Outline: 2/21/07

Today: Chapter 16

Turn in Seminar Reports – to me Jaecker Applications – Chem Dept.

Chemical Equilibrium:

Types of Keq

Manipulating/Calculating Keq

LeChâtelier’s principle

Equilibrium Constant Rules:

Solids and (pure) liquids are left out of the Keq expression

Units of Keq are defined to be 1….

Reactions can be written in either direction at equilibrium

Magnitude of Keq tells you about the extent of reaction

We have defined a constant of the rxn: Keq = [products]/[reactants]

Why is it so important?

Types of Equilibrium Constants:Salt Solubility Product: CuCl(s) Cu+

(aq) + Cl(aq) Keq = [Cu+][Cl] = Ksp

Acid-Base reactions: HA(aq) + H2O(aq) H3O+

(aq) + A(aq)

Keq = [H3O+][A]/ [HA] = Ka

B(aq) + H2O(aq) BH+(aq) + OH

(aq)

Keq = [OH][BH]/ [B] = Kb

Types of Equilibrium Constants:Gas-liquid : (e.g. vapor pressure) H2O() H2O(g) Keq = pH2O

Henry’s Law: Keq = KH

Gas-solid : (e.g. vapor pressure) CO2(s) CO2(g) Keq = pCO2

Gas-aqueous: CO2(g) CO2(aq) Keq = [CO2]/ pCO2

Demo KH

Types of Equilibrium Constants:Formation reactions: Ag+

(aq) + 2NH3(aq) Ag(NH3)2+

(aq)

Keq=[Ag(NH3)2+]/[Ag+][NH3] = Kf

Lots of different names…. Keq, KH , Ksp, Ka , Kb, Kf , Kc, Kp…

All the same idea! (aq)

(g)

Return to Worksheet #6

Do problems B & C & DDo problems B & C & D

Worksheet #6B.B. HOAc + HHOAc + H22O O H H33OO++ + OAc + OAc

0.050 M0.050 M 0 0 0.20 M 0.20 M

KKeqeq = 1.8e = 1.8e = [H = [H33OO++][OAc][OAc] / [HOAc]] / [HOAc]

0.0500.050xx xx 0.200.20+x+x

= (= (xx)(0.20)(0.20+x+x)) / / (0.050(0.050xx))

Assume Assume xx is small… is small…

x x = 4.5e= 4.5eMM

Worksheet #6C.C. NN22 + 3H + 3H22 2NH 2NH33

0.150.15 0.25 0.25 00

KKeqeq = 0.040 = [NH = 0.040 = [NH33]]2 2 / [N/ [N22][H][H22]]33

0.150.15xx 0.250.25xx 2x2x

= (2x)2 / (0.15x)(0.25x)3

x = 4.8e4.8eMM

Is Is xx small compared to 0.15 atm? small compared to 0.15 atm?

Worksheet #6: Last Problem

init: 2 NO(g) + O2(g) 2 NO2(g)

5.0 atm 5.0 atm 0.0 atm

Keq= 4.2 1012Keq= 4.2 1012 = (pNO2)2/(pNO)2(pO2)

= (2x)2/(52x)2(5x) ???

(52x) (5x) +2x

Given this Keq is x small?

NO !

Worksheet #6: A new trick…

init: 2 NO(g) + O2(g) 2 NO2(g)

5.0 atm 5.0 atm 0.0 atm

Keq= 4.2 1012 = (pNO2)2/(pNO)2(pO2)

= (52x)2/(2x)2(2.5+x) ???

(55) (52.5) +5

Given this Keq is x small?

0.0 atm 2.5 atm 5.0 atmnew init:

(+2x) (2.5+x) (52x)

YES !

Worksheet #6: Last Problem

init: 2 NO(g) + O2(g) 2 NO2(g)

0.0 atm 2.5 atm 5.0 atm

Keq= 4.2 1012 = (pNO2)2/(pNO)2(pO2)

= (52x)2/(2x)2(2.5+x)

x = 7.7 10

(+2x) (2.5+x) (52x)

= (5)2/(2x)2(2.5)