The
internal energy of a sample is
the sum of all the kinetic and potential energies
of all the atoms and molecules in a sample
Vaporization
Melting
Solid
Gas
Liquid
Freezing
Condensation
Heating Curves Animation
A plot of temperature vs. time that represents the
process in which energy is added at a constant rate
Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem
140
120
Gas - KE 
Temperature (oC)
100
Boiling - PE 
80
60
40
20
0
-20
Liquid - KE 
Melting - PE 
-40
-60
-80
Solid - KE 
-100
Time
A plot of temperature vs. time that represents
the process in which energy is added at a
constant rate
• The temperature doesn’t change during a
phase change.
• If you have a mixture of ice and water, the
temperature is 0ºC
• At 1 atm, boiling water is 100ºC
• You can’t get the temperature higher until
it boils
Chemical Energy
2 parts of the universe as it relates to a
chemical reaction:
• System
– the reactants and the products
• Surroundings
– everything else in the universe
(such as container, the room, etc.)
Law of Conservation of Energy: the total energy of
the universe is constant and can neither be created
nor destroyed; it can only be transformed.
The First Law of Thermodynamics:
The total energy content of the
universe is constant
surroundings
system
Signs (+/-) will tell you if energy is
entering or leaving a system
+ indicates energy enters a system
- indicates energy leaves a system
Chemical energy lost by combustion =
Energy gained by the surroundings
Chemical Energy
Two types of processes based on energy flow:
• Exothermic
– produces energy (heat flows out of the system)
• Endothermic
– absorbs energy (heat flows into the system)
Conservation of Energy
in a Chemical Reaction
In this example, the energy of the reactants and products decreases,
while the energy of the surroundings increases.
In every case, however, the total energy does not change.
Exothermic Reaction
Energy
Surroundings
mistry, 2004, page 41
System
Before reaction
Surroundings
Reactant
System
After reaction
Product + Energy
Conservation of Energy
in a Chemical Reaction
In this example, the energy of the reactants and products increases,
while the energy of the surroundings decreases.
In every case, however, the total energy does not change.
Surroundings
Reactant + Energy
Energy
Surroundings
mistry, 2004, page 41
Endothermic Reaction
System
System
Before reaction
After reaction
Product
Thermochemistry
• Every reaction has an energy change
associated with it
• Energy is stored in bonds between atoms
• Making bonds gives energy
• Breaking bonds takes energy
20
•To more easily measure and study the energy
changes that accompany chemical reactions,
chemists have defined a property called
enthalpy.
•Enthalpy (H) is the heat content of a system at
constant pressure.
•Although you cannot measure the actual
energy or enthalpy of a substance, you can
measure the change in enthalpy, which is the
heat absorbed or released in a chemical
reaction.
•The change in enthalpy for a reaction is called
the enthalpy (heat) of reaction (∆Hrxn).
•You have already learned that a symbol
preceded by the Greek letter ∆ means a change
in the property.
•Thus, ∆Hrxn is the difference between the
enthalpy of the substances that exist at the end
of the reaction and the enthalpy of the
substances present at the start.
•Because the reactants are present at the
beginning of the reaction and the products are
present at the end, ∆Hrxn is defined by this
equation.
Enthalpy (H) is used to quantify the heat flow into or out of a
system in a process that occurs at constant pressure.
DH = H (products) – H (reactants)
DH = heat given off or absorbed during a reaction at constant pressure
Hproducts > Hreactants
Hproducts < Hreactants
DH > 0
DH < 0
6.4
Energy Change in Chemical
Processes
reaction
Exothermic, heat given off &
temperature of water rises
Exothermic process:
DH < 0 (at constant pressure)
Endothermic, heat taken in &
temperature of water drops
reaction
Endothermic process:
DH > 0 (at constant pressure)
Exothermic process is any process that gives off
heat – transfers thermal energy from the system to
the surroundings.
2H2 (g) + O2 (g)
2H2O (l) + energy
H2O (g)
H2O (l) + energy
Endothermic process is any process in which heat
has to be supplied to the system from the
surroundings.
energy + 2HgO (s)
energy + H2O (s)
2Hg (l) + O2 (g)
H2O (l)
6.2
Endothermic Reactions
Exothermic Reactions
exothermic
endothermic
exothermic
endothermic
endothermic
Effect of Catalyst on Reaction Rate
What is a catalyst? What does it do during a chemical reaction?
Catalyst lowers the activation energy for the reaction.
No catalyst
Energy
activation energy
for catalyzed reaction
reactants
products
Reaction Progress
140
120
DH = mol x DHfus
DH = mol x DHvap
Temperature (oC)
100
80
Heat = mass x Dt x Cp, gas
60
40
20
0
Heat = mass x Dt x Cp, liquid
-20
-40
-60
-80
Heat = mass x Dt x Cp, solid
-100
Time
• The heat that is absorbed by one mole of a
substance in melting at a constant temperature is
the molar heat of fusion DHfus
• The heat lost when one mole of a liquid solidifies
at a constant temperature is the molar heat of
solidification DHsol
H2O (s)
H2O (l)
33
H2O (l)
H2O (g)
DHfus =
6.01 kJ/mol
DHsol = - 6.01 kJ/mol
• The molar heat of vaporization:
– Heat needed to change one mol of a liquid to
gas DHvap
• The molar heat of condensation:
– Heat needed to change one mol of a gas to
liquid DHcon
H2O (l)
H2O (g)
34
H2O (g)
H2O (l)
DHvap =
40.7 kJ/mol
DHcon = - 40.7 kJ/mol
• The heat that is released or absorbed in a
chemical reaction is equivalent to DH
C + O2(g)
C + O2(g)
CO2(g) +394 kJ
CO2(g)
DH = -394 kJ
• In thermochemical equation it is important
to say what state
H2O(g)
H2O(l)
H2(g) + ½ O2 (g)
H2(g) + ½ O2 (g)
DH = 241.8 kJ
DH = 285.8 kJ
Difference = 44.0 kJ
35
“In going from a particular set of reactants to a
particular set of products, the change in
enthalpy is the same whether the reaction takes
place in one step or a series of steps.”
• The change in heat that accompanies the
formation of a mole of a compound from its
elements at standard conditions
• Standard conditions 25°C and 1 atm.
• Symbol is DH◦f
The
standard heat of formation of an element at
its most stable form is 0
This
38
includes the diatomics
• There are tables of heats of formations
(pg. 316)
• For most compounds it is negative
– Because you are making bonds
– Making bonds is exothermic
• The heat of a reaction can be calculated
by subtracting the heats of formation of
the reactants from the products
DH = DH◦f (products) - DH◦f (reactants)
39
1. If a reaction is reversed, the sign of ∆H must
be reversed as well.
– because the sign tells us the direction of heat
flow as constant P
2. The magnitude of ∆H is directly proportional
to quantities of reactants and products in
reaction.
If coefficients are multiplied by an integer,
the ∆H must be multiplied in the same way.
– because ∆H is an extensive property
If H2(g) + 1/2 O2(g)
H2O(l) D H=-285.5 kJ/mol
then
H2O(l)
H2(g) + 1/2 O2(g) D H =+285.5 kJ/mol
If you turn an equation around, you change the sign
2 H2O(l)
2 H2(g) + O2(g)
D H =+571.0 kJ/mol
If you multiply the equation by a number, you multiply
the heat by that number.
– Twice the moles, twice the heat
41
• You make the products, so you need
their heats of formation
• You “unmake” the reactants so you
have to subtract their heats.
ΔHreaction  Σn pΔHf(products)  Σn r ΔHf(reactant s)
https://www.youtube.com/watch?v=_NLAgSnqNOE&noredirect=1
42
Calculate the heat of combustion of methane, CH4
CH4(g) + 2 O2(g)
CO2(g) + 2 H2O(g)
DH◦f CH4 (g) = -74.86 kJ/mol
DH◦f O2(g) = 0 kJ/mol
DH◦f CO2(g) = -393.5 kJ/mol
pg. 316
DH◦ fH2O(g) = - 241.8 kJ/mol
H2 (g) + ½ O2 (g)  H2O(g)
2x(- 241.8)= - 483.6kJ/mol
Step #1: since 2 moles of water are produced by each mole of methane, we
multiply the DH◦ f. of water by 2.
43
Calculate the heat of combustion of methane, CH4
CH4(g) + 2 O2(g)
CO2(g) + 2 H2O(g)
DH◦f CH4 (g) = +74.86 kJ/mol
DH◦f O2(g) = 0 kJ/mol
DH◦f CO2(g) = -393.5 kJ/mol
pg. 316
DH◦ fH2O(g) = -483.6 kJ/mol
DH◦f = [-393.5 kJ/mol + (-483.6 kJ/mol)]- [-74.86 kJ/mol + (0 kJ/mol )]
DH◦f = [-393.5 -483.6] + 74.86 = -877.1 + 74.86 = -802.2 kJ/mol
Step #2: sum up all the DH◦ f. : DHrxn =  DHf(products) -  D Hf(reactants)
44
Specific Heat capacity (J/oC) = heat supplied (J)
temperature (oC)
Specific Heat Capacity = heat required to raise the
temperature of 1 gram of a substance object by
1 oC
Affected by
−What the substance is
−Mass of the object
The amount of heat absorbed or released during a
physical or chemical change can be measured…
…usually by the change in temperature of a known
quantity of water
1 calorie is the heat required to raise the temperature
of 1 gram of water by 1 C
1 BTU is the heat required to raise the temperature of
1 pound of water by 1 F
– A device used to
experimentally determine
the amount of heat
released or absorbed
during a physical or
chemical change
heat gained = heat lost
Most common units of energy
1. S unit of energy is the joule (J), defined as 1
(kilogram•meter2)/second2, energy is also
expressed in kilojoules (1 kJ = 103J).
2. Non-S unit of energy is the calorie.
One cal = 4.184 J or 1J = 0.2390 cal.
Units of energy are the same, regardless of the form of energy
The amount of heat required to raise the temperature of one gram of
substance by one degree Celsius.
Substance
Specific Heat (J/g·K)
Water (liquid)
4.18
Ethanol (liquid)
2.44
Water (solid)
2.06
Water (vapor)
1.87
Aluminum (solid)
0.897
Carbon (graphite,solid)
0.709
Iron (solid)
0.449
Copper (solid)
0.385
Mercury (liquid)
0.140
Lead (solid)
0.129
Gold (solid)
0.129
Page 296
• The higher the specific heat the more
energy it takes to change its temperature.
• Pizza burning the roof of your mouth
• The same amount of heat is released
when an object cools down
50
Q = m. C . DT
Q = Heat lost or gained ( J)
C = Specific Heat (J/ ºC.g)
DT = Temperature change = Tf – Ti (ºC)
Change in energy = mass * specific heat * change in temp.
How much heat is need to raise 5 g of water 10 ̊C?
(Water’s specific heat is 4.18 J/(̊C.g)
Known
m= 5 Kg
DT = 10 ̊C
C = 4.18 J/(g- ̊C)
Unknown
heat needed?
2. Q = m * C5g
* *DT
C * DT
4.18 J/(g- ̊C) * DT
10 ̊C
3. Q = 209 J
1. Q = m * C * DT
28,875 J of energy are added to a 5-kg piece of copper that
has an initial temperature of 293 K. What will be the final
temperature of this piece of copper? (Copper’s specific
heat:385 J/(kg-K))
Known
m= 5 Kg
Ti = 293 K
C = 385 J/(g- C)
Unknown
Final temperature?
1. Q = m * C * DT
2. 28,875 J =
5m
kg ** C
385
C ** DT
J/(kg-K)
DT
* DT
3. DT = 15 K
293 K + 15 K = 308 K
1. Q = m * C * DT
140
120
DH = mol x DHfus
DH = mol x DHvap
Temperature (oC)
100
80
Heat = mass x Dt x Cp, gas
60
40
20
0
Heat = mass x Dt x Cp, liquid
-20
-40
-60
-80
Heat = mass x Dt x Cp, solid
-100
Time
Choose all that apply...
C(s) + 2 S(g)
CS2(l) DH = 89.3 kJ
Which of the following are true?
A) This reaction is exothermic
B) It could also be written
C(s) + 2 S(g) + 89.3 kJ
CS2(l)
C) The products have higher energy than the
reactants
D) It would make the water in the calorimeter colder