Môn Hóa Lý: Chapter 11 solid

Vapor pressure depressionNo solute vapor  total vapor pressure of the solution is the vapor pressure of the solvent No solute vapor  total vapor pressure of the solution is the vapor p

1 Freezing Point – Boiling Point – Osmotic pressure

2 Two- component solid – liquid phase diagrams

3 Three- component solid – liquid phase diagrams

Chapter 11 – Equilibrium involving solids

Content

Vapor pressure depression

No solute vapor  total vapor pressure of the

solution is the vapor pressure of the solvent

No solute vapor  total vapor pressure of the

solution is the vapor pressure of the solvent

Raoult’s law: P = Psolv. = Pvap

solv xsolv. = Pvap

solv (1-x)

Where: P : solution vapor pressure

Pvap

solv. : vapor pressure of the pure solvent

x : total mole fraction of solutes

Raoult’s law: P = Psolv. = Pvap

solv xsolv. = Pvap

solv (1-x)

Where: P : solution vapor pressure

Pvap

solv. : vapor pressure of the pure solvent

x : total mole fraction of solutes

Vapor pressure depression

Higher the concentration of the

solutes, the lower the vapor

pressure

Higher the concentration of the

solutes, the lower the vapor

pressure

x P

P P

P

P

vap solv

vap solv

vap solv

.

Freezing Point & Boiling Point

• Freezing Point depression and Boiling

Freezing Point & Boiling Point

Parameter Boiling point Freezing point

T (K) Elevation Depression

 (cal/mol) Heat of vaporization Heat of fusion

T o (K) Solvent’s boiling point Solvent’s freezing point

R = 1,987 (cal/mol/K) : gas constant

M 2 (g/mol) : Solvent molar weight

Freezing Point & Boiling Point

– 50% NaOH: boiling point  1200C– 22% NaCl in water: freezing point -21 0C

Examples:

Osmotic pressure

( ): is the minimum pressure

which needs to be applied to a

solution to prevent the inward

flow of water across a

semipermeable membrane.

( ): is the minimum pressure

which needs to be applied to a

solution to prevent the inward

flow of water across a

semipermeable membrane.

Osmotic pressure

 (atm): osmotic pressure

V (lit): volume of the solution.

n (mol): moles of solute.

R (= 0,082 atm.lit/mol/K): gas constant

T (K): temperature

C (M) : solute concentration

CRT

RT V

n







In-Class activity 2

Compute the osmotic pressure of the solution:

Sucrose in water at 293 K at:

1 C= 0.1 mol/l; comparing to  (measured)= 262 kPa.

2 C= 0.75 mol/l; comparing to  (measured)= 2,4 MPa

Two-component

Solid – liquid

phase diagrams

Liquid phase’s point

Solid phase’s point

Two-component systems

Amount of phase, example at Q 2 :

Q Q2 2= liquid(l = liquid(l2 2) + solid(s ) + solid(s2 2) )

2 2

2

2

) (

)

(

s Q

Q

l l

g

s g



HR

eH L

g

R

g

 )

(

) (

Two-component systems

• Applications of Eutectic mixtures:

Cooling agent (NaCl-

Two-component systems

• Isothermal process

Reduce the concentration of

A in the solution at constant temeprature (ex By

Reduce the concentration of

A in the solution at constant temeprature (ex By

vacuum):

- Q  l : Starting crystallization, forming SB

g

S

g B

)()(

Adding H2O to Q until the

system point pass “M”

Salt purification

Pure AX

Example:

Filtering

()

(

)(

2

2

O H M

O H

Q Q

)(

)

(

AX B

MB M

g

AX g



At “T”: Adding H2O to Q until

the system point pass “M”