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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