Transcript Chapter 6 6-1 Thermochemistry: Energy Flow and Chemical Change

Chapter 6
Thermochemistry: Energy Flow and Chemical Change
6-1
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 (DHorxn)
6-2
Some Definitions
System: that part of the universe whose change we are going
to observe
Surroundings: everything else relevant to the change
INTERNAL ENERGY, E
Each particle in a system has potential and kinetic energy;
the sum of these energies for all particles in a system is
the internal energy, E.
In a chemical reaction: when reactants are converted to
products, E changes (DE).
6-3
DE = Efinal - Einitial = Eproducts - Ereactants
Energy diagrams for the transfer of internal energy (E)
between a system and its surroundings
Figure 6.2
6-4
DE = q + w
where q = heat and w = work
A system transferring energy as heat only
Figure 6.3
6-5
Sign Conventions
The numerical values of q and w can be either positive or negative,
depending on the change the system undergoes.
Energy coming into the system is positive; energy going out from the
system is negative.
6-6
A system losing energy as work only
Pressure-volume
work (PV work)
Figure 6.4
6-7
Table 6.1
q
Sign Conventions for q, w and DE
+
w
=
DE
+
+
+
+
-
depends on magnitudes of q
and w
-
+
depends on magnitudes of q
and w
-
-
-
For q: (+) means system gains heat, (-) means system loses heat.
For w: (+) means work done on system, (-) means work done by system.
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Law of Conservation of Energy
(First Law of Thermodynamics)
DEuniverse =
DEsystem + DEsurroundings = 0
Units of Energy
joule (J)
1 J = 1 kg m2/s2
calorie (cal)
1 cal = 4.184 J
British Thermal Unit
6-9
1 Btu = 1055 J
Some quantities of energy
Figure 6.5
6-10
Sample Problem 6.1
Determining the Change in Internal Energy E of
a System
PROBLEM:
PLAN:
When gasoline burns in a car engine, the heat released causes
the combustion 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 the system and the surroundings, assign signs to q and w, and
calculate DE. The answer in units of J is converted to kJ and to kcal.
SOLUTION:
q = (-) 325 J (system loses heat)
w = (-) 451 J (system does work)
DE = q + w = -325 J + (-451 J) = -776 J
-776 J x
6-11
kJ
103J
= -0.776 kJ
-0.776 kJ x
kcal
4.184 kJ
= -0.185 kcal
E is a state function - depends on current state of the system,
not on the path taken to reach that state.
q and w are not state functions but DE is path independent
Two different paths for the energy change of a system
Figure 6.6
6-12
Pressure-volume
work
In this case, the system
does PV work on the
surroundings; thus w is
negative in sign.
Figure 6.7
6-13
For reactions that occur at constant pressure….
The Definition of Enthalpy
DH ≈ DE for:
w = - PDV
1. Reactions that do not involve gases.
DH = DE + PDV
2. Reactions in which the number of
moles of gas does not change.
qp = DE + PDV = DH
3. Reactions in which the number of
moles of gas changes but
q is >>> PDV.
H is a state function.
6-14
Enthalpy diagrams for exothermic and endothermic processes
CH4(g) + 2O2(g)
CO2(g) + 2H2O(g) + heat
heat + H2O(s)
Heat is released;
enthalpy decreases.
6-15
H2O(l)
Heat is absorbed;
enthalpy increases.
Figure 6.8
Sample Problem 6.2
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 an
enthalpy diagram.
PROBLEM:
PLAN:
Drawing Enthalpy Diagrams and Determining
(a) H2(g) + 1/2O2(g)
H2O(l) + 285.8 kJ
(b) 40.7 kJ + 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)
DH = -285.8 kJ
EXOTHERMIC
H2O(l)
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(products)
(b) The reaction is endothermic.
H2O(g)
ENDOTHERMIC
H2O(l)
(products)
DH = +40.7 kJ
(reactants)
Some Important Types of Enthalpy Change
heat of combustion (DHcomb)
1C4H10(l) + 13/2O2(g)
4CO2(g) + 5H2O(g)
heat of formation (DHf)
K(s) + 1/2Br2(l)
1KBr(s)
Standard quantity
of either reactant
or product: 1 mol
heat of fusion (DHfus)
1NaCl(s)
NaCl(l)
heat of vaporization (DHvap)
1C6H6(l)
6-17
C6H6(g)
Components of Internal Energy
Contributions to the kinetic energy:
• The molecule moving through space, Ek(translation)
• The molecule rotating, Ek(rotation)
• The bound atoms vibrating, Ek(vibration)
• The electrons moving within each atom, Ek(electron)
Contributions to the potential energy:
• Forces between the bound atoms vibrating, Ep(vibration)
• Forces between nucleus and electrons and between electrons in each
atom, Ep(atom)
• Forces between the protons and neutrons in each nucleus, Ep(nuclei)
• Forces between nuclei and shared electron pair in each bond, Ep(bond)
6-18
Figure 6.9
Components of internal
energy (E)
6-19
H2(g) + F2(g)
2HF(g) + 546 kJ
(exothermic)
Bonds absorb energy when they break and release energy when they form.
Weaker bonds = easier to break = higher in energy (less stable, more reactive)
6-20
Where does the heat of reaction come from?
The energy released or absorbed during a chemical reaction
is due to differences between the strengths of reactant bonds
and product bonds.
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Table 6.2
Heats of Combustion (DHcomb) of Some Carbon Compounds
Two-Carbon Compounds
One-Carbon Compounds
Ethane (C2H6) Ethanol (C2H6O) Methane (CH4) Methanol (CH4O)
structural
formulas
H
H
H C
C
H
H
H
H
H
H C
C
H
H
H
O
H
H C
H
H
H
H C
O
H
sum of C-C and
C-H bonds
7
6
4
3
sum of C-O and
O-H bonds
0
2
0
2
DHcomb(kJ/mol)
-1560
-1367
-890
-727
-51.88
-29.67
-55.5
-22.7
DHcomb(kJ/g)
6-22
H
Table 6.3 Heats of Combustion of Some Fats and Carbohydrates
Substance
DHcomb(kJ/g)
Fats
vegetable oil
-37.0
margarine
-30.1
butter
-30.0
Carbohydrates
6-23
table sugar (sucrose)
-16.2
brown rice
-14.9
maple syrup
-10.4
Ball-and-stick molecular models of a carbohydrate
(sucrose) and a triglyceride (triolean)
A more “oxidized” fuel
6-24
A more “reduced” fuel
Calorimetry = laboratory measurement of heats of reaction
heat capacity = an object’s capacity to absorb heat = the quantity
of heat required to change its temperature by 1 K
q = constant x DT (the constant of proportionality is equal
to the heat capacity; in units of J/K)
specific heat capacity (c) = the quantity of heat required to change
the temperature of 1 gram of a substance by 1 K (in units of J/g.K)
q = c x mass x DT
molar heat capacity (C) = the quantity of heat required to change
the temperature of 1 mole of a substance by 1 K (in units of J/mol.K)
6-25
Table 6.4
Specific Heat Capacities of Some Elements,
Compounds and Materials
Substance
Specific Heat
Capacity (J/g*K)
Substance
Elements
Materials
aluminum (Al)
0.900
graphite (C)
iron (Fe)
0.711
0.450
copper (Cu)
0.387
gold (Au)
0.129
Compounds
6-26
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
wood
cement
1.76
0.88
glass
granite
steel
0.84
0.79
0.45
Sample Problem 6.3
Calculating the Quantity of Heat from the
Specific Heat Capacity
PROBLEM:
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 25 oC to 300. oC? 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 oC and K are the
same.
SOLUTION:
6-27
q=
0.387 J
g.K
x 125 g x (300-25)oC = 1.33 x 104 J
Calorimeter. A device
used to measure the
heat released or absorbed
by a physical or chemical
process.
A coffee-cup
calorimeter
(constant P)
-qsolid = qwater
Figure 6.10
6-28
Sample Problem 6.4
PROBLEM:
PLAN:
Determining the Specific Heat Capacity of a
Solid
A 25.64 g sample of a solid was heated in a test tube to 100.00 oC
in boiling water and carefully added to a coffee-cup calorimeter
containing 50.00 g of water. The water temperature increased from
25.10 oC to 28.49 oC. What is the specific heat capacity of the
solid? (Assume all the heat is gained by the water)
It is helpful to use a table to summarize the data given. Then work the
problem realizing that heat lost by the system must be equal to that
gained by the surroundings.
SOLUTION:
mass (g)
c (J/g.K)
solid
25.64
?
H2O
50.00
4.184
25.64 g x c x
csolid = - 50.00 g x
4.184 J/g.K x 3.39 K
Tfinal
100.00 28.49
25.10
-71.51 K = - 50.00 g x
25.64 g x -71.51 K
6-29
Tinitial
=
DT
-71.51
28.49
3.39
4.184 J/g.K x 3.39 K
0.387 J/g.K
Calorimetry at
Constant Volume
A bomb calorimeter:
used to measure qv
Figure 6.11
6-30
Sample Problem 6.5
PROBLEM:
PLAN:
SOLUTION:
Calculating the Heat of Combustion
A manufacturer claims that its new diet dessert has “fewer than
10 Calories (10 kcal) 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 = 8.151 kJ/K). The temperature increases by 4.937
oC. Is the manufacturer’s claim correct?
- qsample = qcalorimeter
= heat capacity x DT
qcalorimeter
= 8.151 kJ/K x 4.937 K
= 40.24 kJ
40.24 kJ x
kcal
= 9.62 kcal < 10 Calories = 10 kcal
4.184 kJ
The manufacturer’s claim is correct.
6-31
AMOUNT (mol)
Summary of the relationship between
amount (mol) of substance and the heat
(kJ) transferred during a reaction
of compound A
AMOUNT (mol)
molar ratio from
balanced equation
of compound B
HEAT (kJ)
DHrxn (kJ/mol)
Figure 6.12
6-32
gained or lost
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 1.000 x 103 kJ of heat is transferred?
PLAN:
SOLUTION:
heat (kJ)
1.000 x 103 kJ x
2 mol Al
1676 kJ
1676 kJ = 2 mol Al
mol of Al
x M (g/mol)
g of Al
6-33
= 32.20 g Al
x
26.98 g Al
1 mol Al
Hess’s Law of Heat Summation
The enthalpy change of an overall process is the sum of
the enthalpy changes of its individual steps.
Used to predict the enthalpy change (a) of an overall reaction
that cannot be studied directly, and/or (b) of an overall reaction
that can be separated into distinct reactions whose enthalpy
changes can be measured individually.
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Sample Problem 6.7
PROBLEM:
Using Hess’s Law to Calculate an Unknown DH
Two gaseous pollutants that form auto exhaust are CO and NO.
An environmental chemist is studying ways to convert them into
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.
CO2(g) DHA = -283.0 kJ
Equation A: CO(g) + 1/2O2(g)
Equation B: N2(g) + O2(g)
PLAN:
Equations A and B have to be manipulated by reversal and/or multiplied
by factors in order to sum to the target equation.
SOLUTION:
Multiply Equation B by 1/2 and reverse it.
CO(g) + 1/2O2(g)
NO(g)
CO(g) + NO(g)
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2NO(g) DHB = +180.6 kJ
CO2(g)
1/2N2(g) + 1/2O2(g)
CO2(g) + 1/2N2(g)
DHA = -283.0 kJ
DHB = -90.3 kJ
DHrxn = -373.3 kJ
Specifying Standard States
For a gas, the standard state is 1 atm; ideal gas behavior is
assumed.
For a substance in aqueous solution, the standard state is
1 M concentration (1 mol/liter solution).
For a pure substance (element or compound), the standard
state is usually the most stable form of the substance at 1
atm and the temperature of interest (usually 25 oC (298 K).
DHorxn = standard heat of reaction
(enthalpy change determined with all
substances in their standard states)
6-36
Formation Equations
In a formation equation, 1 mol of a compound forms from its
elements. The standard heat of formation (DHof) is the enthalpy
change for the formation equation when all substances are
in their standard states.
C(graphite) + 2H2(g)
CH4(g)
DHof = -74.9 kJ
An element in its standard state is assigned a DHof of 0.
Most compounds have a negative D Hof (i.e., the compound
is more stable than its component elements).
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Selected Standard Heats of Formation at 25 oC (298 K)
Table 6.5
Formula
calcium
Ca(s)
CaO(s)
CaCO3(s)
DHof (kJ/mol)
0
-635.1
-1206.9
carbon
C(graphite)
C(diamond)
CO(g)
CO2(g)
CH4(g)
CH3OH(l)
HCN(g)
CS2(l)
chlorine
Cl(g)
6-38
0
1.9
-110.5
-393.5
-74.9
-238.6
135
87.9
121.0
Formula DHof (kJ/mol)
Formula
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)
DHof (kJ/mol)
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
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
DHof.
(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 (Table 6.5) to calculate DHof
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)
6-39
DHof = -127.0 kJ
AgCl(s)
CaCO3(s)
HCN(g)
DHof = -1206.9 kJ
DHof = 135 kJ
The general process for determining DHorxn from DHof values
Figure 6.13
6-40
Sample Problem 6.9
Calculating the Standard Heat of Reaction from
Standard Heats of Formation
Nitric acid, whose worldwide annual production is about 8 billion kg,
is used to make many products, including fertilizers, 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 DHorxn from DHof values.
PLAN:
Look up DHof values (Table 6.5) and use Hess’s Law to find
DHorxn.
DHorxn = S mDHof (products) - S nDHof
(reactants)
o
DH rxn = [4DHof NO(g) + 6DHof H2O(g)] - [4DHof NH3(g) + 5DHof O2(g)]
SOLUTION:
= [(4 mol)(90.3 kJ/mol) + (6 mol)(-241.8 kJ/mol)] [(4 mol)(-45.9 kJ/mol) + (5 mol)(0 kJ/mol)]
DHorxn = -906 kJ
6-41