Chapter 5: Thermochemistry

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Chapter 5 Thermochemistry
The Study of Energy and its Transformations
The Nature of Energy
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Chapter 5 Thermochemistry
(1) Kinetic Energy (Ek)
Energy of motion a moving object has kinetic
energy

• moving car

• molecules and atoms are constantly moving
  (thermal motion)

speed (velocity)
mass
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Chapter 5 Thermochemistry
(1) Potential Energy (Ep)
Energy an object has due to its position relative
to other objects

• potential energy due to gravity acting on the
  object
• (diver jumping off a diving board)

• potential energy due to electrostatic attraction
  between
• two charged objects

charge
distance between charges

• potential energy due to chemical energy stored
  in bonds of
• molecules

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Chapter 5 Thermochemistry
Different forms of energy are interconvertible
Potential Energy
Kinetic Energy
Chemical Energy
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HW 11a,c
Chapter 5 Thermochemistry
The Units of Energy
A person (50kg) moving at a speed of 1m/s has a
kinetic energy
1kJ 1000 J
1cal 4.184 J
A calorie is an older energy unit
1Cal 1kcal 1000cal
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Chapter 5 Thermochemistry
The System and its Surroundings
Energy/Heat is transferred between regions of
the universe
Surroundings
Region studied system
Molecules cannot escape closed system
System
Everything else surroundings
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Chapter 5 Thermochemistry
The System and its Surroundings
energy
2 H2 O2 ? 2 H2O
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Chapter 5 Thermochemistry
The System and its Surroundings
Energy can be transferred from the system to the
surroundings
Energy can also transferred from the surroundings
to the system
OR
Heat is transferred from the hotter to the colder
object
until
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Chapter 5 Thermochemistry
Energy can be transferred as heat (q)
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Chapter 5 Thermochemistry
First law of Thermodynamics
Energy is conserved

• energy is neither created nor destroyed, it can
  only be converted from one form into another

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Chapter 5 Thermochemistry
Internal Energy

• The internal energy of a system (E) is the sum
  of all kinetic and potential energies

• When a system undergoes change (D), the internal
  energy of the system may change by DE

DE Efinal - Einitial
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Chapter 5 Thermochemistry
Internal Energy
Heat (q) transferred from system to surroundings
Heat (q) transferred from surroundings to system
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Chapter 5 Thermochemistry
Internal Energy
E
final state
Einitial
Efinal
E absorbed
initial state
Einitial
Efinal
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Chapter 5 Thermochemistry
Internal Energy sign conventions
System
Heat (q) transferred TO system or work (w) done
TO system positive sign q, w gt DE gt 0
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Chapter 5 Thermochemistry
Internal Energy sign conventions
DE q w
System
Heat (q) transferred TO surroundings or work (w)
done BY system negative sign -q, -w
gt DE lt 0
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Chapter 5 Thermochemistry
DE q w
Can the sign of DE be predicted for the following
experiment?
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Chapter 5 Thermochemistry
What is the change in internal energy, DE, of a
system that gains 150 J of heat and does 432 J of
work on the surroundings?
-282 J
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Chapter 5 Thermochemistry
Exothermic / Endothermic Processes
Endothermic process

• heat flows into the system

Exothermic process

• heat flows out of the system

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Chapter 5 Thermochemistry
Exothermic / Endothermic Processes
H2O (l)
H2O (g)
Water vapor
Water
H2O (l)
H2O (s)
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Chapter 5 Thermochemistry
Enthalpy (H)
Enthalpy measures the amount of heat exchanged at
constant pressure
DH qp

• Enthalpy is an EXTENSIVE property

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Chapter 5 Thermochemistry
Enthalpy (H) is an extensive property
lots of O2
CO2
heat
H2O
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Chapter 5 Thermochemistry
Exothermic / Endothermic Processes
H2O (l)
H2O (g)
Water vapor
Water
H2O (l)
H2O (s)
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Chapter 5 Thermochemistry
Enthalpies of Reaction
DH Hproducts - Hreactants
2 H2 (g) O2 (g) ? 2 H2O (l)
DH -483.6 kJ
How much heat (q) is given off by reacting 3.4 g
of H2 gas?
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Chapter 5 Thermochemistry
2 H2 (g) O2 (g) ? 2 H2O (l)
DH -483.6 kJ
H
2 H2 (g) O2 (g)
DH lt 0
DH gt 0
2 H2O (l)
DH 483.6 kJ
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Chapter 5 Thermochemistry
2 H2 (g) O2 (g) ? 2 H2O (l)
DH -483.6 kJ
DH 483.6 kJ
DH is equal in magnitude, but opposite in sign,
to DH for the reverse reaction
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Chapter 5 Thermochemistry
Calorimetry
is a way to measure the amount of heat given
off or absorbed in the course of a reaction
dissolves
heat (q)
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Chapter 5 Thermochemistry
Calorimetry
when the reaction occurs in a container that is
thermally insulated
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Chapter 5 Thermochemistry
Calorimetry
The temperature change that a substance undergoes
when it absorbs heat depends on its specific heat
(s)
1 g of H2O 11oC
1 g of H2O 12oC
4.184 J of heat
1 cal of heat
The specific heat (s) is the amount of heat
required to heat 1 g of a substance by 1K (or 1oC)
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Chapter 5 Thermochemistry
Calorimetry
The temperature change that a substance undergoes
when it absorbs heat depends on its specific heat
(s)
temperature change
mass
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Chapter 5 Thermochemistry
Calorimetry
Coffee-cup Calorimeter
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Chapter 5 Thermochemistry
The temperature of the aqueous NaOH solution
(250g) was found to have risen from room
temperature (21oC) to 30oC. How much heat was
transferred? the specific heat of water is 4.184
J/g-K.
DT 30oC 21oC 9oC
m 250 g
9K
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Chapter 5 Thermochemistry
The temperature of the aqueous NaOH solution
(250g) was found to have risen from room
temperature (21oC) to 30oC. How much heat was
transferred? The specific heat of water is 4.184
J/g-K.
were transferred was the reaction exo- or
endothermic?
NaOH (s) ? Na(aq) OH- (aq)
heat
The reaction gives off heat gt exothermic
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Chapter 5 Thermochemistry
NaOH (s) ? Na(aq) OH- (aq)
heat
The reaction gives off heat gt exothermic
qsolution in calorimter -qreaction
DH qreaction -9.41 kJ
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Chapter 5 Thermochemistry
Calorimetry
Temperature increase in calorimeter ? reaction
is exothermic ? DH negative
Temperature decrease in calorimeter ? reaction
is endothermic ? DH positive
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Chapter 5 Thermochemistry
Calorimetry
Where are the system and the surroundings in a
thermally insulated calorimeter?
- both are part of the solution inside the
calorimeter!
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Chapter 5 Thermochemistry
Which of the following substances requires the
smallest amount of energy to increase the
temperature of 50.0 g of that substance by 10K?
Specific Heat
CH4 (g) 2.20 J/g-K
Hg (l) 0.14 J/g-K
H2O (l) 4.18 J/g-K
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Chapter 5 Thermochemistry
Hesss law
You want to convert water vapor into ice at a
constant temperature. You can do this in two
steps first, convert H2O into liquid water,
then, in a second step, into ice
H2O (l)
H2O (g)
DH1 -44 kJ
H2O (l)
H2O (s)
DH1 -6.0 kJ
add
H2O (g)
H2O (l)
H2O (l)
H2O (s)
DH DH1 DH2
H2O (g)
H2O (s)
net
-44kJ (-6.0kJ) -50 kJ
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Chapter 5 Thermochemistry
Hesss law
If a reaction is carried out in a series of
steps, DH for the overall reaction will equal the
sum of the enthalpy changes for the individual
steps
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Chapter 5 Thermochemistry
Hesss law
H
H2O (g)
-44kJ
H2O (l)
-6 kJ
H2O (s)
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Chapter 5 Thermochemistry
From the enthalpies of reaction
2 H2 (g) O2 (g) ? 2 H2O (g)
DH -483.6 kJ
a)
3 O2 (g) ? 2 O3 (g)
DH 284.6 kJ
b)
calculate the heat of the reaction
3 H2 (g) O3 (g) ? 3 H2O (g)
Strategy Find two equations, the addition of
which will give you the final reaction
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Chapter 5 Thermochemistry
2 H2 (g) O2 (g) ? 2 H2O (g)
DH -483.6 kJ
a)
3 H2 (g) O3 (g) ? 3 H2O (g)
(1) start by considering the first reactant find
an equation where H2 appears
- reaction a).
Then, multiply reaction a) so that the
coefficients match the final equation
( 2 H2 (g) O2 (g) ? 2 H2O (g)
DH -483.6 kJ )
3 H2 (g) 3/2 O2 (g) ? 3 H2O (g)
DH -725.4 kJ
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Chapter 5 Thermochemistry
3 O2 (g) ? 2 O3 (g)
DH 284.6 kJ
b)
2 O3 (g) ? 3 O2 (g)
DH -284.6 kJ
b) inverted
3 H2 (g) O3 (g) ? 3 H2O (g)
(2) Consider the second reactant we need to
introduce O3 to get to the final equation.
Find equation where O3 appears rxn. b).
O3 appears on the product side invert it to get
O3 on the reactant side
multiply b) inverted so that the coefficients
match
(2 O3 (g) ? 3 O2 (g)
DH -284.6 kJ) x 1/2
O3 (g) ? 3/2 O2 (g)
DH -142.3 kJ
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Chapter 5 Thermochemistry
3) Now add the two partial reactions - and their
DHs - and check whether you get the overall
reaction
3 H2 (g) 3/2 O2 (g) ? 3 H2O (g)
DH -725.4 kJ
O3 (g) ? 3/2 O2 (g)
DH -142.3 kJ
add
3 H2 (g) O3 (g)
? 3 H2O (g)
DH -867.7 kJ
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Chapter 5 Thermochemistry
Calculate the enthalpy change for the reaction A
2B ? D3 DH ?
Given the following enthalpies of reaction a) A
B ? C DH -125 kJ
b) D3 ? C B DH -41 kJ
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Chapter 5 Thermochemistry
Calculate the enthalpy change for the reaction A
2B ? D3 DH ?
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Chapter 5 Thermochemistry
Calculate the enthalpy change for the reaction A
2B ? D3 DH ?
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Chapter 5 Thermochemistry
Calculate the enthalpy change for the reaction A
2B ? D3 DH ?
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Chapter 5 Thermochemistry
Calculate the enthalpy change for the
reaction P4O6 (s) 2 O2(g) ? P4O10
(s) DH ?
Given the following enthalpies of reaction a)
P4 (s) 3 O2(g) ? P4O6 (s) DH -1640.1
kJ
b) P4 (s) 5 O2(g) ? P4O10 (s) DH
-2940.1 kJ
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Chapter 5 Thermochemistry
P4O6 (s) 2 O2(g) ? P4O10 (s) DH ?
Where in the following reactions does P4O6
appear? a) P4 (s) 3 O2(g) ? P4O6
(s) DH -1640.1 kJ
b) P4 (s) 5 O2(g) ? P4O10 (s) DH
-2940.1 kJ
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Chapter 5 Thermochemistry
P4O6 (s) 2 O2(g) ? P4O10 (s) DH ?
Where in the following reactions does P4O6
appear? a) P4 (s) 3 O2(g) ? P4O6
(s) DH -1640.1 kJ
b) P4 (s) 5 O2(g) ? P4O10 (s) DH
-2940.1 kJ
- it appears on the product side, so a) needs to
be inverted
a) inv. P4O6 (s) ? P4 (s) 3 O2(g)
DH 1640.1 kJ
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Chapter 5 Thermochemistry
P4O6 (s) ? P4 (s) 3 O2(g) DH
1640.1 kJ
II) O2 needs to be introduced on the reactant
side, P4O10 on the product side, and P4 needs to
be eliminated - equation b)
b) P4 (s) 5 O2(g) ? P4O10 (s) DH
-2940.1 kJ
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Chapter 5 Thermochemistry
P4O6 (s) ? P4 (s) 3 O2(g) DH
1640.1 kJ
P4 (s) 5 O2(g) ? P4O10 (s) DH
-2940.1 kJ
add
P4O6 (s) P4(s) 5 O2(g) ? P4(s) 3
O2(g) P4O10 (s)
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Chapter 5 Thermochemistry
P4O6 (s) ? P4 (s) 3 O2(g) DH
1640.1 kJ
P4 (s) 5 O2(g) ? P4O10 (s) DH
-2940.1 kJ
add
P4O6 (s) P4(s) 2 O2(g) ? P4(s) 3
O2(g) P4O10 (s)
P4O6 (s) 2 O2(g) ? P4O10 (s)
DH -1300.0 kJ
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Chapter 5 Thermochemistry
Enthalpy of Formation, DHof,
... is the DH for the reaction that forms one
mole of that compound from its elements, with all
substances in their standard states 25oC, 1atm
H2 (g) ½ O2 (g) ? H2O
(l) DHof -285.8 kJ
½ N2 (g) ½ O2 (g) ? NO (g) DHof
90.4 kJ
2 C (s) 3 H2 (g) ½ O2 (g) ? C2H5OH (l)
DHof -277.7 kJ
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Chapter 5 Thermochemistry
Enthalpy of Formation, DHof
DHof of the most stable form of any element is
zero (it does not need to be formed)
½ O2 (g) ½ O2 (g) ? O2 (g) DHof 0
DHof for O2 (g) , N2 (g) , H2 (g), Br2 (l) etc.
0
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Chapter 5 Thermochemistry
For which of the following reactions would DH
represent a standard enthalpy of formation?
2 Na (s) ½ O2 (g) ? Na2O (s)
K (l) ½ Cl2 (g) ? KCl (s)
CO (g) ½ O2 (g) ? CO2 (g)
2 Na (s) Cl2 (g) ? 2 NaCl (s)
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Chapter 5 Thermochemistry
Enthalpies of formations can be used to calculate
enthalpies of reactions (under standard
conditions)
DHorxn sum of all DHof(products) - sum of all
DHof(reactants)
DHorxn S n DHof(products) S m DHof(reactants)
sum
moles of reactant
moles of product
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Chapter 5 Thermochemistry
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Chapter 5 Thermochemistry
Enthalpies of formations (DHof) can be used to
calculate enthalpies of reactions (DHorxn)
C3H8 (g) 5 O2 (g) ? 3 CO2 (g) 4
H2O (l)
DHorxn 3 x DHof CO2 (g) 4 x DHof H2O
(l)) - (DHof C3H8 (g) 0)
(3 x -394 kJ) (4 x -286 kJ) - -104 kJ
-1182 kJ - 1144 kJ 104 kJ -2222 kJ
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Chapter 5 Thermochemistry
What is DHo for the reaction below?
Mg(OH)2 (s) ? MgO (s) H2O (l)
DHorxn DHof MgO (s) DHof H2O (l) -
DHof Mg(OH)2 (s)
-601.8 kJ (-285.8 kJ) - -924.7 kJ
37.1 kJ
DHof MgO (s) -601.8 kJ DHof H2O (l)
-285.8 kJ DHof Mg(OH)2 (s) -924.7 kJ
DHof values are in Table 5.3 and Appendix C