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).
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%