section 7.3—changes in state

Section 7.3—Changes in State

What’s happening when a frozen ice pack melts?

The energy being put into the system is used for breaking IMF’s, not increasing motion (temperature)

Change in State

To melt or boil, intermolecular forces must be broken

Breaking intermolecular forces requires energy

A sample with solid & liquid will not rise above the melting point until all the solid is gone.

The same is true for a sample of liquid & gas

Melting

When going from a solid to a liquid, some of the intermolecular forces are broken

The Enthalpy of Fusion (Hfus) is the amount of energy needed to melt 1 gram of a substanceThe enthalpy of fusion of water is 80.87 cal/g or 334 J/g

All samples of a substance melt at the same temperature, but the more you have the longer it takes to melt (requires more energy).

fusHmH Energy needed to melt

Mass of the sample

Energy needed to melt 1 g

Phase Changes with a Positive Enthalpy

The energy added during a this type of a phase change is not being used to increase the temperature of a substance, but instead to break the intermolecular forces between molecules to change the state. Energy is absorbed when bonds are broken.

ΔH will be positive State changes

-Melting -Boiling

Example

Example:Find the enthalpy of

fusion of a substance if it takes 5175 J to melt 10.5 g of the

substance.

Example

fusHmH

H = enthalpy (energy)m = mass of sampleHfus = enthalpy of fusion

fusHgJ )5.10(5175

fusHgJ

5.10

5175Hfus = 493 J/g

Example:Find the enthalpy of

fusion of a substance if it takes 5175 J to melt 10.5 g of the

substance.

Vaporization

When going from a liquid to a gas, all of the rest of the intermolecular forces are broken

The Enthalpy of Vaporization (Hvap) is the amount of energy needed to boil 1 gram of a substanceThe Hvap of water is 547.2 cal/g or 2287 J/g

All samples of a substance boil at the same temperature, but the more you have the longer it takes to boil (requires more energy).

vapHmH Energy needed to boil

Mass of the sample

Energy needed to boil 1 g

ExampleExample:

If the enthalpy of vaporization of water is 547.2 cal/g, how many calories are

needed to boil 25.0 g of water?

Example

vapHmH


gcalgH 2.547)0.25(

H = 1.37×104 cal

Example:If the enthalpy of

vaporization of water is 547.2 cal/g, how many calories are

needed to boil 25.0 g of water?

Solid

Liquid

Gas

Melting

Vaporizing or Evaporating

Condensing

Freezing

Incr

easi

ng m

olec

ular

mot

ion

(tem

pera

ture

)

Changes in State go in Both Directions

Going the other way

The energy needed to melt 1 gram (Hfus) is the same as the energy released when 1 gram freezes. If it takes 547 J to melt a sample, then 547 J would be

released when the sample freezes. H will = -547 JThe energy needed to boil 1 gram (Hvap) is the

same as the energy released when 1 gram is condensed. If it takes 2798 J to boil a sample, then 2798 J will be

released when a sample is condensed. H will = -2798 J

Phase Changes with a negative Enthalpy The energy being released by

the system during this type of a phase change is not being used to decrease the temperature of a substance, but instead to re-form the intermolecular forces between the molecules to change the state to a more organized structure. When forces are being re-formed, energy is released.

ΔH will be negative Condensing Freezing

Example

Example:How much energy is

released with 157.5 g of water is condensed?

Hvap water = 547.2 cal/g

Example

vapHmH


gcalgH 2.547)5.157(

H = - 8.6×104 cal

Example:How much energy is

released with 157.5 g of water is condensed?

Hvap water = 547.2 cal/g

Since we’re condensing, we need to “release” energy…H will be negative!

Heating curve of water

-500

50100

150

Energy input

Tem

pera

ture

Heating Curves

Melting & Freezing

Point

Boiling & Condensing

Point

Heating curves show how the temperature changes as energy is added to the sample


-500

50100

150

Energy input

Tem

pera

ture

Going Up & Down

+H

-H

Moving up the curve requires energy, while moving down releases energy


-500

50100

150

Energy input

Tem

pera

ture

States of Matter on the Curve

Gas OnlyEnergy added

increases temp

Liquid & gasEnergy added breaks remaining IMF’s

Liquid OnlyEnergy added

increases temp

Solid & LiquidEnergy added breaks IMF’s

Solid OnlyEnergy added

increases temp


-500

50100

150

Energy input

Tem

pera

ture

Different Heat Capacities

Gas OnlyCp = 0.48 cal/g°C

Liquid OnlyCp = 1.00 cal/g°C

Solid OnlyCp = 0.51 cal/g°C

The solid, liquid and gas states absorb water differently—use the correct Cp!


-500

50100

150

Energy input

Tem

pera

ture

Changing StatesLiquid & gas

Hvap = 547.2 cal/g

Solid & LiquidHfus = 80.87 cal/g


-500

50100

150

Energy input

Tem

pera

ture

Adding steps together

If you want to heat ice at -25°C to water at 75°C, you’d have to first warm the ice to zero before it could melt.

Then you’d melt the ice

Then you’d warm that water from 0°C to your final 75°

You can calculate the enthalpy needed for each step and then add them together

Example:How many calories are

needed to change 15.0 g of ice at -12.0°C to steam at

137.0°C?

ExampleUseful information:Cp ice = 0.51 cal/g°C

Cp water = 1.00 cal/g°CCp steam = 0.48 cal/g°C

Hfus = 80.87 cal/gHvap = 547.2 cal/g

TCpmH

fusHmH

vapHmH

Example:How many calories are

needed to change 15.0 g of ice at -12.0°C to steam at

137.0°C?

Example

gcalgH 2.5470.15

CCgcalgH 010000.10.15

CCgcalgH 12051.00.15

Useful information:Cp ice = 0.51 cal/g°C



TCpmH

fusHmH

CCCg

calgH 1000.13748.00.15

Warm ice from -12.0°C to 0°C

Melt ice

Warm water from 0°C to 100°C

Boil water

Warm steam from 100°C to 137°C

gcalgH 87.800.15

91.8 cal

1500 cal

1213 cal

8208 cal

266 cal

vapHmH

Total energy = 11279 cal

Example:How many needed to

change 40.5 g of water at 25°C to steam at 142°C?

Let’s PracticeUseful information:Cp ice = 0.51 cal/g°C



TCpmH

fusHmH

vapHmH

Example:How many needed to

change 40.5 g of water at 25°C to steam at 142°C?

Let’s Practice

gcalgH 2.5475.40

CCgcalgH 2510000.15.40

Useful information:Cp ice = 0.51 cal/g°C



TCpmH

fusHmH

CCCg

calgH 1000.14248.05.40

Warm water from 25°C to 100°C

Boil water

Warm steam from 100°C to 142°C

3038 cal

22162 cal

816 cal

vapHmH

Total energy = 26016 cal

section 7.3—changes in state

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