Transcript Document

Chapter 6
Thermochemistry
Contents and Concepts
Understanding Heats of Reaction
The first part of the chapter lays the groundwork
for understanding what we mean by heats of
reaction.
1.
2.
3.
4.
5.
6.
Energy and Its Units
Heat of Reaction
Enthalpy and Enthalpy Changes
Thermochemical Equations
Applying Stoichiometry to Heats of Reaction
Measuring Heats of Reaction
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Using Heats of Reaction
Now that we understand the basic properties of
heats of reaction and how to measure them, we
can explore how to use them.
7. Hess’s Law
8. Standard Enthalpies of Formation
9. Fuels—Foods, Commercial Fuels, and Rocket
Fuels
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Learning Objectives
Understanding Heats of Reaction
1. Energy and Its Units
a. Define energy, kinetic energy, and internal
energy.
b. Define the SI unit of energy (joule) as well
as the common unit of energy (calorie).
c. Calculate the kinetic energy of a moving
object.
d. State the law of conservation of energy.
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2. Heat of Reaction
a. Define a thermodynamic system and its
surroundings.
b. Define heat and heat of reaction.
c. Distinguish between an exothermic process
and an endothermic process.
3. Enthalpy and Enthalpy Changes
a. Define enthalpy and enthalpy of reaction.
b. Explain how the terms enthalpy of reaction
and heat of reaction are related.
c. Explain how enthalpy and internal energy
are related.
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4. Thermochemical Equations
a. Define a thermochemical equation.
b. Write a thermochemical equation given
pertinent information.
c. Learn the two rules for manipulating
(reversing and multiplying) thermochemical
equations.
d. Manipulate a thermochemical equation
using these rules.
5. Applying Stoichiometry to Heats of
Reaction
a. Calculate the heat absorbed or evolved
from a reaction given its enthalpy of
reaction and the mass of a reactant or
product.
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6. Measuring Heats of Reaction
a. Define heat capacity and specific heat.
b. Relate the heat absorbed or evolved to the
specific heat, mass, and temperature
change.
c. Perform calculations using the relationship
between heat and specific heat.
d. Define a calorimeter.
e. Calculate the enthalpy of reaction from
calorimetric data (temperature change and
heat capacity).
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Using Heats of Reaction
7. Hess’s Law
a. State Hess’s law of heat summation.
b. Apply Hess’s law to obtain the enthalpy
change for one reaction from the enthalpy
changes of a number of other reactions.
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8. Standard Enthalpies of Formation
a. Define standard state and reference form.
b. Define standard enthalpy of formation.
c. Calculate the heat of a phase transition
using standard enthalpies of formation for
the different phases.
d. Calculate the heat (enthalpy) of reaction
from the standard enthalpies of formation
of the substances in the reaction.
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9. Fuels—Foods, Commercial Fuels, and Rocket
Fuels
a. Define a fuel.
b. Describe the three needs of the body that are
fulfilled by foods.
c. Give the approximate average values quoted
(per gram) for the heat values (heats of
combustion) for fats and for carbohydrates.
d. List the three major fossil fuels.
e. Describe the processes of coal gasification
and coal liquefaction.
f. Describe some fuel-oxidizer systems used in
rockets.
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Thermodynamics
The science of the relationship between heat and
other forms of energy.
Thermochemistry
An area of thermodynamics that concerns the
study of the heat absorbed or evolved by a
chemical reaction
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Energy
The potential or capacity to move matter.
One form of energy can be converted to another
form of energy: electromagnetic, mechanical,
electrical, or chemical.
Next, we’ll study kinetic energy, potential energy,
and internal energy.
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Kinetic Energy, EK
The energy associated with an object by virtue of
its motion.
1
EK  mv 2
2
m = mass (kg)
v = velocity (m/s)
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The SI unit of energy is the joule, J, pronounced
“jewel.”
kg  m 2
J
s2
The calorie is a non-SI unit of energy commonly
used by chemists. It was originally defined as the
amount of energy required to raise the
temperature of one gram of water by one degree
Celsius. The exact definition is given by the
equation:
1 cal  4.184 J (exact)
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?
A person weighing 75.0 kg (165 lbs)
runs a course at 1.78 m/s (4.00 mph).
What is the person’s kinetic energy?
m = 75.0 kg
V = 1.78 m/s
EK = ½ mv2
1
m

E K  (75.0 kg) 1.78 
2
s

2
kg  m 2
E K  119
 119 J
2
s
(3 significant figures)
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Potential Energy, EP
The energy an object has by virtue of its position in
a field of force, such as gravitaitonal, electric or
magnetic field.
Gravitational potential energy is given by the
equation
E P  mgh
m = mass (kg)
g = gravitational constant (9.80 m/s2)
h = height (m)
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Internal Energy, U
The sum of the kinetic and potential energies of
the particles making up a substance.
Total Energy
Etot = EK + EP + U
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Law of Conservation of Energy
Energy may be converted from one form to
another, but the total quantity of energy remains
constant.
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Thermodynamic System
The substance under study in
which a change occurs is
called the thermodynamic
system (or just system).
Thermodynamic
Surroundings
Everything else in the vicinity
is called the thermodynamic
surroundings (or just the
surroundings).
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Heat, q
The energy that flows into or out of a system
because of a difference in temperature between
the thermodynamic system and its surroundings.
Heat flows spontaneously from a region of higher
temperature to a region of lower temperature.
• q is defined as positive if heat is absorbed by
the system (heat is added to the system)
• q is defined as negative if heat is evolved by a
system (heat is subtracted from the system)
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Heat of Reaction
The value of q required to return a system to the
given temperature at the completion of the reaction
(at a given temperature)
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Endothermic Process
A chemical reaction or process in which heat is
absorbed by the system (q is positive). The
reaction vessel will feel cool.
Exothermic Process
A chemical reaction or process in which heat is
evolved by the system (q is negative). The reaction
vessel will feel warm.
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In an endothermic reaction:
The reaction vessel cools.
Heat is absorbed.
Energy is added to the system.
q is positive.
In an exothermic reaction:
The reaction vessel warms.
Heat is evolved.
Energy is subtracted from the system.
q is negative.
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Enthalpy, H
An extensive property of a substance that can be
used to obtain the heat absorbed or evolved in a
chemical reaction.
Extensive Property
A property that depends on the amount of
substance. Mass and volume are extensive
properties.
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A state function is a property of a system that
depends only on its present state, which is
determined by variables such as temperature and
pressure, and is independent of any previous
history of the system.
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The altitude of a campsite is
a state function.
It is independent of the path
taken to reach it.
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Enthalpy of Reaction
The change in enthalpy for a reaction at a given
temperature and pressure:
DH = H(products) – H(reactants)
Note: D means “change in.”
Enthalpy change is equal to the heat of reaction at
constant pressure:
DH = qP
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The diagram illustrates the enthalpy change for the
reaction
2Na(s) + 2H2O(l)  2NaOH(aq) + H2(g)
The reactants are at
the top. The products
are at the bottom.
The products have less
enthalpy than the
reactants, so enthalpy
is evolved as heat.
The signs of both q and
DH are negative.
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Enthalpy and Internal Energy
The precise definition of enthalpy, H, is
H = U + PV
Many reactions take place at constant pressure, so
the change in enthalpy can be given by
DH = DU + PDV
Rearranging:
DU = DH – PDV
The term (–PDV) is the energy needed to change
volume against the atmospheric pressure, P. It is
called pressure-volume work.
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For the reaction
2Na(s) + 2H2O(l)  2NaOH(aq) + H2(g)
DV
–P
The H2 gas had to do work to raise the piston.
For the reaction as written at 1 atm, -PDV = -2.5 kJ.
In addition, 368.6 kJ of heat are evolved.
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Thermochemical Equation
The thermochemical equation is the chemical
equation for a reaction (including phase labels) in
which the equation is given a molar interpretation,
and the enthalpy of reaction for these molar
amounts is written directly after the equation.
For the reaction of sodium metal with water, the
thermochemical equation is:
2Na(s) + 2H2O(l) 
2NaOH(aq) + H2(g); DH = –368.6 kJ
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?
Sulfur, S8, burns in air to produce sulfur
dioxide. The reaction evolves 9.31 kJ of
heat per gram of sulfur at constant
pressure. Write the thermochemical
equation for this reaction.
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We first write the balanced chemical equation:
S8(s) + 8O2(g)  8SO2(g)
Next, we convert the heat per gram to heat per mole.
ΔH  
9.31 kJ 256.56 g S 8

1 g S8
1 mol S 8
ΔH  2.39 10 3 kJ
Note: The negative sign indicates that heat is
evolved; the reaction is exothermic.
Now we can write the thermochemical equation:
S8(s) + 8O2(g)  8SO2(g); DH = –2.39 × 103 kJ
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Manipulating a Thermochemical Equation
• When the equation is multiplied by a factor, the
value of DH must be multiplied by the same
factor.
• When a chemical equation is reversed, the sign
of DH is reversed.
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CH4(g) + H2O(g)  CO(g) + 3H2(g); DH = 206 kJ
a.
CH4(g) + H2O(g)  CO(g) + 3H2(g)
This reaction is identical to the given reaction.
It is endothermic.
DH = 206 kJ
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CH4(g) + H2O(g)  CO(g) + 3H2(g); DH = 206 kJ
b. 2 CH4(g) + 2H2O(g)  2CO(g) + 6H2(g)
This reaction is double the given reaction.
It is endothermic.
DH = 412 kJ
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CH4(g) + H2O(g)  CO(g) + 3H2(g); DH = 206 kJ
c. CO(g) + 3H2(g)  CH4(g) + H2O(g)
This reaction is the reverse of the given reaction.
It is exothermic.
DH = -206 kJ
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CH4(g) + H2O(g)  CO(g) + 3H2(g); DH = 206 kJ
d.
2CO(g) + 6H2(g)  2CH4(g) + 2H2O(g)
This reaction is reverse and double the given
reaction.
It is exothermic.
DH = -412 kJ
Equations c and d are exothermic.
Equation d is the most exothermic reaction.
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?
When sulfur burns in air, the following
reaction occurs:
S8(s) + 8O2(g)  8SO2(g);
DH = – 2.39 x 103 kJ
Write the thermochemical equation for the
dissociation of one mole of sulfur dioxide into
its elements.
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S8(s) + 8O2(g)  8SO2(g); DH = –2.39 × 103 kJ
We want SO2 as a reactant, so we reverse the
given reaction, changing the sign of DH:
8SO2(g)  S8(g) + 8O2(g) ; DH = +2.39 × 103 kJ
We want only one mole SO2, so now we divide
every coefficient and DH by 8:
SO2(g)  1/8S8(g) + O2(g) ; DH = +299 kJ
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Applying Stoichiometry to Heats of Reaction
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?
You burn 15.0 g sulfur in air. How much
heat evolves from this amount of
sulfur? The thermochemical equation is
S8(s) + 8O2(g)  8SO2(g); DH = -2.39 x 103 kJ
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S8(s) + 8O2(g)  8SO2(g); DH = -2.39 x 103 kJ
Molar mass of S8 = 256.52 g
3
1 mol S 8
 2.39 x 10 kJ
q  15.0 g S 8 

256.5 g S 8
1 mol S 8
q = –1.40 × 102 kJ
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The daily energy requirement for a 20year-old man weighing 67 kg is 1.3 x
104 kJ. For a 20-year-old woman
weighing 58 kg, the daily requirement is
8.8 x 103 kJ. If all this energy were to be
provided by the combustion of glucose,
C6H12O6, how many grams of glucose would
have to be consumed by the man and the
woman per day?
C6H12O6(s) + 6O2(g)  6CO2(g) + 6H2O(l);
DH = -2.82 x 103 kJ
?
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C6H12O6(s) + 6O2(g)  6CO2(g) + 6H2O(l);
DH = -2.82 x 103 kJ
For a 20-year-old man weighing 67 kg:
mglucose
1 mol glucose 180.2 g glucose
 1.3x10 kJ 

3
1 mol glucose
2.82x10 kJ
4
= 830 g glucose required
(2 significant figures)
For a 20-year-old woman weighing 58 kg:
mglucose
1 mol glucose 180.2 g glucose
 8.8x10 kJ 

3
1 mol glucose
2.82x10 kJ
3
= 560 g glucose required
(2 significant figures)
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Measuring Heats of Reaction
We will first look at the heat needed to raise the
temperature of a substance because this is the
basis of our measurements of heats of reaction.
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Heat Capacity, C, of a Sample of Substance
The quantity of heat needed to raise the
temperature of the sample of substance by one
degree Celsius (or one Kelvin).
Molar Heat Capacity
The heat capacity for one mole of substance.
Specific Heat Capacity, s (or specific heat)
The quantity of heat needed to raise the
temperature of one gram of substance by one
degree Celsius (or one Kelvin) at constant
pressure.
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The heat required can be found by using the
following equations.
Using heat capacity:
q = CDt
Using specific heat capacity:
q = s x m x Dt
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A calorimeter is a device used to measure the
heat absorbed or evolved during a physical or
chemical change. Two examples are shown below.
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?
A piece of zinc weighing 35.8 g was
heated from 20.00°C to 28.00°C. How
much heat was required? The specific
heat of zinc is 0.388 J/(g°C).
m = 35.8 g
s = 0.388 J/(g°C)
Dt = 28.00°C – 20.00°C = 8.00°C
 0.388
q  35.8 g  
 gC
q = m  s  Dt
J
8.00 C

q = 111 J
(3 significant figures)
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?
Nitromethane, CH3NO2, an organic
solvent burns in oxygen according to
the following reaction:
CH3NO2(g) + 3/4O2(g) 
CO2(g) + 3/2H2O(l) + 1/2N2(g)
You place 1.724 g of nitromethane in a
calorimeter with oxygen and ignite it. The
temperature of the calorimeter increases
from 22.23°C to 28.81°C. The heat capacity
of the calorimeter was determined to be
3.044 kJ/°C. Write the thermochemical
equation for the reaction.
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We first find the heat evolved for the 1.724 g of
nitromethane, CH3NO2.
q rxn   Ccal Δt
q rxn
 3.044 kJ 
 
28.81 C  22.23 C   20.03 kJ
C 

Now, covert that to the heat evolved per mole by
using the molar mass of nitromethane, 61.04 g.
q rxn
61.04 g CH3NO 2
- 20.03 kJ


1.724 g CH3NO 2
1 mol CH3NO 2
DH = –709 kJ
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We can now write the thermochemical equation:
CH3NO2(l) + ¾O2(g)  CO2(g) + 3/2H2O(l) + ½N2(g);
DH = –709 kJ
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Hess’s Law of Heat Summation
For a chemical equation that can be written as the
sum of two or more steps, the enthalpy change for
the overall equation equals the sum of the
enthalpy changes for the individual steps.
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Suppose we want DH for the reaction
2C(graphite) + O2(g)  2CO(g)
It is difficult to measure directly.
However, two other reactions are known:
C(graphite) + O2(g)  CO2(g); DH = -393.5 kJ
2CO2(g)  2CO(g) + O2(g); DH = – 566.0 kJ
In order for these to add to give the reaction we
want, we must multiply the first reaction by 2. Note
that we also multiply DH by 2.
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2C(graphite) + O2(g)  2CO(g)
2C(graphite) + 2O2(g)  2CO2(g); DH = -787.0 kJ
2CO2(g)  2CO(g) + O2(g); DH = – 566.0 kJ
2 C(graphite) + O2(g)  2 CO(g); DH = –1353.0 kJ
Now we can add the reactions and the DH values.
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DHsub = DHfus + DHvap
6 | 58
?
What is the enthalpy of reaction, DH,
for the reaction of calcium metal with
water?
Ca(s) + 2H2O(l)  Ca2+(aq) + 2OH-(aq) + H2(g)
This reaction occurs very slowly, so it is
impractical to measure DH directly. However,
the following facts are known:
H+(aq) + OH-(aq)  H2O(l); DH = –55.9 kJ
Ca(s) + 2H+(aq)
 Ca2+(aq) + H2(g); DH = –543.0 kJ
6 | 59
Ca(s) + 2H2O(l)  Ca2+(aq) + 2OH-(aq) + H2(g)
First, identify each reactant and product:
H+(aq) + OH-(aq)  H2O(l); DH = –55.9 kJ
Ca(s) + 2H+(aq)  Ca2+(aq) + H2(g); DH = –543.0 kJ
Each substance must be on the proper side.
Ca(s), Ca2+(aq), and H2(g) are fine.
H2O(l) should be a reactant.
OH-(aq) should be a product.
Reversing the first reaction and changing the sign of
its DH accomplishes this.
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Ca(s) + 2H2O(l)  Ca2+(aq) + 2OH-(aq) + H2(g)
H2O(l)  H+(aq) + OH-(aq); DH = +55.9 kJ
Ca(s) + 2H+(aq)  Ca2+(aq) + H2(g); DH = –543.0 kJ
The coefficients must match those in the reaction
we want.
The coefficient on H2O and OH- should be 2.
We multiply the first reaction and its DH by 2 to
accomplish this.
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Ca(s) + 2H2O(l)  Ca2+(aq) + 2OH-(aq) + H2(g)
2H2O(l)  2H+(aq) + 2OH-(aq);
DH = +111.8 kJ
Ca(s) + 2H+(aq)  Ca2+(aq) + H2(g); DH = –543.0 kJ
We can now add the equations and their DH’s.
Note that 2H+(aq) appears as both a
reactant and a product.
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Ca(s) + 2H2O(l)  Ca2+(aq) + 2OH-(aq) + H2(g)
2H2O(l)  2H+(aq) + 2OH-(aq); DH = +111.8 kJ
Ca(s) + 2H+(aq)  Ca2+(aq) + H2(g); DH = –543.0 kJ
Ca(s) + 2H2O(l)  Ca2+(aq) + 2OH-(aq) + H2(g);
DH = –431.2 kJ
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Standard Enthalpies of Formation
The term standard state refers to the standard
thermodynamic conditions chosen for substances
when listing or comparing thermodynamic data:
1 atm pressure and the specified temperature
(usually 25°C). These standard conditions are
indicated with a degree sign (°).
When reactants in their standard states yield
products in their standard states, the enthalpy of
reaction is called the standard enthalpy of
reaction, DH°. (DH° is read “delta H zero.”)
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Elements can exist in more than one physical
state, and some elements exist in more than one
distinct form in the same physical state. For
example, carbon can exist as graphite or as
diamond; oxygen can exist as O2 or as O3 (ozone).
These different forms of an element in the same
physical state are called allotropes.
The reference form is the most stable form of the
element (both physical state and allotrope).
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The standard enthalpy of formation, DHf°, is the
enthalpy change for the formation of one mole of
the substance from its elements in their reference
forms and in their standard states. DHf° for an
element in its reference and standard state is zero.
For example, the standard enthalpy of formation
for liquid water is the enthalpy change for the
reaction
H2(g) + 1/2O2(g)  H2O(l)
DHf° = –285.8 kJ
Other DHf° values are given in Table 6.2 and
Appendix C.
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Standard enthalpies of formation can be used to
calculate the standard enthalpy for a reaction, DH°.
CH4(g) + 4Cl2(g)  CCl4(l) + 4HCl(g); DH° = ?
Table 6.2 shows the DHf° values:
C(graphite) + 2Cl2(g)  CCl4(l); DHf° = –135.4 kJ
1/
2
H2(g) + 1/2 Cl2(g)  HCl(g); DHf° = -92.3 kJ
CH4(g)  C(graphite) + 2H2(g); DHf° = +74.9 kJ
We first identify each reactant and product from
the reaction we want.
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Standard enthalpies of formation can be used to
calculate the standard enthalpy for a reaction, DH°.
CH4(g) + 4Cl2(g)  CCl4(l) + 4HCl(g); DH° = ?
Table 6.2 shows the DHf° values:
C(graphite) + 2Cl2(g)  CCl4(l); DHf° = –135.4 kJ
1/
2
H2(g) + 1/2 Cl2(g)  HCl(g); DHf° = -92.3 kJ
CH4(g)  C(graphite) + 2H2(g); DHf° = +74.9 kJ
Each needs to be on the correct side of the arrow
and is. Next, we’ll check coefficients.
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Standard enthalpies of formation can be used to
calculate the standard enthalpy for a reaction, DH°.
CH4(g) + 4Cl2(g)  CCl4(l) + 4HCl(g); DH° = ?
Table 6.2 shows the DHf° values:
C(graphite) + 2Cl2(g)  CCl4(l); DHf° = –135.4 kJ
1/
2
H2(g) + 1/2 Cl2(g)  HCl(g); DHf° = -92.3 kJ
CH4(g)  C(graphite) + 2H2(g); DHf° = +74.9 kJ
Cl2 and HCl need a coefficient of 4. Multiplying the
second equation and its DH by 4 does this.
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Standard enthalpies of formation can be used to
calculate the standard enthalpy for a reaction, DH°.
CH4(g) + 4Cl2(g)  CCl4(l) + 4HCl(g); DH° = ?
Table 6.2 shows the DHf° values:
C(graphite) + 2Cl2(g)  CCl4(l); DHf° = –135.4 kJ
2H2(g) + 2Cl2(g)  4HCl(g); DHf° = -369.2 kJ
CH4(g)  C(graphite) + 2H2(g); DHf° = +74.9 kJ
Now, we can add the equations.
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Standard enthalpies of formation can be used to
calculate the standard enthalpy for a reaction, DH°.
CH4(g) + 4Cl2(g)  CCl4(l) + 4HCl(g); DH° = ?
Table 6.2 shows the DHf° values:
C(graphite) + 2Cl2(g)  CCl4(l); DHf° = –135.4 kJ
2H2(g) + 2Cl2(g)  4HCl(g); DHf° = -369.2 kJ
CH4(g)  C(graphite) + 2H2(g); DHf° = +74.9 kJ
CH4(g) + 4Cl2(g)  CCl4(g) + 4HCl(g); DH° = –429.7 kJ
6 | 71
?
What is the heat of vaporization of
methanol, CH3OH, at 25°C and 1 atm?
Use standard enthalpies of formation
(Appendix C).
6 | 72
We want DH° for the reaction:
CH3OH(l)  CH3OH(g)

ΔH reaction


nΔH f 
products
For liquid methanol :
ΔH f
For gaseous methanol :

nΔH f
reactants
kJ
 238.7
mol
ΔH f
kJ
 200.7
mol

kJ   
kJ  


ΔH v ap  1 mol   200.7
   1 mol   238.7

mol   
mol  



DHvap= +38.0 kJ
6 | 73
Methyl alcohol, CH3OH, is toxic
because liver enzymes oxidize it to
formaldehyde, HCHO, which can
coagulate protein. Calculate DHo for the
following reaction:
2CH3OH(aq) + O2(g)  2HCHO(aq) + 2H2O(l)
?
Standard enthalpies of formation, ΔH fo :
CH3OH(aq): -245.9 kJ/mol
HCHO(aq):
-150.2 kJ/mol
H2O(l):
-285.8 kJ/mol
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We want DH° for the reaction:
2CH3OH(aq) + O2(aq)  2HCHO(aq) + 2H2O(l)

ΔH reaction


nΔH f 
products
o
ΔH reacton

nΔH f
reactants

kJ 
kJ  


 2 mol   150.2
  2 mol - 285.8

mol 
mol  




kJ 

 kJ  
 2 mol   245.9
  1 mol 0

mol 

 mol  

o
ΔH reaction
   300.4 kJ   571.6 kJ   491.8 kJ
o
ΔH reaction
  872.0 kJ   491.8 kJ
o
ΔH reaction
 380.2 kJ
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Foods fuels three needs of the body:
• They supply substances for the growth and
repair of tissue.
• They supply substances for the synthesis of
compounds used in the regulation processes.
• They supply energy.
Foods do this by a combustion process.
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For glucose, a carbohydrate:
C6H12O6(s) + 6O2(g)  6CO2(g) + 6H2O(l);
DHf° = –2803 kJ
For glycerol trimyristate, a fat:
C45H86O6(s) + 127/2O2(g)  45CO2(g) + 43H2O(l);
DHf°= –27,820 kJ
The average value for carbohydrates is 4.0 kcal/g
and for fats is 9.0 kcal/g.
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Fossil fuels originated millions of years ago when
aquatic plants and animals were buried and
compressed by layers of sediment at the bottoms
of swamps and seas.
Over time this organic matter was converted by
bacterial decay and pressure to petroleum (oil),
gas, and coal.
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Coal, which accounts for 22.9% of total U.S.
energy consumption, varies in terms of the amount
of carbon it contains and so varies in terms of the
amount of energy it produces in combustion.
Anthracite (hard coal) was laid down as long as
250 million years ago and can contain as much as
80% carbon. Bituminous coal, a younger variety,
contains between 45% and 65% carbon.
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Natural gas, which accounts for 22.7% of total U.S.
energy consumption, is convenient because it is
fluid and can be easily transported. Purified natural
gas is primarily methane, CH4, plus small amounts
of ethane, C2H6; propane, C3H8; and butane,
C4H10.
Petroleum is a mixture of compounds. Gasoline,
which is obtained from petroleum, is a mixture of
hydrocarbons (compounds of carbon and
hydrogen).
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The main issue with natural gas and petroleum is
their relatively short supply. It has been estimated
that petroleum deposits will be 80% depleted by
2030. Natural gas deposits may be depleted even
sooner.
Coal deposits, however, are expected to last for
several more centuries. This has led to the
development of commercial methods for
converting coal to the more easily handled liquid
and gaseous fuels.
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Coal gasification is one way. Steam is passed over
hot coal:
C(s) + H2O(g)  CO(g) + H2(g)
The mixture containing carbon monoxide can be
converted by various methods into useful products.
The mixture of carbon monoxide and hydrogen
can be converted by various methods into useful
products such as methane, CH4.
CO(g) + 3H2(g)  CH4(g) + H2O(g)
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Rockets are self-contained missiles propelled by
the ejection of gases from an orifice. Usually these
are hot gases expelled from the rocket from the
reaction of a fuel with an oxidizer.
One factor in determining the appropriate
fuel/oxidizer combination is its mass of the mixture.
6 | 83