Chapter 11 - Thermochemistry
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Transcript Chapter 11 - Thermochemistry
Chapter 16 - Energy and
Chemical Change
Energy and Heat
Thermochemistry
- concerned with
heat changes that occur during
chemical reactions
Energy - capacity for doing work or
supplying heat
• weightless, odorless, tasteless
• if within the chemical substancescalled chemical potential energy
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Energy and Heat
Gasoline
contains a significant
amount of chemical potential energy
Heat - represented by “q”, is energy
that transfers from one object to
another, because of a temperature
difference between them.
• only changes can be detected!
• flows from warmer cooler object
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Exothermic and Endothermic
Processes
Essentially
all
chemical reactions,
and changes in
physical state, involve
either:
• release of heat, or
• absorption of heat
Exo- and Endothermic Processes
In
studying heat changes,
think of defining these
two parts:
• the system - the part of
the universe on which
you focus your attention
• the surroundings includes everything
else in the universe
Exo- and Endothermic Processes
Together,
the system and
its surroundings
constitute the universe
Thermochemistry is
concerned with the flow
of heat from the system
to its surroundings, and
vice-versa.
Exothermic and Endothermic
Processes
The
Law of Conservation of Energy
states that in any chemical or
physical process, energy is neither
created nor destroyed.
• All the energy is accounted for as
work, stored energy, light, or
heat.
Exothermic and Endothermic
Processes
Heat
flowing into a system from its
surroundings:
• defined as positive
• q has a positive value
• called endothermic
–system gains heat as the
surroundings cool down
Exo- and Endothermic Processes
Heat
flowing out of a system
into its surroundings:
• defined as negative
• q has a negative value
• called exothermic
–system loses heat as
the surroundings heat up
Exothemic and Endothermic
Every
reaction has an energy
change associated with it
Exothermic reactions release energy,
usually in the form of heat.
Endothermic reactions absorb
energy
Energy is stored in bonds between
atoms
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Heat Capacity and Specific Heat
A calorie
is defined as the quantity of
heat needed to raise the temperature
of 1 g of pure water 1 oC.
• Used except when referring to food
• a Calorie, written with a capital C,
always refers to the energy in food
• 1 Calorie = 1 kilocalorie = 1000 cal.
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Heat Capacity and Specific Heat
The
calorie is also related to the
joule, the SI unit of heat and energy
• named after James Prescott Joule
• 4.184 J = 1 cal
Heat Capacity - the amount of heat
needed to increase the temperature
of an object exactly 1 oC
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Heat Capacity and Specific Heat
Specific
Heat Capacity - the
amount of heat it takes to raise the
temperature of 1 gram of the
substance by 1 oC (abbreviated “c”)
• often called simply “Specific Heat”
has a HUGE value,
compared to other chemicals
Movie
Water
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Specific Heats of
Common Substances
are provided in a
table in your book.
Heat Capacity and Specific Heat
water, c = 4.184 J/(g oC), and
also c = 1.00 cal/(g oC)
Thus, for water:
• it takes a long time to heat up, and
• it takes a long time to cool off!
Water is used as a coolant!
For
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Heat Capacity and Specific Heat
To
calculate, use the formula:
q = c x mass (g) x T
heat abbreviated as “q”
T = change in temperature
c = Specific Heat
Units are either J/(g oC) or cal/(g oC)
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Energy Conversion
Factors are given in a
table in your book.
Section 16.2
Heat in Chemical Reactions and
Processes
OBJECTIVES:
• Calculate heat
changes in
chemical and
physical
processes.
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Calorimetry
Calorimetry
- the accurate and
precise measurement of heat
change for chemical and physical
processes.
The device used to measure the
absorption or release of heat in
chemical or physical processes is
called a Calorimeter
Calorimetry
Foam
cups are
excellent heat
insulators, and are
commonly used as
simple calorimeters
• Fig. 16.5, page 497
Calorimetry
Calorimetry
experiments
can be performed at a
constant volume using a
device called a “bomb
calorimeter”
For systems at constant
pressure, the heat
content is the same as a
property called Enthalpy
(H) of the system
Calorimetry
in enthalpy = H
q = H These terms will
be used interchangeably
in this textbook
Thus, q = H = m x c x T
H is negative for an
exothermic reaction
H is positive for an
endothermic reaction
Changes
In terms of bonds
C
O
O
O
C
O
Breaking this bond will require energy.
O
C
O C O
O
Making these bonds gives you energy.
In this case making the bonds gives you
more energy than breaking them.
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Exothermic
The
products are
lower in energy than
the reactants
Releases energy
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Energy
C + O2 CO2+ 395 kJ
C + O2
395kJ
C O2
Reactants
Products
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Endothermic
The
products are higher in energy
than the reactants
Absorbs energy
Movie
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Energy
CaCO
CaO
CaCO
CaO
+ CO+2 CO2
3 + 176
3 kJ
CaO + CO2
176 kJ
CaCO3
Reactants
Products
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Sec. 16.2&3 Chemistry Happens in
MOLES
An equation that includes energy is
called a thermochemical equation
CH4 + 2O2 CO2 + 2H2O + 802.2 kJ
1 mole of CH4 releases 802.2 kJ of
energy.
When you make 802.2 kJ you also
make 2 moles of water
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Thermochemical Equations
A heat
of reaction is the heat
change for the equation.
• The physical state of reactants
and products must also be given.
• Standard conditions for the
reaction are 101.3 kPa (1 atm.)
and 25 oC
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CH4 + 2 O2 CO2 + 2 H2O + 802.2 kJ
If 10. 3 grams of CH4 are burned
completely, how much heat will be
produced?
10. 3 g CH4
1 mol CH4
16.05 g CH4
802.2 kJ
1 mol CH4
= 515 kJ
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CH4 + 2 O2 CO2 + 2 H2O + 802.2 kJ
How
many liters of O2 at STP would
be required to produce 23 kJ of
heat?
How many grams of water would be
produced with 506 kJ of heat?
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Summary, so far...
Enthalpy
The
heat content a substance has at a
given temperature and pressure
Can’t be measured directly because
there is no set starting point
The reactants start with a heat content
The products end up with a heat
content
So we can measure how much
enthalpy changes
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Enthalpy
Symbol
is H
Change in enthalpy is H (delta H)
If heat is released, the heat content of
the products is lower
H is negative (exothermic)
If heat is absorbed, the heat content
of the products is higher
H is positive (endothermic)
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Energy
Change is down
H is <0
Reactants
Products
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Energy
Change is up
H is > 0
Reactants
Products
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Heat of Reaction
The heat that is released or absorbed in a
chemical reaction
Equivalent to H
C + O2(g) CO2(g) + 393.5 kJ
C + O2(g) CO2(g)
H = -393.5 kJ
In thermochemical equation, it is important
to indicate the physical state
H2(g) + 1/2O2 (g) H2O(g) H = -241.8 kJ
H2(g) + 1/2O2 (g) H2O(l) H = -282.5 kJ
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Heat of Combustion
The
heat from the
reaction that
completely burns 1
mole of a
substance.
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Section 16.3
Heat in Changes of State
OBJECTIVES:
• Calculate heat
changes that occur
during melting,
freezing, boiling, and
condensing.
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Heats of Fusion and
Solidification
Molar
Heat of Fusion (Hfus) - the
heat absorbed by one mole of a
substance in melting from a solid to
a liquid
Molar Heat of Solidification (Hsolid)
- heat lost when one mole of liquid
solidifies Movie
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Heats of Fusion and
Solidification
Heat
absorbed by a melting solid is
equal to heat lost when a liquid
solidifies
• Thus, Hfus = -Hsolid
Note Table 16.6, page 502
Sample Problem 16-4, page 504
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Heats of Vaporization and
Condensation
When
liquids absorb heat at their
boiling points, they become vapors.
Molar Heat of Vaporization (Hvap) the amount of heat necessary to
vaporize one mole of a given liquid.
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Heats of Vaporization and
Condensation
Condensation
is the opposite of
vaporization.
Molar Heat of Condensation (Hcond)
- amount of heat released when one
mole of vapor condenses
Hvap = - Hcond
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Heats of Vaporization and
Condensation
large values for Hvap and
Hcond are the reason hot vapors
such as steam is very dangerous
• You can receive a scalding burn
from steam when the heat of
condensation is released!
The
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Heats of Vaporization and
Condensation
H20(g)
H20(l)
Hcond = - 40.7kJ/mol
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Heat of Solution
Heat
changes can also occur when
a solute dissolves in a solvent.
Molar Heat of Solution (Hsoln) heat change caused by dissolving
one mole of a substance
Sodium hydroxide provides a good
example of an exothermic molar
heat of solution:
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Heat of Solution
NaOH(s)
H2O(l)
Na1+(aq) + OH1-(aq)
Hsoln = - 445.1 kJ/mol
The heat is released as the ions
separate and interact with water,
releasing 445.1 kJ of heat as Hsoln
thus becoming so hot it steams!
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Section 16.4
Calculating Heat Changes
OBJECTIVES:
• Apply Hess’s law of heat
summation to find heat changes
for chemical and physical
processes.
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Hess’s Law
If
you add two or more
thermochemical equations to give a
final equation, then you can also
add the heats of reaction to give
the final heat of reaction.
Called Hess’s law of heat summation
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Why Does It Work?
If you turn an equation around, you change
the sign:
If H2(g) + 1/2 O2(g) H2O(g) H=-285.5 kJ
then,
H2O(g) H2(g) + 1/2 O2(g) H =+285.5 kJ
also,
If you multiply the equation by a number,
you multiply the heat by that number:
2 H2O(g) H2(g) + O2(g) H =+571.0 kJ
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Why does it work?
You make the products, so you need
their heats of formation
You “unmake” the products so you have
to subtract their heats.
How do you get good at this?
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Applying Hess’s Law
2H2O2(l) → 2H2O(l) + O2(g)
Can be shown in two equations
a) 2H2(g)+O2(g) → 2H2O(l) ΔH = -572 kJ
b) H2(g) + O2(g) → H2O2(l) ΔH = -188 kJ
Reverse the equation b) to make H2O2 a
reactant and then double it.
c) 2H2O2(l)→2H2(g)+2O2(g) ΔH = 376 kJ
Note how ΔH= -188 kJ was doubled and its
sign was reversed.
Applying Hess’s Law
Adding equations a) and c) reactants and
products together yields:
2H2O2(l) + 2H2(g) + O2(g) →
2H2(g) + 2O2(g) + 2H2O(l)
Canceling like items of both sides reduces
the equation to:
2H2O2(l) → 2H2O(l) + O2(g)
Adding equations a) and c) enthalpies
yields:
376 kJ + -572 kJ = -196 kJ
Standard Heats of Formation
The H for a reaction that produces one mol of
a compound from its elements at standard
conditions
Standard conditions are 25°C and 1 atm.
0
The symbol is H f
The
standard heat of formation of an
element is zero H 0f element = 0
This
includes the diatomics like O2
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Using Standard Heat of Formation
Table 16.7, page 510 has standard heats of
formation for numerous compounds
The heat of a reaction can be calculated by
subtracting the heats of formation of the reactants
from the products
H
0
=
rxn
0
0
(H f Products) - (H f Reactants)
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Example Standard Heat of Formation
CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g)
0 CH (g) = - 74.81 kJ/mol
H f
4
0 O (g) = 0 kJ/mol
H f 2
0 CO (g) = - 393.5 kJ/mol
H f
2
0 H O(g) = - 241.8 kJ/mol
H f 2
H
0
=
rxn
[-393.5 + 2(-241.8)] - [-74.81 +2 (0)]
= - 802.3 kJ
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