Thermochemistry: Energy Flow and Chemical Change

1
Chapter 6
Thermochemistry Energy Flow and Chemical Change
2
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 Hesss Law of Heat Summation
6.6 Standard Heats of Reaction (DHorxn)
3
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).
4
DE Efinal - Einitial Eproducts -
Ereactants
Figure 6.2
5
DE q w
where q heat and w work
A system transferring energy as heat only
Figure 6.3
6
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.
7
A system losing energy as work only
Pressure-volume work (PV work)
Figure 6.4
8
Table 6.1 Sign Conventions for q, w and DE
q
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.
9
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
1 Btu 1055 J
10
Some quantities of energy
Figure 6.5
11
Determining the Change in Internal Energy E of a
System
Sample Problem 6.1
PLAN
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
-0.776 kJ
-0.185 kcal
12
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
13
Pressure-volume work
In this case, the system does PV work on
the surroundings thus w is negative in sign.
Figure 6.7
14
For reactions that occur at constant pressure.
The Definition of Enthalpy
DH DE for
w - PDV
1. Reactions that do not involve gases.
2. Reactions in which the number of moles of gas
does not change.
DH DE PDV
3. Reactions in which the number of moles of gas
changes but q is gtgtgt PDV.
qp DE PDV DH
H is a state function.
15
Enthalpy diagrams for exothermic and endothermic
processes
Heat is released enthalpy decreases.
Heat is absorbed enthalpy increases.
Figure 6.8
16
Sample Problem 6.2
Drawing Enthalpy Diagrams and Determining the
Sign of DH
PLAN
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.
(b) The reaction is endothermic.
EXOTHERMIC
DH -285.8 kJ
DH 40.7 kJ
ENDOTHERMIC
17
Some Important Types of Enthalpy Change
heat of combustion (DHcomb)
heat of formation (DHf)
Standard quantity of either reactant or product
1 mol
heat of fusion (DHfus)
heat of vaporization (DHvap)
18
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)

19
Figure 6.9
Components of internal energy (E)
20
Bonds absorb energy when they break and release
energy when they form.
Weaker bonds easier to break higher in energy
(less stable, more reactive)
21
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.
22
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
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
DHcomb(kJ/g)
-51.88
-29.67
-55.5
-22.7
23
Table 6.3 Heats of Combustion of Some Fats and
Carbohydrates
Fats
Carbohydrates
24
Ball-and-stick molecular models of a carbohydrate
(sucrose) and a triglyceride (triolean)
A more oxidized fuel
A more reduced fuel
25
Calorimetry laboratory measurement of heats of
reaction
heat capacity an objects 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)
26
Table 6.4 Specific Heat Capacities of Some
Elements, Compounds and Materials
27
Sample Problem 6.3
Calculating the Quantity of Heat from the
Specific Heat Capacity
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
1.33 x 104 J
q
125 g
(300-25)oC
x
x
28
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
29
Sample Problem 6.4
Determining the Specific Heat Capacity of a Solid
PLAN
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
c x
25.64 g x
-71.51 K

-
csolid
30
Calorimetry at Constant Volume
A bomb calorimeter used to measure qv
Figure 6.11
31
Sample Problem 6.5
Calculating the Heat of Combustion
PLAN
- qsample qcalorimeter
SOLUTION
qcalorimeter
heat capacity x DT
8.151 kJ/K x 4.937 K
40.24 kJ
9.62 kcal lt 10 Calories 10 kcal
The manufacturers claim is correct.
32
Summary of the relationship between amount (mol)
of substance and the heat (kJ) transferred during
a reaction
molar ratio from balanced equation
DHrxn (kJ/mol)
Figure 6.12
33
Sample Problem 6.6
Using the Heat of Reaction (DHrxn) to Find Amounts
SOLUTION
PLAN
heat (kJ)
1.000 x 103 kJ x
x
1676 kJ 2 mol Al
32.20 g Al
mol of Al
x M (g/mol)
g of Al
34
Hesss 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.
35
Sample Problem 6.7
Using Hesss Law to Calculate an Unknown DH
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.
DHB -90.3 kJ
DHrxn -373.3 kJ
36
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)
37
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 DHof
(i.e., the compound is more stable than its
component elements).
38
Table 6.5 Selected Standard Heats of
Formation at 25 oC (298 K)
39
Sample Problem 6.8
Writing Formation Equations
(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
DHof -127.0 kJ
DHof -1206.9 kJ
DHof 135 kJ
40
The general process for determining DHorxn from
DHof values
Figure 6.13
41
Sample Problem 6.9
Calculating the Standard Heat of Reaction from
Standard Heats of Formation
PLAN
Look up DHof values (Table 6.5) and use Hesss
Law to find DHorxn.
SOLUTION
DHorxn S mDHof (products) - S nDHof
(reactants)
DHorxn 4DHof NO(g) 6DHof H2O(g)
- 4DHof NH3(g) 5DHof 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)
DHorxn -906 kJ