Chapter 6
Thermochemistry
Jozsef Devenyi
Department of Chemistry, UTM
Chapter 5
1
The Nature of Energy
Recall:
force: a push or pull on an object
work: the product of force applied to an object over a
distance
w= F
x
d
energy: the work done to move an object against a force
(energy is the capacity to do work or transfer heat)
• kinetic energy is the energy of motion:
Chapter 5
2
The Nature of Energy
• potential energy is the energy an object possesses by
virtue of its position
• potential energy can be converted
into kinetic energy
example: a ball is falling from a balcony
Chapter 5
3
The Nature of Energy
Units of Energy
SI Unit for energy is the joule, J:
for example, an object with a mass of 2 kg that moves at
a speed of 1 m/s; its kinetic energy is
We sometimes use the calorie instead of the joule:
1 cal = 4.184 J (exactly)
Chapter 6
4
The Nature of Energy
Systems and Surroundings
• system: part of the universe we are interested in
• surroundings: the rest of the universe
Chapter 6
5
First Law of Thermodynamics
Internal Energy
• Internal energy: total energy of a system
(cannot measure absolute internal energy)
• Change in internal energy:
Chapter 6
6
First Law of Thermodynamics
Relating DE to Heat and Work:
• Energy cannot be created or destroyed.
first law of thermodynamics:
energy of (system + surroundings) is constant
• Any energy transferred from a system must be
transferred to the surroundings (and vice versa).
• internal energy of a system:
when a system undergoes a physical or chemical change,
the change in internal energy is given by the heat
released or absorbed by the system plus the work done
on or by the system
Chapter 6
7
First Law of Thermodynamics
Relating DE to Heat and Work:
Chapter 6
8
First Law of Thermodynamics
Chapter 6
9
First Law of Thermodynamics
Exothermic and Endothermic Processes
• endothermic: absorbs heat from the surroundings
(an endothermic reaction feels cold)
example:
Chapter 6
10
First Law of Thermodynamics
Exothermic and Endothermic Processes
• exothermic: transfers heat to the surroundings
(an exothermic reaction feels warm/hot)
example:
Chapter 6
11
First Law of Thermodynamics
State Functions
DE is a state function; that is, the value
of DE depends only on the initial and
final states of
system, not on
how change
occurred
Chapter 6
12
Enthalpy
enthalpy (H):
heat transferred between the system and its
surroundings while pressure is constant;
we can measure the change in enthalpy:
DH = Hfinal – Hinitial = qp
enthalpy is a
state function
Chapter 6
13
Enthalpies of Reaction
For a reaction:
enthalpy is an extensive property (magnitude DH is
directly proportional to amount):
CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O(g)
DH = - 802 kJ
2 CH4 (g) + 4 O2 (g)  2 CO2 (g) + 4 H2O(g)
Chapter 6
DH =
14
Enthalpies of Reaction
When reaction is reversed the sign of DH is reversed:
CO2 (g) + 2 H2O(g)  CH4 (g) + 2 O2 (g)
DH =
Change in enthalpy also depends on physical state:
H2O(g)  H2O(l)
DH = - 88 kJ
Chapter 6
15
Enthalpies of Reaction
Example:
Chapter 6
16
Enthalpies of Reaction
Example:
Chapter 6
17
Calorimetry
Heat Capacity and Specific Heat
calorimetry = measurement of heat flow
calorimeter = apparatus that measures heat flow by
measuring the change in temperature
heat capacity (C) = the amount of energy required to raise
the temperature of an object by one degree Celsius (J / oC)
Chapter 6
18
Calorimetry
Heat Capacity and Specific Heat
molar heat capacity = heat capacity of 1 mol of a substance
[J/(mol . oC)]
Chapter 6
19
Calorimetry
Heat Capacity and Specific Heat
specific heat (s) = specific heat capacity = heat capacity
of 1 g of a substance
unit: J/(g . oC)
that is, the amount of energy required to raise the temperature of
1 g of substance by one degree Celsius (J /g oC)
Chapter 6
20
Calorimetry
Heat Capacity and Specific Heat
heat released/absorbed:
q = (specific heat) x (grams of substance) x Dt =
= s.h.
X
m
x
Dt
where Dt = tfinal - tinitial
Chapter 6
21
Calorimetry
Examples:
A)
Chapter 6
22
Calorimetry
Examples:
B)
Chapter 6
23
Calorimetry
Constant-Pressure Calorimetry
• at constant atmospheric
pressure:
DH = qp
• in such system, we assume that
no heat is lost to surroundings
qrxn + qwater = 0
therefore
qrxn = - qwater
Chapter 6
24
Calorimetry
Example:
When a student mixes 50.0 mL of 1.0 M NaOH solution
and 50.0 mL of 1.0 M HCl solution in a coffee cup
calorimeter, the temperature of the resultant solution
increases from 21.3 oC to 27.8 oC.
Calculate the enthalpy change for this neutralization
reaction.
Chapter 6
25
Calorimetry
Example:
Assume that
i. the calorimeter loses only negligible quantity of heat,
ii. the total volume of the solution is 100 mL,
iii. the density and the specific heat of the solution are
the same as those of water, 1.00 g/mL and 4.184 J/g oC,
respectively.
(these assumptions are usually true, unless stated
otherwise)
Chapter 6
26
Calorimetry
Example:
Chapter 6
27
Calorimetry
Example:
Chapter 6
28
Calorimetry
Constant-Volume Calorimetry (Bomb Calorimetry)
• rxn carried out under constant volume
• uses a device called bomb calorimeter
• usually used to study
combustion rxns
qrxn + qwater + qcal = 0
qwater + qcal = - qrxn
Note: since pressure is
not constant under
these conditions, q
measured this way is
not equal to DH.
Chapter 5
29
Calorimetry
Example:
In a laboratory test 9.20 g of ethanol, C2H5OH was burned
in a bomb calorimeter that contained 500.0 g of water and
the heat capacity of the calorimeter is 4.821 kJ/oC. The
temperature increased from 22.9 oC to 24.85 oC.
A) Calculate the heat of combustion per gram ethanol.
Chapter 6
30
Calorimetry
Example:
Chapter 6
31
Calorimetry
Example:
B)
Chapter 6
32
Hess’s Law
Hess’s law: if a reaction is carried out in a number of
steps, DH for the overall reaction is the sum of DH for
each individual step.
For example:
CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O(g)
DH = - 802 kJ

DH = - 88 kJ
2 H2O(g)
2 H2O(l)
Chapter 6
33
Hess’s Law
Note that:
Chapter 6
DH1 = DH2 + DH3
34
Hess’s Law
From the two reactions:
N2 (g) + 2 O2 (g)
2 NO(g)
+
O2 (g)
2 NO2 (g)
DH1 = + 67.6 kJ
2 NO2 (g)
DH2 = - 521 kJ
calculate the heat of reaction (enthalpy of reaction, DHrxn)
for the following rxn: N2 (g) + O2 (g)
2 NO(g) .
Chapter 6
35
Hess’s Law
Chapter 6
36
Hess’s Law
Chapter 6
37
Enthalpies of Formation
• If 1 mol of compound is formed from its
constituent elements, then the enthalpy change for the
reaction is called the enthalpy of formation, DHf .
• Standard conditions (standard state): 1 atm and 25 oC
(298 K).
• Standard enthalpy, DHo, is the enthalpy measured when
everything is in its standard state (every component).
Chapter 6
38
Enthalpies of Formation
• Molar enthalpy of formation: 1 mol of compound is
formed from substances in their standard states.
• If there is more than one state for a substance under
standard conditions, the more stable one is used.
Chapter 6
39
Enthalpies of Formation
See Table 6.5
• Standard
enthalpy of
formation
of the most
stable form
of an
element
is zero.
Chapter 6
40
Enthalpies of Formation
Chapter 6
41
Enthalpies of Formation
Using a set of DHof values to calculate DHrxn .
We use Hess’ Law to calculate the enthalpy of any
reaction using the enthalpies from the table of enthalpies
of formation.
where
- “S“ means “the sum”
- n and m are stoichiometric coefficients for the
each product and reactant, respectively
Chapter 6
42
Enthalpies of Formation
Example:
Consider the following combustions reaction of methane:
CH4 (g) + O2 (g)  CO2 (g) +
H2O(g)
DHrxn = ??
Using Hess’ Law and the relevant standard enthalpies of
formation, calculate DHrxn for this reaction.
Chapter 6
43
Enthalpies of Formation
Example:
Chapter 6
44
Enthalpies of Formation
Example:
Chapter 6
45
Enthalpies of Formation
Example:
Chapter 6
46
Enthalpies of Formation
Example:
Chapter 6
47
End of Chapter 6
Thermochemistry
Chapter 6
48