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