Transcript Chapter 11 - Thermochemistry

Chapter 17 - Thermochemistry
Heat and Chemical Change
Energy and Heat
 Thermochemistry
- concerned
with heat changes that occur
during chemical reactions.
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Energy and Heat
Energy
- capacity for doing work
or supplying heat
• weightless, odorless, tasteless
• if within the chemical
substances - called chemical
potential energy
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Energy and Heat
 Gasoline
contains a significant
amount of chemical potential
energy
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Energy and 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
 Heat
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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
Exothermic and Endothermic
 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
Exothermic and Endothermic
 Together,
the system and it’s
surroundings constitute the
universe
 Thermochemistry is concerned
with the flow of heat from the
system to it’s surroundings, and
vice-versa.
• Figure 17.2, page 506
Law of Conservation of Energy
 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, or heat.
Endothermic
 Fig.
17.2a, p.506 - heat flowing
into a system from it’s
surroundings:
• defined as positive
• q has a positive value
• called endothermic
–system gains heat as the
surroundings cool down
Endothermic
Exothermic
 Fig.
17.2b, p.506 - heat flowing out of
a system into it’s surroundings:
• defined as negative
• q has a negative value
• called exothermic
–system loses heat as the
surroundings heat up
Exothermic
Exothermic 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.
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Heat Capacity and Specific Heat
• 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
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Heat Capacity and Specific Heat
 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”
• Note Table 17.1, page 508
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Heat Capacity and Specific Heat
Water has a HUGE value,
compared to other
chemicals
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Heat Capacity and Specific Heat
water, C = 4.18 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
q = mass (g) x T x C
 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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Example
When 435J of heat is added to 3.4 g of
olive oil at 21 °C, the temperature
increases to 85 °C. What is the specific
heat of olive oil?
q = m x T x C
C= q
m x T
C=
C=
q
m x T
435J
3.4g x (85°C - 21°C)
C = 2 J/g°C
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
 For
systems at constant pressure,
the heat content is the same as a
property called Enthalpy (H) of the
system.
Calorimeter
Calorimetry
in enthalpy = H
 q = H These terms will be
used interchangeably in this
textbook.
 Changes
Calorimetry
 Thus,
q = H = m x C x T
H
is negative for an
exothermic reaction.
H
is positive for an
endothermic reaction.
Calorimetry
 Calorimetry
experiments can
be performed at a constant
volume using a device called
a “bomb calorimeter” - a
closed system
Bomb Calorimeter
Calorimeter Problem
50ml of HCl
(in H2O)
Ti= 22.5°C
Calorimeter Problem
Add 50ml of NaOH
(in H2O)
Ti= 22.5°C
Calorimeter Problem
Chemical
Reaction
Calorimeter Problem
100ml of HCl and NaOH
(in H2O)
Tf= 26°C
Calorimeter Problem
1ml of H2O = 1g
100ml of H2O = 100g
m = 100g
Ti= 22.5°C
Tf= 26°C
CH20= 4.18J/g°C
Calorimeter Problem
H = m x C x T
H = m x C x (Tf –Ti)
= 100g x 4.18J/g°C x (26°C - 22.5°C)
H = 1460J = 1.46kJ
Energy
C + O2  CO2 + 395 kJ
C + O2
395kJ
C O2
Reactants

Products
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In terms of bonds
O
O
C
O
C
O
Breaking this bond will require energy.
O
C
O
O C 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
CaCO
 CaO
CaCO
CaO
+ CO+2 CO2
3 + 176
3 kJ
CaO + CO2
176 kJ
CaCO3
Reactants

Products
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Endothermic
The
products are higher in
energy than the reactants
Absorbs energy
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Chemistry Happens in
MOLES
 An
equation that includes
energy is called a
thermochemical equation
CH4+2O2CO2+2H2O + 802.2 kJ
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CH4+2O2CO2+2H2O + 802.2kJ
1
mole of CH4 releases 802.2kJ
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, exactly as
written
• The physical state of reactants
and products must also be given.
• Standard conditions for the
reaction is 101.3 kPa (1 atm.)
and 25 oC
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CH4+2O2  CO2 + 2 H2O + 802.2 kJ
 If
10. 3 grams of CH4 are burned
completely, how much heat will be
produced?
10.3g CH4
1 mol CH4
802.2 kJ
16.05g CH4 1mol CH4
ΔH = 514 kJ
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CH4+2O2  CO2+2H2O + 802.2 kJ
 How many liters of O2 at STP
would be required to produce 23
kJ of heat?
22.4L x 2 = 44.8L produces 802.2kJ
23kJ
= 0.029 0.029 x 44.8L =
802.2kJ
VO2 = 1.30 L
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CH4+2O2  CO2+2H2O+802.2 kJ
 How
many grams of water would
be produced with 506 kJ of heat?
2 mol of H2O produced with 802.2kJ
2 mol of H2O is 36g
506kJ
x 36g =
802.2kJ
22.71g of H2O
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Heat of Combustion
 The
heat from the reaction that
completely burns 1 mole of a
substance
 Note Table 17.2 page 517
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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
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Heat of Reaction
 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 = -285.8 kJ
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Section 17.3
Heat in Changes of State
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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
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Heats of Fusion and
Solidification
 Molar
Heat of Solidification
(Hsolid) - heat lost when one
mole of liquid solidifies
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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 17.3, page 522
 Sample Problem 17-4, page 421
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Sample Problem
How many kJs of heat are required to
melt a 10.0g popsicle at 0°C? Assume
the popsicle has the same molar mass
and heat of fusion as water.
It takes 6.01kJ to melt 1 mol of ice.
There are 18g in 1 mol of ice.
10g
x 6.01kJ/mol = 3.34 kJ
18g/mol
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.
 Table 17.3, page 521
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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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ΔHfusion
ΔHvaporization
Heats of Vaporization and
Condensation
 The 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!
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Heats of Vaporization and
Condensation
H20(l)  H20(g) m = 63.7g of liquid
Hvap = ?
63.7g
x 40.7kJ/mol
18g/mol
= 144kJ
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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 dissolution of one
mole of substance
 Sodium hydroxide provides a good
example of an exothermic molar heat
of solution:
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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!
 Sample Problem 17-7, page 526
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Section 17.4
Calculating Heat Changes
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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
 Example shown on page 529 for
graphite and diamonds
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Diamond to Graphite
C(s,diamond)

C(s,graphite)
C(s,diamond) + O2  CO2 ΔH=-393.5kJ
C(s,graphite) + O2  CO2 ΔH=-395.4kJ
CO2  C(s,graphite) + O2 ΔH=395.45kJ
C(s,diamond) + O2  CO2 ΔH=-393.5kJ
CO2  C(s,graphite) + O2 ΔH=395.45kJ
+
C(s,diam)  C(s,graph)
ΔH=-1.9kJ
Why Does It Work?
 If
you turn an equation around,
you change the sign:
H2(g)+1/2 O2(g)H2O(g) H=-285.5 kJ
H2O(g)H2(g)+1/2 O2(g)H =+285.5 kJ
 If
you multiply the equation by a
number, you multiply the heat by
that number:
2H2O(g)H2(g)+O2(g) H =+571.0 kJ
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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Standard Heats of Formation
H for a reaction that
produces 1 mol of a compound
from its elements at standard
conditions
 Standard conditions: 25°C and 1
atm.
 The
 Symbol
is ΔHf
0
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Standard Heats of Formation
The
standard heat of
formation of an element = 0
This
includes the diatomics
O2 N2
Cl2 etc……
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What good are they?
 Table
17.4, page 530 has
standard heats of formation
 The heat of a reaction can be
calculated by:
• subtracting the heats of
formation of the reactants
from the products
Ho = (H 0f Products) - (H 0f Reactants)
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CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)
0
H f CH4 (g) = - 74.86 kJ/mol
0
H f O2(g) = 0 kJ/mol
0 CO (g) = - 393.5 kJ/mol
H f
2
0 H2O(g) = - 241.8 kJ/mol
H f
H= [-393.5+2(-241.8)]-[-74.68+2 (0)]
H= - 802.4 kJ
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Examples
 Sample
Problem 17-8, page 531
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