1. Thermochemistry

2. Definitions
Thermochemistry – study of energy changes that occur during phase changes and chemical reactions.
Energy – capacity for doing work or supplying heat.
Chemical (Potential) Energy – energy stored in chemical bonds.

3. Example Lots of energy stored in bonds! Energy difference. 5473 kJ/mol
Little energy stored in bonds.

4. Heat
Heat - Energy that transfers from one object to another because of a temperature difference between them.
Represented by the symbol q
Heat flows from warm  cool until the two objects are at the same temperature
heat = amount of energy
temperature = average kinetic energy

5. Exothermic vs. Endothermic
Exothermic – energy is released
Surroundings are warmed up
q is negative because the system is losing heat.
Endothermic – energy is absorbed
surroundings are cooled down.
q is positive because the system is gaining heat.

6. Potential Energy Diagram of Ice Melting at 0ºC.
Time 
Potential Energy 
Ice
Water
Is the melting of ice an endothermic or an exothermic process? How can you tell?

7. Measuring Heat Flow SI Unit of heat flow: Joule (J)
Common unit used in chemistry: calorie (cal) -amount of heat needed to raise 1 gram of water by 1ºC.
1 cal = J
Food Calorie (capital “C”) = 1000 cal, or 1 kilocalorie = 4184 J

8. Heat Capacity
Amount of heat needed to raise an object’s temperature by 1°C.
Depends on the chemical composition and the mass(grams) of the object.
EXAMPLE: 1 gram of water requires 1 cal to raise its temperature by 1°C.
100. g of water require 100. cal to raise the temp. by 1°C.

9. Specific Heat (c)
Amt. of heat needed to raise 1 gram of a substance’s temperature by 1ºC.
Expressed in J/g ºC
The higher a substance’s specific heat, the more energy it takes to heat it.
Water: 4.18 J/gºC

10. Energy of Temperature Change
q = mcΔT
q = heat energy in Joules
m = mass in grams
c = specific heat(for water, it’s 4.18 J/gºC)
ΔT = change in temperature

11. Example Problem
What is the energy required to heat a gram glass of water from 25.0ºC to ºC?
SOLUTION: q = m c ΔT
q = (120.0 g) (4.18 J/gºC) (100.0ºC – 25.0ºC)
q = (120.0 g) (4.18 J/gºC) (75.0ºC)
q = J

12. Example Problem
1000. Joules of energy is added to a gram sample of water at 25.0ºC. What is the change in temperature?
SOLUTION: q = m c ΔT
1000. J = (50.0 g) (4.18 J/gºC) ΔT
1000. J = (209 J) ΔT
ΔT = 4.78 ºC
______ ______
209 J J

13. Thermochemical Equations and Stoichiometry
The combustion of 5.00 grams of methane releases 250. kJ of heat.
Now we’ll calculate how hot the water in the container will get if it absorbs all of the heat.
First convert 25.0 kJ to J
q = m x c x T
2.50x104 J = (100. g) (4.18 J/gºC) T
2.50x104 J = (418 J/ºC) T
T = 59.8ºC
The water will get 59.8ºC warmer.
The final temperature will be 20.0ºC ºC = 79.8ºC.
25.0 kJ
= 2.50x104 J