Chapter 5 Thermochemistry

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Chapter 5Thermochemistry
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Thermochemistry
Very broad field combustion of fossil fuels,
development of biomass technologies,
biochemistry, fuel cells Began during
Industrial Revolution to enhance performance of
steam engines.
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The Nature of Energy
Energy (i.e., move a mass through some
distance w F x d). It is interconvertible with
heat.
Two important types of energy
energy due to motion Depends on mass and
velocity Ex Moving car.
energy due to position in a gravitational
field (stored energy) Ex Standing on hill
m mass v velocity (speed) g grav. const.
9.8m/s2 h height
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due to
the position of a charge in an electric field
k 8.99 ? 109 J m/C2 Q and q electrical
Charges on two objects
eg e- in the electric field of the nucleus of an
atom
This is the energy that gets converted during
chemical reactions.
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In a chemical reaction
Rearrangement of electrons causes energy to flow
in or out of a chemical system.
Thermal energy
Energy a substance possess due to its temperature
above the absolute zero of temperature.
Associated with the kinetic energy of atoms and
molecules.
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SI Unit for energy is the joule, J
James Joule 1818-1889 work and heat
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Commit to memory! J kg m2 s-2
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System and Surroundings
Thermodynamic system Surroundings
For example C(s) CO2(g) ? 2CO(g)
System Surroundings
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Types of Systems
Closed systems
Open systems
Isolated systems
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Example
Identify each of the following systems as open,
closed or isolated
Mercury in a thermometer A living plant Coffee in
a tightly sealed thermos flask
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Transferring Energy Work and Heat
Sir Benjamin Thompson, Count Rumford Demonstrated
the equivalence of mechanical work and heat
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Transferring Energy Work and Heat
SYSTEM C2H5OH(g) 3O2(g) ? 2CO2(g) 2H2O(g)
energy
mechanical work
heat flows
ENERGY ability to do WORK and/or to TRANSFER HEAT
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The First Law of Thermodynamics
The proper definition is
Or simply, ENERGY IS CONSERVED.
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Internal Energy, U
(textbook uses E)
Sum of all Ek and Ep of a system
Motion of molecules Motion of electrons Position
of electrons in electrostatic field of the
nucleus Interactions between molecules, etc.
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Virtually impossible to determine or calculate
internal energy
During a chemical reaction can only determine
changes in U
?U
delta U, delta implies change.
A positive ?U results when Ufinal gt Uinitial,
system gained energy from surroundings. Bricks
warmed from the sun.
A negative ?U results when Uinitial gt Ufinal,
surroundings gained energy from system. A
cooling cup of tea.
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Two ways for energy to be transferred
Heat (q) and work (w)

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Example
A battery does 555 kJ of work every hour driving
a pump. The system loses 124 kJ of energy (in
the form of heat) per hour to the
surroundings. What is ?Usystem per hour?
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Exothermic and Endothermic Processes
Endothermic Exothermic

• Note
• An exothermic reaction (qlt0) feels
• An endothermic reaction (qgt0) feels

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U is a state function
State functions depend only on
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Many chemical systems expand or contract when
they react. These systems are doing expansion
(PV) work.
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Work done by the system against opposing force
distance moved x opposing force
Piston of area A
Expanding system
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Example
A 10.0 W heater heats gas in a cylinder for 1
min. The cylinders volume expands from 1.00 l
to 1.50 l. Calculate ?U for the gas if Patm
0.965 atm.
Analysis
Heat is transferred to the gas ? q is ve (heat
IN)
Gas expands and does work ? w is ve (work on
surr)
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Solve for q
Need Joules, and 1 Watt (W) 1 J/s energy (q or
w) Pt
q 600 J
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Solve for w
Again, need Joules Recall 101 kPa/atm and 1 Pa
1 N/m2
-48.9 Nm (1Nm 1J)
OR, can use this conversion1 L atm 101.3 J
w - 48.9 J
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Therefore, ?U q w
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Measuring ?U
?U q w q - P?V
If the reaction is carried out in a sealed, rigid
container, ?V 0
The change in internal energy of a system is
equal to the heat transferred to or from the
system at constant volume.
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Enthalpy
Most reactions carried our under constant
pressure.
Define a new state function called the ENTHALPY
of the system, H, such that
As for U, it is virtually impossible to measure
or to calculate H, only ?H. Under constant P
conditions
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Measuring ?H
?H ?U P?V
The change in the enthalpy of a system is equal
to the heat transferred to or from the system at
constant pressure
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Examples
For a reaction at constant P, ?U -65 kJ and 28
kJ expansion work is done by the system. What is
?H?
?H ?U P?V
For a reaction ?H -15 kJ and 22 kJ expansion
work is done on the system. What is ?U?
?H ?U P?V
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Calorimetry
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Need to relate q (heat) to ?T
Heat capacity, C, is the temp change of an object
after absorbing an amount of energy.
?T Tf - Ti
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Recall that size of 1 degree Kelvin size of 1
degree centigrade
?T will depend on the quantity of water (in grams
or moles) so more common to use the intensive
property SPECIFIC HEAT CAPACITY Cs or MOLAR
HEAT CAPCITY Cm
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J K-1 g-1
J K-1 mol-1
Cs
Cm
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Example
The molar heat capacity of water is 75.4 J K-1
mol-1. How long must a 1.000 kW heater run to
raise the temperature of 500.0 g of water from
20.0 oC to 80.0 oC?
Strategy
Mass
n m/MW
q m C ?T
P w/t ? t w/P
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Coffee cup calorimeter
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Bomb calorimeter
qV ?U
We measure change in internal energy.
p. 161 8th Ed p. 172 9th Ed
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Example
When 100 ml of an aqueous solution of certain
reactants are mixed, it is known that their
reaction releases 1.78 kJ of energy.
When this reaction occurred in an open
calorimeter, the temperature rose by 3.65 oC.
Next 50 ml HCl was mixed with 50 ml NaOH in the
same calorimeter in this case the temperature
rise was 1.26 oC.
What is the heat output of the neutralization
reaction?
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When 100 ml of an aqueous solution of certain
reactants are mixed, it is known that their
reaction releases 1.78 kJ of energy.
When this reaction occurred in an open
calorimeter, the temperature rose by 3.65 oC.
Next 50 ml HCl was mixed with 50 ml NaOH in the
same calorimeter in this case the temperature
rise was 1.26 oC.
What is the heat output of the neutralisation
reaction?
First reaction calibrates the calorimeter
Ccal q/?T
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When 100 ml of an aqueous solution of certain
reactants are mixed, it is known that their
reaction releases 1.78 kJ of energy.
When this reaction occurred in an open
calorimeter, the temperature rose by 3.65 oC.
Next 50 ml HCl was mixed with 50 ml NaOH in the
same calorimeter in this case the temperature
rise was 1.26 oC.
What is the heat output of the neutralization
reaction?
Second reaction uses same volume of aqueous
solution so dont have to account for any mass
difference.
q Ccal ?T
Since T has increased (exothermic), q must be ve.
?H qp
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Example
A 2.200 g sample of quinone (C6H4O2) is burned in
a bomb calorimeter with a heat capacity of 7.854
kJ K-1.
The temperature of the calorimeter increases from
23.44 to 30.57 oC.
What is the molar heat of combustion of quinone?
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A 2.200 g sample of quinone (C6H4O2) is burned in
a bomb calorimeter with a heat capacity of 7.854
kJ K-1.
The temperature of the calorimeter increases from
23.44 to 30.57 oC.
What is the molar heat of combustion of quinone?
?T
How many moles?
Mr 108.09 n m/MW
Heat transfer (q ve as T incr, exothermic)
q C?T
Converting to a per mol basis
-2752 kJ mol-1
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Enthalpies of Reaction
Enthalpy is an extensive property (magnitude of
?H is directly proportional to amount) CH4(g)
2O2(g) ? CO2(g) 2H2O(g) ?H -802 kJ
mol-1 2CH4(g) 4O2(g) ? 2CO2(g) 4H2O(g)
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For the reverse reaction the sign of ?H
changes CO2(g) 2H2O(g) ? CH4(g) 2O2(g) ?H
802 kJ mol-1
Values of ?H depend on conditions such as
pressure and physical state of reactants and
products.
Define the standard enthalpy of reaction, ?Ho, as
Pressure Temp
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If a reaction is carried out in a number of
steps, ?H for the overall reaction is

.
This is known as Hess Law. It is merely a
consequence of the First Law of Thermodynamics.
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Most useful application of Hess Law
Use extensive data available for heats of
combustions to determine ?H for other reactions
Example
Given
A. C3H8(g) 5O2(g) ? 3CO2(g) 4H2O(l) ?Ho
-2220 kJ mol-1
B. C(s) O2(g) ? CO2(g) ?Ho -394 kJ mol-1
C. H2(g) ½O2(g) ? H2O(l) ?Ho -286 kJ mol-1
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A. C3H8(g) 5O2(g) ? 3CO2(g) 4H2O(l) ?Ho
-2220 kJ mol-1
B. C(s) O2(g) ? CO2(g) ?Ho -394 kJ mol-1
C. H2(g) ½O2(g) ? H2O(l) ?Ho -286 kJ mol-1
Calculate ?Ho for
3C(s) 4H2(g) ? C3H8(g)
Step 1
Select one of the reactants and write down an
equation in which it also appears as a reactant.
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C3H8(g) 5O2(g) ? 3CO2(g) 4H2O(l) ?Ho
-2220 kJ mol-1
C(s) O2(g) ? CO2(g) ?Ho -394 kJ mol-1
H2(g) ½O2(g) ? H2O(l) ?Ho -286 kJ mol-1
Calculate ?Ho for
3C(s) 4H2(g) ? C3H8(g)
Step 2
Select one of the products and do the same thing
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C3H8(g) 5O2(g) ? 3CO2(g) 4H2O(l) ?Ho
-2220 kJ mol-1
C(s) O2(g) ? CO2(g) ?Ho -394 kJ mol-1
H2(g) ½O2(g) ? H2O(l) ?Ho -286 kJ mol-1
Calculate ?Ho for
3C(s) 4H2(g) ? C3H8(g)
Step 3
Add Steps 1 and 2 and simplify
3C(s) 3O2(g) 3CO2(g) 4H2O(l) ? 3CO2(g)
C3H8(g) 5O2(g) ?Ho
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C3H8(g) 5O2(g) ? 3CO2(g) 4H2O(l) ?Ho
-2220 kJ mol-1
C(s) O2(g) ? CO2(g) ?Ho -394 kJ mol-1
H2(g) ½O2(g) ? H2O(l) ?Ho -286 kJ mol-1
Calculate ?Ho for
3C(s) 4H2(g) ? C3H8(g)
Step 4
3C(s) 4H2O(l) ? C3H8(g) 2O2(g) ?Ho
1038 kJ mol-1
Cancel any unwanted species by adding an equation
with the same species on the other side of the
arrow
3C(s) 4H2O(l) ? C3H8(g) 2O2(g) ?Ho
1038 kJ mol-1
3C(s) 4H2(g) ? C3H8(g) ?Ho
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Standard Enthalpies of Formation
The standard enthalpy of formation of the
most stable form of any element is zero no
formation needed when in its standard state.
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p. 167 in 8th Ed p. 177 in 9th Ed p. 192 in
10th Ed. See also Appendix C.
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Usefulness?
We use Hess Law to calculate enthalpies of a
reaction from enthalpies of formation
For any reaction
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Example
Calculate ?Ho for the combustion of 1 mol of
acetylene C2H2.
Step 1
Write a balanced equation for the process
Step 2
Use Appendix C to find Sn?Hof (products)
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Step 3
Use Appendix C to find Sn?Hof (reactants)
Step 4
Subtract reactants total from the products
total
This is for combustion of 2 mols so for 1 mol
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Lets see how we have used Hess Law to get this
result
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End of Chapter 5Thermochemistry