1. 10.5 Water

2. POINT > Define ∆Hv and ∆Hf for water
POINT > Define density of water and ice
POINT > Calculate the energy needed to boil or melt volumes/masses of water

3. POINT > Define ∆Hv and ∆Hf for water
The extensive hydrogen bonding that occurs between water molecules gives water unique properties
∆Hv and ∆Hf for water are both relatively high compared to other compounds of similar molar mass

4. ∆Hv = 40.7kJ/mol ∆Hf = 6.01kJ/mol
POINT > Define ∆Hv and ∆Hf for water
∆Hv = 40.7kJ/mol
∆Hf = 6.01kJ/mol

5. POINT > Define ∆Hv and ∆Hf for water
Ex. How much energy is absorbed when 47.0g of ice melts? (at STP)
Energy = 47.0g x 1 mol x kJ
18g mol
= 15.7 kJ

6. WB CHECK:
How much energy is needed to boil 47.0g of water? (at STP) Energy = 47.0g x 1 mol x 40.7kJ 18g 1 mol = 106 kJ

7. POINT > Define ∆Hv and ∆Hf for water
What mass of water requires 5.97 x 105kJ of energy to boil?
5.97 x 105kJ = kJ = 1.47 x 104 mol
1 mol
1.47 x 104 mol x g mol
= 2.65 x 105 g H2O
(x mol)

8. WB CHECK:
What mass of ice requires 3.00 x 103kJ of energy to melt?
3.00 x 103kJ = kJ = 4.99 x 102 mol
1 mol
x g
1 mol = 8.99 x 103 g
(x mol)

9. POINT > Define density of water and ice
The density of liquid water is greater than that of ice. This is unique and important!
Density of liquid water = 1.00g/cm3
Density of ice = 0.917g/cm3

10. WB CHECK:
How much energy is required to melt 5.00 liters of ice? 5 x 103 cm3 x 0.917g = 4590g x 1mol = 255mol cm3 18.0g 255mol x 6.01kJ = 1.53 x 103 kJ mol
(1.00L = 103cm3)

11. Homework:
Read
Practice #1-2 page 333
F.A. #1-6 page 333
Practice #1-2 page 334