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Chapter 6
Thermochemistry:
Energy Flow and Chemical Change
6-1
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Thermochemistry: Energy Flow and Chemical Change
6.1 Forms of Energy and Their Interconversion
6.2 Enthalpy: Heats of Reaction and Chemical Change
6.3 Calorimetry: Laboratory Measurement of Heats of Reaction
6.4 Stoichiometry of Thermochemical Equations
6.5 Hess’s Law of Heat Summation
6.6 Standard Heats of Reaction (DH0rxn)
6-2
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Thermodynamics is the study of heat and its transformations.
Thermochemistry is a branch of thermodynamics that deals with
the heat involved with chemical and physical changes.
Fundamental premise
When energy is transferred from one object to another,
it appears as work and/or as heat.
For our work we must define a system to study; everything else
then becomes the surroundings.
The system is composed of particles with their own internal energies (E or U).
Therefore the system has an internal energy. When a change occurs, the
internal energy changes.
6-3
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Figure 6.1
Energy diagrams for the transfer of internal energy (E)
between a system and its surroundings.
DE = Efinal - Einitial = Eproducts - Ereactants
6-4
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Figure 6.2
6-5
A system transferring energy as heat only.
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Figure 6.3
A system losing energy as work only.
Zn(s) + 2H+(aq) + 2Cl-(aq)
Energy, E
DE<0
work done on
surroundings
H2(g) + Zn2+(aq) + 2Cl-(aq)
6-6
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Table 6.1 The Sign Conventions* for q, w, and DE
q
+
w
=
DE
+
+
+
+
-
depends on sizes of q and w
-
+
depends on sizes of q and w
-
-
-
* For q: + means system gains heat; - means system loses heat.
* For w: + means word done on system; - means work done by system.
6-7
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DEuniverse = DEsystem + DEsurroundings
Units of Energy
Joule (J)
1 J = 1 kg*m2/s2
Calorie (cal)
1 cal = 4.18J
British Thermal Unit
6-8
1 Btu = 1055 J
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Sample Problem 6.1
PROBLEM:
PLAN:
System
When gasoline burns in a car engine, the heat released causes
the products CO2 and H2O to expand, which pushes the pistons
outward. Excess heat is removed by the car’s cooling system.
If the expanding gases do 451 J of work on the pistons and the
system loses 325 J to the surroundings as heat, calculate the
change in energy (DE) in J, kJ, and kcal.
Define system and surroundings, assign signs to q and w and calculate
DE. The answer should be converted from J to kJ and then to kcal.
SOLUTION:
q = - 325 J
DE = q + w =
-776J
6-9
Determining the Change in Internal Energy of a
kJ
103J
w = - 451 J
-325 J + (-451 J) = -776 J
= -0.776kJ
-0.776kJ
kcal
4.18kJ
= -0.185 kcal
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Figure 6.4
6-10
Two different paths for the energy change of a system.
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Figure 6.5
Pressure-volume
work.
6-11
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The Meaning of Enthalpy
w = - PDV
H = E + PV
DH ≈ DE in
1. Reactions that do not involve gases.
where H is enthalpy
DH = DE + PDV
qp = DE + PDV = DH
6-12
2. Reactions in which the number of
moles of gas does not change.
3. Reactions in which the number of
moles of gas does change but q is >>>
PDV.
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Figure 6.6
Enthalpy diagrams for exothermic and endothermic processes.
CH4(g) + 2O2(g)
CO2(g) + 2H2O(g)
H2O(l)
CH4 + 2O2
H2O(g)
DH < 0
heat out
Enthalpy, H
Enthalpy, H
Hinitial
DH > 0
CO2 + 2H2O
H2O(l)
Hfinal
A
6-13
Exothermic process
B
H2O(g)
Hfinal
heat in
Hinitial
Endothermic process
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Sample Problem 6.2
PROBLEM:
PLAN:
Drawing Enthalpy Diagrams and Determining
the Sign of DH
In each of the following cases, determine the sign of DH, state
whether the reaction is exothermic or endothermic, and draw
and enthalpy diagram.
(a) H2(g) + 1/2O2(g)
H2O(l) + 285.8kJ
(b) 40.7kJ + H2O(l)
H2O(g)
Determine whether heat is a reactant or a product. As a reactant,
the products are at a higher energy and the reaction is
endothermic. The opposite is true for an exothermic reaction
SOLUTION: (a) The reaction is exothermic.
H2(g) + 1/2O2(g) (reactants)
EXOTHERMIC
H2O(l)
6-14
DH = -285.8kJ
(products)
(b) The reaction is endothermic.
H2O(g)
ENDOTHERMIC
H2O(l)
(products)
DH = +40.7kJ
(reactants)
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Table 6.2 Specific Heat Capacities of Some Elements, Compounds, and
Materials
Substance
Specific Heat
Capacity (J/g*K)
Elements
Substance
Materials
aluminum, Al
0.900
wood
1.76
graphite,C
0.711
cement
0.88
iron, Fe
0.450
glass
0.84
copper, Cu
0.387
granite
0.79
gold, Au
0.129
steel
0.45
Compounds
6-15
Specific Heat
Capacity (J/g*K)
water, H2O(l)
4.184
ethyl alcohol, C2H5OH(l)
2.46
ethylene glycol, (CH2OH)2(l)
2.42
carbon tetrachloride, CCl4(l)
0.864
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Sample Problem 6.3
PROBLEM:
Finding the Quantity of Heat from Specific Heat
Capacity
A layer of copper welded to the bottom of a skillet weighs 125 g.
How much heat is needed to raise the temperature of the
copper layer from 250C to 300.0C? The specific heat capacity
(c) of Cu is 0.387 J/g*K.
PLAN: Given the mass, specific heat capacity and change in temperature, we
can use q = c x mass x DT to find the answer. DT in 0C is the same as for
K.
SOLUTION:
6-16
q=
0.387 J
g*K
x 125 g x (300-25)0C = 1.33x104 J
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Figure 6.7
Coffee-cup calorimeter.
6-17
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Sample Problem 6.4
Determining the Heat of a Reaction
PROBLEM: You place 50.0 mL of 0.500 M NaOH in a coffee-cup calorimeter at
25.000C and carefully add 25.0 mL of 0.500 M HCl, also at 25.000C.
After stirring, the final temperature is 27.210C. Calculate qsoln (in J) and
DHrxn (in kJ/mol). (Assume the total volume is the sum of the individual
volumes and that the final solution has the same density and specfic
heat capacity as water: d = 1.00 g/mL and c = 4.18 J/g*K)
PLAN:
6-18
We need to determine the limiting reactant from the net ionic equation.
The moles of NaOH and HCl as well as the total volume can be
calculated. From the volume we use density to find the mass of the
water formed. At this point qsoln can be calculated using the mass, c, and
DT. The heat divided by the M of water will give us the heat per mole of
water formed.
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Sample Problem 6.4
Determining the Heat of a Reaction
continued
SOLUTION:
HCl(aq) + NaOH(aq)
H+(aq) + OH-(aq)
NaCl(aq) + H2O(l)
H2O(l)
For NaOH
0.500 M x 0.0500 L = 0.0250 mol OH-
For HCl
0.500 M x 0.0250 L = 0.0125 mol H+
HCl is the limiting reactant.
0.0125 mol of H2O will form during the rxn.
total volume after mixing = 0.0750 L
0.0750 L x 103 mL/L x 1.00 g/mL = 75.0 g of water
q = mass x specific heat x DT
= 75.0 g x 4.18 J/g*0C x (27.21-25.00)0C
= 693 J
(693 J/0.0125 mol H2O)(kJ/103 J) = 55.4 kJ/ mol H2O formed
6-19
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Figure 6.8
A bomb calorimeter
6-20
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Sample Problem 6.5
A manufacturer claims that its new dietetic dessert has “fewer
than 10 Calories per serving.” To test the claim, a chemist at the
Department of Consumer Affairs places one serving in a bomb
calorimeter and burns it in O2(the heat capacity of the
calorimeter = 8.15 kJ/K). The temperature increases 4.9370C.
Is the manufacturer’s claim correct?
PROBLEM:
PLAN:
Calculating the Heat of Combustion
- q sample = qcalorimeter
SOLUTION:
qcalorimeter
= heat capacity x DT
= 8.151 kJ/K x 4.937 K
= 40.24 kJ
40.24 kJ
kcal
= 9.63 kcal or Calories
4.18 kJ
The manufacturer’s claim is true.
6-21
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Figure 6.9
AMOUNT (mol)
of compound A
Summary of the relationship between
amount (mol) of substance and the heat
(kJ) transferred during a reaction.
AMOUNT (mol)
of compound B
molar ratio from
balanced equation
HEAT (kJ)
DHrxn (kJ/mol)
6-22
gained or lost
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Sample Problem 6.6
PROBLEM:
Using the Heat of Reaction (DHrxn) to Find
Amounts
The major source of aluminum in the world is bauxite (mostly
aluminum oxide). Its thermal decomposition can be represented by
Al2O3(s)
2Al(s) + 3/2O2(g)
DHrxn = 1676 kJ
If aluminum is produced this way, how many grams of aluminum can
form when 1.000x103 kJ of heat is transferred?
PLAN:
SOLUTION:
heat(kJ)
1676kJ=2mol Al
mol of Al
xM
g of Al
6-23
1.000x103 kJ x
2 mol Al
26.98 g Al
1676 kJ
1 mol Al
= 32.20 g Al
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Sample Problem 6.7
PROBLEM:
Using Hess’s Law to Calculate an Unknown DH
Two gaseous pollutants that form in auto exhaust are CO and
NO. An environmental chemist is studying ways to convert them
to less harmful gases through the following equation:
CO(g) + NO(g)
CO2(g) + 1/2N2(g)
DH = ?
Given the following information, calculate the unknown DH:
Equation A: CO(g) + 1/2O2(g)
Equation B: N2(g) + O2(g)
PLAN:
2NO(g) DHB = 180.6 kJ
Equations A and B have to be manipulated by reversal and/or
multiplication by factors in order to sum to the first, or target, equation.
SOLUTION:
Multiply Equation B by 1/2 and reverse it.
CO(g) + 1/2O2(g)
NO(g)
CO(g) + NO(g)
6-24
CO2(g) DHA = -283.0 kJ
CO2(g) DHA = -283.0 kJ
1/2N2(g) + 1/2O2(g)
CO2(g) + 1/2N2(g)
DHB = -90.3 kJ
DHrxn = -373.3 kJ
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Table 6.3 Selected Standard Heats of Formation at 250C(298K)
Formula
calcium
Ca(s)
CaO(s)
CaCO3(s)
DH0f(kJ/mol)
0
-635.1
-1206.9
carbon
C(graphite)
C(diamond)
CO(g)
CO2(g)
CH4(g)
CH3OH(l)
HCN(g)
CSs(l)
chlorine
Cl(g)
6-25
0
1.9
-110.5
-393.5
-74.9
-238.6
135
87.9
121.0
Formula DH0f(kJ/mol)
Formula DH0f(kJ/mol)
0
-92.3
silver
Ag(s)
AgCl(s)
hydrogen
H(g)
H2(g)
218
0
sodium
nitrogen
N2(g)
NH3(g)
NO(g)
0
-45.9
90.3
oxygen
O2(g)
O3(g)
H2O(g)
0
143
-241.8
H2O(l)
-285.8
Cl2(g)
HCl(g)
Na(s)
Na(g)
NaCl(s)
0
-127.0
0
107.8
-411.1
sulfur
S8(rhombic) 0
S8(monoclinic) 2
SO2(g)
-296.8
SO3(g)
-396.0
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Sample Problem 6.8
PROBLEM:
Writing Formation Equations
Write balanced equations for the formation of 1 mol of the following
compounds from their elements in their standard states and include
DH0f.
(a) Silver chloride, AgCl, a solid at standard conditions.
(b) Calcium carbonate, CaCO3, a solid at standard conditions.
(c) Hydrogen cyanide, HCN, a gas at standard conditions.
PLAN:
Use the table of heats of formation for values.
SOLUTION:
(a) Ag(s) + 1/2Cl2(g)
(b) Ca(s) + C(graphite) + 3/2O2(g)
(c) 1/2H2(g) + C(graphite) + 1/2N2(g)
6-26
DH0f = -127.0 kJ
AgCl(s)
CaCO3(s)
HCN(g)
DH0f = -1206.9 kJ
DH0f = 135 kJ
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Figure 6.10
Elements
-DH0f
formation
Reactants
decomposition
Enthalpy, H
The general process for determining DH0rxn from DH0f values.
DH0f
Hinitial
DH0rxn
Products
DH0rxn = S mDH0f(products) - S nDH0f(reactants)
6-27
Hfinal
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Sample Problem 6.9
Calculating the Heat of Reaction from Heats of
Formation
Nitric acid, whose worldwide annual production is about 8 billion
kilograms, is used to make many products, including fertilizer, dyes,
and explosives. The first step in the industrial production process is
the oxidation of ammonia:
PROBLEM:
4NH3(g) + 5O2(g)
4NO(g) + 6H2O(g)
Calculate DH0rxn from DH0f values.
PLAN:
Look up the DH0f values and use Hess’s Law to find DHrxn.
SOLUTION:
DHrxn = S mDH0f (products) - S nDH0f (reactants)
DHrxn = [4(DH0f NO(g) + 6(DH0f H2O(g)] - [4(DH0f NH3(g) + 5(DH0f O2(g)]
= (4 mol)(90.3 kJ/mol) + (6 mol)(-241.8 kJ/mol) [(4 mol)(-45.9 kJ/mol) + (5 mol)(0 kJ/mol)]
DHrxn = -906 kJ
6-28
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Figure 6.11
The trapping of heat by the atmosphere.
6-29