Intermolecular Forces

Chapter 11

Intermolecular Forces: Introduction

• Forces between separate molecules and dissolved ions (not bonds)

• “Van der Waals Forces”

• 15% as strong as covalent or ionic bonds

Intermolecular Forces: Introduction

• Low temperature – strong

• High temperature – kinetic energy of motion overcomes the IMF

• Boiling point is a good indicator– Stronger IMF = higher boiling point– Weaker IMF = lower boiling point

Ion-Ion Full to Full Charge

Ion-Dipole Full to Partial Charge

Hydrogen Bonds Partial to Partial Charge (H involved)

Dipole-Dipole Partial to Partial Charge (No H)

London Dispersion Forces

Non-polar to Non-Polar

Predict what type of IMF would form between:

a. Br2 and I2

b.KCl and water

c. Water and ammonia

d.Two SO2 molecules

e. NaCl ions in a crystal

Type 1: Ion-Ion Forces

• Full Charges to full charges

• High melting points (ionic solids)

Ex: Melting Point

Na ~98 oC

NaCl ~800 oC

Type 2: Ion-Dipole Forces

• Full Charges to partial charges

• Very Strong

Type 3: Hydrogen Bonds

• Stronger than dipole-dipole that do not have hydrogen (no inner electrons, strong + pull)

• Generally involves hydrogen and O, N or F

R-H · · · · O-R

R-H · · · · N-R

R-H · · · · F-R

Miscibility

Miscibility

• “Like dissolves like”

• Substances that can hydrogen bond dissolve in one another.

Glucose and other sugars

Water Beading

Ice

Ice

DNA

DNA

DNA is TWO molecules

that are hydrogen

bonded (like a zipper)

Boiling Point

• Generally increases with increasing molar mass

• H2O unusually high - H-bonding

Type 4: Dipole-Dipole

• Slightly weaker IMF

• Involve + and - charges other than those in hydrogen bonding

Draw Lewis Dot Structures to explain the following boiling points

MM (amu) BP (oC)

CH3CH2CH3 44 -42

CH3CHO 44 21

CH3CN 41 82

Type 5: London Dispersion Forces

• Very weak IMF• Caused by temporary imbalances in electrons

London Forces: Inorganic Molecules

• More electrons, more chance for temporary dipoleBoiling Point Table

Halogen Molar Mass

BP(oC) Noble Gas

Atomic Mass

BP(oC)

F2 (g) 38.0 -188 He 4.0 -268

Cl2 (g) 71.0 -35 Ne 20.2 -246

Br2 (l) 159.8 59 Ar 39.9 -186

I2 (s) 253.8 185 Kr 83.8 -152

Explain the differences in boiling point between Cl2 (-35oC) and Ar (-186oC)

London Forces: Organic Compounds• The longer the carbon chain, the higher the

London Dispersion Forces (the higher the melting point and boiling point)

• Chainlike molecules greater London Forces than “bunched up” molecules (branched)

Ex 1

Rank the following compounds in terms of increasing melting points: NaCl, CF4, CH3OH.

Ex 2

Separate the following compounds by whether they have dipole-dipole attractions (including H-bonding) or London Forces. Which should have the highest dipole-dipole attraction? Which should have the strongest London Force?

Br2, Ne, HCl, N2, HF

Ex 3

Rank the following in order of increasing boiling point:

BaCl2, H2, CO, HF, Kr

Ex 4

Rank the following in order of increasing boiling point:

N2, KBr, O2, CH3CH2OH, HCN

Properties of Liquid

• Viscosity – resistance of a liquid to flow

• Oil is more viscous than water– Water has H-bonds (stronger)– Oil has London forces (weaker, but there are many

more of them, long carbon chain)

40 W oil 10 W oil

Miscibility

Surface Tension

Surface Tension

Capillary Action

• Water is attracted to the glass

• Mercury more attracted to itself

Heating Curves

1. Changes of state do not have a temperature change.

1. Melting/Freezing

2. Boiling/Condensing

2. A glass of soda with ice will stay at 0oC until all of the ice melts.

3. Graph “flattens out” during changes of state

Heating Curves

Heat (Joules)

Temperature (oC) Boiling

Melting

Ice warms up

Water warms up

Steam heats up

Heating Curve

No phase change is occurring (heating ice, water, or steam):

q = mCpT

Melting or boiling:

q = mLf or q = mLv

Lf and Lv

Latent Heat - heat for phase changes. No temperature change.

Lf –latent heat of freezing/melting

Lv –latent heat of boiling/condensing

Heating Curves

Heat (calories)

Temperature (oC) Boiling

Melting

Use q = mCpT

Use q = mLf

Use q = mLv

Important Values

Substance Cp

Steam 2.01 J/goC

Water 4.18 J/goC

Ice 2.09 J/goC

Latent Heat of fusion (water) Lf = 334.7 J/g

Latent Heat of vaporization(water) Lv = 2259.4 J/g

Heating Curves: Example 1

How much energy must be removed to cool 100.0 grams of water at 20.0oC to make ice at –10.0oC?

Heating Curves: Example 1

Heat

Temperature (oC)

Melting (q=mLf)

Ice cools (q=mCpT)

Water cools (q=mCpT)

Heating Curves: Example 1

1. Cooling the water

q = mCpT = (100 g)(4.18 J/goC)(0oC-20oC)

q = 8360 J (8.36 kJ) (ignore the negative sign for now)

2. Freezing the water

q = mLf = (100.0 g)(334.7 J/g)= 33.47 kJ

3. Cooling the ice down to –10.0oC

q = mCpT = (100 g)(2.09 J/goC)(-10oC-0oC)

q = 2.09 kJ (we will ignore the negative sign for now)

8.36 kJ+ 33.47 kJ+ 2.09 kJ= 43.92 kJ

Ex 2

How much energy is needed to convert 18.0 grams of ice at -25oC to steam at 125oC?

ANS: 56.0 kJ

Ex 3

How much energy must be used to convert 100.0 grams of water from steam at 110.0oC to ice at -25.0oC ?

(309 kJ)

Vapor Pressure

• Pressure of a gas above a liquid caused by that liquid

• Temperature – measure of the average kinetic energy of molecules

• At any given moment, some molecules have enough energy to escape

• EX: Even cold water will evaporate

Volatility

• Volatile – liquids that evaporate easily– Acetone– Often weak intermolecular forces

• Boiling point – point at which the vapor pressure of a liquid = vapor pressure of the atmosphere

• Normal Boiling Point – vapor pressure = 1 atm– Steam pressure cookers – Forces water to boil at a

higher temperature – High Altitude – water boils at a lower temperature

Four Types of Solids

1. Molecular Solids (single molecules)

2. Covalent Network Solids (one large molecule)

3. Ionic Solids

4. Metallic Solids

1. Molecular Solids

• Held together by IMF

• Ice

• Plastics

2. Covalent Network Solids

• Basically one big molecule

• Held together by covalent bonds

• Diamond

• Graphite

• Quartz (SiO2)

3. Ionic Solids

• Held together by electrostatic attraction (Ion:Ion)

• Usually crystalline (unit cells)

4. Metallic Solids

• Atoms share electrons very freely

• Positive nuclei in a “sea of electrons”

• Electrons held loosely– Conducts electricity– Photoelectric effect– Malleable and ductile

Rank by boiling point (low to high). Below each, tell me which IMF is important:

CH3OH

Cl2

N2

CH3Cl

CH3CH2CH2CH3

CH3CH2CH2OH Draw Lewis Dots for these

CH3OCH3

CH3CH2CH3

How much energy must be removed to convert 100.0 grams of steam at 120.0oC to ice at -5.00oC?

(100.0g)(2.01 J/gK)(20.0oC) = 4.02 kJ

(100.0g)(2259.4 J/g) = 225.9 kJ

(100.0g)(4.18 J/gK)(100.0oC) = 41.8 kJ

(100.0g)(334.7 J/g) = 33.47 kJ

(100.0g)(2.09 J/gK)(5.0oC) = 1.05 kJ

306.2 kJ

The heat of vaporization of ammonia is 23.35 kJ/mol. How many grams of ammonia must evaporate to freeze 100.0 grams of water initially at 10.00oC? Assume the ice stays at 0oC when frozen, and is used to cool a cup of too hot tea.

(Ans: 27.4 g)

Water

qcool (100)(4.18)(10) = 4.18 kJ

qfreeze (100)(334.7) = 33.47 kJ

37.65 kJ

Ammonia

qv = mLv

m =qv/Lv

m = 37.65 kJ/23.35 kJ/mol = 1.61 mol

mass = (1.61 mol)(17.0 g/mol) = 27.4 g

2. a) H-bonding, (b)London (c) Ion-dipole (d) dipole-dipole. Ion-dipole and h-bonding are stronger

10. a) Solids = attractive forces (IMF) win

Liquids = Balance

Gases = Kinetic energy wins

b) Increasing T increase KE, eventually overcoming IMF

c) High pressure forces gas molecules clsoe together and IMF’s can win

12.a) Distance greater in liquid state

b) More movement, more volume, lower density

14. Overall, net forces are attractive

16.a) CH3OH has h-bonding, CH3SH does not

b) Xe is heavier, greater London Forces

c) Cl2 more polarizable than Kr

d) Acetone has dipole-dipole forces

18. a) True b) False c) False d) True

20. a) Br2 b) C5H11SH c) CH3CH2CH2Cl

22. Propyl alcohol is longer and more polarizable

24. a) HF has hydrogen bonds, HCl dipole/dipole

b) CHBr3 higher molar mass, more dispersion

c) ICl has dipole-dipole, Br2 only dispersion

26.a) Dispersion, C8H18 higher boiling point

b) C3H8(dispersion) CH3OCH3 (dip-dip)

c) HOOH (h-bonding) HSSH (dip-dip)

d) NH2NH2 (h-bonding) CH3CH3 (dispersion)

32. H2NNH2, HOOH, H2O can all h-bond

34.a) Exo b) Endo c) Endo d) Exo

38. 275 g CCl2F2

40. 10.3 kJ