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Chapter 8
Thermochemistry
8.1 Principles of Heat Flow
8.2 Calorimetry
8.3 Enthalpy
8.4 Thermochemical Equations. Hess’s Law
8.5 Enthalpies of Formation
8.6 Bond Enthalpy
8.7 The First Law of Thermodynamics
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.
A chemical system and its surroundings.
the surroundings
the system
Energy is the capacity to do work.
Potential Energy
energy due to the position of an object
or energy due to composition which can be
released in a chemical reaction
Kinetic Energy
energy due to the motion of the object
Potential and kinetic energy can be interconverted.
Energy is the capacity to do work.
Figure A
less stable, Ep= mgh
change in potential energy
EQUALS
kinetic energy
more stable, Ek= 1/2 mv2
A gravitational system. The potential energy of a lifted weight
is converted to kinetic energy as the weight falls.
Energy is the capacity to do work.
Figure B
less stable
change in potential energy
EQUALS
kinetic energy
more stable
A system of two balls attached by a spring. The potential energy
gained by a stretched spring is converted to kinetic energy when the
moving balls are released.
Energy is the capacity to do work.
Figure C
less stable
change in potential energy
EQUALS
kinetic energy
more stable
A system of oppositely charged particles. The potential energy gained
when the charges are separated is converted to kinetic energy as the
attraction pulls these charges together.
Energy is the capacity to do work.
Figure D
less stable
change in potential energy
EQUALS
kinetic energy
more stable
A system of fuel and exhaust. A fuel is higher in chemical potential
energy than the exhaust. As the fuel burns, some of its potential energy is
converted to the kinetic energy of the moving car.
Energy diagrams for the transfer of internal energy (E)
between a system and its surroundings.
DE = Efinal - Einitial = Eproducts - Ereactants
Internal Energy, E, of the system consists of potential
(composition, state of subdivision), and kinetic (vibrational,
rotational, and translational) energies.
A system transferring energy as heat only.
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)
Changes in Internal Energy
• If DE < 0, Efinal < Einitial
– Therefore, the system
released energy to the
surroundings.
– This energy change is
called exergonic, or
exothermic.
Changes in Internal Energy
• When energy is
exchanged between
the system and the
surroundings, it is
exchanged as either
heat (q) or work (w),
or both q, and w.
• That is, DE = q + w.
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.
DEuniverse = DEsystem + DEsurroundings
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
Two different paths for the energy change of a system.
State Functions
Usually we have no way of knowing the internal
energy of a system; finding that value is simply
too complex a problem.
– In the system below, the water could have reached
room temperature from either direction.
However, we do know that the internal energy of a
system is independent of the path by which the system
achieved that state.
State Functions
• A state function depends only on the current state
of a system. The change in a state function
between two states is independent of the
pathway between them.
• Therefore, internal energy is a state function.
• And so, DE depends only on Einitial and Efinal.
State Functions
• However, q and w are
not state functions.
• Whether the battery is
shorted out (a) or is
discharged by running
the fan (b), its DE is the
same.
 But q and w are different
in the two cases!
Sample Problem
PROBLEM:
PLAN:
Determining the Change in Internal Energy of a
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
kJ
103J
w = - 451 J
-325 J + (-451 J) = -776 J
= -0.776kJ
-0.776kJ
kcal
4.18kJ
= -0.185 kcal
Pressure-volume
work.
Sample Problem
1. A system expands against a constant pressure of 1.50
atm from 10.0 L to 25.0 L, while absorbing 150 J of
heat. Determine the change in internal energy, ∆E.
[1 L·atm = 101.3 J]
Plan: Find work done by the system, add to heat change.
w = -P∆V = -(1.50 atm)(25.0 L - 10.0 L)(101.3 J/L·atm)
= - 2279.25 J = - 2280 J
∆E = q + w = 150 J - 2280 J = -2130 J
Enthalpy
• If a process takes place at constant
pressure (as the majority of processes we
study do) and the only work done is this
pressure-volume work, we can account for
heat flow during the process by measuring
the enthalpy of the system.
• Enthalpy is the internal energy plus the
product of pressure and volume:
H = E + PV
Enthalpy
• When the system changes at constant
pressure, the change in enthalpy, DH, is
DH = D(E + PV)
• This can be written
DH = DE + PDV
Enthalpy
• Since DE = q + w and w = - PDV, we
can substitute these into the enthalpy
expression:
DH = DE + PDV
DH = (q+w) - w
DH = qp
• So, at constant pressure the change in
enthalpy is the heat gained or lost.
The Truth about Enthalpy
1. Enthalpy is an extensive property.
2. DH for a reaction in the forward
direction is equal in size, but opposite
in sign, to DH for the reverse reaction.
3. DH for a reaction depends on the state
of the products and the state of the
reactants, that is ∆H is a state
function.
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
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
>>> (much, much, much greater) PDV
When gases are involved, ∆H = ∆E + ∆nRT
Where ∆n = moles gaseous products - moles
gaseous reactants.
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
Exothermic process
B
H2O(g)
Hfinal
heat in
Hinitial
Endothermic process
Some Important Types of Enthalpy Change
heat of combustion (DHcomb)
C4H10(l) + 13/2O2(g)
4CO2(g) + 5H2O(g)
heat of formation (DHf)
K(s) + 1/2Br2(l)
KBr(s)
heat of fusion (DHfus)
NaCl(s)
NaCl(l)
heat of vaporization (DHvap)
C6H6(l)
C6H6(g)
Sample Problem
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
(Sp) of Cu is 0.387 J/g·˚C.
PLAN: Given the mass, specific heat capacity and change in temperature, we
can use q = Sp x mass x DT to find the answer. DT in 0C is the same as
for K.
SOLUTION:
q=
0.387 J
g·˚C
x 125 g x (300-25)0C = 1.33x104 J
Coffee-cup calorimeter.
Sample Problem
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 Sp = 4.184 J/g·˚C)
PLAN:
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.
Sample Problem
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 4.184 J/g·0C x (27.21-25.00)0C
= - 693.50 J
(-693.50 J/0.0125 mol H2O)(kJ/103 J) = -55.5 kJ/ mol H2O formed
Figure 5.10
A bomb calorimeter
Sample Problem
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/˚C). The temperature increases 4.9370C.
Is the manufacturer’s claim correct?
PROBLEM:
PLAN:
Calculating the Heat of Combustion
Energy conservation; qsample + qcal = 0 => - q sample = qcalorimeter
SOLUTION:
qcalorimeter
= heat capacity x DT
= 8.151 kJ/˚C x 4.937 ˚C
= 40.24 kJ
40.24 kJ
kcal
= 9.63 kcal or Calories
4.18 kJ
The manufacturer’s claim is true.
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)
gained or lost
Sample Problem
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
1.000x103 kJ x
2 mol Al
26.98 g Al
1676 kJ
1 mol Al
= 32.20 g Al
Using Hess’s Law to Calculate an Unknown DH
Sample Problem
PROBLEM:
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:
CO2(g) DHA = -283.0 kJ
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)
CO2(g)
1/2N2(g) + 1/2O2(g)
CO(g) + NO(g)
CO2(g) + 1/2N2(g)
DHA = -283.0 kJ
DHB = -90.3 kJ
DHrxn = -373.3 kJ
Sample Problem
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)
DH0f = -127.0 kJ
AgCl(s)
(b) Ca(s) + C(graphite) + 3/2O2(g)
(c) 1/2H2(g) + C(graphite) + 1/2N2(g)
CaCO3(s)
HCN(g)
DH0f = -1206.9 kJ
DH0f = 135 kJ
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)
Hfinal
Sample Problem
PROBLEM:
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:
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
State Functions
Usually we have no way of knowing the
internal energy of a system; finding that value
is simply too complex a problem.