Transcript Chapter 6

Chapter 6
Principles of Reactivity:
Energy and
Chemical Reactions
Dr. S. M. Condren
Thermite Reaction
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Terminology
Energy
• capacity to do work
Kinetic Energy
• energy that something has because it is moving
Potential Energy
• energy that something has because of its
position or its chemical bonding
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Kinetic Energy
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Chemical Potential Energy
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Chemical Potential Energy
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Internal Energy
• The sum of the individual energies of all
nanoscale particles (atoms, ions, or
molecules) in that sample.
• E = 1/2mc2
• The total internal energy of a sample of
matter depends on temperature, the type
of particles, and how many of them there
are in the sample.
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Energy Units
• calorie - energy required to heat 1-g of
water 1oC
• Calorie - unit of food energy;
– 1 Cal = 1-kcal = 1000-cal
• Joule - 1-cal = 4.184 J = 1-kg*m2/sec2
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Law of Conservation of Energy
• energy can neither be created nor
destroyed
• the total amount of energy in the universe
is a constant
• energy can be transformed from one form
to another
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First Law of Thermodynamics
• the amount of heat transferred into a
system plus the amount of work done on
the system must result in a corresponding
increase of internal energy in the system
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Thermochemistry Terminology
system => that part of the universe under
investigation
surroundings => the rest of the universe
universe = system + surroundings
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System and Surroundings
• SYSTEM
–The object under
study
• SURROUNDINGS
–Everything outside
the system
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Announcement
Learn@UW will be unavailable
on Thursday July 19
from 5:00am until 12:00 noon.
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Energy & Chemistry
2 H2(g) + O2(g) -->
2 H2O(g) + heat and light
This can be set up to provide
ELECTRIC ENERGY in a fuel cell.
Oxidation:
2 H2 ---> 4 H+ + 4 eReduction:
4 e- + O2 + 2 H2O ---> 4 OH-
H2/O2 Fuel Cell
Energy, page 288
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Energy & Chemistry
ENERGY is the capacity to do work or
transfer heat.
HEAT is the form of energy that flows
between 2 objects because of their
difference in temperature.
Other forms of energy —
• light
• electrical
• kinetic and potential
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Potential Energy in the Atomic Scale
• Positive and negative
particles (ions) attract
one another.
• Two atoms can bond
• As the particles attract
they have a lower
potential energy
http://mrsec.wisc.edu/Edetc/pmk/NaCl_alt.html
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NaCl — composed
of Na+ and Cl- ions.
Internal Energy (E)
PE + KE = Internal energy (E or U)
Int. E of a chemical system
depends on
–number of particles
–type of particles
–temperature
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Energy Transfer
Energy is always transferred from the hotter
to the cooler sample
Heat – the energy that flows into or out of a
system because of a difference in
temperature between the thermodynamic
system and its surroundings
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Thermochemistry Terminology
state properties => properties which depend
only on the initial and final states
=> properties which are path independent
non-state properties => properties which are
path dependent
state properties => E
non-state properties => q & w
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Thermochemistry Terminology
exothermic - reaction that gives off energy
endothermic - reaction that absorbs
energy
chemical energy - energy associated with
a chemical reaction
thermochemistry - the quantitative study
of the heat changes accompanying
chemical reactions
thermodynamics - the study of energy and
its transformations
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Exothermic Reaction
First-Aid Hotpacks, containing either calcium chloride or
magnesium sulfate, plus water
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Endothermic Reaction
First-aid cold packs, containing ammonium nitrate and
water in separate inner pouches
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Enthalpy
• heat at constant pressure
qp = DH = Hproducts - Hreactants
Exothermic Reaction
DH = (Hproducts - Hreactants) < 0
H2O(l) -----> H2O(s)
DH < 0
Endothermic Reaction
DH = (Hproducts - Hreactants) > 0
H2O(l) -----> H2O(g)
DH > 0
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Enthalpy
H = E + PV
DH = DE + PDV
DE = DH – PDV
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First Law of Thermodynamics
heat => q
internal energy => E
internal energy change =>DE
work => w = - P*DV
DE = q + w
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Specific Heat-Specific Heat Capacity
• the amount of heat necessary to raise the
temperature of 1 gram of the substance
1oC
• independent of mass
• substance dependent
• s.h.
• Specific Heat of Water = 4.184 J/goC
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Heat
q = m * s.h. * Dt
where q => heat, J
m => mass, g
s.h. => specific heat, J/g*oC
Dt = change in temperature, oC,
(always tf – ti)
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Molar Heat Capacity
• the heat necessary to raise the
temperature of one mole of substance by
1oC
• substance dependent
• C
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Heat Capacity
• the heat necessary to raise the
temperature 1oC
• mass dependent
• substance dependent
• C
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Heat Capacity
C = m X s.h.
where
C => heat capacity, J/oC
m => mass, g
s.h. => specific heat, J/goC
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Plotted are graphs
of heat absorbed
versus
temperature for
two systems.
Which system has
the larger heat
capacity?
A, B
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Heat Transfer
qlost = - qgained
(m X s.h. X Dt)lost = - (m X s.h. X Dt)gained
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Heat Transfer
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EXAMPLE
If 100. g of iron at 100.0oC is
placed in 200. g of water at 20.0oC in an insulated
container, what will the temperature, oC, of the iron
and water when both are at the same temperature?
The specific heat of iron is 0.106 cal/goC.
(100.g*0.106cal/goC*(Tf - 100.)oC) = qlost
- qgained = (200.g*1.00cal/goC*(Tf - 20.0)oC)
10.6(Tf - 100.oC) = - 200.(Tf - 20.0oC)
10.6Tf - 1060oC = - 200.Tf + 4000oC
(10.6 + 200.)Tf = (1060 + 4000)oC
Tf = (5060/211.)oC = 24.0oC
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Melting of Ice
http://mrsec.wisc.edu/Edetc/pmk/ice.html
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EXAMPLE: How much heat is required to
heat 10.0 g of ice at -15.0oC to steam at
o
127.0 C?
q = DHice
+ DHfusion
+ DHwater
+ DHboil.
+ DHsteam
q = DHice + DHfusion + DHwater + DHboil. + DHsteam
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Heat Transfer
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EXAMPLE: How much heat is required to
heat 10.0 g of ice at -15.0oC to steam at
127.0oC?
q = DHice + DHfusion + DHwater + DHboil. + DHsteam
{
q = (10.0g*2.09J/goC*((0.0 – (-15.0))oC))
Mass of the ice
specific heat of ice
Temperature change
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EXAMPLE: How much heat is required to
heat 10.0 g of ice at -15.0oC to steam at
127.0oC?
q = DHice + DHfusion + DHwater + DHboil. + DHsteam
q = (10.0g*2.09J/goC*15.0oC)
+ (10.0g*333J/g) Melting of ice occurs at a
constant temperature
Mass of ice Heat of fusion
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EXAMPLE: How much heat is required to
heat 10.0 g of ice at -15.0oC to steam at
127.0oC?
q = DHice + DHfusion + DHwater + DHboil. + DHsteam
q = (10.0g*2.09J/goC*15.0oC)
+ (10.0g*333J/g)
+ (10.0g*4.18J/goC*((100.0-0.00)oC))
Mass of water Specific heat of
liquid water
Temperature change
of the liquid water
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EXAMPLE: How much heat is required to
heat 10.0 g of ice at -15.0oC to steam at
127.0oC?
q = DHice + DHfusion + DHwater + DHboil. + DHsteam
q = (10.0g*2.09J/goC*15.0oC)
+ (10.0g*333J/g)
+ (10.0g*4.18J/goC*100.0oC)
+ (10.0g*2260J/g)
Boiling of water occurs at a
constant temperature
Mass of water Heat of vaporization
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EXAMPLE: How much heat is required to
heat 10.0 g of ice at -15.0oC to steam at
127.0oC?
q = DHice + DHfusion + DHwater + DHboil. + DHsteam
q = (10.0g*2.09J/goC*15.0oC)
+ (10.0g*333J/g)
+ (10.0g*4.18J/goC*100.0oC)
+ (10.0g*2260J/g)
+ (10.0g*2.03J/goC*((127.0-100.0)oC))
Mass of steam Specific heat
of steam
Temperature change for the steam
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EXAMPLE: How much heat is required to
heat 10.0 g of ice at -15.0oC to steam at
127.0oC?
q = DHice + DHfusion + DHwater + DHboil. + DHsteam
q = (10.0g*2.09J/goC*15.0oC)
+ (10.0g*333J/g)
+ (10.0g*4.18J/goC*100.0oC)
+ (10.0g*2260J/g)
+ (10.0g*2.03J/goC*27.0oC)
q = (314 + 3.33X103 + 4.18X103 + 2.26X104 + 548 )J
= 30.96 kJ
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Spreadsheet of Previous Problem
10
10
10
10
10
2.09
333
4.18
2260
2.03
15
1
100
1
27
313.5
3330
4180
22600
548.1
314
3330
4180
22600
548
31
333
418
2260
54
x10
x10
x10
x10
x10
3096 x10
30.96 x10^3
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Coffee Cup Calorimeter
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Bomb Calorimeter
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EXAMPLE
A 1.000g sample of a particular compound
produced 11.0 kJ of heat. The temperature
of the calorimeter and 3000 g of water was
raised 0.629oC. How much heat is gained
by the calorimeter?
heat gained = - heat lost
heatcalorimeter + heatwater = heatreaction
heatcalorimeter = heatreaction - heatwater
Dr. S. M. Condren
EXAMPLE
A 1.000g sample of a particular compound
produced 11.0 kJ of heat. The temperature
of the calorimeter and 3000. g of water was
raised 0.629oC. How much heat is gained
by the calorimeter?
heatcalorimeter = heatreaction - heatwater
heat = 11.0 kJ - ((3.000kg)(0.629oC)(4.184kJ/kgoC))
= 3.10 kJ
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Example
What is the mass of water equivalent of the
heat absorbed by the calorimeter?
#g = (3.10 kJ/0.629oC)(1.00kg*oC/4.184kJ)
= 6.47 x 102 g
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Example
A 1.000 g sample of ethanol was burned in
the sealed bomb calorimeter described
above. The temperature of the water rose
from 24.284oC to 26.225oC. Determine the
heat for the reaction.
m = (3000. + 647)g H2O
q = m X s.h. X Dt
= (3647g)(4.184J/goC)(1.941oC) = 29.61 kJ
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When graphite is burned to yield CO2, 394 kJ of
energy are released per mole of C atoms
burned. When C60 is burned to yield CO2
approximately 435 kJ of energy is released per
mole of carbon atoms burned. Would the
buckyball-to-graphite conversion be exothermic
or endothermic?
exothermic, endothermic
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Laws of Thermochemistry
1.
The magnitude of DH is directly
proportional to the amount of reactant or
product.
s --> l
l --> g
DH => heat of fusion
DH => heat of vaporization
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Laws of Thermochemistry
2. DH for a reaction is equal in magnitude
but opposite in sign to DH for the reverse
reaction.
H2O(l) -----> H2O(s)
DH < 0
H2O(s) -----> H2O(l)
DH > 0
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Laws of Thermochemistry
3. The value of H for the reaction is the
same whether it occurs directly or in a
series of steps.
DHoverall = DH1 + DH2 + DH3 + · · ·
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Hess' Law
• a relation stating that the heat flow in a
reaction which is the sum of a series of
reactions is equal to the sum of the heat
flows in those reactions
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EXAMPLE
CH4(g) + 2 O2(g) -----> CO2(g) + 2 H2O(l)
CH4(g) -----> C(s) + 2 H2(g)
DH1
2 O2(g) -----> 2 O2(g)
DH2
C(s) + O2(g) -----> CO2(g)
DH3
2 H2(g) + O2(g) -----> 2 H2O(l)
DH4
--------------------------------------------CH4(g) + 2 O2(g) -----> CO2(g) + 2 H2O(l)
DHoverall = DH1 + DH2 + DH3 + DH4
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Standard Enthalpy of Formation
the enthalpy associated with the formation of
a substance from its constituent elements
under standard state conditions
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Calculation of DH
DH =
o
o
Sc*DHf products
-
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o
o
Sc*DHf reactants
Example
What is the value of DHrx for the reaction:
2 C6H6(l) + 15 O2(g) --> 12 CO2(g) + 6 H2O(g)
from Appendix L Text
C6H6(l) DHfo = + 48.95 kJ/mol
O2(g) DHfo = 0
CO2(g) DHfo = - 393.509
H2O(g) DHfo = - 241.83
D Hrx = [S c* D Hfo]products - [S c* D Hfo]reactants
Dr. S. M. Condren
Example
What is the value of DHrx for the reaction:
2 C6H6(l) + 15 O2(g) --> 12 CO2(g) + 6 H2O(g)
from Appendix L Text
C6H6(l) DHfo = + 48.95 kJ/mol; O2(g) DHfo = 0
CO2(g) DHfo = - 393.509; H2O(g) DHfo = - 241.83
D Hrx = [S c* D Hfo]products - [S c* D Hfo]reactants
DHrx = [(12mol)(- 393.509kJ/mol)
+ (6mol)(- 241.83kJ/mol)]products
- [(2mol)(+ 48.95kJ/mol)
+ (15mol)(0kJ/mol)]reactants
= - 6.271 x 103 kJ
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Fossil Fuels
coal
petroleum
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natural gas
Based on 1998 Data
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