5e 618 table 13.1
• Z5e 619 Section 13.3
• Section 13.4 Z5e 622
• Section 13.5 Z53 634 Applications of =m constant
• Z5e 627
• Section 13.6 Z5e 634 RICE: reaction initial [ ] change in [ ] equilibrium [ ]
• Make table to show initial, change and final moles - use mole ratios Convert to molarity by dividing by volume (since 1.00 flask, moles = molarity) Plug into =m expression to get K = 4.36 Use K p = K (RT)  n to get K p = 6.09 x 10 -2 (mol 2 atm -1 ) Remember  n is change of gaseous moles only, but can be ±!!
• Same temp so K is same (4.36)! Convert to [ ] Make initial, change, final mole table
• Same temp so K is same (4.36)! Convert to [ ] Make initial, change, final mole table
• Z5e 633 SE 13.11
• Z5e 633 SE 13.11
• Z5e 633 SE 13.11
• Z5e 633 SE 13.11
• Z5e 633 SE 13.11
• This is the Green method.
• This is the Green method.
• This is the Green method.
• Remember to divide by 4 L to get molarity. Q = 1 = products/reactants. Since less than K, reaction shifts to right. Use table of molarities (divide by 4 to get 0.025M ClO, .25 M O 2 and .0025 M Cl 2 Since shift to right, make [=m] Cl 2 = x (since LR) Cl 2 O 2 Cl 2 Init .0025 .25 .025 Chg x-.0025 x-.0025 .0050-2x Fin x x+.2475 .0300-2x - add above Simplify &amp; solve for x = 2.33 x 10 -5 = [ Cl 2 ]
• Remember to divide by 4 L to get molarity. Q = 1 = products/reactants. Since less than K, reaction shifts to right. Use table of molarities (divide by 4 to get 0.025M ClO, .25 M O 2 and .0025 M Cl 2 Since shift to right, make [=m] Cl 2 = x (since LR) Cl 2 O 2 Cl 2 Init .0025 .25 .025 Chg x-.0025 x-.0025 .0050-2x Fin x x+.2475 .0300-2x - add above Simplify &amp; solve for x = 2.33 x 10 -5 = [ Cl 2 ]
• Remember to divide by 4 L to get molarity. Q = 1 = products/reactants. Since less than K, reaction shifts to right. Use table of molarities (divide by 4 to get 0.025M ClO, .25 M O 2 and .0025 M Cl 2 Since shift to right, make [=m] Cl 2 = x (since LR) Cl 2 O 2 Cl 2 Init .0025 .25 .025 Chg x-.0025 x-.0025 .0050-2x Fin x x+.2475 .0300-2x - add above Simplify &amp; solve for x = 2.33 x 10 -5 = [ Cl 2 ]
• Q = 1, &gt; K so shift to left Chlorine = LR so make it X 2ClO(g) Cl 2 (g) + O 2 (g) Init .025 &lt;--&gt; .0025 .25 Chg ,0050 - x x-.0025 x-.0025 Fin .030 - x x x+.2475 X = .000023 5% rule .000023/.025 = .09%
• .361 &amp; .820
• .361 &amp; .820
• Z5e 640 Section 13.7 Le Chatelier’s Principle
• Vonderbrink PL 17
• ### Ch13 z5e equilibrium

1. 1. 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. 13.1 Reactions are reversible <ul><li>A + B C + D ( forward) </li></ul><ul><li>C + D A + B (reverse) </li></ul><ul><li>Initially there is only A and B so only the forward reaction is possible </li></ul><ul><li>As C and D build up, the reverse reaction speeds up while the forward reaction slows down. </li></ul><ul><li>Eventually the rates are equal </li></ul>
3. 3. Reaction Rate Time Forward Reaction Reverse reaction Equilibrium
4. 4. What is equal at Equilibrium? <ul><li>Rates are equal </li></ul><ul><li>Concentrations are not! </li></ul><ul><li>Rates are determined by concentrations and activation energy. </li></ul><ul><li>The concentrations do not change at equilibrium. </li></ul><ul><li>Or, if the reaction is verrrry slooooow. </li></ul>
5. 5. 13.2 Law of Mass Action pp <ul><li>For any solution or gaseous reaction </li></ul><ul><li>j A + k B l C + m D </li></ul><ul><li>K c = [C] l [D] m PRODUCTS power [A] j [B] k REACTANTS power </li></ul><ul><li>K c is called the equilibrium constant. </li></ul><ul><li>K c often is written as just K </li></ul><ul><li>is how we indicate a reversible reaction </li></ul><ul><li>Does not apply to solids or pure liquids ! </li></ul>
6. 6. Playing with K pp <ul><li>If we write the reaction in reverse. l C + m D j A + k B </li></ul><ul><li>Then the new equilibrium constant is K ’ = [A] j [B] k = 1/K [C] l [D] m </li></ul>
7. 7. Playing with K pp <ul><li>If we multiply the equation by a constant </li></ul><ul><li>nj A + nk B nl C + nm D </li></ul><ul><li>Then the equilibrium constant is </li></ul><ul><li>K = [C] nl [D] nm = ( [C] l [D] m ) n = K n [A] nj [B] nk ([A] j [B] k ) n </li></ul>
8. 8. Calculate K - all are gases, given N 2 + 3H 2 2NH 3 where K = 1.3 x 10 -2 <ul><li>1/2N 2 + 3/2H 2 NH 3 K = ? . . . </li></ul><ul><li> K 1/2 = 0.11 </li></ul><ul><li>2NH 3 N 2 + 3H 2 K = ? . . . </li></ul><ul><li> K -1 = 77 </li></ul><ul><li>NH 3 1/2N 2 + 3/2H 2 K = ? . . . </li></ul><ul><li> (K -1 ) 1/2 = (1/K) 1/2 = 8.8 </li></ul><ul><li>2N 2 + 6H 2 4NH 3 K = ? . . . </li></ul><ul><li> (K) 2 = 1.7 x 10 -4 </li></ul>
9. 9. The units for K <ul><li>Are determined by the various powers and units of concentrations. </li></ul><ul><li>They depend on the reaction. </li></ul><ul><li>K values usually written without units </li></ul>
10. 10. K is CONSTANT <ul><li>At any constant temperature. </li></ul><ul><li>Temperature affects rate. </li></ul><ul><li>The equilibrium concentrations don’t have to be the same; only K. </li></ul><ul><li>Equilibrium position is a set of concentrations at equilibrium. </li></ul><ul><li>Indicates reaction shift to left or right. </li></ul><ul><li>There are an unlimited number. </li></ul>
11. 11. Equilibrium Constant One for each Temperature
12. 12. Calculate K c <ul><li>N 2 + 3H 2 2NH 3 </li></ul><ul><li>Initial At Equilibrium [N 2 ] 0 =1.000 M [N 2 ] = 0.921 M [H 2 ] 0 =1.000 M [H 2 ] = 0.763 M [NH 3 ] 0 =0 M [NH 3 ] = 0.157 M </li></ul><ul><li>Now do the calculation to get . . . </li></ul><ul><li>K = 6.02 x 10 -2 </li></ul>
13. 13. Calculate K <ul><li>N 2 + 3H 2 2NH 3 </li></ul><ul><li>Initial At Equilibrium [N 2 ] 0 = 0 M [N 2 ] = 0.399 M [H 2 ] 0 = 0 M [H 2 ] = 1.197 M [NH 3 ] 0 = 1.000 M [NH 3 ] = 0.203 M </li></ul><ul><li>Your answer . . . </li></ul><ul><li>K c is the same (6.02 x 10 -2 ) no matter what the amount of starting materials </li></ul>
14. 14. 13.3 Equilibrium and Pressure <ul><li>Some reactions are gaseous </li></ul><ul><li>PV = nRT </li></ul><ul><li>P = (n/V)RT where n/V = [ ] of gas </li></ul><ul><li>P = CRT C is a concentration in moles/Liter </li></ul><ul><li>C = P/RT </li></ul>
15. 15. Equilibrium and Pressure <ul><li>2SO 2 (g) + O 2 (g) 2SO 3 (g) </li></ul><ul><li>K p = ( P SO3 ) 2 ( P SO2 ) 2 ( P O2 ) </li></ul><ul><li>K c = [ SO 3 ] 2 [ SO 2 ] 2 [ O 2 ] </li></ul>
16. 16. Deriving a relationship between K c & K p <ul><li>2SO 2 (g) + O 2 (g) 2SO 3 (g) </li></ul><ul><li>Since C = [ ] = P/RT </li></ul><ul><li>K c = (P SO3 /RT) 2 (P SO2 /RT) 2 (P O2 /RT) </li></ul><ul><li>K c = (P SO3 ) 2 (1/RT) 2 (P SO2 ) 2 (P O2 ) (1/RT) 3 </li></ul><ul><li>K c = K p (1/RT) 2 = K p RT (1/RT) 3 </li></ul>
17. 17. General Equation - Deriving: <ul><li>j A + k B l C + m D </li></ul><ul><li>K p = (P C ) l (P D ) m = (C C xRT ) l (C D xRT ) m (P A ) j (P B ) k (C A xRT ) j (C B xRT ) k </li></ul><ul><li>K p = (C C ) l (C D ) m x(RT) l+m (C A ) j (C B ) k x(RT) j+k </li></ul><ul><li>K p = K (RT) ( l+m)-(j+k) = K c (RT)  n </li></ul><ul><li> n = (l+m)-(j+k) = Change in moles of gas </li></ul>
18. 18. General Equation pp <ul><li>K p = K c (RT)  n (R = .08206 l atm/K mol & T in Kelvin) </li></ul><ul><li>Where  n = change in moles of gas. </li></ul><ul><li>Watch for mixed equilibria (solids & liquids in reaction) </li></ul><ul><li>Do not count their moles in calculating  n. </li></ul><ul><li>Be sure to convert C to K for temp. </li></ul>
19. 19. Homogeneous Equilibria <ul><li>So far every example dealt with reactants and products where all were in the same phase. </li></ul><ul><li>So, we can use K in terms of either concentration or pressure if gas phase. </li></ul><ul><li>Units depend on the reaction. </li></ul><ul><li>Try # 29, 39 & 56, p 615 (5’ then WB). </li></ul>
20. 20. 13.4 Heterogeneous Equilibria <ul><li>If the reaction involves pure solids or pure liquids the concentration of the solid or the liquid does not change. </li></ul><ul><li>As long as they are not used up we can leave them out of the equilibrium expression. </li></ul><ul><li>For example: </li></ul>
21. 21. For Example <ul><li>H 2 (g) + I 2 ( s ) 2HI(g) </li></ul><ul><li>K = [HI] 2 [ H 2 ][ I 2 ] </li></ul><ul><li>But the [ ] of I 2 does not change since it’s a solid (treat as = to 1). So . . . </li></ul><ul><li>K [ I 2 ] = [HI] 2 = K’ [ H 2 ] </li></ul>
22. 22. E.g. , Write the K p Expression: <ul><li>2Fe (s) + 3/2 O 2(g) Fe 2 O 3(s) </li></ul><ul><li>K p = 1 (P O2 ) 3/2 </li></ul>
23. 23. 13.5 Applications of the Equilibrium Constant
24. 24. The Extent of a Reaction <ul><li>Size of K ( i.e , extent of rxn) and time to reach =m are not directly related. </li></ul><ul><li>Time to reach =m determined by kinetics; i.e., activation energy. </li></ul><ul><li>Large K means strong shift to the right </li></ul><ul><li>I.e., at =m have mostly products </li></ul>
25. 25. The Reaction Quotient <ul><li>Tells you the direction the reaction will go to reach equilibrium (lt vs. rt) </li></ul><ul><li>Calculated the same as the equilibrium constant, but for a system not at equilibrium (e.g., time = 0) </li></ul><ul><li>Q = [Products] coefficient [Reactants] coefficient </li></ul><ul><li>Compare value to equilibrium constant </li></ul>
26. 26. What Q tells us pp <ul><li>If Q < K </li></ul><ul><ul><li>Not enough products, so adjusts by </li></ul></ul><ul><ul><li>Shift to right (to make more) </li></ul></ul><ul><li>If Q > K </li></ul><ul><ul><li>Too many products, so adjusts by </li></ul></ul><ul><ul><li>Shift to left (to reduce products) </li></ul></ul><ul><li>If Q = K system at =m and no shift </li></ul>
27. 27. Example <ul><li>For the reaction 2 NOCl (g) 2 NO (g) + Cl 2 (g) </li></ul><ul><li>K = 1.55 x 10 -5 M at 35ºC </li></ul><ul><li>In an experiment 0.10 mol NOCl , 0.0010 mol NO and 0.00010 mol Cl 2 are mixed in 2.0 L flask. </li></ul><ul><li>Which direction will the reaction proceed to reach equilibrium? (Use initial concentrations) Answer . . . </li></ul>
28. 28. Example <ul><li>Q is not 1.0 x 10 -8 . Why? </li></ul><ul><li>You were given moles in a 2 liter flask </li></ul><ul><li>So [ ]s are halved </li></ul><ul><li>Q = 5.0 x 10 -9 </li></ul><ul><li>Since Q < K (which was 1.55 x 10 -5 ) </li></ul><ul><li>Reaction shifts to the right . </li></ul>
29. 29. 13.6 Solving Equilibrium Problems pp <ul><li>Given the starting concentrations and one equilibrium concentration. </li></ul><ul><li>Use stoichiometry to figure out other concentrations and K. </li></ul><ul><li>Learn to create a table of initial and final conditions. </li></ul><ul><li>Use “RICE” </li></ul>
30. 30. <ul><li>Consider the following at 600ºC pp </li></ul><ul><li>2 SO 2 (g) + O 2 (g) 2SO 3 ( g) </li></ul><ul><li>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: </li></ul><ul><li>The equilibrium concentrations of O 2 and SO 2 , K c and K P (procedure follows . . .) </li></ul>
31. 31. Strategy pp <ul><li>Make RICE table to show initial, change and final molarity (convert moles to M by dividing by volume) </li></ul><ul><li>Use stoich & mole ratios as needed. </li></ul><ul><li>Plug into =m expression to get K c </li></ul><ul><li>Use K p = K c (RT) ∆n </li></ul><ul><li>Remember: ∆n is change in gaseous moles only!!! </li></ul>
32. 32. Solving pp <ul><li> 2SO 2 (g) + O 2 (g 2SO 3 (g) initial 2.00 M 1.5 M 3.00 M change equil. 3.50 M </li></ul><ul><li>K c = [SO 3 ] 2 [SO 2 ] 2 [O 2 ] </li></ul>
33. 33. Solving pp <ul><li> 2SO 2 (g) + O 2 (g 2SO 3 (g) initial 2.00 M 1.5 M 3.00 M change 0.50 M 0.25 M 0.50 M final 3.50 M </li></ul><ul><li>K c = [SO 3 ] 2 [SO 2 ] 2 [O 2 ] </li></ul>
34. 34. Solving pp <ul><li> 2SO 2 (g) + O 2 (g 2SO 3 (g) initial 2.00 M 1.5 M 3.00 M change -0.50 M -0.25 M +0.50 M final 1.50 M 1.25 M 3.50 M </li></ul><ul><li>K c = [SO 3 ] 2 = (3.50) 2 [SO 2 ] 2 [O 2 ] (1.50) 2 (1.25) </li></ul><ul><li>K c = 4.36 </li></ul>
35. 35. Solving pp <ul><li>600º C 2SO 2 (g) + O 2 (g 2SO 3 (g) initial 2.00 M 1.5 M 3.00 M change -0.50 M -0.25 M +0.50 M final 1.50 M 1.25 M 3.50 M </li></ul><ul><li>K c = 4.36 </li></ul><ul><li>K p = K c (RT) ∆n and ∆n = -1 </li></ul><ul><li>K p = 4.36(0.08206 x 873) -1 = 6.09 x 10 -2 atm (or 3.80 torr). </li></ul>
36. 36. <ul><li>2SO 2 (g) + O 2 (g) 2SO 3 (g) @ 600ºC </li></ul><ul><li>In a different experiment .500 mol SO 2 and .350 mol SO 3 were placed in a 1.000 L container. When the system reaches equilibrium 0.045 mol of O 2 are present. </li></ul><ul><li>Calculate the final concentrations of SO 2 and SO 3 and K. Do on own -- your answers are . . . </li></ul>
37. 37. <ul><li>In a different expt at 600 ºC .500 mol SO 2 & .350 mol SO 3 placed in a 1.000 L container. At =m 0.045 mol of O 2 are present. </li></ul><ul><li>2SO 2 (g) + O 2 (g) 2SO 3 (g) </li></ul><ul><li>Same temp. , so K c is the same!! (4.36) </li></ul><ul><li>Since no O 2 initially, its change is +0.045 and the reaction shifted to the left . </li></ul><ul><li>Plug in to get SO 2 and SO 3 =m [ ]s (derived from the O 2 change in [ ]) </li></ul><ul><li>Your answer for their concentrations . . . </li></ul><ul><li>[SO 2 ] = .590 M & [SO 3 ] = .260 (mole ratios) </li></ul>
38. 38. What if not given equilibrium concentration ? pp <ul><li>Easy if you get a “perfect square.” </li></ul><ul><li>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 . . . </li></ul>
39. 39. Example continued pp <ul><li>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 . Write balanced equation, determine Q and direction of shift, set up RICE </li></ul><ul><li>Steps on next slides. </li></ul>
40. 40. Example cont. where K = 1.15 x 10 2 pp <ul><li>3.00 mol each of H 2 , F 2 , HF added to a 1.50 L flask. So, M = 2.00 for each. </li></ul><ul><li>H 2 (g) + F 2 (g) 2HF(g) </li></ul><ul><li>Q = (2.00) 2 = 1.00 < K eq (2.00)(2.00) </li></ul><ul><li>So, shifts to right . So, Left side is losing stuff and right side is gaining. </li></ul>
41. 41. Example continued pp <ul><li>H 2 (g) + F 2 (g) 2HF(g) initial 2.00 M 2.00 M 2.00 M change -x -x + 2x final 2.00-x 2.00-x 2.00+2x </li></ul><ul><li>K = 1.15 x 10 2 = (2.00 + 2x) 2 (2.00 - x)(2.00 - x) </li></ul><ul><li>K = 1.15 x 10 2 = (2.00 + 2x) 2 (2.00 - x) 2 </li></ul><ul><li>Perfect square, so take square root of both sides. </li></ul>
42. 42. Example continued pp <ul><li>√ K = 10.74 = (2.00 + 2x) (2.00 - x) </li></ul><ul><li>x = 1.528 </li></ul><ul><li>H 2 (g) + F 2 (g) 2HF(g) </li></ul><ul><li>final 2.00-x 2.00-x 2.00+2x </li></ul><ul><li>2.00 - x = [H 2 ] eq & [F 2 ] eq = 0.472 M </li></ul><ul><li>(2.00 + 2 x) = [HF] eq = 5.056 M </li></ul>
43. 43. What if you don’t get a “perfect square”? pp <ul><li>We are then stuck with having to solve a quadratic expression unless we can make simplifying assumptions . </li></ul><ul><li>To avoid a quadratic, we will take different approaches depending on the size of K c . </li></ul>
44. 44. General Rules to avoid quadratics pp <ul><li>When K c is large - find limiting reactant and set it as “x” in the “equilibrium” line of RICE. (Example follows). </li></ul><ul><li>When K c is small - easiest is to put “x” in the “change” line or RICE (see example). </li></ul><ul><li>If putting “x” in “change line then requires a quadratic or you get a negative [ ], put it in the “equilibrium line” (see example). </li></ul><ul><li>When K c is “middle” then must do quadratic. </li></ul>
45. 45. General Rules to avoid quadratics pp <ul><li>Why don’t we just always put “x” in the “equilibrium” line since it always works? </li></ul><ul><li>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. </li></ul><ul><li>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 -[ ]. </li></ul>
46. 46. Large K c example pp <ul><li>H 2 (g) + I 2 (g) 2HI(g) </li></ul><ul><li>K = 7.1 x 10 2 at 25ºC </li></ul><ul><li>Calculate the equilibrium concentrations if a 5.00 L container initially contains 15.7 g of H 2 and 294 g I 2 . </li></ul><ul><li>What is Q? . . . (can do without a calculator) </li></ul><ul><li>Q = 0 because initially no product. </li></ul><ul><li>[ H 2 ] 0 = (15.7g/2.02)/5.00 L = 1.56 M </li></ul><ul><li>[I 2 ] 0 = (294g/253.8)/5.00L = 0.232 M </li></ul><ul><li>[HI] 0 = 0 </li></ul><ul><li>Problem: Need to know limiting reactant . </li></ul>
47. 47. <ul><li>H 2 (g) + I 2 (g) 2HI(g) pp </li></ul><ul><li>Q = 0 < K so more product formed. </li></ul><ul><li>Assumption : Since K is large, reaction will go to completion ( I.e., forward arrow only). </li></ul><ul><li>But, it doesn’t quite go to completion. </li></ul><ul><li>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. </li></ul><ul><li>Set up RICE table in concentrations (use moles with stoich,but M in RICE) </li></ul>
48. 48. <ul><li>Choose X so it is small (so use I 2 since it was limiting reactant. </li></ul><ul><li>For I 2 the change in X must be X - 0.232 M </li></ul><ul><li>Final must = initial + change </li></ul>H 2 (g) + I 2 (g) 2HI(g) pp initial 1.56 M 0.232 M 0 M change Equilibrium X
49. 49. <ul><li>H 2 (g) + I 2 (g) 2HI(g) </li></ul><ul><li>Using stoichiometry we can find: </li></ul><ul><li>Change in H 2 = X-0.232 M </li></ul><ul><li>Change in HI = - twice change in H 2 </li></ul><ul><li>Change in HI = 0.464-2X </li></ul>H 2 (g) I 2 (g) 2HI(g) pp initial 1.56 M 0.232 M 0 M change X-0.232 final X
50. 50. <ul><li>Now we can determine the final concentrations by adding . </li></ul>H 2 (g) I 2 (g) 2HI(g) pp initial 1.56 M 0.232 M 0 M change X-0.232 X-0.232 0.464-2X final X
51. 51. <ul><li>Now plug these values into the equilibrium expression </li></ul><ul><li>K = (0.464-2X) 2 = 7.1 x 10 2 (1.328+X)(X) </li></ul><ul><li>Now you can see why we made sure X was small (by using the LR). </li></ul>H 2 (g) I 2 (g) 2HI(g) pp initial 1.56 M 0.232 M 0 M change X-0.232 X-0.232 0.464-2X final 1.328+X X 0.464-2X
52. 52. Why We Chose x to be Small pp <ul><li>K = (0.464-2X) 2 = 7.1 x 10 2 (1.328+X)(X) </li></ul><ul><li>Since x is going to be small, we can ignore it in relation to 0.464 and 1.328 </li></ul><ul><li>So we can rewrite the equation </li></ul><ul><li>7.1 x 10 2 = (0.464) 2 (1.328)(X) </li></ul><ul><li>Makes the algebra easy </li></ul>
53. 53. <ul><li>When we solve for X we get 2.3 x 10 -4 </li></ul><ul><li>So we can find the other concentrations </li></ul><ul><li>I 2 = 2.3 x 10 -4 M </li></ul><ul><li>H 2 = 1.328 M </li></ul><ul><li>HI = 0.464 M </li></ul><ul><li>Note: If had set “x” in change line we would have got (-) =m concentrations for H 2 & I 2 . </li></ul>H 2 (g) I 2 (g) 2HI(g) pp initial 1.56 M 0.232 M 0 M change X-0.232 X-0.232 0.464-2X final 1.328+X X 0.464-2X
54. 54. Checking the assumption pp <ul><li>Rule of thumb: If the value of X is less than 5% of all the other concentrations, assumption is valid. </li></ul><ul><li>If not we would have had to use the quadratic equation </li></ul><ul><li>More on this later. </li></ul><ul><li>Our assumption was valid. </li></ul>
55. 55. Why we couldn’t put x in “change” line pp <ul><li>(2x) 2 = 7.1 x 10 2 (flies on Matt) (1.56-x)(.232-x) </li></ul><ul><li>x = 8.01, and plugging into “final” we get: </li></ul><ul><li> -6.45 -7.78 +16.02 ( - ) concentrations for H 2 & I 2 !!! </li></ul>H 2 (g) I 2 (g) 2HI(g) initial 1.56 M 0.232 M 0 M change -x -x +2x final 1.56-x 0.232-x 0+2x
56. 56. Practice (put “x” in final line) <ul><li>For the reaction Cl 2 + O 2 2ClO(g) K = 156 </li></ul><ul><li>In an experiment 0.100 mol ClO, 1.00 mol O 2 and 0.0100 mol Cl 2 are mixed in a 4.00 L flask. </li></ul><ul><li>If the reaction is not at equilibrium, which way will it shift? </li></ul><ul><li>Calculate the equilibrium concentrations. </li></ul><ul><li>Try first, then we’ll review (next slides) </li></ul>
57. 57. Practice <ul><li>Cl 2 + O 2 2ClO(g) K = 156 </li></ul><ul><li>Remember to divide by 4 L to get molarity. </li></ul><ul><li>Q = 1 = products/reactants. Since less than K, reaction shifts to right. </li></ul><ul><li>Use RICE table of molarities (divide by 4 to get 0.025M ClO , .25 M O 2 and .0025 M Cl 2 </li></ul><ul><li>Since shift to right, make [=m] Cl 2 = x (since it is the LR) </li></ul>
58. 58. Practice <ul><li>K = 156 </li></ul><ul><li> Cl 2 + O 2 2ClO </li></ul><ul><li>Init . 0025 .25 .025 Chg x-.0025 x-.0025 .0050-2x Final x x+.2475 .0300-2x </li></ul><ul><li>add above </li></ul><ul><li>Simplify & solve for x. Your answer . . . </li></ul><ul><li>X = 2.33 x 10 -5 = [ Cl 2 ] </li></ul>
59. 59. Problems with small K pp K< .01
60. 60. Example of small K c pp <ul><li>90% of the time you can x in “change” line with a small K c , </li></ul><ul><li>For the reaction 2 NOCl (g) 2 NO (g) + Cl 2 (g) </li></ul><ul><li>K = 1.6 x 10 -5 M at 35ºC </li></ul><ul><li>In an experiment 1.00 mol NOCl , is placed in a 2.0 L flask. </li></ul><ul><li>Find the =m concentrations </li></ul>
61. 61. Example of small K c pp <ul><li>1 mol NOCl in 2 L, K = 1.6 x 10 -5 2NOCl (g) 2 NO (g) + Cl 2 (g) 0.50 M 0 0 -2x +2x +x 0.5 - 2x +2x +x </li></ul><ul><li>K c = 1.6 x 10 -5 = (2x) 2 (x) (0.50 - 2x) 2 </li></ul><ul><li>K c ≈ (2x) 2 (x) ≈ 4x 3 (0.50) 2 (0.50) 2 </li></ul><ul><li>x = 1.0 x 10 -2 </li></ul>
62. 62. Example of small K c pp <ul><li>1 mol NOCl in 2 L, K = 1.6 x 10 -5 2NOCl (g) 2 NO (g) + Cl 2 (g) 0.50 M 0 0 -2x +2x +x 0.5 - 2x +2x +x </li></ul><ul><li>Since x = 1.0 x 10 -2 , =m [ ]s are 0.48 0.02 0.01 </li></ul>
63. 63. If can’t solve with “x” in change line pp <ul><li>Set up table of initial, change, and final concentrations (RICE). </li></ul><ul><li>Choose X to be small (find LR) and put in “final” line of RICE. </li></ul><ul><li>For this case it will be a product because K c is small. </li></ul><ul><li>The following problem must be done this way to avoid a quadratic. </li></ul>
64. 64. For example pp <ul><li>For the reaction 2NOCl 2NO + Cl 2 </li></ul><ul><li>K= 1.6 x 10 -5 </li></ul><ul><li>If 1.20 mol NOCl , 0.45 mol of NO and 0.87 mol Cl 2 are mixed in a 1 L container </li></ul><ul><li>What are the equilibrium concentrations </li></ul><ul><li>Q = [NO] 2 [Cl 2 ] = (0.45) 2 (0.87) = 0.15 M [NOCl] 2 (1.20) 2 </li></ul><ul><li>Since Q > K, shift to left </li></ul>
65. 65. <ul><li>Choose X to be small </li></ul><ul><li>NO will be LR </li></ul><ul><li>Choose NO to be X </li></ul>2NOCl 2NO + Cl 2 pp Initial 1.20 0.45 0.87 Change Final
66. 66. <ul><li>Figure out change in NO </li></ul><ul><li>Change = final - initial </li></ul><ul><li>change = X-0.45 </li></ul>2NOCl 2NO Cl 2 pp Initial 1.20 0.45 0.87 Change Final X
67. 67. <ul><li>Now figure out the other changes </li></ul><ul><li>Use stoichiometry </li></ul><ul><li>Change in Cl 2 is 1/2 change in NO </li></ul><ul><li>Change in NOCl is - change in NO </li></ul>2NOCl 2NO Cl 2 pp Initial 1.20 0.45 0.87 Change X-.45 Final X
68. 68. <ul><li>Now we can determine final concentrations </li></ul><ul><li>Add </li></ul>2NOCl 2NO Cl 2 pp Initial 1.20 0.45 0.87 Change 0.45-X X-.45 0.5X - .225 Final X
69. 69. <ul><li>Now we can write equilibrium constant </li></ul><ul><li>K = (X) 2 (0.5X+0.645) (1.65-X) 2 </li></ul><ul><li>Now we can test our assumption X is small so ignore it in (+) and (-) </li></ul>2NOCl 2NO Cl 2 pp Initial 1.20 0.45 0.87 Change 0.45-X X-.45 0.5X - .225 Final 1.65-X X 0.5X + 0.645
70. 70. <ul><li>K = (X) 2 (0.645) = 1.6 x 10 -5 (1.65) 2 </li></ul><ul><li>X= 8.2 x 10 -3 </li></ul><ul><li>Figure out final concentrations </li></ul>2NOCl 2NO Cl 2 pp Initial 1.20 0.45 0.87 Change 0.45-X X-.45 0.5X - .225 Final 1.65-X X 0.5X +0.645
71. 71. <ul><li>X= 8.2 x 10 -3 </li></ul><ul><li>[NOCl] = 1.6 5 </li></ul><ul><li>[Cl 2 ] = 0.64 9 </li></ul><ul><li>Check assumptions (x < all other [ ]s) </li></ul><ul><li>.0082/.45 = 1.8 % OKAY!!! </li></ul>2NOCl 2NO + Cl 2 pp Initial 1.20 0.45 0.87 Change 0.45-X X-.45 0.5X - .225 Final 1.65-X X 0.5X +0.645
72. 72. Quadratic if put x in “change” pp <ul><li>(0.45+2x (0.87+x) 2 = 1.6 x 10 -5 (1.20-2x) 2 </li></ul><ul><li>So, here we need “x” in the “final” line </li></ul>2NOCl 2NO + Cl 2 Initial 1.20 0.45 0.87 Change -2x +2x +x Final 1.20-2X 0.45+2X 0.87+x
73. 73. Practice Problem - put “x” in final <ul><li>For the reaction 2ClO (g) Cl 2 (g) + O 2 (g) </li></ul><ul><li>K = 6.4 x 10 -3 </li></ul><ul><li>In an experiment 0.100 mol ClO(g) , 1.00 mol O 2 and 1.00 x 10 -2 mol Cl 2 are mixed in a 4.00 L container. </li></ul><ul><li>What are the equilibrium [ ]s? (Remember to change to molarity .) </li></ul>
74. 74. 2ClO(g) Cl 2 (g) + O 2 (g) <ul><li>Q = 1, > K so shift to left </li></ul><ul><li>Chlorine = LR so make it X </li></ul><ul><li> 2ClO(g) Cl 2 (g) + O 2 (g) Init . .025 .0025 .25 Chg 0050 - 2x x-.0025 x-.0025 Fin .030 - 2x x x+.2475 </li></ul><ul><li>X = 2.3 x 10 -5 </li></ul><ul><li>5% rule (.000023/.025)(100%) = 0.09% </li></ul>
75. 75. Mid-range K’s .01<K<10 2
76. 76. No Simplification Scenario <ul><li>Choose X to be small. </li></ul><ul><li>Can’t simplify so we will have to solve the quadratic (we hope) </li></ul><ul><li>H 2 (g) + I 2 (g) HI(g) K=38.6 </li></ul><ul><li>What is the equilibrium concentrations if 1.800 mol H 2 , 1.600 mol I 2 and 2.600 mol HI are mixed in a 2.000 L container? </li></ul>
77. 77. Problems Involving Pressure <ul><li>Solved exactly the same, with same rules for choosing “x” depending on K P </li></ul><ul><li>For N 2 O 4 (g) 2NO 2 (g) K p = .131atm What are the equilibrium pressures if a flask initially contains 1.000 atm N 2 O 4 ? </li></ul><ul><li>Assume no quadratic, put “x” in “change” column. Your Answers are . . . </li></ul><ul><li>0.819 atm NO 2 & .362 atm NO 2 </li></ul>
78. 78. Super Hints pp <ul><li>Always look first for a perfect square!! </li></ul><ul><li>Large K - put “x” in “final row </li></ul><ul><li>Small K - 98% of time can put “x” in change row, but . . . </li></ul><ul><li>If with small K you get (-) [ ]s or must use quadratic then put “x” in final row. </li></ul><ul><li>Putting “x” in final row always works, but is difficult to use when we get to acid/base/buffer equilibrium. </li></ul><ul><li>So for small K try “x” in change row first. </li></ul>
79. 79. 13.7 Le Chatelier’s Principle <ul><li>If a stress is applied to a system at equilibrium, the position of the equilibrium will shift to reduce the stress. </li></ul><ul><li>3 Types of stress </li></ul><ul><li>Change in amounts, pressure or temperature </li></ul>
80. 80. Changing the amounts of reactants and/or products <ul><li>Adding product makes Q > K </li></ul><ul><li>Removing reactant makes Q > K </li></ul><ul><li>Adding reactant makes Q < K </li></ul><ul><li>Removing product makes Q < K </li></ul><ul><li>Determining the effect on Q will tell you the direction of shift </li></ul>
81. 81. Can Change Pressure <ul><li>Because partial pressures (and concentrations) change a new equilibrium position must be reached. </li></ul><ul><li>By reducing volume </li></ul><ul><li>System will move in the direction that has the least moles of gas. </li></ul><ul><li>System tries to minimize the moles of gas. </li></ul><ul><li>By increasing volume , system moves in direction that has the most moles of gas. </li></ul>
82. 82. Can Change Total Pressure <ul><li>By adding an inert gas. </li></ul><ul><li>BUT partial pressures of reactants and products are not changed </li></ul><ul><li>No effect on equilibrium position (test concept and problem questions) </li></ul>
83. 83. Change in Temperature <ul><li>Affects the rates of both the forward and reverse reactions. </li></ul><ul><li>So, doesn’t just change the equilibrium position , but also changes the equilibrium constant . </li></ul><ul><li>The direction of the shift depends on whether it is exo- or endothermic </li></ul>
84. 84. Exothermic <ul><li> H < 0 </li></ul><ul><li>Releases heat </li></ul><ul><li>Think of heat as a product in exothermic rxns </li></ul><ul><li>Raising temperature pushes toward reactants . </li></ul><ul><li>Shifts to left . </li></ul>
85. 85. Endothermic <ul><li> H > 0 </li></ul><ul><li>Requires heat </li></ul><ul><li>Think of heat as a reactant </li></ul><ul><li>Raising temperature push toward products . </li></ul><ul><li>Shifts to right . </li></ul>
86. 86. Review <ul><li>6CO 2(g) + 6H 2 O (l) C 6 H 12 O 6(s) + 6O 2(g) ∆H = 2820 kJ </li></ul><ul><li>Some CO 2(g is added and causes . . . </li></ul><ul><li>Shift to right (Q < K) so get more C 6 H 12 O 6(s) . </li></ul><ul><li>Temperature is raised . . . </li></ul><ul><li>Q < K Shift to right, more C 6 H 12 O 6(s) forms. </li></ul><ul><li>Volume is decreased </li></ul><ul><li>Equal moles of gaseous reactants and products, so no change. </li></ul>
87. 87. Review <ul><li>6CO 2(g) + 6H 2 O (l) C 6 H 12 O 6(s) + 6O 2(g) ∆H = 2820 kJ </li></ul><ul><li>Some O 2(g) is removed . . . </li></ul><ul><li>Q < K shift to right, moreC 6 H 12 O 6(s) forms. </li></ul><ul><li>Some C 6 H 12 O 6(s) is removed . . . </li></ul><ul><li>No change (glucose is a solid so is not in the equilbrium expression). No more C 6 H 12 O 6(s) . </li></ul><ul><li>A catalyst is added . . . </li></ul><ul><li>No effect on equilibrium, only allows it be attained more quickly. No more C 6 H 12 O 6(s) . </li></ul>
88. 88. Review <ul><li>6CO 2(g) + 6H 2 O (l) C 6 H 12 O 6(s) + 6O 2(g) ∆H = 2820 kJ </li></ul><ul><li>Some H 2 O is removed . . . </li></ul><ul><li>No change, water is a liquid. No additional C 6 H 12 O 6(s) forms. </li></ul><ul><li>End, Chapter 13. </li></ul>