section 7.3—changes in state
DESCRIPTION
Section 7.3—Changes in State. Change in State. During a change in state:. We are breaking intermolecular forces. . Breaking intermolecular forces requires energy. - PowerPoint PPT PresentationTRANSCRIPT
Section 7.3—Changes in State
• During a change in state, the energy being put into the system is used to breaking IMF’s, not increasing motion (temperature). WE CAN’T USE mc∆T TO GET THIS ENERGY!!
Change in State
• We are breaking intermolecular forces.
• Breaking intermolecular forces requires energy
A sample with solid & liquid will not rise above the melting point until all the solid is gone. The temp stays the same! WE CAN’T USE mc∆T TO GET THIS ENERGY!!
During phase changes we need ANOTHER equation!
During a change in state:
Melting - at 1 atm of Pressure
Enthalpy of Fusion (Hfus) - the amount of energy needed to melt 1 gram of a substance
The enthalpy of fusion of water is 80.87 cal/g or 334 J/g
Example: All samples of ice will melt at 0°C and 1atm BUT the more ice you melt, the more energy you need!!!
fusHmH Energy needed to melt
Mass of the sample
Energy needed to melt 1 g
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 – at 1 atm of Pressure
Enthalpy of Vaporization (Hvap) - the amount of energy needed to boil 1 gram of a substance
The Hvap of water is 547.2 cal/g or 2287 J/g
Example: All samples of water boil at 100°C BUT the more you have the more energy it takes to boil!
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
H = enthalpy (energy)m = mass of sampleHfus = enthalpy of fusion
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
Melting only breaks some IMF’s!!!
Boiling Breaks All IMF’s!!!
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 vaporize a sample, then 547 J would be
released when the sample condenses. H will = -547 J
The 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
Example
Example:How much energy is
released when 157.5 g of water is condensed? Hvap water = 547.2 cal/g
Example
vapHmH
H = enthalpy (energy)m = mass of sampleHfus = enthalpy of fusion
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
Heating curve of water
-500
50100
150
Energy input
Tem
pera
ture
Going Up & Down
+H
-H
Moving up the curve requires energy, while moving down releases energy
Heating curve of water
-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
Heating curve of water
-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!
Heating curve of water
-500
50100
150
Energy input
Tem
pera
ture
Changing StatesLiquid & gas
Hvap = 547.2 cal/g
Solid & LiquidHfus = 80.87 cal/g
Heating curve of water
-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
Cp water = 1.00 cal/g°CCp steam = 0.48 cal/g°C
Hfus = 80.87 cal/gHvap = 547.2 cal/g
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