Ch 6

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Chapter 6 Thermochemistry Energy Flow and
Chemical Change
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Thermochemistry Energy Flow and Chemical Change
6.1 Forms of Energy and Their Interconversion
6.2 Enthalpy Heats of Reaction and Chemical
Change
6.3 Calorimetry Laboratory Measurement of
Heats of Reaction
6.4 Stoichiometry of Thermochemical Equations
6.5 Hesss Law of Heat Summation
6.6 Standard Heats of Reaction (DH0rxn)
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Thermochemistry Energy Flow and Chemical Change
In the observation and measurement of a change in
energy, we talk about the System and the
Surroundings. The System is what we are
studying, e.g. a reaction taking place in a
flask. The Surroundings are the rest of the
universe, but generally only the nearby portion
that is relevant to the system.
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A change in energy in the System is always
accompanied by an opposite change in energy of
the Surroundings.
Change in energy, DE Efinal - Einitial
Eproducts - Ereactants
Energy diagrams for the transfer of internal
energy (E) between a system and its surroundings.
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Heat (thermal energy) and Work, Two Forms of D E
The symbol for heat is q. The symbol for work is
w. They can be positive or negative
values. Energy flowing into a system is defined
as positive. Energy flowing out of a system is
negative.
DE q w
DE q 0
In a system transferring energy as heat only.
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A system losing energy as work only.
Work is done on the surroundings, so work on the
system is negative.
DE 0 w
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The Sign Conventions for q, w and DE
DE
q
w






-
depends on sizes of q and w
-

depends on sizes of q and w
-
-
-
For q means system gains heat - means
system loses heat.
For w means word done on system -
means work done by system.
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Law of Conservation of Energy (First Law of
Thermodynamics)
The total energy of the universe is constant.
Energy transfers can occur between the system and
surroundings in the form of heat and/or work. But
the energy of the system and the energy of the
surroundings remains constant energy is
conserved.
DEuniverse DEsystem DEsurroundings 0
Units of Energy
Joule (J)
1 J 1 kgm2/s2 (mass x accel x dist)
Calorie (cal)
1 cal 4.18J
British Thermal Unit
1 Btu 1055 J
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Sample Problem 6.1
Determining the Change in Internal Energy of a
System
PLAN
Define system and surrounds, assign signs to q
and w and calculate DE. The answer should be
converted from J to kJ and then to kcal.
SOLUTION
q - 325J
w - 451J
DE q w
-325J (-451J)
-776J
0.776kJ
0.186kcal
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State functions such as internal energy of a
system are dependent only upon the initial state
and the final state. It is independent upon the
pathway to get between the two states.
Two different paths for the energy change of a
system.
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Enthalpy, H
Pressure-volume work.
In chemistry the work is usually electrical (Ch
21) or pressure-volume (PV) work.....work done by
expanding gas (change in volume, ?V) ?V Vfinal -
Vinitial So this work done on surroundings is
negative.
DE q w
DE q - PDV
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The Meaning of Enthalpy, H
In many reactions little or no work done so DH
DE in
DE q - PDV
1. Reactions that do not involve gases.
qp DE PDV
2. Reactions in which the number of
moles of gas does not change.
(constant pressure)
DH DE PDV
3. Reactions in which the number of moles of gas
does change but q is gtgtgt PDV.
DH is the change in enthalpy - change in internal
energy plus PDV work
qp DE PDV DH
DH qp , equals the heat gained or lost at
constant pressure.
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?H Hfinal - Hinitial
Change of Enthalpy, ? H
Exothermic / Endothermic
If the products have less energy than the
reactants, i.e. Hfinal lt Hinitial , then ?H is
negative. Heat has been released. Exothermic
processes have a negative ?H.
Enthalpy diagram for exothermic process.
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?H Hfinal - Hinitial
Change of Enthalpy, ? H
Exothermic / Endothermic
If the products have more energy than the
reactants, i.e. Hfinal gt Hinitial , then ?H is
positive. Heat has been absorbed. Endothermic
processes have a positive ?H.
Enthalpy diagram for endothermic process.
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Sample Problem 6.2
Drawing Enthalpy Diagrams and Determining the
Sign of DH
PLAN
Determine whether heat is a reactant or a
product. As a reactant, the products are at a
higher energy and the reaction is endothermic.
The opposite is true for an exothermic reaction
SOLUTION
(a) The reaction is exothermic.
(b) The reaction is endothermic.
EXOTHERMIC
DH -285.8kJ
DH 40.7kJ
ENDOTHERMIC
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Some Important Types of Enthalpy Change
heat of combustion (DHcomb)
heat of formation (DHf)
heat of fusion (DHfus)
heat of vaporization (DHvap)
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Table 6.3 Heats of Combustion of Some Fats and
Carbohydrates
Fats
vegetable oil
-37.0 (x 1/4.18) -8.85 Cal / g
Carbohydrates
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Components of Internal Energy
Contributions to the kinetic energy

• The molecule moving through space,
  Ek(translation)
• The molecule rotating, Ek(rotation)
• The bound atoms vibrating, Ek(vibration)
• The electrons moving within each atom,
  Ek(electron)

Contributions to the potential energy

• Forces between the bound atoms vibrating,
  Ep(vibration)
• Forces between nucleus and electrons and
  between electrons in each atom, Ep(atom)
• Forces between the protons and neutrons in each
  nucleus, Ep(nuclei)
• Forces between nuclei and shared electron pair
  in each bond, Ep(bond)

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Components of internal energy (E)
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Breaking and Forming Chemical Bonds
During a chemical reaction, chemical bonds are
broken and new chemical bonds are formed.
Breaking each chemical bond requires energy,
Endothermic
Forming each chemical bond releases energy,
Exothermic
The ?H for a reaction is the sum of the energy
required and released during the bond breaking
and forming process.
The enthalpy for any reverse process has the same
numerical value but a change in sign.
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Specific Heat Capacity
When an object absorbs heat, it gets hotter. The
more heat it absorbs the hotter it gets, i.e. q
DT or q constant x DT or q /DT
constant This capacity of an object to absorb
heat is called itsheat capacity q / DT
(units of J/K)The amount of heat to change the
temperature by 1 K A related property is
Specific Heat Capacity (c)This is the amount of
heat to change the temperature of 1 gram of
substance by 1 K c q / mass x DT (units of
J/g-K) Another property is Molar Heat Capacity
(C)This is the amount of heat to change the
temperature of 1 mole of substance by 1 K C q
/ moles x DT (units of J/mol-K)
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Sample Problem 6.3
Calculating the Quantity of Heat from the
Specific Heat Capacity
PLAN
Given the mass, specific heat capacity and change
in temperature, we can use q c x mass x DT to
find the answer. DT in 0C is the same as for K.
SOLUTION
1.33x104J
q
125g
(300-25)0C
x
x
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Calorimetry
Calorimetry is a way to measure heat from a
process by directing the heat into surroundings
where the increase of heat can be measured. The
surroundings will be a container called a
calorimeter. The heat lost by the system equals
the heat gained by the calorimeter. - q sample
qcalorimeter (minus sign for heat lost)
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Coffe-cup calorimeter
- q sample qcalorimeter
- (mass x heat capacity x DT)samp (mass x
heat capacity x DT)cal.
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Sample Problem 6.4
Determining the Specific Heat Capacity of a Solid
PLAN
It is helpful to use a table to summarize the
data given. Then work the problem realizing that
heat lost by the system must be equal to that
gained by the surroundings.
SOLUTION
c x
25.64g x
-71.51K

-
csolid
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A bomb calorimeter
- q sample qcalorimeter
qcalorimeter
heat capacity x DT
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Sample Problem 6.5
Calculating the Heat of Combustion
PLAN
- q sample qcalorimeter
SOLUTION
qcalorimeter
heat capacity x DT
8.151kJ/K x 4.937K
40.24kJ
9.62kcal or Calories
The manufacturers claim is true.
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Summary of the relationship between amount (mol)
of substance and the heat (kJ) transferred during
a reaction.
molar ratio from balanced equation
DHrxn (kJ/mol)
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Sample Problem 6.6
Using the Heat of Reaction (?Hrxn) to Find Amounts
SOLUTION
PLAN
heat(kJ)
1.000x103kJ x
1676kJ2mol Al
mol of Al
32.20g Al
x M
g of Al
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Hesss Law
Since enthalpy is a state function, the enthalpy
for an overall process is equal to the sum of the
enthalpies of all individual steps.
Procedure Identify the target equation whose ?H
is unknown. Manipulate contributing equations by
multiplying coefficients and/or reversing to get
the correct amounts of compounds as is target
equation. Remembering to change the sign of ?H
when reversing equations and multiplying ?H when
multiplying coefficients. The sum of all
equations must be the same as the target
equation. The sum of the individual ?H values
will then equal the target ?H .
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Sample Problem 6.7
Using Hesss Law to Calculate an Unknown DH
PLAN
Equations A and B have to be manipulated by
reversal and/or multiplication by factors in
order to sum to the first, or target, equation.
SOLUTION
Multiply Equation B by 1/2 and reverse it.
DHB -90.6kJ
DHrxn -373.6kJ
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Formation Equations
The chemical equation showing the formation of 1
mole of a compound from the elements in their
standard state. The enthalpy change for the
formation is called the Standard Heat of
Formation, DHof . Element in their standard
state have a DHof 0 Most compounds have a
negative DHof , i.e. The reaction is
exothermic........the compound is lower in energy
than the elements from which it was formed.
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Table 6.5 Selected Standard Heats of Formation
at 250C(298K)
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Sample Problem 6.8
Writing Formation Equations
(a) Silver chloride, AgCl, a solid at standard
conditions.
(b) Calcium carbonate, CaCO3, a solid at standard
conditions.
(c) Hydrogen cyanide, HCN, a gas at standard
conditions.
PLAN
Use the table of heats of formation for values.
SOLUTION
DH0f -127.0kJ
DH0f -1206.9kJ
DH0f 135kJ
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The general process for determining DH0rxn from
DH0f values
Using Hesss Law to Calculate DH0rxn
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Sample Problem 6.9
Calculating the Heat of Reaction from Heats of
Formation
PLAN
Look up the DH0f values and use Hesss Law to
find DHrxn.
SOLUTION
DHrxn S mDH0f (products) - S nDH0f (reactants)
DHrxn 4(DH0f NO(g) 6(DH0f H2O(g)
- 4(DH0f NH3(g) 5(DH0f O2(g)
(4mol)(90.3kJ/mol) (6mol)(-241.8kJ/mol) -
(4mol)(-45.9kJ/mol) (5mol)(0kJ/mol)
DHrxn -906kJ
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End of Chapter 6