Outline:2/23/07

Outline: 2/23/07

Today: Finish Chapter 16

Chem Dept Seminar – today@4pm CAPA 10 – deadline moved to Mon.

Chemical Equilibrium:

LeChâtelier’s principle

G and Keq relationship

Types of Equilibrium Constants:

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

All the same idea: Extent of rxn…

Keq is a number that describes the ratio of products/reactants…

LeChâtelier’s Principle: A system reacts to change in the

direction that minimizes the change.

Change in concentrations Change in pressures Change in temperatures

Demo

LeChâtelier’s Principle Demo:

Which way will the reaction shift when water is added?

Co(H2O)62+

(aq) + 4Cl CoCl4 2

(aq) + 6H2O

Toward reactants!!

LeChâtelier’s Principle Demo:

Which way will the reaction shift when it is heated?

Toward products!!

H > 0 (endothermic)

Co(H2O)62+

(aq) + 4Cl CoCl4 2

(aq) + 6H2O

Co(H2O)62+

(aq) + heat CoCl4 2

(aq)

Practice problem:

Keq = [Pb2+][Cl]2

More PbCl2 is added:

More H2O is added:

Solid NaCl is added: Solid KNO3 is added:

PbClPbCl2(s)2(s) PbPb2+2+(aq)(aq) + 2 Cl + 2 Cl(aq)(aq)

More dissolves More precip.

Nothing

Nothing

One last link to thermodynamics: G = Go

+ RTlnQ At equilibrium: G = 0 ; Q =Keq

0 = Go + RTlnKeq

or Go = RTlnKeq

or Keq= eGº/RT

There is a relationship between Go and Keq! (see CAPA-11)

Example:

Use the thermodynamic tables to find Keq for the following rxn:

N2 + 3 H2 2 NH3 at 298 K

Go(kJ/mol) 0 0 16.45

Gorxn = 32.9 kJ/mol

Keq= eGo/RT = e32,900/(8.315*298) = 585000Watch units!

Example (non std. temp):


N2 + 3 H2 2 NH3 at 773 KHo(kJ/mol) 0 0 46.1So(J/molK) 192 131 192o

rxn = 92.2 kJ/molSo

rxn = 198.8 J/mol K

GTrxn = o

rxn T Sorxn = 61.5 kJ

Keq= eGT/RT = e61,500/(8.315*773) = 7.0105

Example (non std. temp):


N2 + 3 H2 2 NH3 Ho(kJ/mol) 0 0 46.1So(J/molK) 192 131 192o

rxn = 92.2 kJ/molSo

rxn = 198.8 J/mol K

Exothermic reaction

+ heat

Keq@ 773K= 7.010-5

Keq@ 298K= 585000

Example: Keq to Go

You can also find Go given Keq

N2 + 3 H2 2 NH3 at 773 KKeq= 7.0105

Go = RTlnKeq

Go = (8.314 J/k mol)(773 K) ln(7.0 105)

=61.5 kJ/mol

Let’s practice some more:

CAPA-11: problem #3CAPA-11: problem #3

Some reaction…

calculate Keq from G… at 220 K!

Keq= eGº/RT

GT = HºTSº

Keq= eHº/RT+Sº/R

Let’s practice some more: CAPA-11: problem #7CAPA-11: problem #7

2A B + C

If 1.00 atm of A initially, and 0.24 atm of C at equilibrium…whats Keq?

0.24 atm C means 0.24 atm B at equil.

Keq= (0.24)(0.24)/ (1.00.48)2

0.24 atm C means 0.48 atm A decays

Let’s practice some more: CAPA-11: problems #9 & #10CAPA-11: problems #9 & #10

A + B C Kc = 710

calculate [C]/[B] at equilibrium if:

0.01 M A is added to 0.20 M B

What must you do first?

K is big: take reactants over to products


A + B C Kc = 7100.01 0.20 0.00 Initial

0.00 0.19 0.01 Initial0.01 0.01 +0.01

+ x + x x Changex (0.19+x) (0.01x) Equilibrium

K=710 = (0.01x) / x (0.19+x)

x is small…x = 7.41e-5


A + B C Kc = 7107.41e-5 0.19 0.01 Initial

Does reaction quotient change?Dilute everything by 10.0…

K=710 = (0.01) / (7.41e-5)(0.19)

= (0.001) / (7.41e-6)(0.019) ?

Q = 7100 Yes!


A + B C Kc = 7107.41e-6 0.019 0.001 Initial

Towards more reactants…

Q > Keq…which way does it go?

+ x + x x Change

(7.41e-6+x) (0.019+x) (0.001x) Equil

Q = 7100

Let’s test out those keypads…

What is the equilibrium expression for: BaS(s) + 2 O2(g) BaSO4(s)

11 22 33 44 55

20%

20%

20%

20%

20% 1.1. [O[O22]]22

2.2. [BaSO[BaSO44]/[BaS][O]/[BaS][O22]]22

3.3. [BaS][O[BaS][O22]]22/[BaSO/[BaSO44]]

4.4. 1/[O1/[O22]]22

5.5. None of the aboveNone of the above

The Kp expression equals 400.0 for: CO(g) + H2O(g) CO2 (g) + H2(g)

If 4 atm each of CO & H2O are put into a container what’s the final pressure of CO ?

11 22 33 44 55

20%

20%

20%

20%

20% 1.1. 0.04 atm0.04 atm

2.2. 0.2 atm0.2 atm

3.3. 5 atm5 atm

4.4. 80 atm80 atm

5.5. None of the aboveNone of the above

Have a great week-end!

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Outline:2/23/07