Embed Size (px)
DESCRIPTION
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. - PowerPoint PPT Presentation
Citation preview
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
Solids and liquids are left out
[C][A] [B]Keq=
[D]dc
a b
Chemical Concentration before and after equilibrium
Experimental Data:
If the Temperature is the same, K will be the same.
[NO2][N2O4]______Keq=
2 [0.0125][0.0337]_______2
=
TrialNumber
Initial Concentration
Initial Concentration
EquilibriumConstant
[N2O4] [NO2] [N2O4] [NO2] [NO2] 2/[N2O4]
1
2
3
4
5
0.0400 0.0000 0.0337 0.0125 4.64x10-3
0.0000 0.0800 0.0337 0.0125 4.64x10-3
0.0600 0.0000 0.0522 0.0156 4.66x10-3
0.0000 0.0600 0.0246 0.0107 4.65x10-3
0.0200 0.0600 0.0429 0.0141 4.63x10-3
[0.0125][0.0337]_______2
=[0.0156][0.0522]_______2
=
Mass Action Expression
• 4NH3(g) + 7O2(g)4NO2(g) + 6H2O(g)
• CaCO3(s)CaO(s) + CO2(g)
• PCl3(l) + Cl2(g) PCl5 (s)
• H2(g) + F2(g) 2HF(g)
[NO2][NH3] [O2]__________
Keq=[H2O]
64
4 7
[CO2]Keq=
[Cl2]Keq=
[HF]
[H2] [F2]_______
Keq=2
____1
Calculating K
2) Mass Action Equation2) Mass Action Equation
1) Balanced Equation1) Balanced Equation
3) Calculate 3) Calculate K
Calculate the Keq at 740C, if [CO] = 0.012 M, [Cl2] = 0.054 M, and [COCl2] = 0.14 M.
CO(g) + Cl2(g)→ COCl2(g)
[COCl2][CO][Cl2]________
Keq=
[0.14 M1][0.012M1] [0.054M1]_________________
Keq= 220M-1Keq=
Calculating Concentration From K
2) Mass Action Equation2) Mass Action Equation
3) Solve for [O3) Solve for [O22]?]?
1) Balanced Equation1) Balanced Equation
4) Calculate 4) Calculate 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?
2 12NO2(g) NO(g) + O2(g)
[NO]
[NO2][O2]________K=
2
2
[NO][NO2] =[O2]
_______K x2
2
(0.270atm)
(0.400atm)=[O2 ]
_________(158atm)2
2[O2 ]= 347atm
K[NO] [O2][NO2]
2 K=
Kx[NO2]
2[NO2] 2 x
2
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
Co(H2O)6+2(aq) + 4Cl-1(aq) CoCl4
+2(aq) +6H2O(l) In step 3, hydrochloric acid is used as a source of Cl-1 ions.
We see more blue!
Co(H2O)6+2(aq) + 4Cl-1(aq) CoCl4
+2(aq) +6H2O(l) In step 5, why did adding H2O cause the change that it did?
We see more red!
Co(H2O)6+2(aq) + 4Cl-1(aq) CoCl4
+2(aq) +6H2O(l)
In step 6, silver ions from the AgNO3 react with Cl- ions to produce an insoluble precipitate.
We see more red!
Co(H2O)6+2(aq) + 4Cl-1(aq) CoCl4
+2(aq) +6H2O(l) In step 7, acetone has an attraction for H2O.
We see more blue!
Writing Solubility Reactions
Ag3PO4 (s) Ag+1 + PO4-33
ScF3 (s) Sc+3 + F-13
Sn3P4 (s) Sn+4 + P-33 4
Ag3PO4 Dissolves:
ScF3 Dissolves:
Sn3P4 Dissolves:
1 cation 1 ion
1 cation 1 ion
1 cation 1 ion
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
CdCd33(AsO(AsO44))22(s) CdCd+2+2(aq)(aq) + AsOAsO44-3-3(aq)(aq)223
[AsOAsO44-3-3][CdCd+2+2] 23
00 00
3X3X++ 2X2X++3X3X 2X2X
[2X2X][3X3X] 23
108108XX55
(2.2×10(2.2×10-33-33))________________108108
^(1/5)^(1/5) X=1.2x10X=1.2x10-7-7MM11
[Cd[Cd+2+2] =3(] =3(1.2x101.2x10-7-7M M ))[AsO[AsO44
-3-3] =2(] =2(1.2x101.2x10-7-7M M ))X=X=
Ksp =Ksp =2) Mass Action Equation2) Mass Action Equation
3) What do we know?3) What do we know?Before Eq Before Eq ChangeChange
At Eq At Eq
1) Balanced Equation1) Balanced Equation
4) Calculate X4) Calculate X2.2×102.2×10-33-33==
2.2×102.2×10-33-33==
What is the concentration of the cation and anion for What is the concentration of the cation and anion for cadmium arsenate,Cdcadmium arsenate,Cd33(AsO(AsO44))22, if Ksp=2.2×10, if Ksp=2.2×10-33 -33 M5??
Calculating Ksp from Concentration
00 00
XX++ 3X3X++XX 3X3X
Ksp =Ksp =2) Mass Action Equation2) Mass Action Equation
3) What do we know?3) What do we know?Before Eq Before Eq ChangeChange
At Eq At Eq
1) Balanced Equation1) Balanced Equation
4) Calculate 4) Calculate Ksp
Ksp ==
Ksp ==
If the molar solubility of BiI3 is 1.32 x 10-5, find its Ksp.
BiI3(s) Bi+3 + I-13
[I-1][Bi+3] 3
[I-1][Bi+3] 3
[3X][X] 3 27X4=
(1.32 x 10-5)27 4 8.20 x 10-19= M4Ksp ==
Calculating K from Initial Conditions
2) Mass Action Equation2) Mass Action Equation
3) What do we know?3) What do we know?
Before Eq Before Eq ChangeChange
At Eq At Eq
1) Balanced Equation1) Balanced Equation
4) Calculate 4) Calculate K
In a flask 1.50M H2 and 1.50M N2 is allowed to reach equilibrium. At equilibrium [NH3] =0.33M. Calculate K.
3 21H2(g) + N2 (g)→ NH3(g)
[NH3][H2] [N2]________K=
2
3
H2 N2 NH31.50 1.50 00.501.00
0.17 0.331.33 0.33
- - +
[0.33][1.00] [1.33]___________
K=2
3K= 0.082M-2
