Transcript Notes One Unit Ten

Notes One Unit Eleven
• Dynamic Equilibrium
• Demo Siphon
• Demo Dynamic Equilibrium
• Rate of reaction Forward = Rate of Reaction Backward
• Chemical Concentration before and after equilibrium
• Mass Action Expression
• Calculating Equilibrium Constant K from Concentration.
Demo Siphon
Demo Dynamic Equilibrium
Demo Dynamic Equilibrium
Dynamic Equilibrium
• Reversible reactions.
• One reaction going forward.
• One reaction going backward.
• Temperature can change the Equilibrium
Chemical Concentration before and after equilibrium
•(a) Only 0.04 M N2O4 present initially
•(b) Only 0.08 M NO2 present initially
Mass Action Expression
Mass Action Expressions
c
d
[C] [D]
Keq=
b
a
[A] [B]
Solids and liquids are left out
Chemical Concentration before and after equilibrium
2
2
2
2
[NO
]
[0.0125]
[0.0125]
[0.0156]
______
_______
_______
_______
2
Keq=
=
=
=
[N2O4]
[0.0337] [0.0337] [0.0522]
Experimental Data:
Initial
Initial
Equilibrium
Trial Concentration Concentration
Constant
Number [N O ] [NO ] [N O ] [NO ] [NO ] 2/[N O ]
2 4
2
2 4
2
2
2 4
1
2
3
4
5
0.0400 0.0000 0.0337 0.0125
0.0000 0.0800 0.0337 0.0125
4.64x10-3
0.0600 0.0000 0.0522 0.0156
0.0000 0.0600 0.0246 0.0107
0.0200 0.0600 0.0429 0.0141
4.66x10-3
4.64x10-3
4.65x10-3
4.63x10-3
If the Temperature is the same, K will be the same.
Mass Action Expression
• 4NH3(g) + 7O2(g)4NO2(g) + 6H2O(g)
4
6
[NO2] [H2O]
__________
Keq=
4
7
[NH3] [O2]
• CaCO3(s)CaO(s) + CO2(g)
Keq= [CO2]
• PCl3(l) + Cl2(g) PCl5 (s)
• H2(g) + F2(g)  2HF(g)
1
____
Keq=
[Cl2]
2
[HF]
_______
Keq=
[H2] [F2]
Calculating K
Calculate the Keq at 740C, if [CO] = 0.012 M,
[Cl2] = 0.054 M, and [COCl2] = 0.14 M.
1) Balanced Equation
CO(g) + Cl2(g)→ COCl2(g)
2) Mass Action Equation
[COCl2]
________
Keq=
[CO][Cl2]
3) Calculate K
1]
[0.14
M
_________________
Keq=
[0.012M1] [0.054M1]
Keq= 220M-1
Calculating Concentration From K
The equilibrium constant K for the reaction is 158 atm at
1000K. What is the equilibrium pressure of O2, if the PNO2 =
0.400 atm and PNO = 0.270 atm?
1) Balanced Equation
2 NO2(g) 2 NO(g) + 1O2(g)
2) Mass Action Equation
2
[NO] [O2]
________
K=
[NO2]2
2
[NO] [O2]
2
[NO2] x K =
x[NO
]
2
2
[NO
]
K
K
2
2
3) Solve for [O2]?
2
[NO2]
_______
=[O2]
Kx
2
[NO]
4) Calculate K
2
(0.400atm)
(158atm) _________2 =[O2 ]
(0.270atm)
[O2 ]= 347 atm
Notes Two Unit Eleven Chapter Fourteen

Le Chatelier's Principle

Silver Chloride Demo

Calculating K from Initial Conditions

Calculating Concentration From Ksp
Le Chatelier's Principle
• If an external stress is applied, the system adjusts
• An increased stress is reduced.
• A decreased stress is increased.
Le Chatelier's Lab
In step 3, hydrochloric acid is used as a source of Cl-1 ions.
Co(H2O)6+2(aq) + 4Cl-1(aq) CoCl4+2(aq) +6H2O(l)
We see more blue!
In step 5, why did adding H2O cause the change that it did?
Co(H2O)6+2(aq) + 4Cl-1(aq) CoCl4+2(aq) +6H2O(l)
We see more red!
In step 6, silver ions from the AgNO3 react with Cl- ions to
produce an insoluble precipitate.
Co(H2O)6+2(aq) + 4Cl-1(aq) CoCl4+2(aq) +6H2O(l)
We see more red!
In step 7, acetone has an attraction for H2O.
Co(H2O)6+2(aq) + 4Cl-1(aq) CoCl4+2(aq) +6H2O(l)
We see more blue!
Writing Solubility Reactions
Ag3PO4 Dissolves:
1 cation 1 ion
Ag3PO4 (s)3 Ag+1 + PO4-3
ScF3 Dissolves:
1 cation
ScF3 (s)
Sc+3 +
Sn3P4 Dissolves:
1 ion
3 F-1
1 cation
1 ion
Sn3P4 (s)  3 Sn+4 + 4P-3
Silver Chloride Demo
AgCl(s) Ag+1(aq) + Cl-1(aq)
NaCl is added
We see a cloudy solid:AgCl(s)!
Le Chatelier's Principle!
Writing Solubility Reactions
NaCl Dissolves:
NaCl(s) Na+1 + Cl-1
CaF2 Dissolves:
CaF2(s) Ca+2 + 2F-1
Calculating Concentration From Ksp
What is the concentration of the cation and anion for
cadmium arsenate,Cd3(AsO4)2, if Ksp=2.2×10-33 M5?
1) Balanced Equation
Cd3(AsO4)2(s)  3Cd+2(aq) + 2 AsO4-3(aq)
2) Mass Action Equation
+2] 3 [AsO -3]2
[Cd
Ksp =
4
3) What do we know?
Before Eq
0
0
Change
+ 3X + 2X
At Eq
3X
2X
4) Calculate X
2.2×10-33= [3X] 3 [2X] 2
2.2×10-33= 108X5
^(1/5)
-33
(2.2×10 )
________
X=
108
X=1.2x10-7M1
[Cd+2] =3(1.2x10-7M )
[AsO4-3] =2(1.2x10-7M )
Calculating Ksp from Concentration
If the molar solubility of BiI3 is 1.32 x 10-5, find its Ksp.
1) Balanced Equation
BiI3(s)  Bi+3 + 3 I-1
2) Mass Action Equation
Ksp = [Bi+3] [I-1]3
3) What do we know?
Before Eq
0
0
Change
+ X + 3X
At Eq
X
3X
4) Calculate Ksp
Ksp = [Bi+3][I-1]3
Ksp = [X][3X]3 = 27 X 4
Ksp = 27(1.32 x
4
-5
10 ) =
8.20 x 10-19 M4
Calculating K from Initial Conditions
In a flask 1.50M H2 and 1.50M N2 is allowed to reach
equilibrium. At equilibrium [NH3] =0.33M. Calculate K.
1) Balanced Equation
3 H2(g) +1 N2 (g)→ 2 NH3(g)
2) Mass Action Equation
2
[NH
]
________
3
K=
3
[H2] [N2]
3) What do we know?
Before Eq
Change
At Eq
H2
1.50
-0.50
1.00
N2
1.50
- 0.17
1.33
NH3
0
+0.33
0.33
4) Calculate K
2
[0.33]
___________
K= [1.00]3 [1.33]
K= 0.082M-2