![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).](https://image.slidesharecdn.com/ch13z5eequilibrium-110115231237-phpapp01/85/ch13-z5e-equilibrium-1-320.jpg)
![13.1 Reactions are reversible ,[object Object],[object Object],[object Object],[object Object],[object Object]](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
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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).
- 3. Reaction Rate Time Forward Reaction Reverse reaction Equilibrium
- 11. Equilibrium Constant One for each Temperature
- 23. 13.5 Applications of the Equilibrium Constant
- 59. Problems with small K pp K< .01
- 75. Mid-range K’s .01<K<10 2
Notes de l'éditeur
- 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 & 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 & 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 & solve for x = 2.33 x 10 -5 = [ Cl 2 ]
- Q = 1, > K so shift to left Chlorine = LR so make it X 2ClO(g) Cl 2 (g) + O 2 (g) Init .025 <--> .0025 .25 Chg ,0050 - x x-.0025 x-.0025 Fin .030 - x x x+.2475 X = .000023 5% rule .000023/.025 = .09%
- .361 & .820
- .361 & .820
- Z5e 640 Section 13.7 Le Chatelier’s Principle
- Vonderbrink PL 17