Transcript Document

H2O (g)
H2O (s)
H2O ()
Heat & Changes of State
Heat & Changes of State
sublimation
sublimation
melting
freezing
boiling
vaporization
condensation
deposition
Heat & Changes of State
Molar Heat of:
Latent Heat of:
Fusion
Heat involved
per mole
Heat involved
per gram
Vaporization
0 C
100 C
80 cal/g
540 cal/g
Energy Requirements for changing state:
In ice the water molecules are held together by strong intermolecular
forces.
The energy required to melt 1 gram of a substance is called the
latent heat of fusion
For ice it is 80.0
cal/g
The energy required to change 1 gram of a liquid to its vapor is
called the latent heat of vaporization
For water it is 540
cal/g
It takes more energy to vaporize water than to melt it.
This is because in melting you weaken the intermolecular forces.
Here about 1/6 of the hydrogen bonds are broken.
In vaporization you totally break them.
All the hydrogen bonds are broken
Fusion is when a solid melts to form a liquid
Vaporization is when a liquid evaporates to form a gas.
Heating and cooling curve for water heated at a constant rates.
A-B = Solid ice, temperature is increasing.
Particles gain kinetic energy, vibration of particles increases.
Ice
B-C = Solid starts to change state from solid to liquid.
Temperature remains constant as energy is used to break intermolecular bonds.
H2O (s)  H2O () energy required  80 cal/g
0ºC
C-D = temperature starts to rise once all the solid has melted.
Particles gain kinetic energy.
Liquid
water
D-E = Liquid starts to vaporize, turning from liquid to gas. The
temperature remains constant as energy is used to break intermolecular forces.
H2O ()  H2O (g)
energy required  540 calg
100ºC
E-F = temperature starts to rise once all liquid is vaporized. Gas
particles gain kinetic energy.
steam
Heating Diagram
Calculating Energy Changes
2.
4.
3.
5.
1.
DHf = 80 cal/g Te
DHv= 540 cal/g m
p
o
cl = 1.00 cal/g • eC
cv = 0.50 cal/g • oraC
cs = 0.50 cal/g • oC
t
u
r
e
(oC)
Heat Added
Calculating Energy Changes
Phase change:
Q(gained or lost) = m x L. H.(fusion/vaporization)
Temperature change:
Q(gained or lost) = m • c
heat
= mass
specific
heat
• DT
(Tf - Ti )
Problem
How much energy is required to heat 25 g of liquid water from 25C
to 100C and change it to steam?
Step 1: Calculate the energy needed to heat the water from 25C to
100C
Q = m  c  DT
Q = 25g
 1.0 cal g-1 C –1  75 C =
Step 2: Vaporization: Use the Latent Heat to calculate the energy
required to vaporize 25g of water at 100C
Q = 25.0 g  540 cal/g =
.25g  1mol H2O / 18g mol-1 H2O = 1.4 mol H2O
Dvap H (H2O) = 1.4 mol H2O  40.6kJ/mol = 57 kJ
Total energy change is:
Calculating Energy Changes
Calculate the total amount of heat needed to change 1 mole
of ice at -7oC to steam at 125oC.
18.0 g x 0.5 cal/g•oC x 25oC = 225 cal
Dphase
DT
Dphase
DT
18.0 g x 540.0 cal/g
DT
= 9720 cal
18.0 g x 1.0 cal/g•oC x 100oC = 1800 cal
18.0 g x 80.0 cal/g
= 1440 cal
18.0 g x 0.5 cal/g•oC x 7oC = 63 cal
= 13248 cal
Intermolecular Forces
Intra-molecular forces are (within the molecule) while inter-molecular
forces are (between molecules)
Types of inter-molecular forces
dipole-dipole (1% as strong as covalent bonds)
POLAR MOLECULES
A special type of dipole-dipole force is the hydrogen bond. These form
between molecules that contain a hydrogen atom bonded to a very
electronegative element like N, O or F. Hydrogen bonds are very strong
compared to an ordinary dipole-dipole bond.
E.g HF, NH3, H2O all form hydrogen bonds
Hydrogen bonding 10% as strong as covalent bonds
London dispersion forces (instantaneous and induced dipoles)
NON-POLAR MOLECULES
Non-polar molecule
This instantaneous dipole will
effect any nearby molecules
Movement of electrons causes
an instantaneous dipole
This induces a dipole in a
nearby molecule
Water molecules are polar molecules. The  - oxygen forms
intermolecular bonds with the  + hydrogen of another water
molecules. Water has a special type of intermolecular bond called a
hydrogen bond.
Inter-molecular forces
Ice molecules are locked in
fixed positions, held by
intermolecular-bonds.
Ice is less dense than liquid
water because the molecules are
further apart than in liquid
water.
Other properties of Liquids: Many properties are due to the
forces between the particles.
Why do liquids on a surface form
droplets?
Why do some liquids exhibit capillary
action?
Hg
H2O
Why are some liquids more viscous than
others?
This allows insects to walk on water!
The inward force or pull which tends to
minimize the surface area of any liquid is
surface tension.
Surface tension is caused by hydrogen bonding between water
molecules. The more polar a liquid the stronger its surface
tension.
Hg
pure H2O
H2O with detergent
Surfactants are chemicals that decrease the surface tension of
water, detergents and soaps are examples.
The smallest surface area a liquid can form is a sphere.
Viscosity is the resistance to motion of a liquid.
Maple syrup is more viscous than water.
But water is much more viscous than
gasoline or alcohol.
The stronger the attraction between
molecules of a liquid, the greater its
resistance to flow and so the more
viscous it is.
Capillary action is the spontaneous rising of a liquid in a
narrow tube.
Two forces are responsible for this action:
Cohesive forces,the intermolecular forces between molecules of
the liquid
Adhesive forces, between the liquid molecules and their container
If the container is made of a substance
that has polar bonds then a polar liquid
will be attracted to the container.
This is why water forms a concave
meniscus while mercury forms
convex meniscus
Hg
H2O
The fact that water has both strong cohesive (intermolecular)
forces and strong adhesive forces to glass, it pulls itself up a glass
capillary tube.
This also allows it to be drawn up high into trees like giant
redwoods.
If you place a liquid in a container, then some of the particles
will have enough kinetic energy to evaporate. You will notice
the amount of liquid decreasing.
Dynamic - at the same time some of these gaseous molecules
condense to reform liquid.
In an open container all the liquid
will eventually evaporate out if they
have enough kinetic energy.
Volatility – the higher the vapor
pressure, the more volatile the liquid.
Temperature – vapor pressure
increases in a non-linear fashion with
increasing temperature.
Closed system - In a sealed
container, molecules will start to
evaporate and the liquids
volume will decrease.
But some of these molecules will then condense and after a short
time the volume of the liquid will not change. Has evaporation
and condensation stopped?
No, both evaporation and condensation continue.
But an equilibrium has been reached.
The rate of evaporation = the rate of condensation
When water is heated bubbles of
vapor form within it. The vapor
pressure in the bubble is the same
as the vapor pressure of the water
at that temperature.
As long as this vapor pressure is less
than atmospheric pressure the bubbles
collapse.
When the temperature of the water reaches a
point that the vapor pressure of the bubble
equals atmospheric pressure, the bubbles
don’t collapse, they get larger and more form
and escape as steam. The water begins to boil.
The normal boiling point
of a liquid occurs at 1 atm.
Calculating Energy Changes: Solid to liquid
How much energy is required to melt 8.5 g of ice at 0C?
The molar heat of fusion for ice is 6.02 kJmol-1
Step 1: How many moles of ice do we have?
n = m/M
n = 8.5g / 18gmol-1
= 0.47 mol H2O
Step 2: Use the equivalence statement to work the energy (6.02 kJ is
required for 1 mol H2O)
kJ = 0.47 mol H2O  6.02 kJ / mol H2O
= 2.8kJ
What is specific heat capacity?
The amount of energy required to change the temperature of one
gram of a substance by 1C .
10 C
11 C
Another name for specific heat is a calorie
(1 calorie = 4.184 Joules)
Specific heat capacity of liquid water (H2O (L) ) is 4.18 J g-1C–1.
Water (s) = 2.03 J g-1 C –1
Water (g) = 2.0 J g-1 C –1
 0.5 cal/g to break up ice
Calculating the energy to increase the temperature of liquid water.
Calculating specific heat using the equation:
Q = ms (tf  ti)
or
Q = energy (heat) required
Q = ms DT
or
s = specific heat capacity
Heat (H) = ms (tf  ti)
m = mass of the sample
DT = change in temperature in C
EXAMPLE:
How much energy does it take to heat 10g of water from 50 to 100 C ?
Specific heat capacity of water = 4.184 J g-1C–1
Q = m  s  DT
Q=
(10g)  (4.184 J g-1 C -1)  (50 C) = 2.1  10 3 J
Dvap H (H2O) = 1.4 mol  40.6kJ/mol = 57 kJ
Explain these trends in Boiling points
Boiling point is effected by the strength of the inter-molecular
forces between liquid molecules.
The general trend is an increase in B.P. due the greater size of the
molecules and hence the greater intermolecular forces
The anomalous B.P. for H2O, HF, and NH3 are explained by the
fact that they exhibit hydrogen bonding.
Heating Diagram