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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).
13.1 Reactions are reversible ,[object Object],[object Object],[object Object],[object Object],[object Object]
Reaction Rate Time Forward Reaction Reverse reaction Equilibrium
What is equal at  Equilibrium? ,[object Object],[object Object],[object Object],[object Object],[object Object]
13.2 Law of Mass Action   pp ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Playing with K  pp ,[object Object],[object Object]
Playing with K  pp ,[object Object],[object Object],[object Object],[object Object]
Calculate K - all are gases, given    N 2  + 3H 2     2NH 3   where K = 1.3 x 10 -2 ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
The units for K ,[object Object],[object Object],[object Object]
K is CONSTANT ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Equilibrium Constant One for each Temperature
Calculate K c ,[object Object],[object Object],[object Object],[object Object]
Calculate K ,[object Object],[object Object],[object Object],[object Object]
13.3 Equilibrium and  Pressure ,[object Object],[object Object],[object Object],[object Object],[object Object]
Equilibrium and Pressure ,[object Object],[object Object],[object Object]
Deriving a relationship between K c  & K p ,[object Object],[object Object],[object Object],[object Object],[object Object]
General Equation - Deriving: ,[object Object],[object Object],[object Object],[object Object],[object Object]
General Equation  pp ,[object Object],[object Object],[object Object],[object Object],[object Object]
Homogeneous Equilibria ,[object Object],[object Object],[object Object],[object Object]
13.4 Heterogeneous Equilibria ,[object Object],[object Object],[object Object]
For Example ,[object Object],[object Object],[object Object],[object Object]
E.g. , Write the K p  Expression: ,[object Object],[object Object]
13.5 Applications of the Equilibrium Constant
The Extent of a Reaction ,[object Object],[object Object],[object Object],[object Object]
The Reaction Quotient ,[object Object],[object Object],[object Object],[object Object]
What Q tells us  pp ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Example ,[object Object],[object Object],[object Object],[object Object]
Example ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
13.6 Solving Equilibrium Problems  pp ,[object Object],[object Object],[object Object],[object Object]
[object Object],[object Object],[object Object],[object Object]
Strategy  pp ,[object Object],[object Object],[object Object],[object Object],[object Object]
Solving  pp ,[object Object],[object Object]
Solving  pp ,[object Object],[object Object]
Solving  pp ,[object Object],[object Object],[object Object]
Solving  pp ,[object Object],[object Object],[object Object],[object Object]
[object Object],[object Object],[object Object]
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
What if not given equilibrium  concentration ?   pp ,[object Object],[object Object]
Example continued   pp ,[object Object],[object Object]
Example cont. where K = 1.15 x 10 2   pp ,[object Object],[object Object],[object Object],[object Object]
Example continued  pp ,[object Object],[object Object],[object Object],[object Object]
Example continued  pp ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
What if you don’t get a “perfect square”?  pp ,[object Object],[object Object]
General Rules to avoid quadratics   pp ,[object Object],[object Object],[object Object],[object Object]
General Rules to avoid quadratics   pp ,[object Object],[object Object],[object Object]
Large K c  example   pp ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
[object Object],[object Object],[object Object],H 2 (g)  +   I 2 (g)  2HI(g)   pp  initial 1.56  M 0.232  M   0  M   change   Equilibrium   X
[object Object],[object Object],[object Object],[object Object],[object Object],H 2 (g)    I 2 (g)  2HI(g)   pp  initial 1.56  M 0.232  M   0  M   change X-0.232 final   X
[object Object],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
[object Object],[object Object],[object Object],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
Why We Chose x to be Small   pp ,[object Object],[object Object],[object Object],[object Object],[object Object]
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],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
Checking the assumption  pp ,[object Object],[object Object],[object Object],[object Object]
Why we couldn’t put x in “change” line  pp ,[object Object],[object Object],[object Object],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
Practice (put “x” in final line) ,[object Object],[object Object],[object Object],[object Object],[object Object]
Practice ,[object Object],[object Object],[object Object],[object Object],[object Object]
Practice ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Problems with small K  pp K< .01
Example of small K c   pp ,[object Object],[object Object],[object Object],[object Object],[object Object]
Example of small K c   pp ,[object Object],[object Object],[object Object],[object Object]
Example of small K c   pp ,[object Object],[object Object]
If can’t solve with “x” in change line   pp ,[object Object],[object Object],[object Object],[object Object]
For example  pp ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
[object Object],[object Object],[object Object],2NOCl   2NO + Cl 2   pp   Initial  1.20   0.45 0.87 Change Final
[object Object],[object Object],[object Object],2NOCl   2NO Cl 2   pp Initial  1.20   0.45 0.87 Change Final   X
[object Object],[object Object],[object Object],[object Object],2NOCl   2NO Cl 2   pp   Initial  1.20   0.45 0.87 Change   X-.45 Final   X
[object Object],[object Object],2NOCl   2NO Cl 2   pp   Initial  1.20   0.45 0.87 Change 0.45-X   X-.45  0.5X - .225 Final   X
[object Object],[object Object],[object Object],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
[object Object],[object Object],[object Object],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
[object Object],[object Object],[object Object],[object Object],[object Object],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
Quadratic if put x in “change”  pp   ,[object Object],[object Object],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
Practice Problem - put “x” in final ,[object Object],[object Object],[object Object],[object Object]
2ClO(g)    Cl 2  (g) + O 2  (g) ,[object Object],[object Object],[object Object],[object Object],[object Object]
Mid-range K’s .01<K<10 2
No Simplification Scenario ,[object Object],[object Object],[object Object],[object Object]
Problems Involving Pressure ,[object Object],[object Object],[object Object],[object Object]
Super Hints  pp ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
13.7 Le Chatelier’s Principle ,[object Object],[object Object],[object Object]
Changing the amounts of reactants and/or products ,[object Object],[object Object],[object Object],[object Object],[object Object]
Can Change Pressure ,[object Object],[object Object],[object Object],[object Object],[object Object]
Can Change Total Pressure ,[object Object],[object Object],[object Object]
Change in  Temperature ,[object Object],[object Object],[object Object]
Exothermic ,[object Object],[object Object],[object Object],[object Object],[object Object]
Endothermic ,[object Object],[object Object],[object Object],[object Object],[object Object]
Review ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Review ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Review ,[object Object],[object Object],[object Object],[object Object]

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Ch13 z5e equilibrium

  • 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).
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  • 3. Reaction Rate Time Forward Reaction Reverse reaction Equilibrium
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  • 11. Equilibrium Constant One for each Temperature
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  • 23. 13.5 Applications of the Equilibrium Constant
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Notas del editor

  1. Ch 13 Section 13.1 Z5e
  2. Z5e 13.1 p611 Analogy: Flow of cars across a bridge connecting two island cities
  3. Z5e 13.2 p615 Law of Mass Action does NOT apply to solids or pure liquids.
  4. Z5e 650 #21
  5. Z5e p617
  6. Z5e 618 Table 13.1
  7. Rf Z5e 618 table 13.1
  8. Z5e 619 Section 13.3
  9. Section 13.4 Z5e 622
  10. Section 13.5 Z53 634 Applications of =m constant
  11. Z5e 627
  12. Section 13.6 Z5e 634 RICE: reaction initial [ ] change in [ ] equilibrium [ ]
  13. 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 ±!!
  14. Same temp so K is same (4.36)! Convert to [ ] Make initial, change, final mole table
  15. Same temp so K is same (4.36)! Convert to [ ] Make initial, change, final mole table
  16. Z5e 633 SE 13.11
  17. Z5e 633 SE 13.11
  18. Z5e 633 SE 13.11
  19. Z5e 633 SE 13.11
  20. Z5e 633 SE 13.11
  21. This is the Green method.
  22. This is the Green method.
  23. This is the Green method.
  24. 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 ]
  25. 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 ]
  26. 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 ]
  27. 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%
  28. .361 &amp; .820
  29. .361 &amp; .820
  30. Z5e 640 Section 13.7 Le Chatelier’s Principle
  31. Vonderbrink PL 17