Objectives: Students will be able to: -describe The structure of a water
molecule
-describe Hydrogen bonding
-
describe
the
physical properties of
water
A. Structure
A single water molecule consists of 2 hydrogen atoms and 1 oxygen atom. Each H-atom is attached by a single covalent bond to the oxygen atom. The water molecule is nonlinear and has a V-shape structure.
(figure available in print form)
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electron
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bond angle &
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molecular
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dipole
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distribution
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bond length
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orbital
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representation
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structure
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The bond angle between the 2 H-atoms is approximately
105° Oxygen is the second most electronegative element, as a result, the two covalent 0-H bonds in water are polar. The two polar covalent bonds and the bent structure result in a partial negative charge on the oxygen atom and a partial positive charge on each hydrogen atom which repel each other 105°. The polar nature of water is responsible for many of its properties, including its behavior as a solvent.
B. The Hydrogen Bond
A
hydrogen
bond
is a chemical bond that is formed between polar molecules that contain a hydrogen atom covalently bonded to a small, highly electronegative atom such as fluorine, oxygen or nitrogen. The high electronegative atom tends to have a negative charge while the hydrogen is positively charged. When a negatively charged atom from one molecule is attracted to a positively charge H-atom of another molecule. The hydrogen bond is actually a dipoledipole attraction of polar molecules.
The hydrogen bonding between water molecules give water physical properties that do not fit the trend relative to the molecular weights of similar compounds from the same chemical family. Water tends to have unusually high melting point, boiling point, heat of fusion and heat of vaporization. This is because hydrogen bonds are stronger than the intermolecular forces between other molecules.
Below is a table and graph that compares the physical properties of H20, H2S, H2Se and H2Te at STP.
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Molecular
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Melting
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Boiling
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for
mula
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color
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weight
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Point C
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Point
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H20
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colorless
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18.0 g
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0.0 C
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100 C
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H
22S
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colorless
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34.1 g
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-85.5 C
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-60 C
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H2Se
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colorless
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81.0 g
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-65.7 C
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-41 C
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H2Te
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colorless
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129.6 g
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-51.0 C
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-2 C
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Heat of fusion
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Heat of vaporization
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80 cal/g
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540 cal/g
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17 cal/g
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131 cal/g
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7 cal/g
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57 cal/g
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43 cal/g
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Temperature as a function of Molecular Weights in the Melting and boiling points of H 0, H2S, H2Se & H2Te
(figure available in print form)
C. Physical Properties
Water is a colorless, odorless, tasteless liquid with a melting point of 0° C and a boiling point of 100°C at 1 atm pressure. Two additional physical properties of matter are introduced with the study of water: Heat of fusion and heat of vaporization.
Heat of fusion
is the amount of heat required to change one gram of a solid into a liquid at its melting point. The heat of fusion of water is 80 cal/g. The heat of fusion is the energy needed to breakdown the crystalline lattice of ice from a solid to a liquid.
Heat of vaporization
is the amount of heat required to change one gram of a liquid into a gas at its boiling point. The heat of vaporization for water is 540 cal/g. The heat of vaporization is the energy needed to breakdown the inter- molecular forces (or hydrogen bonds) between molecules in a liquid to be unattracted to each other in a gas.
The
Specific Heat
of water is 1 cal/g°C. The specific heat is the amount of energy needed to raise one gram of water one degree Celsius.
ACTIVITY: (may be a demonstration or lab exercise, 30 min.)
Materials: hot plate, thermometer, 250 grams of ice, 500 ml beaker
Measure the temperature of the ice as time zero. Then begin heating ice and take temperature every 30 seconds. Once the ice starts to melt, the temperature no longer rises even though the hot plate continues to supply heat. Once all the ice has melted, the temperature again rises as the water heats up. The temperature of the water stops rising, once the water starts to boil even though the hot plate continues to supply heat at a constant rate.
DATA TABLE
time (seconds) temperature Heating Curves
(figure available in print form)
Problem Solving:
-
1. How many calories are required to change 20 grams of ice at 0.0° C to liquid water at 0.0° C? (the heat of fusion is 80.0 cal/gram)
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answer:
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Q
=
Hm
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Q
= 80.0
cal/g x 20 g
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Q= 1600
cal
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2. Calculate the number of calories required to vaporize 20 grams of liquid water at 100 C to gaseous water(steam) at 100 C. (The heat of vaporization for water is 540 cal/g)
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answer:
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Q= Hm
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Q=54O cal/g x 20 g
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Q= 10,800 cal
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3. How much heat would be needed to raise the temperature of a 20 gram sample of water from 50° to 80° C?
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answer:
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Q
=
m Tc
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Q= 20 g x 30
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C x 7 cal/g C
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Q = 600 cal
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4. Calculate the number of calories required to melt 20 grams of ice at 0.0 C and convert the resulting water to steam at 100 C.
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answer:
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Q= Hfm + m
Tc +
Hvap
m
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Q=
80
cal x 20 g + 20 g x 700 C x 7
cal
+ 540
cal
x
2Og
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g g C g
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Q= 1600
cal + 2000 cal + 70,000 cal
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Q= 74,400
cal
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