Chapter 12-13

Chapter 12-13

States of Matter

States of Matter• Types of atoms (composition) and arrangement (structure) determine chemical properties of matter

States of Matter• Composition and structure also affect the physical properties of matter

Solids, Liquids and Gases

• It is easy to see the difference between solids and liquids but gases are different

Gases• Why are physical properties of gases different from solids and liquids?

Kinetic Molecular Theory

• A model that describes the behavior of gases in terms of particles in motion

• The model makes several assumptions about the size motion and energy of gas particles

Assumption 1: Size• Gas particles are small and separated from one another

• The volume of the particle is small compared to the space in between

• There are no significant attractions or repulsions among gas particles

Assumption 2: Motion

• Gas particles are in constant random motion

• They move in a straight line until they run into something

• Collisions are elastic=no kinetic energy is lost

Assumption 3: Energy

• Mass and velocity of particle determine kinetic energy

• The bigger the mass the bigger the KE. KE=1/2 mv2

• Temperature is a measure of KE

Explanation of Gas Behavior

• Diffusion is the movement of one material through another.

• Because of the space between particles gases diffuse easily

Effusion: Graham’s Law

• Similar to diffusion, effusion is the rate that a gas moves through a tiny opening.

• Graham determined that the rate that the gas effused was related to its molar mass

Rate of effusion is proportional to 1 molar mass

• To compare the diffusion of two gases

• Ratea = molar massb• Rateb molar massa

Explanation of Gas Behavior

KMT explains• Density is mass per volume

• Gases occupy large volumes with very little mass

Explanation of Gas Behavior

• Compression and Expansion the empty space found between gas particles allows for compression and expansion

Gas Pressure• Pressure is defined as the force per unit area. The earth is surrounded by the atmosphere which exerts pressure on the surface of the earth.

• Units of atmospheric pressure are1 atm = 101.3kPa = 760 mm Hg = 760

torr

Measuring Pressure• Barometer Measures the force exerted on the surface of a pool of mercury

Measuring PressureManometers measure the pressure in a vessel relative to the atmosphere

Converting Pressure Units

1 atm = 101.3kPa = 101325 Pa = 760 mm Hg = 760 torr

The pressure of N2 gas was measured to be 750 mm Hg. What is the pressure of the gas in kPa?

750 mm Hg 101.3kPa = 760 mm Hg

Daltons Law of Particle Pressure

• The pressure total is equal to the sum of the individual pressure that make up the gas mixture

• Ptotal = P1 + P2 +P3 + … Pn

Daltons Law of Particle Pressure

• A mixture of O2, CO2, and N2 has a total pressure of 0.97 atm. What is the partial pressure of O2 if CO2 has a pressure of 0.70 atm and N2 has a pressure of 0.12 atm?

0.97 atm = 0.70atm + 0.12atm + xatm

x= .15atm

Forces of Attraction• Intramolecular forces: Occur between atoms to form molecules. Examples include Ionic, Covalent and metallic bonds

• Intermolecular forces: Occur between molecules. Examples include dispersion, dipole-dipole and hydrogen bonding.

Forming a dipole• A dipole is a charged region of a

molecule. (some regions of the molecule have a partially positive or negative charge.) This occurs when there is a shift in the electron density

Neutral Dipole

OH H

d-

d+ d+

Intermolecular Forces

Intermolecular Forces

• Dispersion Forces Involve temporary dipole-dipole interactions

d- d+ d- d+

Intermolecular Forces

• Dipole- Dipole forces involve molecules with permanent dipoles. An example would be HCl:

Intermolecular Forces

• Hydrogen Bonding is a dipole-dipole force that involves H bonded to nitrogen, oxygen or a halogen.

• Hydrogen bonding is very strong. An example of this is water

Effects of Intermolecular Forces

• Solids and Liquid tend to have strong molecular forces.

• Water is composed of 2 gases. When combined to make a molecule with strong molecular forces water exists at room temperature as a liquid.

• Phase changes involve forming or breaking intermolecular forces.

The Liquid State• Intermolecular forces play a part in:Surface TensionCapillary ActionViscosity

Surface Tension• All particles are not

attracted equally in a solution.

• Particles in the middle are attracted by those above and below while those on the surface are attracted to those below.

Surface Tension• So the surface stretches over the top

• ST is a measure of the inward pull by particles in the interior

Capillary Action• Capillary

action is a result of the different degrees of attraction between the container and the liquid.

Capillary Action• If the

attraction to the container is greater than the attraction to other water particle the water will travel up the capillary

Viscosity• The resistance of a liquid to

flow. • Particles in a liquid are close

enough to each other to have intermolecular forces involved.

• The stronger the forces the higher the viscosity.

• The size and shape of the molecules and the temperature effect viscosity.

Viscosity• Viscosity of motor oil increases in the summer.

Surface Tension Activity

• Obtain a 250 or 400mL beaker with water in it

• Float the paper clip on the surface of the water

• Use a dropper to add one drop of water containing detergent to the beaker. Observe what happens

Answer these questions for Homework

• Is a paper clip likely to be more or less dense than water?

• How does the shape of the pin help it float

• Hypothesize about the reason for the pin’s behavior before and after you added the detergent.

Change of State/Phase Changes

• NO bonds are broken when a compound changes state.

Phase Changes That REQUIRE energy

Melting, Evaporation,

Sublimation and Boiling

Vaporization• Vaporization is the process by which gas particles escape the surface of a solid or a liquid. This process can occur through:–Evaporation–Sublimation–Boiling

Melting• The energy

absorbed by the ice is used to break hydrogen bonds that held the ice together. The temp at which this happens is the melting point

Evaporation• Requires energy to change from a liquid to a gas

• This is how we cool ourselves

Sublimation

Boiling

Vapor Pressure• Vapor pressure is the

pressure of the gas over a liquid in equilibrium.

• The rate of evaporation = rate of condensation

Vapor Pressure• Vapor pressure is

determined principally by the size of the intermolecular forces in the liquid.

• Vapor pressure increases significantly with temperature.

Boiling• If the

temperature of a liquid increases, the molecules of water gain kinetic energy and the vapor pressure increases.

Boiling• When the vapor pressure of the liquid = the pressure of the surrounding atmosphere then the boiling will occur.

BoilingAt high altitudes, the boiling point of

liquids is lower than at sea level. In Denver, Colorado, water will boil at about 94°C. Do not confuse boiling with cooking. Cooking pasta in Denver is a slower process because the water is at a lower temperature. Also, realize that water boiling rapidly is no hotter than water boiling slowly.

Boiling• The temperature of a boiling

liquid never rises above its boiling point. No matter how much heat is applied, the liquid only boils faster, not hotter.

Boiling• Compounds with a high degree of intermolecular forces will have high boiling points.

Phase Changes that RELEASE energy

Condensation, Depositon, Freezing

Condensation• When a water vapor molecule

loses energy, it’s velocity is reduced

• Then it is more likely to interact and form a hydrogen bond when it collides with another water vapor molecule

Deposition• Some substances can change directly into a solid without first forming a liquid

• Frost

Freezing• AS the heat is removed from the liquid the molecules release energy and velocity of molecules decreases

• When enough energy is removed hydrogen bonds form

• Freezing point-temp at which liquid converts into a solid

Freezing• Scanning Electron Microscope picture of snow

Phase Diagrams• There are 2 variables that

control the phase of a substance: Temperature and Pressure

• These 2 have opposite effect on substance

• Phase diagram is a graph that predicts the phase of a substance at a given temperature and pressure

Phase Diagram

Phase Diagram• Represents

phases as a function of temperature and pressure.

• critical temperature: temperature above which the vapor can not be liquefied.

Phase Diagram• critical pressure: pressure required to liquefy AT the critical temperature.

Phase Diagram• critical point: critical temperature and pressure (for water, Tc = 374°C and 218 atm).

Phase Diagram• Triple point: the temperature and pressure where all three phases can exist

Phase diagrams

Energy in a Phase Change

Copyright© by Houghton Mifflin Company. All rights reserved. 12

Figure 14.7: The heating/cooling curve for water heated or coolFigure 14.7: The heating/cooling curve for water heated or cooled at a constant rate. ed at a constant rate.

Solids

Energy in a phase change

• Class work: due at end of period

• Comment on this graph: Write a paragraph about what data was collected and what the graph shows. You should include what things are on each axis and why.

Vocabulary Flash Cards

• Make a Flash Card for each Vocabulary Word

• Design them like this:

Definition:What it is:What it isn’t:

Vocabulary Word

Vocabulary Flash Cards

• Intramolecular forces

• Intermolecular forces

• Dipole• Dispersion force

• Hydrogen Bond

• Phase Change

• Surface Tension

• Capillary Action

• Viscosity• Adhesion• Cohesion

• Evaporation• Sublimation• Boiling• Vapor Pressure

• Condensation

• Deposition• Freezing• Phase Diagram

• Critical Temperature

• Critical Pressure

• Critical Point

• Triple Point

Heat• Heat is represented in

equations as H . • It is the energy that flows

from a warm object to a cold object.

• When a warmer object loses heat the temperature goes down. When a colder object gains H it’s temperature goes up

Measuring Heat• A calorie is the amount of

heat required to raise the temperature of 1 gram of water 1 degree Celsius.

• Joules are the SI units for heat.

• 1 calorie = 4.184 J• 1kilocalorie (kcal) = 1 Calorie

= 1000 calories• 1 kilojoules (kJ) = 1000 Joules

Specific Heat• The specific heat of a substance is the amount of energy required to raise 1 gram of the substance 1oCelsius.

• C is the symbol for specific heat.

• The specific heat of water is 1 cal/(g oC) or 4.184 J/(g oC)

Specific Heat• Specific heats can be listed on data tables–Smaller the specific heat the less energy it takes the substance to feel hot

–Larger the specific heat the more energy it takes to heat a substance up

SUBSTANCE SPECIFIC HEAT CAPACITY, CP

WATER 4.18J/g°C OR 1cal/g°CICE 2.10 J/g°C OR .502cal/g°C

STEAM 1.87J/g°C OR .447cal/g°CMERCURY, Hg .139 J/g°C OR .033cal/g°C

ALCOHOL (Ethyl) 2.40 J/g°C OR .580cal/g°C

CALCIUM, Ca .647 J/g°C OR .155cal/g°CALUMINUM, Al .992J/g°C OR .237cal/g°CTABLE SALT,

NaCl .865 J/g°C OR .207cal/g°CAMMONIA,

NH32.09 J/g°C OR .500cal/g°C

SILVER, Ag .235 J/g°C OR .056cal/g°CLEAD, Pb .129J/g°C OR .031cal/g°C

Specific Heat• The energy required to raise the

temperature of a compound is related to the magnitude of the change, the mass of the compound, and the specific heat of the compound.

DH = M * C * DT Change in Heat=Mass* Sp Heat*Change in

temp

Specific Heat• How much heat energy would be required to raise the temperature of a glass of liquid water (400 g) from room temperature 20oC to 50oC.

DH = M * C * DT= 400 g * (50-20 oC)* 4.184 kJ

g oC = 50208 kJ

Calorimeter• A calorimeter is an insulated

device used to measure the amount of heat absorbed or released.

• The law of conservation of energy states that energy cannot be gained or lost. So if one object heats up another object must have lost that heat.

COFFEE CUP CALORIMETER

Calorimeter

• If hot metal is added to cold water the amount of energy lost by the hot metal is gained by the water.

Calorimetry• Calorimetry is the process of measuring heat energy – Uses the heat absorbed by H2O to measure the heat given off by an object

• The amount of heat soaked up by the water is equal to the amount of heat released by the object

DHSYS = -DHSUR+ SIGN MEANS

HEAT WASABSORBED

BY THE object

- SIGN MEANSHEAT WAS

RELEASED BYWATER

• With calorimetry we use the sign of what happens to the water– When the water loses heat into the system it obtains a (-) sign

HEATHEAT

• With calorimetry we use the sign of what happens to the water– When the water gains heat from the system it obtains a (+) sign

-DHSYS = DHSUR- SIGN MEANS

HEAT WASRELEASED BY

THE object

+ SIGN MEANS

HEAT WASABSORBED BY

WATER

HEATHEAT

Calorimetry• You calculate the amount of heat absorbed by the water (using DH= mCDT)

• Which leads to the amount of heat given off by the object

Calorimetry–you know the mass of the water (by weighing it)

–you know the specific heat for water (found on a table)

–and you can measure the change in the temp of water (using a thermometer)

A chunk of Al that weighs 72.0g is heated to 100°C is dropped in a calorimeter containing 120ml of

water at 16.6°C. the H2O’s temp rises to 27°C.- mass of Al = 72g- Tinitial of Al = 100°C- Tfinal of Al = 27°C- CAl = .992J/g°C (from table)

DHSYS

DH = 72g .992J/g°C 27°C-100°CDH = -5214J

We can do the same calc with the water info

DHSUR

DH=5216JEqual but opposite, means that the Al

decreased in temp, it released its stored heat into the H2O, causing the

temp of the H2O to increase.

DH=120g 4.18J/g°C27°C-16.6°C

– Mass of H2O= 120g– Tinitial of H2O= 16.6°C– Tfinal of H2O = 27°C– CH2O= 4.18J/g°C (from table)

Calorimetery Calculations

• DH metal = DH water

• DHmetal=mCDT • DH water = mCDT• So….• mCDT= mCDT

Specific heat of a metal

• You heat 50.0g of metal to 100oC

• You put the 50.0g of metal in a calorimeter containing 100mL of water.

• The water temperature changes from 25oC to 30oC

• What is the specific heat of the metal?

What metal is it?• DH metal = DH water

• mCDT = mCDT50g x C x (100oC-30oC) = 100g x 1cal/goC x 5oC

C =0.125 cal goC

What metal is it?• Let’s go to a table of the specific heats of metals and find out what metal we were using

Homework: • When a 4.25 g sample of solid NH4NO3 dissolves in 60.0 g of water in a calorimeter, the temperature drops from 21.0°C to 16.9°C. Calculate the energy involved in the dissolving of the NH4NO3.