Unit 11 - Intermolecular Forces.pptx

Unit 11: Intermolecular Forces

11.1: Employ the kinetic molecular model to explain the changes in motion or structure of particles in gases,

liquids, and solids.

11.2: Describe the various types of intermolecular and

interatomic attractive forces and state the kinds of forces expected for a given substance given its chemical

formula.

11.3: Calculate the heat absorbed or evolved when a given quantity of a chemical changes from one

phase/temperature to another.

11.4: Draw a phase diagram of a substance given

Phase

States of Matter (phase)

The fundamental difference between

States of Matter

Because in the solid and liquid states particles are closer together, we refer to them as condensed

States of Matter

• Since the phase of a substance is determined by how far apart the particles are, we can change the phase by making the

particles move more/less (temperature) or by giving them more/less room to move in (pressure.)

• _{But what really makes particles stay in one phase or another?} • _{What dictates how much of a temperature/pressure change is}

needed to cause one phase or another?

• _{Why is this amount (melting point/boiling point) different for} different substances?

• We already know that INTRAmolcular forces (bonding) dictates much of a substances chemical/physical properties.

• _{While INTERmolecular forces (attractive forces between whole}

Intermolecular forces

• _{Intermolecular Forces: Forces of attraction} (typically caused by polarity) between

molecules.

• Typically much weaker forces (typically less

than 10%) than INTRAmolecular bonding forces such as covalent/ionic/metallic.

• There are 4 main types of intermolecular forces (Also called Van Der Waals Forces, named ofter Johannes Diderik van der Waals, 1910 Nobel

Prize winner in Physics): -London forces

-Ion-Dipole forces

London

Dispersion

London Dispersion Forces

While the electrons in the 1s orbital of helium would repel each other

(and, therefore, tend to stay far away from each other), it does

London Dispersion Forces

London Dispersion Forces

Another helium nearby, then,

would have a dipole induced in it,

as the electrons on the left side

of helium atom 2 repel the

London Dispersion Forces

London dispersion forces, or dispersion forces, are attractions between an

instantaneous dipole (random dipole caused by moving e-) and an induced dipole (dipole caused by one atom pushing e- around in

London Dispersion Forces

• These forces are always present in all molecules, whether they are polar or nonpolar.

• _{The ONLY I.M. Forces present in nonpolar}

substances

• _{The tendency of an electron cloud to distort in this}

Factors Affecting London Forces

• _{The shape of the molecule affects the}

strength of dispersion forces:

• long, skinny molecules (like n -pentane) tend to have stronger dispersion forces

• _{than short, fat ones (like neopentane).} • This is due to the increased surface

area in n-pentane.

• _{They can line up and “spoon” better,}

Factors Affecting London Forces

•

The

strength of LDF also tends to increase

with increased molecular weight.

•

Larger atoms have larger/dispersed

Dipole-Dipole

(Polarity)

Dipole-Dipole Interactions

• Molecules that have

permanent dipoles

(dipoles that are always present, not

instantaneous or

induced) are attracted to each other.

• _{The positive end of}

one is attracted to the negative end of the

other and vice-versa.

Dipole-Dipole

• _{D-D interactions are probably the easiest to}

understand conceptually.

• _{+ and – are attracted to each other}

• _{The more polar a permanent molecule dipole}

the more the molecules are attracted to each other.

• _{This is why we spent most of the last two units}

Ion-Dipole

Ion-Dipole Interactions

• _{Ion-dipole interactions are an}

important force in ionic

solutions (typically aqueous.) • The strength of these forces

are what makes it possible for ionic substances to dissolve in polar solvents.

• Remember for solubility:

“Like dissolves like”

Polar substances dissolve in polar solvents, nonpolar substances

Hydrogen

Bonding

Hydrogen Bonding

• _{Hydrogen Bonding: The strongest possible} dipole-dipole interaction that happens in

molecules containing Hydrogen bonded directly to oxygen, nitrogen, or Fluorine.

• Typically the strongest of the intermolecular forces. So strong that it is called Hydrogen “Bonding” because it is closest to the

strength of a chemical bond.

• _{BUT H-bonds are NOT chemical bonds,}

just a really strong intermolecular force.

• _{Remember, IMF are only 10% of bonding}

Hydrogen Bonding

• Hydrogen is very exposed by O’s high EN pulling H’s bonding e- towards it

• _{The exposed nucleus is VERY attracted to an adjacent}

molecules lone pair

Intermolecular Forces

Roundup

• _{So really we only have two types:} • _{Dipole-Dipole: Polarity attraction}

-Only occur if the molecule has polarity

-if it involves ions acting alone then it is ion-dipole (but basically dipole)

-If it involves H bonded to O, N, or F then H-bonding (extreme dipole-dipole)

• London Forces:

-ALWAYS occur in polar OR non-polar substances -dominant force in non-polar substance

Now you try



• Predict the dominant intermolecular force found in each substance (it may help to draw the Lewis Dot Diagrams):

1. Hydrofluoric acid 2. CO₂ 3. H₂O

4. NaCl_(aq) 5. CH₄ 6. CCl₃Br

1. H-Bonding 2. London 3. H-Bonding

Intermolecular

IM Force Properties

• _{Many properties of a substance are caused by}

its chemical bonding: color, conductivity, solubility, texture, etc…

• IMF are usually correlated to four properties: -viscosity

-surface tension -vapor pressure

Viscosity

•

Resistance of a liquid to flow is called

viscosity.

•

It is related to the

ease with which

molecules can move past each other.

•

Viscosity increases with stronger

Surface Tension

•

Surface tension results from

the net inward force

experienced by the molecules

on the surface of a liquid.

•

Enables floating of certain

objects.

•

Causes “meniscus” effect in

liquids like water…

http://apod.nasa.gov/apod/ap130424.html

Cool nasa water surface tension video

https://

Vapor Pressure

• At any temperature, some molecules in a liquid have enough energy to escape as a gas.

• _{As the temperature rises, the fraction of}

Vapor Pressure

As temperature increases or pressure decreases,

more molecules escape the liquid and the

Vapor Pressure

In a closed system like this, the liquid and vapor reach a state of dynamic equilibrium: liquid

molecules evaporate and vapor molecules condense

Vapor Pressure

•

Liquids with high IMF make it more difficult for

particles to escape as gas.

•

So liquids with high IMF have low vapor

pressures

Boiling

• _{The boiling point of a liquid}

is the temperature at which its vapor pressure equals

atmospheric pressure.

• _{The ”normal” boiling point is}

the temperature at which its vapor pressure is 760 torr.

• _{Do not confuse with}

evaporation!

• _{A boiling diagram like this}

only shows part of the picture, we need phase

Phase Diagrams

Phase diagrams display the phase of a substance at

• The AB line is the liquid-vapor interface.

It ends at the ”critical point” (B); above this critical

• The AD line is the interface between liquid and solid.

• Below the triple point pressure, the substance cannot exist in the liquid state.

Phase Diagram of Water

• _{Note the high critical}

temperature and critical pressure:

• _{These are due to the}

strong hydrogen bonding forces between water

molecules.

• _{The slope of the solid–liquid}

line is negative.

• _{This means that water}

uniquely prefers to be a liquid at high pressures due to H-Bonding

Phase Diagram of

Carbon Dioxide

• _{Carbon dioxide cannot}

exist in the liquid state at pressures below

5.11 atm; CO₂

sublimes at normal pressures.

• _{Carbon dioxide has a} “normal” positive

melting point curve

Now you try



• What phase changes would occur if…

Liquid water was heated from

-100 °C to 40 °C at 1 atm pressure.

Ice  liquid (melting)

Water vapor in a vacuum at 2 torr is cooled from 0 °C to -265 °C

Vapor  Ice (deposition)

Steam at 140 °C is pressurized from 1 atm to 200 atm, then cooled to -130 °C

Phase Change

Energetics of Phase Change

• Heat of Fusion: Energy required to change a solid at its melting point to a liquid.

Energetics of Phase Change

• The energy required to undergo phase change can be calculated for a given mass of matter using the heat of fusion or heat of vaporization, ΔH_fus and ΔH_vap

respectively.

• _{These values are usually given in a problem or looked}

up in table.

• _{Example: to calculate the energy needed to melt 100}

g of ice.

q = ΔH_fus x m = 0.333 kJ/g x 100 g = 33.3 kJ

Or to calculate the energy needed to vaporize the same amount.

q = ΔH_vap x m = 2.26 kJ/g x 100 g = 226 kJ

Phase Change Energetics

• _{To calculate the full energy change for a given}

quantity of matter that you have to account for:

• _{1. Heating/cooling matter: The energy needed}

to heat the substance to or beyond the melting and/or boiling temperature. (the ol’ q

equation)

• _{2. Changing the phase of the matter: The}

energy needed to overcome the intermolecular forces once the melting and/or boiling

Heating/Cooling Curve

energetics can be summarized in a “Heat Curve”

• The diagonal sections

correspond to heating a

substance to the next phase

change temp.

• _{The flat sections} correspond to the substance

Heating/Cooling Curve

energy to make ice

 steam requires 5 calculations:

1. E to heat ice 2. E to phase

change solid 

liquid

3. E to heat liquid 4. E to phase

change liquid 

gas

5. E to heat steam Steam  ice would be the reverse

Heating/Cooling Curve

• Steam  ice would be the reverse

(exotherm)

• Since energy is conserved, you just reverse your signs.

• _{ΔHfus = -ΔHmelting} • _{ΔHvap = -}

ΔH_condensation

Example

• _{How much energy is needed to heat 100g ice at -10 °C to steam} at 250 °C

• _{Watch your units!}

1.) heat ice: q = Cs_ice x m x Δt = 2.1 x 100 x 10 = 2100 J

2.) Melting: q = ΔH_fus x m = 0.333 kJ/g x 100 g = 33.3 kJ = 33,300 J

3.) heat water: q = Cs x m x Δt = 4.18 x 100 x 100 = 41,800 J

4.) Boiling: q = ΔH_vap x m = 2.26 kj/g x 100 g = 226 kj = 226,000 J

5.) heat steam: q = Cs_steam x m x Δt = 1.996 x 100 x 150 = 29,940 J

• Add up each q:

2100 J + 33,300 J + 41,800 J +

Now you try



• Calculate the amount of energy RELEASED when 400.0

grams of steam at 180 °C is cooled at 1 atm to ice at -35 °C Cs ice = 2.09 J/gk Cs water = 4.18 J/gk Cs steam = 1.99 J/gk

ΔH_fus = 0.334 kJ/g ΔH_vap = 2.26 kJ/g

• Cooling steam: 400g x 1.99 x (100-180) = -63680 J = -637 kJ

• _Steamwater: 400g x -2.26 kJ/g = -904 kJ

• Cooling water: 400g x 4.18 x (0-100) = -167200 J = -167 kJ

• _Waterice: 400g x -0.334 kJ/g = -133.6 kJ

Solid

Structures

Solid Specific Properties

Two general types of solids:

• _{Crystalline: Highly ordered/structured atomic}

arrangement.

Attractions in Ionic Crystals

In ionic crystals, ions pack themselves so as to maximize the attractions and minimize

Crystalline Solids

Because of the order in a

crystal, we can focus on

the repeating pattern of

arrangement called the

unit cell.

Crystalline Solids

There are several types of basic arrangements in

crystals, but we will focus on three for this class:

Now you try



• _{Gallium crystallizes in a primitive cubic unit}

cell. The length of the unit cell edge is 3.70 The radius of a Ga atom is __________ Å.

3.70 / 2 = 1.85 A

• _{Chromium crystallizes in a body-centered cubic}

unit cell. There are __________ chromium atoms per unit cell.