1. Intermolecular forces of attraction include ionic, dipole-dipole, hydrogen bonding, and dispersion forces. These forces determine physical properties like boiling point. 2. Stronger intermolecular forces require more kinetic energy to overcome, resulting in higher boiling points. Polar substances have stronger intermolecular forces than nonpolar substances. 3. The boiling point of a liquid is reached when the vapor pressure of the escaping molecules equals the pressure of the air above the liquid. Stronger intermolecular forces make it harder for molecules to break away from the liquid, increasing the boiling point.

1. INTERMOLECULAR FORCES OF ATTRACTION
Outside and
weak
attraction
Physical
properties
Physical
states
Boiling
point
compounds
Partial
charges
Farther
distance
solubility
are
between
explains

2. KIND OF IMFA COMPOUNDS THAT FORMED THEM
electrostatic Ionic & ionic
Ion dipole Ionic & polar
H bond
Polar H terminal & lone pair of small
Electronegative atom
Dipole dipole Polar & polar
Ion induced dipole Ionic & non polar
Dipole induced diipole Polar & non polar
dispersion Non polar & non polar
Decreasing
strenght

3. electrostatic
Ionic with ionic
NaCl NaCl
electrostatic

5. Ion dipole
Ion with polar
Na
+
H2O
iondipole

6. H bond
Polar H terminal
with lone pairs
of small
electronegative
atoms
H2O H2O
H-bond

7. Dipole-dipole
Polar with polar
CH3 C = O
CH3
CH3
CH3 C=O dipole-dipole

8. Ion-induced
dipole
Ion with non
polar
O = O Na
+
ion induced dipole
O = O Cl
-
e'movement
e'movement

9. Dipole induced
diploe
Dipole with non
polar
O = O H2O
dipoleinduceddipole

10. Dispersion
Non polar with
non polar
O = O O = O
dispersion

11. IMFA and Solubility
1. Dissolved substance
(solute) must separate
(IMFA breaking-
endothermic heat)
2. Dissolving substance
(solvent) must separate
(IMFA breaking-
endothermic heat)
3. Solute and solvent must
mix (IMFA forming-
exothermic heat)
For dissolving to
happen, 3
processes must
occur

12. IMFA between solute and solvent > IMFA among solute and or IMFA among solvent
_
• Summarized as : LIKE DISSOLVES LIKE ;
• polar solvent dissolves polar solute
• Non polar solvent dissolves non polar solute

13. IMFA and Physical State

14. PHYSICAL STATE
IMFA KINETIC ENERGY
MOLECULES TOGETHER MOLECULES APART
Interplay between
draws
separates

15. Kinds
Solid
IMFA
>>>
KE
Gas IMFA
<<<
KE
liquid IMFA is continuously being
formed and continuously
being broken

16. PHYSICAL
STATES
Solid
Gas
Liquid
IMFA >>> KE
KE >>> IMFA
IMFA is continously being formed
And continously being broken

17. Phase Changes are

18. MELTING FREEZINGEVAPORATION CONDENSATION
S to L L to SG to LL to G
Temp > IMFA Temp >>> IMFA
Temp < IMFA Temp <<<IMFA
ENDO EXO
Sorrounding
cold
Sorrounding
Warm
IS IS IS IS
WHENWHEN WHEN WHEN
makes makes
is is

19. ENERGY CURVE DURING PHASE
CHANGE (OF WATER)

20. TE MP
E NE R GY
-4
0
s
l
l - g
s - l
g
80 cal/g
540 cal/g
100
25
Heat of Fusion- amount of
energy needed to melt 1 gram
of a substance at its melting
point
H fusion water = 80 cal/g
Q= mass X H fusion
Heat of Vaporization – amount
of energy needed to evaporate
1 gram of a substance at its
boiling point
H vap water = 540 cal/g
Q = mass X H vap
Heat of Freezing = energy
released to change 1 gram of
liquid to solid
Heat of Condensation= energy
released to convert 1 gram of gas to
liquid
SPECIFIC HEATS
Energy involved to change the
temperature of 1 gram of a
substance, 10Celsius

21. LATENT HEATS
• Heat of Fusion- amount of energy needed to
melt 1 gram of a substance at its melting
point
• H fusion water = 80 cal/g
• Q= mass X H fusion

22. LATENT HEATS
• Heat of Freezing = energy released to change
1 gram of liquid to solid
•
• Heat of Fusion (endo) = Heat of Freezing (exo)

23. LATENT HEATS
• Heat of Vaporization – amount of energy
needed to evaporate 1 gram of a substance at
its boiling point
• H vap water = 540 cal/g
• Q = mass X H vap

24. LATENT HEATS
• Heat of Condensation= energy released to
convert 1 gram of gas to liquid
•
• Heat of Vaporization (endo) = Heat of Condensation(exo)

25. SPECIFIC HEATS
• Energy involved to change the temperature of
1 gram of a substance, 10Celsius
• For water: Sp. Heat = 1 cal/g-0C

26. UNIQUE PROPERTIES OF LIQUIDS
1. Surface Tension
2. Capillarity
3. Viscosity
•

27. IMFA & Boiling Point
Because of the KE energy of
evaporating liquids, they break their
IMFA and result in some of them going
into the gaseous state
and exert a pressure called vapor
pressure.
Under a given temperature each liquid
has its own vapor pressure.
When the vapor pressure of escaping
liquid molecules become equal to the
pressure of the air above it. The
temperature at that point is called the
boiling point of the liquid.
The stronger the IMFA, the harder to
break, the longer time the liquid to
evaporate the higher the boiling point
Vapor pressure
Air pressure
• Vapor pressure = air pressure : Boiling point

28. IMFA & Boiling Point
• IMFA increases, boiling point increases
• MWt increases, IMFA increases, boiling point
increases
• Branching increases, IMFA decreases, boiling
point decreases