Thermochemistry

Thermochemistry
The study of the transfer of
heat energy
Let’s Revisit Intensive and
Extensive Properties
• Intensive Extensive Video Clip
• Restate in your own words
intensive and extensive
properties
Energy Flow
• Thermodynamics is the study of
the flow of energy.
• Thermodynamic Energy Flow
Definitions
• Thermochemistry—the study of heat or
thermal energy transfer and its application to
chemical reactions and physical processes
• Enthalpy—a measure of the total energy of a
system
• Heat capacity—the amount of heat that is
required to change the temperature of that
object by 1.00 degree Celsius (or Kelvin)
• Specific heat capacity (specific heat)—the
amount of heat energy needed to change the
temperature of 1.00 gram of the substance by
1.00 degree Celsius
Different Types of Heat
with Physical Processes
• There is also a certain amount of energy associated
with phase changes.
• There is a difference in the sensible heat (the heat
that is associated with temperature changes) and the
latent heat (the heat that is associated with a change
in phase of a substance while the temperature
remains constant)
• The heat associated with the phase change of solid
to liquid is the latent heat of fusion (Hf). The heat
required to change from a liquid to a vapor is called
the latent heat of vaporization (Hv).
Heat with Temperature
Change
• The specific heat of a substance can be
determined experimentally by
determining the quantity of heat
transferred by a definite mass of the
substance as its temperature rises or
falls at constant pressure
•
Specific heat (cp) = quantity of heat transferred (Q)
mass of material (m) x temperature change (DT)
• Better known as:
Q = m cp DT
Heating and Cooling
Curve
• Heating Curve Animation
• Sketch a heating and cooling
curve and label according to the
animation
Example 1: Temp
Change
q = m c DT
How many joules of energy must be
transferred from a cup of coffee to your
body if the temperature of the coffee
drops from 60.0 oC to 37.0 oC? Assume
that the cup holds 250. mL and that the
specific heat of coffee is the same as
that of water, 4.18 J/goC.
Example 1 Solution
Knowns:
DT = 60.0oC – 37.0oC
= 23.0oC
V = 250. mL
cp = 4.18 J/goC
Unknowns:
Q (heat energy)
Formula:
Q = mcpDT
What else do you need to
solve it?
Mass!
Example 1 Solution
Q = mcpDT
Density = m/V and the D of H2O is 1.0 g/mL
DxV=m
1.0 g/mL x 250. mL = 250. g
Q = (250. g) (4.18 J/goC) (23.0oC) = 24035 J
= 24.0 kJ
Example 2: Phase
Change
• Q = m Hf
• Q = m Hv
for solid to liquid
for liquid to a vapor
Example 2:
You have four ice cubes at 0oC. If each cube
has a mass of 9.63 g, how much heat is
required to melt these ice cubes to liquid
water at 0oC? The latent heat of fusion of ice
is 333 J/g.
Example 2 Solution
Knowns:
4 ice cubes
m of 1 cube = 9.63 g
Hf of ice = 333 J/g
Unknowns:
Q
Formula:
Q = m Hf
Example 2 Solution
q = m Hf
m = 4 x 9.63 g
= 38.5 g
q = (38.5 g) (333 J/g) = 12820.5 J
= 12.8 kJ
Heat Transfer
• When heat is transferred into the system, it is an
endothermic process
• When heat is transferred out of the system, the
process is exothermic
• All energy transfers follow the first law of
thermodynamics, the law of energy conservation.
All of the energy transferred between a system and
its surroundings must be conserved
• The heat transferred into (or out of) a system at a
constant pressure is the enthalpy change (DH)
• Enthalpies are experimentally determined, and can
be calculated for individual reactions.
Enthalpy of Formation
• One way to determine the DH of a reaction is to use a
value called the enthalpy of formation (Hof)
• The enthalpy of formation is defined as the energy needed
to form one mole of the compound from its elements
• Example: The energy for H2 & O2 to form H2O
• The values are always in units of kJ/mol
• The Hof of an element is 0 kJ/mol because element cannot
be formed from anything simpler than itself
• Enthalpies of formations are usually found in a reference
table and are not something students memorize
• The enthalpy of an entire reaction can be calculated from
the Hof of the reactants and products
• The formula is: DH = DHof (products) – DHof(reactants)
Example 3
DH = DHof (products) – DHof(reactants)
• Given: DHof CO2(g)
= -393.5 kJ/mol
DHof H2O(l)
= -285.8 kJ/mol
DHof C5H12 (g) = -145.7 kJ/mol
Calculate the heat of the combustion
reaction:
C5H12(g) + 8O2(g)  5CO2 (g) + 6H2O(l)
Example 3 Solution
DH = DHof (products) – DHof(reactants)
Given:
DHof CO2(g) = -393.5 kJ/mol
DHof H2O(l) = -285.8 kJ/mol
DHof C5H12 (g) = -145.7 kJ/mol
C5H12(g) + 8O2(g)  5CO2 (g) + 6H2O(l)
DH = [5(-393.5 kJ) + 6(-285.8 kJ)]
– [1(-145.7 kJ) + 0]
= -3536.6 kJ
Graphing Enthalpy of
Reactions
• An endothermic reaction means that the
enthalpies of formation of the reactants
is lower than the enthalpy of formations
of the products
• An exothermic reaction that the
enthalpies of formation of the products
is lower than the enthalpy of formations
of the reactants
Reaction Energy
Diagrams
• Note that the energy of each reaction does not go
directly from reactants to products
• There is an energy “hump” that must be overcome in
order for a reaction to occur
• This energy barrier is also called the energy of
activation (or activation energy, Ea)
• It is the amount of energy needed for a reaction to go
from reactants to the transition complex (or activated
complex) that must occur before the products can
form. The energy associated with this activation step
must be invested before the reaction will go to
completion, no matter whether the net reaction is
endothermic or exothermic.
Endothermic Pathway
• (a) = Energy of
Activation (Ea)
• (c) = Enthalpy of
reaction (DHrxn)
• Note that
Enthalpy of
reaction DHrxn is a
positive number
Exothermic Pathway
• (a) = Energy of
Activation (Ea)
• (c) = Enthalpy of
reaction (DHrxn)
• Note that
Enthalpy of
reaction DHrxn is a
negative number
Enthalpy Signs
A positive enthalpy of reaction indicates that the
reaction is endothermic, and the energy term can be
written as if it is a reactant into the chemical equation.
Any reaction that contains an energy term is known
as a thermochemical equation
A negative enthalpy value indicates that the reaction is
exothermic. The energy term is written as a product
in the thermochemical reaction. Note that when the
exothermic enthalpy is written into the products, it
does not carry its negative sign with it
The positive or negative sign in front of an enthalpy
term simply indicates whether the reaction is
endothermic or exothermic. By convention, the math
is not performed with negative numbers.
Example 4
Carbon dioxide can be decomposed into carbon
monoxide and oxygen.
2CO2  2 CO + O2
The DH for this reaction is 43.9 kJ
• Is the reaction endothermic or exothermic?
• If there is only 22.2 kJ of heat transferred,
how many moles of CO2 can be
decomposed?
Example 4
2CO2  2 CO + O2
The DH for this reaction is 43.9 kJ
• Is the reaction endothermic or exothermic?
• Endothermic because it is a positive value
• Therefore it can be written into the reactants
side of the equation
43.9 kJ + 2CO2  2 CO + O2
Example 4 Solution
43.9 kJ + 2CO2  2 CO + O2
Known: 22.2 kJ
Unknown: mol CO2
22.2 kJ x
=
Example 4 Solution
43.9 kJ + 2CO2  2 CO + O2
Known: 22.2 kJ
Unknown: mol CO2
22.2 kJ x 2 mol CO2 =
43.9 kJ
Example 4 Solution
43.9 kJ + 2CO2  2 CO + O2
Known: 22.2 kJ
Unknown: mol CO2
22.2 kJ x 2 mol CO2 = 1.01 mol CO2
43.9 kJ
Measuring Heat Transfer
• Calorimetry is the process by which we
measure heat transfer and its
associated energy changes in the lab.
• Calorimetry
Example 5
• Suppose you heat a 55.0-g piece of silver in the
flame of a Bunsen burner to 425.0oC and then you
plunge it into a beaker of water. The beaker holds
600. g of water, and its temperature before you drop
in the hot silver is 25.0oC. The final temperature of
the silver and water mixture is 44.0oC. What is the
specific heat of the silver? The specific heat of water
is 4.18 J/goC.
Example 5
• Suppose you heat a 55.0-g piece of silver in the
flame of a Bunsen burner to 425.0oC and then you
plunge it into a beaker of water. The beaker holds
600. g of water, and its temperature before you drop
in the hot silver is 25.0oC. The final temperature of
the silver and water mixture is 44.0oC. What is the
specific heat of the silver? The specific heat of water
is 4.18 J/goC.
• Before solving, here is a hint: Keep the data about
silver separate from the data about water
Example 5 Solution
• Knowns:
• Ag: m = 55.0 g; Ti = 425.0oC; Tf =
44.0oC
• H2O: m = 600. g; cp = 4.18 J/goC; Ti =
25.0oC; Tf = 44.0oC
• Unknown: cp for Ag
• Formula: Q = mcpDT
Example 5 Solution
• Notice that you do NOT have enough
information to solve for the cp of Ag
because you do not know Q for Ag.
• So what is Q? The heat the silver lost.
Where did silver lose its heat? To the
water! The water got the heat energy to
warm up from the hot silver.
• Because of heat transfer, if we can
calculate the q of the water, we will
know q for silver!
Example 5 Solution
• Solve for Q of water:
• Q = mcpDT (*Be sure to use the data for
WATER!)
DT (H2O) = 44.0oC – 25.0oC = 19.0oC
• Q (H2O) = 600 g x 4.18 J/goC x 19.0oC
• Q (H2O) = 47652 J
Example 5 Solution
• From the first law of Thermodynamics:
• Heat loss = Heat gained
• Therefore Q (Ag) = Q (H2O)
• Q (H2O) = 47652 J = Q (Ag)
• So, for Ag, Q = mcpDT and now we
have everything we need to solve for cp
of Ag
Example 5 Solution
• Q = mcpDT & our knowns for Ag are
m = 55.0 g; Ti = 425.0oC; Tf = 44.0oC;
Q = 47652 J
DT = 425.0oC – 44.0oC = 381.0oC
• 47652 J = 55.0 g x cp x 381.0oC
• cp = 2.27 J/goC
Example 5 Solution
• So the specific heat of Ag is about half
that of H2O. This means that it takes
twice as much heat to change the
temperature of 1 g of H2O by 1oC as it
does Ag metal. The high specific heat
of water is one of its special properties!