Chapter 7
Thermochemistry
7.1 Terminology
System- the part you are
investigating
Surroundings- the rest of the
universe
Open System: freely exchanges matter and
energy
Closed System: exchanges energy but not
matter
Isolated System:
does not interact
with surroundings
Energy
The ability to do work or transfer heat.
–Work: Energy used to cause an object
that has mass to move.
–Heat: Energy used to cause the
temperature of an object to rise.
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Work
• Energy used to move an
object over some
distance.
• w = F  d,
where w is work, F is
the force, and d is the
distance over which the
force is exerted.
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Kinetic Energy
Energy an object possesses by virtue of its motion.
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KE =  mv2
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Kinetic energy increases
as the speed and mass of
an object increases.
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Potential Energy
Energy an object possesses by virtue of its position or chemical
composition.
PE= m x g x h
m=mass=kg
g=gravitational constant=9.8m/s2
h= height (or distance)
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7.2 Heat
Energy transferred between a system and its
surroundings as a result of temperature
difference
Heat flows
from a warm
body to a
cooler body.
The quantity of heat, q, required to change a
substances temperature is dependent on:
– Temperature change
– Quantity of substance
– Nature of substance
calorie: heat to change 1 g of water 1°C
1 cal = 4.184 J
Specific Heat Capacity: heat to change 1 g of a
substance 1°C
Example 1A
How much heat in kJ is required to raise the
temperature of 237 g of water from 4.0°C to
37.0°C?
q = mCΔT
qsystem +qsurroundings = 0
Example 2A
• When 1.00 kg of lead (c = 0.13J/g°C) at
100.0°C is added to a quantity of water and
the final temperature of the lead/water is
32.5°C. What is the mass of water present?
Specific Heat values
given on page 247 of
textbook.
7.3 Calorimetry
Chemical Energy: the energy associated with
chemical bonds and intermolecular forces
Chemical Reaction: breaking and forming bonds
Heat of Reaction: heat exchange during a
chemical reaction
Exchange of Heat
• When heat is absorbed by the system from the
surroundings, the process is endothermic.
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Exchange of Heat
• When heat is absorbed by the system from the
surroundings, the process is endothermic.
• When heat is released by the system to the
surroundings, the process is exothermic.
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Bomb Calorimetry
Reactions can be
carried out in a sealed
“bomb” and measure
the heat absorbed by
the water.
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Example 3A
• Vanillin in a natural constituent of vanilla. It is
also manufactured for use in artificial vanilla
flavor. The combustion of 1.013 g of vanillin,
C8H8O3, in a bomb calorimeter causes the
temperature to raise from 28.89°C to 30.09°C.
What is the heat of combustion of vanillin in
kJ/mol?
Constant Pressure Calorimetry
By carrying out a reaction
in aqueous solution in a
simple calorimeter such as
this one, one can indirectly
measure the heat change
for the system by
measuring the heat
change for the water in
the calorimeter.
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Example 4A
Two solutions, 100.0 mL of 1.00M AgNO3(aq)
and 100.0 mL of 1.00M NaCl (aq), both initially
at 22.4°C are added to a Styrofoam cup allowed
to react. The temperature rises to 30.2°C.
Determine the qrxn per mole of AgCl(s) in the
reaction
Ag+(aq) + Cl-(aq)  AgCl(s)
7.4 Work
We can measure the work done by the gas if the
reaction is done in a vessel that has been fitted
with a piston.
w = −PV
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Example 5A
How much work, in Joules, is involved when
0.255 mol N2 at constant temperature of 23°C is
allowed to expand by 1.50L in volume against an
external pressure of 0.750 atm?
7.5 First Law of Thermodynamics
• Energy is neither created nor destroyed.
• In other words, the total energy of the universe is a
constant; if the system loses energy, it must be gained
by the surroundings, and vice versa.
qsystem = -qsurroundings
U = q + w
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E, q, w, and Their Signs
q
+ (system gains heat)
- (system loses heat)
w
+ (work done on system)
- ( work done by system)
U
+ (net gain of energy by system)
- ( net loss of energy by system)
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Example 6A
In compressing a gas, 355J of work is done on
the system. At the same time 185 J of heat
escapes from the system. What is the U for
the system?
7.6 Enthalpies of Reaction
This quantity, H, is called the enthalpy of reaction,
or the heat of reaction.
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Example 7
How much heat is associated with the complete
combustion of 1.00kg of sucrose?
ΔH = -5.65 x 103 kj/mol
Example 7A
What mass of sucrose must be burned to
produce 1,000 kJ of heat? ΔH = -5.65 x 103 kj/mol
Changing States
Example 8A
What is the enthalpy change when a cube of ice
2.00 cm on edge is brought from -10.0°C to a
final temperature of 23.2°C? For ice, use d =
0.917 g/mL, a specific heat of 2.01 J/g°C and
enthalpy of fusion of 6.01kJ/mol.
7.7 Hess’s Law
Change in enthalpy is the same if the reaction
took one step or many steps
• If you can add balanced equations, you can
add ΔH
• If you alter the equation, alter ΔH
– Double reactant/products, double ΔH
– Reverse equation, change sign of ΔH
– Cancel items if they are in reactants and products
Hess’s Law Example
7.8 Heat of Formation
ΔHf°: enthaply of formation
• Defined as heat change that results when one
mole of compound is formed from its
constituent elements
• Pure element ΔHf°= 0
Example 10A
The standard enthalpy of formation for the
amino acid leucine is -637.3 kJ/mol C6H13O2N (s).
Write the chemical equation to which this value
applies.
Finding ΔHrxn
ΔHrxn = Σ Hf°(products) - Σ Hf°(reactants)
Values in back of textbook
Example 11A
Use data table 7-2 to calculate the enthalpy for
the combustion of ethanol, CH3CH2OH (l), at
298.15K.
Example 12A
Determine the standard enthalpy for the
formation of glucose C6H12O6.
ΔHrxn = 2803 kJ
Example 13A
Given that Hf°[AgI(s)] = -61.84 kJ/mol, what is
the standard enthanpy change for the
precipitation of silver iodide?
Calculation of H
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
• Imagine this as occurring
in 3 steps:
C3H8 (g)  3 C(graphite) + 4 H2 (g)
3 C(graphite) + 3 O2 (g)  3 CO2 (g)
4 H2 (g) + 2 O2 (g)  4 H2O (l)
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Calculation of H
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
• Imagine this as occurring
in 3 steps:
C3H8 (g)  3 C(graphite) + 4 H2 (g)
3 C(graphite) + 3 O2 (g)  3 CO2 (g)
4 H2 (g) + 2 O2 (g)  4 H2O (l)
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Calculation of H
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
• Imagine this as occurring
in 3 steps:
C3H8 (g)  3 C(graphite) + 4 H2 (g)
3 C(graphite) + 3 O2 (g)  3 CO2 (g)
4 H2 (g) + 2 O2 (g)  4 H2O (l)
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Calculation of H
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
• The sum of these
equations is:
C3H8 (g)  3 C(graphite) + 4 H2 (g)
3 C(graphite) + 3 O2 (g)  3 CO2 (g)
4 H2 (g) + 2 O2 (g)  4 H2O (l)
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
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Calculation of H
We can use Hess’s law in this way:
H = nHf(products) - mHf(reactants)


where n and m are the stoichiometric
coefficients.
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Calculation of H
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
H =
=
=
=
[3(-393.5 kJ) + 4(-285.8 kJ)] - [1(-103.85 kJ) + 5(0 kJ)]
[(-1180.5 kJ) + (-1143.2 kJ)] - [(-103.85 kJ) + (0 kJ)]
(-2323.7 kJ) - (-103.85 kJ)
-2219.9 kJ
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