Thermochemistry is the study of energy changes that occur during chemical reactions and phase changes. Heat (q) is energy that transfers from one object to another due to a temperature difference. Exothermic processes release heat (negative q value), while endothermic processes absorb heat (positive q value). Enthalpy (H) is the heat content of a system at constant pressure, and enthalpy changes (ΔH) quantify the heat absorbed or released during chemical reactions. Thermochemical equations relate the enthalpy change to the heat of reaction. Calorimetry problems use these equations and concepts to calculate energy changes involved in physical and chemical processes.

1. Thermochemistry

2. Definitions
• Energy – capacity for doing work or
supplying heat.
• Thermochemistry – study of energy
changes that occur during phase changes
and chem. rxns.
• Chem. Potential Energy – energy stored in
chemical bonds.

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

4. Heat
• Represented by q.
• Energy that transfers from one object to
another because of a Temp. difference
between them.
• Heat flows from warm  cool until the
two objects are at the same Temp.

5. Exothermic vs. Endothermic
• In exothermic processes, the system loses
heat as its surroundings warm up.
– q has a negative value b/c the system is losing
heat.
• In endothermic processes, the system gains
heat as its surroundings cool down.
– q has a positive value b/c the system is gaining
heat.

6. Potential Energy Diagram of Ice
Melting at 0ºC.
Time 
PotentialEnergy
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)
– Amt. of heat needed to raise 1 gram of water
by 1ºC.
– 1 cal = 4.184 J
• Food Calorie (capital “C”) = 1000 cal, or 1
kilocalorie = 4184 J

8. What do Calories mean in food?
• 10 grams of sugar has 41 Calories.
– When 10 grams of sugar are burned, 41 kcal
(170 kJ) of energy are released.
– Your body “burns” food for energy.
– In order to use the energy available in 10
grams of sugar, you must do 41 kcal worth of
work.

9. Heat Capacity
• Amount of heat needed to raise an object’s
temperature by 1°C.
– Depends on the chemical composition and the
mass 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.

10. Heat Capacity
1 g H2O
10 g H2O
Same temperature change

11. Specific Heat (c)
• Amt. of heat needed to raise 1 gram of a
substance’s temperature by 1ºC.
– Expressed in J/g ºC, or cal/g ºC
• The higher a substance’s specific heat, the
more energy it takes to heat it.
• Substance’s with low specific heats heat up
and cool down quickly (most metals, e.g.)

12. Some Specific Heats
Substance Specific Heat
J/gºC cal/g ºC
Water 4.18 1.00
Grain alcohol 2.4 0.58
Ice 2.1 0.50
Steam 1.7 0.40
Chloroform 0.96 0.23
Aluminum 0.90 0.21
Iron 0.46 0.11
Silver 0.24 0.057
Mercury 0.14 0.033

13. Specific Heat (c)
• c = heat / (mass x change in Temp.)
• c = q / (m x ΔT)
• q = m x c x ΔT

14. Example Problem
• The temperature of a 95.4-g piece of Cu
increases from 25.0ºC to 48.0ºC when the
Cu absorbs 849 J of heat. What is the
specific heat of Cu?
– SOLUTION: q = m x c x ΔT
• 849 J = (95.4 g) c (48.0ºC – 25.0ºC)
• 849 J = (95.4 g) c (23.0ºC)
• 849 J = (2190 gºC) c
• c = 0.388 J/gºC
• Based on what you know about metals,
does this answer make sense?

15. Example Problem
• When 435 J of heat is added to 3.4 g of
olive oil at 21ºC, the temperature increases
to 85ºC. What is the specific heat of olive
oil?
– SOLUTION: q = m x c x ΔT
• 435 J = (3.4 g) c (85ºC – 21ºC)
• 435 J = (3.4 g) c (64ºC)
• 435 J = (220 gºC) c
• c = 2.0 J/gºC

16. Example Problem
• How much heat is required to raise the
temperature of 250.0 g of mercury by
52ºC? The specific heat of mercury is 0.14
J/gºC.
– SOLUTION: q = m x c x ΔT
– q = (250.0 g)(0.14 J/gºC)(52ºC)
– q = 1800 J = 1.8 kJ

17. Enthalpy Changes
• Enthalpy (H) – the heat content of a system
at constant pressure.
• Enthalpy change (ΔH) – the heat that
enters or leaves a system at constant
pressure.
• q = ΔH
• Neg. ΔH = exothermic process
• Pos. ΔH = endothermic process

18. Thermochemical Equations
• Enthalpy change can be written as a
reactant or a product.
– Reactant  endothermic
– Product  exothermic
• Example: The reaction of calcium oxide
with water is exothermic.
– It produces 65.2 kJ of heat per mole of CaO
reacted.
– CaO(s) + H2O(l)  Ca(OH)2(s) + 62.5 kJ

19. An Exothermic Reaction
CaO(s) + H2O(l)
Ca(OH)2(s)
ΔH = -65.2 kJ
CaO(s) + H2O(l)  Ca(OH)2(s) + 62.5 kJ

20. Thermochemical Equations
2NaHCO3(s) + 129 kJ  Na2CO3(s) + H2O(g) + CO2(g)
2NaHCO3(s)
Na2CO3(s) + H2O(g) + CO2(g)
ΔH = +129 kJ

21. Thermochemical Equations and
Stoichiometry
• You can use thermochemical equations in
stoichiometry.
– How much heat energy is produced when 55.0 grams
of ethanol is burned completely?
• C2H5OH(l) + 3O2(g)  2CO2(g) + 3H2O(g) + 1300. kJ
• Given: 55.0 g C2H5OH
• Want: kJ
• Conversion factors:
– 1 mol C2H5OH produces 1300. kJ when burned
– 1 mol C2H5OH = 46.07 g/mol
55.0 g C2H5OH
OHHCg46.07
OHHCmol1
x
52
52
OHHCmol1
kJ1300.
x
52
= 1550 kJ

22. Thermochemical Equations and
Stoichiometry
• 0.500 grams of methane gas are burned completely beneath a
container that holds 100. grams of water, originally at 20.0º. If
all of the heat from the combustion reaction goes into the
water, what will the water’s final temperature be?
– CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) + 803 kJ
– First find out how much total heat is released.
– Given: 0.500 g CH4(g)
– Want: kJ
– Conversion factors:
• 1 mol CH4 = 16.05 g CH4
• 1 mol CH4 produces 803 kJ when completely burned
0.500 g CH4
4
4
CHg16.05
CHmol1
x
4CHmol1
kJ803
x = 25.0 kJ

23. 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 + 59.8ºC = 79.8ºC.
25.0 kJ
kJ1
J1000
x = 2.50x104
J