© University of South Carolina Board of Trustees Adding a nonvolatile solute to a pure solvent causes: lower vapor pressure P = - P x sol higher

© University of South Carolina Board of Trustees

Adding a nonvolatile solute to a pure solvent causes:

lower vapor pressure DP = - P x csol

higher boiling point DT = +kb x msol

lower freezing point DT = - kf x msol

osmotic pressure

Colligative Properties



lower vapor pressure DP = - P x csolute

higher boiling point DT = +kb x msolute

lower freezing point DT = - kf x msolute

osmotic pressure



● Solvent flow initially in balance.

● Adding solute blocks flow right left.


● The solvent level on the right rises due to osmotic pressure.


● External pressure can restore equilibrium.

P






osmotic pressure P = RT x Msol







osmotic pressure P = RT x Msol


moles of solute

Solvent properties

Each can measure molecular mass


Last Lecture

Find the molar mass of a solute, if a solution of 1.33 g of the compound dissolved in 25.0 g of benzene has a boiling point of 81.22 oC.

SolventTb

(oC)kb

(oC/molal)

Aceticacid

117.90 3.07

Benzene 80.10 2.53

Water 100.0 0.512


Student Example: Molar Mass by Osmotic Pressure

A 5.70 mg sample of protein is dissolved in water to give 1.00 mL of solution. Calculate the molar mass of the protein, if the solution has an osmotic pressure of 6.52 Torr at 20 C.

R = 0.0821 L-atm/K-mol

1 atm = 760 Torr


Chapt. 12Solutions

Sec. 5Colligative Properties

with Ions


Ions and Colligative Properties

NaCl(s) Na+(aq) + Cl-(aq)

Ca(NO3)2(s) Ca2+(aq) + 2 NO3

-(aq)

nsolute = i nformulafound by

colligative propertiesusually desired

van’t Hoff factor

1 mole byformula wt.

2 moles ofsolute i = 2

1 mole byformula wt.

3 moles ofsolute i = 3


Student Example

Arrange the following aqueous solutions in order of

increasing boiling points:

0.03 molal urea(s) (a nonelectrolyte)

0.01 molal NaOH

0.02 molal BaCl2

0.01 molal Fe(NO3)3


Student Example



0.03 molal urea(s) urea(aq) 0.03 m solute

0.01 molal NaOH

0.02 molal BaCl2

0.01 molal Fe(NO3)3


Student Example




0.01 molal NaOH Na+ + OH- 0.02 m ions

0.02 molal BaCl2

0.01 molal Fe(NO3)3


Student Example





0.02 molal BaCl2 Ba2+ + 2 Cl- 0.06 m ions

0.01 molal Fe(NO3)3


Student Example






0.01 molal Fe(NO3)3 Fe3+ + 3 NO3- 0.04 m ions


Student Example






0.01 molal Fe(NO3)3 Fe3+ + 3 NO3- 0.04 m ions


Chapt. 12Solutions

Sec. 6Vapor Pressure of Liquid-Liquid

Solutions(skip)


Chapt. 13Kinetics

(Later)


Chapt. 14Chemical Equilibrium

Sec. 1What is Chemical Equilibrium?


What is Chemical Equilibrium?

● Both products and reactants remain

● Macroscopically Static: System has finished changing (not just slow)


Macroscopically Unchanging

2 NO2(g) N2O4(g)

brown colorless

equilibrium

Next Day









● Molecularly Dynamic: Forward and backward rates are in balance


Molecules Dynamic

2 NO2(g) N2O4(g)

brown colorless

equilibrium

Next Day





● Molecularly Dynamic: Forward and backward rates are in balance

● Equilibrium concentrations independent of starting conditions


Independent of Starting Conditions

final: 2 NO2(g) N2O4(g) 2 NO2(g) N2O4(g)

initial: 100% NO2(g)100 % N2O4(g)

equilibriumconcentrations


Chapt. 14Chemical Equilibrium

Sec. 1How are the Equilibrium

Concentrations Calculated?


Experiments:Equilibrium concentrations from

several initial concentrations

Initial concentration, M

Equilibriumconcentration, M

[NO2] [N2O4]

0.0200 0.0000 0.0103 4.86 x 10-3

0.0000 0.0100 0.0103 4.86 x 10-3

2 NO2(g) N2O4(g)

junk

0.0300 0.0000 0.0134 8.29 x 10-3 0.62

0.0400 0.0000 0.0161 11.9 x 10-3 0.74

[N2O4][NO2]

Ratio I

0.47

0.47

[N2O4]

[NO2]


Experiments:Equilibrium concentrations from

several initial concentrations

Initial concentration, M

Equilibriumconcentration, M

Ratio I

[NO2] [N2O4]

0.0200 0.0000 0.0103 4.86 x 10-3 0.47 45.8

0.0000 0.0100 0.0103 4.86 x 10-3 0.47 45.8

0.0300 0.0000 0.0134 8.29 x 10-3 0.62 46.2

0.0400 0.0000 0.0161 11.9 x 10-3 0.74 45.9

2 NO2(g) N2O4(g)

Ratio II

[N2O4][NO2] [N2O4] [N2O4]

[NO2] [NO2]2

constant !


Fast Forward

>>>


Equilibrium Constant Expression

2 NO2(g) N2O4(g)

2

45.9 2 4N O

NO

Example:

a A + b B c C + d D

c d

eq a bK

C D

A B

In General:

products – top

reactants – bottom

concentration


Equilibrium Constant Expression

2 NO2(g) 1 N2O4(g)

2

45.9 1

2 4

2

N O

NO

Example:

a A + b B c C + d D

eqK C D

A B

c d

a b

In General:

products – top

reactants – bottomEquilibrium Constant• a number• independent of starting amounts


Examples

32

23

eqK O

O

2 2

2eqK

H CO

CO H O

2

2eqK

Cl

Cl

CO(g) + H2O(g) H2(g) + CO2(g)

Cl2(g) 2 Cl•(g)

2 O3(g) 3 O2(g)

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© University of South Carolina Board of Trustees Adding a nonvolatile solute to a pure solvent causes: lower vapor pressure P = - P x sol higher