Transcript Chapter 5 Thermochemistry

Chemistry, The Central Science, 11th edition
Theodore L. Brown; H. Eugene LeMay, Jr.;
and Bruce E. Bursten
Chapter 5
Thermochemistry
John D. Bookstaver
St. Charles Community College
Edited by Debbie Amuso
Thermochemistry
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Thermochemistry
Thermodynamics = study of energy and
its transformations
Thermochemistry = study of chemical
reactions involving changes in heat
energy
Thermochemistry
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Energy
Energy = the ability to do work or transfer
heat energy.
Work = energy used to cause an object
with mass to move (w = f x d)
Heat = energy used to cause the
temperature of an object to increase
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Major Types of Energy
Potential energy = energy an object
possesses by virtue of its position or
chemical composition.
Kinetic energy = energy an object
possesses by virtue of its motion.
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Kinetic Energy
1
KE =  m v
2
2
m = mass in kilograms (kg)
v = velocity in meters per second (m/s)
KE = kinetic energy in joules (J)
1 Joule = 1 kg-m2/s2
A mass of 2 kg moving at a speed of one meter per second
possesses a kinetic energy of 1 Joule.
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Potential Energy
PE = m g h
m = mass in kilograms (kg)
g = acceleration due to gravity (9.8 m/s2)
h = height (m)
PE = potential energy in joules (J)
1 Joule = 1 kg-m2/s2
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Units of Energy
• The SI unit of energy is the joule (J).
• An older, non-SI unit is still in
widespread use: the calorie (cal).
1 cal = 4.184 J
• 1000 calories = one nutritional Calorie
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First Law of Thermodynamics
• Energy is neither created nor destroyed, but
it can undergo a transformation from one
type to another. (Law of Conservation of
Energy)
• The total energy of the universe is a constant.
• The energy lost by a system must equal the
energy gained by its surroundings, and vice
Thermochemistry
versa.
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System and Surroundings
System = the molecules
we want to study (here,
the hydrogen and
oxygen molecules).
Surroundings = everything
else (here, the cylinder
and piston).
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Internal Energy
The internal energy of a system is the sum of all
kinetic and potential energies of all components
of the system; we call it E.
E = Efinal − Einitial
(It’s a state function)
• If E is positive, the system absorbed energy
from the surroundings.
• If E is negative, the system released energy
Thermochemistry
to the surroundings.
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E = q + w
• When energy is exchanged between the
system and the surroundings, it is
exchanged as either heat (q) or work (w).
• That is, E = q + w.
Thermochemistry
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q, w, and their signs
+ q = system gains or takes in heat
- q = system loses or gives off heat
+ w = work is done on the system by the
surroundings (piston pushed in)
- w = work is done by the system on its
Thermochemistry
surroundings (piston moves out)
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Example
As hydrogen and oxygen gas are ignited in a cylinder,
the system loses 550 J of heat to its surroundings.
The expanding gases move a pistion to do 240 J of
work on its surroundings. E for system = ?
Answer:
E = q + w
E = (-550 J) + (-240 J)
E = - 790 J
What does it mean?
The system gave off 790 J of energy to its surroundings
Thermochemistry
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Enthalpy & H
• The symbol for enthalpy is H.
• Enthalpy is the internal energy plus the
product of pressure and volume:
H = E + PV
• At constant pressure:
H = E = q
• So at constant pressure,
H = heat lost or gained by the system.
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Recall: Endothermic
• When heat is absorbed (taken in) by the
system from the surroundings, the process is
endothermic.
H = Hfinal − Hinitial
H = Hproducts − Hreactants
H = positive value
for endothermic
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Recall: Exothermic
When heat is released (given off) by the
system into the surroundings, the process is
exothermic.
H = Hfinal − Hinitial
H = Hproducts − Hreactants
H = negative value
for exothermic
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Enthalpy of Reaction
1. This quantity, H, is called the
enthalpy of reaction, or the heat of
reaction.
2. Enthalpy is an extensive property.
3. Reverse Rx  Negate H value.
4. Phase (state) matters.
Thermochemistry
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Enthalpies of Formation
An enthalpy of formation, Hf, is defined
as the enthalpy change for the
reaction in which a compound is
made from its constituent elements
in their elemental forms.
Example:
2 Al (s) + 3 O2 (g)  2 Al2O3 (s)
Thermochemistry
Hf = -3340 kJ
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Standard Enthalpies of
Formation
Standard enthalpies of formation, Hf°, is
defined as the enthalpy change for the
reaction in which one mole of a compound
is made from its constituent elements in
their standard states at 25o C (298 K) and
1.00 atm pressure.
Example:
2 Al (s) + 3/2 O2 (g)  Al2O3 (s)
Hf°= - 1670 kJ/mol
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Calculation of H
H =  n Hf°products –  m Hf° reactants
where n and m are the coefficients.
Hf°Values are listed in Appendix C of
your Textbook.
Thermochemistry
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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
H =  n Hf
products
–  m Hf° reactants
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Calorimetry is Another Way to
Measure H values
We measure H
through calorimetry,
which measures
heat flow.
But first, some
definitions.
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Heat Capacity
The amount of energy (joules) required to raise
the temperature of a substance by 1 K (1C) is its
heat capacity. Units are J/K or J/ C
q = C * T
or
C = q / T
q = heat in Joules
C = heat capacity in J/K
T = change in temperature
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Specific Heat
Specific heat capacity (or simply specific
heat) is the amount of energy (joules)
required to raise the temperature of 1 g of a
substance by 1 K. Units are J/g-K
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Molar Heat Capacity
• Molar heat capacity is the amount of
energy (joules) required to raise the
temperature of 1 mole of a substance
by 1 K. Units are J/mol-K
Also helpful: # mol = mass/gfm
Thermochemistry
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Use Units
What is the molar heat capacity of CH4 (g) if
its specific heat capacity is 2.20 J/g-K?
Answer:
2.20 J x 16.0 g =
g–K
1 mol
35.2 J/mol-K
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Constant Pressure Calorimetry
By carrying out a reaction
in aqueous solution in a
simple calorimeter such
as this one, one can
indirectly measure the H
for the system by
measuring the heat
change for the water in
the calorimeter.
q = m  sh  T
Thermochemistry
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Calorimetry
q = m * sh * T
q = Joules of heat
m = total mass in calorimeter
(If a solution, use m = D * V to obtain the mass)
Water has a density of 1.00 g/mL
Most water solutions have a density of 1.02 g/mL
sh = specific heat (for water, 4.18 J/g-K)
T = Tf - Ti
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Calorimetry
Σq=0
In other words, the heat released (or absorbed) by the reaction of interest = the
sum of the heat gained (or released) by the resulting solution & calorimeter.
qcalorimeter + qreaction+ qresulting solution = 0
If we assume qcalorimeter is negligible the equation reduces to the following:
qreaction = - qsolution = -m*sh*ΔT
where T = Tf - Ti
and m = D * V
H = qreaction /# moles
Thermochemistry
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Specific Heat of a Metal
How to determine specific heat of a metal:
1) Mass metal
2) Put metal in HOT water & measure initial temp of
hot metal
3) Measure temp of 100.0 mL (100.0 g) of COLD water
4) Put hot metal in cold water
5) Record temp of water with metal in it (that temp is
the final temp for both the metal & water)
6) Calculate the sh of the metal
(m * sh * T)metal = (m * 4.18 * T)water
Thermochemistry
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Bomb Calorimetry
• Reactions can be
carried out in a sealed
“bomb” such as this
one.
• The heat absorbed
(or released) by the
water is a very good
approximation of the
enthalpy change for
the reaction.
Thermochemistry
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Bomb Calorimetry
• Because the volume
in the bomb
calorimeter is
constant, what is
measured is really the
change in internal
energy, E, not H.
• For most reactions,
the difference is very
small.
Thermochemistry
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H via Heat Capacity
q = C * T
q = heat in Joules
C = heat capacity in J/K
T = change in temperature
where T = Tf – Ti
H = q / # moles
Thermochemistry
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Stoichiometric Determination of H
Given the H for a reaction, you can use
mole ratios to determine H values of
different sample sizes.
See examples.
Thermochemistry
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Sample Problem #1
H2 (g) + I2 (g)  2 HI (g)
H = +53.0 kJ
How many joules of heat are required to form 5
moles of HI (g)?
5 mol HI
2 mol HI
=
__x___
+53.0 kJ
x = +133 kJ
Thermochemistry
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Sample Problem #2
H2 (g) + I2 (g)  2 HI (g)
H = +53.0 kJ
How many moles of HI (g) are formed by the
expenditure of 235 kJ of heat energy?
__x___
2 mol HI
=
+235 kJ
+53.0 kJ
x = 8.87 mol HI
Thermochemistry
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Sample Problem #3
H2 (g) + I2 (g)  2 HI (g)
H = +53.0 kJ
How many joules of heat are required to form 109 grams of HI?
# mol HI = mass/gfm
# mol HI = 109 g / 128 g/mol = 0.852 mol HI
0.852 mol HI
2 mol HI
=
__x___
+53.0 kJ
Thermochemistry
x = + 22.6 kJ
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Hess’s Law
H is well known for many reactions,
and it is inconvenient to measure H
for every reaction in which we are
interested.
However, we can estimate H using
published H values and the
properties of enthalpy.
Thermochemistry
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Hess’s Law
Hess’s law states that “If a reaction is carried
out in a series of steps, H for the overall
reaction will be equal to the sum of the
enthalpy changes for the individual steps.”
H = H1 + H2 + H3
Thermochemistry
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Calculation of H
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
• Imagine this as occurring
in three 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)
Thermochemistry
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Calculation of H
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
• Imagine this as occurring
in three 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)
Thermochemistry
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Calculation of H
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
• Imagine this as occurring
in three 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)
Thermochemistry
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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)
Thermochemistry
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Energy in Foods
Most of the fuel in the
food we eat comes
from carbohydrates
and fats.
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Energy in Fuels
The vast
majority of the
energy
consumed in
this country
comes from
fossil fuels.
Thermochemistry
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Summary: Ways to Calculate H
H =  n Hf°products –  m Hf° reactants
Using Hfo values from Appendix C
Or
By Using a Calorimeter
qreaction = - qsolution = - m*sh*ΔT
where T = Tf - Ti
and m = D * V
H = qreaction /# moles
Thermochemistry
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Summary: Ways to Calculate H
Using Heat Capacity
q = C * T
q = heat in Joules
C = heat capacity in J/K
T = change in temperature
where T = Tf – Ti
Thermochemistry
H = q / # moles
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Summary: Ways to Calculate H
Hess’s Law
H = H1 + H2 + H3
Or
By Stoichiometric Determination
Given the H for a reaction, you can use mole
ratios to determine H values of different
sample sizes.
Thermochemistry
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