Chemistry 101 : Chap. 5
Thermochemistry
(1) The Nature of Energy
(2) The First Law of Thermodynamics
(3) Enthalpy
(4) Enthalpy of Reaction
(5) Calorimetry
(6) Hess’s Law
(7) Enthalpy of Formation
NOTE: We will discuss chapter 19, “Chemical Thermodynamics”,
after this chapter
Nature of Energy
 Thermodynamics : The study of energy and its transformation
Energies can be in many different forms. These energy changes can
be as simple as those associated with a falling object to complexes
processes like metabolism
 Thermochemistry : Thermodynamics associated with chemical
processes.
The study of energy changes accompanying chemical reactions
Nature of Energy
 Energy : The capacity to do work or transfer heat
 Work : Energy used to move an object with mass
work = force  distance
 Heat : Energy used to increase the temperature of an object
An object can do work against the environment or transfer heat to the
environment. In such case, the energy of an object needs to be
converted to work or heat.
An object can possess two fundamentally different kinds of energy
: Kinetic energy and Potential energy
Nature of Energy
 Kinetic Energy : The energy an object has due to its motion
1 2
Ek  mv
2
velocity
mass
Ex. (1) A moving car (macroscopic object)
(2) Atoms and molecules are constantly moving.
 This is the source of thermal energy
(temperature increase or decrease)
Nature of Energy
 Potential Energy : The energy an object has due to its position
relative to other objects
All objects in the universe are interacting with each other. In other words,
they push or pull surrounding objects (exerting force). Potential energy
arises when there is any kind of force acting on an object.
m1m2
(or gravitational potential energy)
r
m1m2
Gravitational Force : Fg  G
r2
Ex. (1) Gravitation Energy : E g  G
For the gravitational energy between an object (m1 = m) on earth and
earth (m2 = mearth) itself, r ~ radius of earth = rearth.
Fg, earth = m  (Gmearth/rearth2) = mg
Gravitation potential energy (on earth) = mgh
This is the formal
definition of “weight”
Nature of Energy
(2) Electrostatic Energy :
Q1Q2
Eel  
d
electrical charge (in C)
Electrostatic interaction is either attractive (charges with opposite
signs) or repulsive (charges with same signs).
(3) Chemical Energy : Potential energy stored in the arrangement of
the atoms in the substance of interest (mostly electrostatic)
One of the important goal in chemistry is to relate the energy change
that we see in macroscopic world (i.e. in laboratory experiment) to the
kinetic energy and potential energy of substances at the atomic and
molecular level.
The Units of Energy
 joule (J) :
1 J = 1 kgm2/s2
The kinetic energy of a 2kg object with
the speed of 1 m/s is equal to 1J:
Ek = (1/2)mv2 = (1/2)2 kg  (1 m/s)2
= 1 kgm2/s2 = 1 J
James Joule (1818 ~ 1889)
NOTE: kg, m and s are the SI units of mass, distance and time.
Therefore, joule is the SI unit of energy.
 calorie (cal) : This is an older unit of energy. But, it is still
frequently used in chemistry.
1 cal = 4.184 J
The Units of Energy
 Example
: If a hot dog is burned, it produces 180 dietary
Calories (1 Cal = 1000 cal). How many joules is this?
System and Surrounding
We divide the universe into two parts: system and surrounding
System = The portion of the universe that we are interested in
Surrounding = Everything else
surrounding
system
* No exchange of particles between
system and surrounding
(closed system)
* Energy (heat and work) exchange
is allowed between system and
surrounding
Transferring Energy
Energy can be transferred between a system and its surrounding
in the form of heat (q) and/or work (w)
 Work = Force Distance ( w = f  d )
Energy (work) is transferred from surrounding to an object (system)
when the object is moved by external force.
 Heat :
Energy (heat) is transferred from an object at higher temperature
to one at a lower temperature
Transferring Energy
Heat Transfer: Whenever two objects at different temperature are brought
into contact, heat (thermal energy) will flow from the hot object to the cold
object.
hot
cold
heat flow
Heat is transferred
from system to surrounding
Heat is transferred
from surrounding to system
Transferring Energy
Work Transfer: Work can be done on a system by surrounding or work
can be done by the system to surrounding.
Work done on the system
Work done by the system
The First Law of Thermodynamics
Energy is conserved
Energy can neither be created nor destroyed, only converted
from one form to another (Law of conservation of energy)
Energy that is lost by the system must be gained by the
surrounding, and visa versa.
 Internal Energy of a system (E): The sum of all the kinetic and
potential energy of all its components.
e.g, translation, vibrational and rotational motions of atoms,
molecules
The First Law of Thermodynamics
 Internal Energy change (E) : The change in E that accompanies
a change in the system
E = Efinal - Einitial
q
q
Heat (q) transferred from
system to surroundings:
Efinal < Einitial
E < 0
Heat (q) transferred from
surroundings to system:
Efinal > Einitial
E > 0
The First Law of Thermodynamics
 First
Law of Thermodynamics and internal energy change
E = q + w
heat (thermal energy)
transferred in
+q
endothermic
work performed by
surrounding on system
+w
heat (thermal energy)
transferred out
-q
exothermic
work performed by
system on surrounding
-w
The First Law of Thermodynamics
w = +500 J
work
E
heat
q = -1200J
A(g) + B(g)
C (s)
E = q + w = -1200 J + 500 J = -700 J
The system lost 700 J of energy to surrounding
The First Law of Thermodynamics
Most of the work involved in chemical or physical changes
are associated with volume change.
V > 0  w < 0
w done by the system
V < 0  w > 0
w done on the system
State Functions
 State Function : A property of a system that is determined by the
system’s present condition, or state.
The value of a state function depends only on the present state
of the system, not on the path the system took to reach that state.
Example of state function : Internal Energy (E).
State Functions
Because E is a state function, E is also a state function
NOTE: Although E is
a state function,
q and w are NOT
state functions
Enthalpy
Most of chemical reactions we study take place in open container
under constant pressure condition (atmospheric pressure)
 Enthalpy Change (H) : A measure of the amount of heat
exchanged when a reaction takes place under constant pressure
H = qp
NOTE: Enthalpy is also a state function (H = Hfinal – Hinitial) and it
is defined mathematically as H = E + PV where P is
the pressure.
Enthalpy
 Properties of enthalpy change :
H2O (l)
H2O (g)
H = 44kJ (endothermic)
H2O (g)
H2O (l)
H = -44kJ (exothermic)
2H2O (l)
2H2O (g)
H = 88kJ (endothermic)
Enthalpy changes, like all energy changes, are extensive properties.
The amount of heat (or enthalpy change) liberated or absorbed
depends on the amount of material undergoing reaction
Enthalpy
 Example: What is the enthalpy change (in kJ) for the sublimation
of 15 g of iodine
I2 (s)
I2(g)
Hsub = 62.4kJ
MW=254
Enthalpy of Reactions
 Example : How much heat is given off by burning 3.4 g of H2 gas?
2 H2 (g) + O2 (g)
2 H2O (l)
H = - 483.6 kJ
Enthalpy of Reactions
 Example : What is the enthalpy change for the formation 12 g
of CO2(g) from the combustion of CH4 (g)?
CH4(g) + 2O2 (g)
CO2(g) + 2H2O(g)
MW=44
H = - 806 kJ
Enthalpy of Reactions
2 H2 (g) + O2 (g)
→
2 H2O (l)
H = - 483.6 kJ
H
2 H2 (g) + O2 (g)
H > 0
H < 0
q absorbed
q given off
2 H2O (l)
2 H2O (l)
→
2 H2 (g) + O2 (g)
H = +483.6 kJ
Calorimetry
 Calorimetry : The measure of heat flow during physical and
chemical changes
+ NaOH
dissolves
heat (q)
How can we
measure q ?
 Calorimeter : A device for measuring heat flow during physical
and chemical processes
Calorimetry
We need to carry out the reaction in a container
that is thermally insulated
simple calorimeter
NaOH dissolves
heat
(q)
This is under constant pressure condition
Calorimetry
When NaOH is dissolved in water, the temperature of the solution
will increase. But, how much?
 Heat Capacity : The amount of energy required to raise the
temperature of an object by 1 oC
Heat capacity is an extensive property
 Specific Heat Capacity : The amount of energy required to raise
the temperature of 1 g of a substance by 1 oC
Specific heat capacity is an unique property of each substance
Calorimetry
 Unit of specific heat capacity
1 g of H2O
+ 4.184 J of heat
1 g of H2O
11oC
12oC
heat
q
C
m  T
mass
q
T 
mC
J

gK
unit of specific
heat capacity
temperature
change
q  T  m  C
Calorimetry
 Example :Which of the following substances requires the smallest
amount of energy to increase the temperature of 50.0 g
of that substance by 10K?
specific heat
capacity (J/gK)
CH4(g)
Hg(l)
H2O(l)
2.20
0.14
4.18
Calorimetry
 Example :Which of the following substances requires the smallest
amount of energy to increase the temperature by 1K?
CH4(g)
Hg(l)
H2O(l)
C (J/gK)
2.20
0.14
4.18
mass (g)
10
100
2
Calorimetry
 Example : Determine the specific heat of Al if it takes 16 J of
thermal energy to raise the temperature 6.0g of Al
by 3.0 oC
Calorimetry
 Example : What is the change in thermal energy (in kJ) for
300 g iron rod when it is cooled from 250oC to 50oC ?
(C for iron is 0.451 J/gK)
Calorimetry
 Molar Heat Capacity (Cm) : The amount of energy required to raise
the temperature of 1 mol of substance by 1 oC
heat
q
J

Cm 
mol  T mol  K
mol
q
T 
mol  Cm
unit of molar
heat capacity
temperature
change
q  T  mol  Cm
Calorimetry
 Example : The specific heat capacity of iron is 0.451 J/goC.
Determine the molar heat capacity of iron.
Calorimetry
 Example : A 5.00 g pellet of copper at 75.0oC is placed in a beaker
containing 35.0g of water at 25.0oC. What is the final temperature
of the water and copper? (Cu = 0.385 J/goC, H2O = 4.18 J/goC)
Calorimetry
 Example : A 2.00 g of sodium is placed in 150 g of water at 22.0oC
in a calorimeter. When the reaction is complete, the temperature
of the solution is found to be 41.0oC. Calculate H for the reaction
between Na and water: 2Na(s) + 2H2O(l)  H2(g) + 2NaOH(aq)
Strategy to solve the problem:
(1) compute the total amount of heat gained by water
(2) That is the -H value for 2.00 g of sodium
(3) We need H value for 2 mol of sodium
(conversion from 2.00 g to 2 mol of sodium for H )
Hess’s Law
If a reaction is carried out in a series of steps, H for the overall
reaction will equal the sum of the enthalpy changes for the
individual steps
H
H2O (g)
-44kJ
-50kJ
H2O (l)
-6 kJ
H2O (s)
Hess’s Law
Consider the following imaginary reactions.
(A)
A + B
(B)
D3
→
→
H = -125 kJ
C
H = -41 kJ
C + B
How can we calculate the H for the following reaction?
(C) A
+
2B
→
D3
To use Hess’s law, we have to rearrange reaction (A) and (B)
so that the overall reaction becomes (C) A+2B  D3. Be sure
that you make appropriate changes for H’s when you rearrange
chemical equations.
Hess’s Law
This is how we can rearrange the reactions.
(1) Compare (A) and (C):
(A) A +
B →
(C) A + 2B →
C
D3
This is what we have.
This is what we want.
We need one more B on the left and one D3 on the right and remove
C on the right
(2) Consider (B) : D3
→
C + B
H = -41 kJ
If we reverse (B), it pretty much satisfies our need: one B on the left
and one D3 on the right
(B’)
C + B  D3
H = +41 kJ
Hess’s Law
(3) Add (A) and (B’)
A
+
B
 C
H = -125kJ
C
+
B
 D3
H = + 41kJ
2B  C + D3
H = -84 kJ
A
+ C
A +
+
2B

D3
H = -84 kJ
Hess’s Law
 Example : Given the enthalpy changes of following two reactions,
calculate the enthalpy change for the combustion of
methane to form liquid water and CO2
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
H2O(g)  H2O(l)
H = -802 kJ
H = -44 kJ
Hess’s Law
 Example : Given the enthalpy changes of following two reactions,
calculate the enthalpy change for making diamond from
graphite : C(graphite)  C(diamond)
C(graphite) + O2(g)  CO2(g)
H = -393.5 kJ
 CO2(g)
H = -395.4 kJ
C(diamond) + O2(g)
graphite
diamond
Hess’s Law
 Example : Given the enthalpy changes of following two reactions,
calculate the enthalpy change for the reaction between
hydrogen and ozone : 3 H2 (g) + O3 (g) → 3 H2O (g)
2 H2 (g) + O2 (g) → 2 H2O (g)
3 O2 (g) →
2 O3 (g)
H = -483.6 kJ
H = +284.6 kJ
Standard Enthalpy of Formation
H for the formation of one mole of compound from elements,
with all substances in their standard states [25oC, 1atm]
H2 (g)
+
½ N2 (g)
½ O2 (g)
→
H2O (l)
Hof = -285.8 kJ
+ ½ O2 (g)
→
NO (g)
Hof = 90.4 kJ
2 C (s) + 3 H2 (g) + ½ O2 (g) → C2H5OH (g)
Hof = -277.7 kJ
graphite
NOTE : Hof of the most stable form of any element is zero
i.e. Hof for O2 (g) , N2 (g) , H2 (g), Br2 (l) etc. = 0
Standard Enthalpy of Formation
 Example : For which of the following reactions would H represent
a standard enthalpy of formation?
2 Na (s)
K (l)
→
Na2O (s)
→
KCl (s)
+ ½ O2 (g)
→
CO2 (g)
+ Cl2 (g)
→
2 NaCl (s)
+
½ O2 (g)
+ ½ Cl2 (g)
CO (g)
2 Na (s)
Standard Enthalpy of Formation
Enthalpies of formations can be used to calculate enthalpies
of reactions (under standard conditions)
Horxn = sum of all Hof(products) - sum of all Hof(reactants)
Horxn = Σ n Hof(products) – Σ m Hof(reactants)
sum
moles of
products
moles of
reactants
Standard Enthalpy of Formation
Standard Enthalpy of Formation
 Example : Determine the standard enthalpy change for the
following reaction
C2H4(g) + H2O(g)  C2H5OH(l)
Hof [C2H4(g)] = 52.30kJ, Hof [H2O(g)] = -241.82 kJ, Hof[C2H6O(l)] =-277.7kJ
Standard Enthalpy of Formation
 Example : What is the enthalpy change of the following reaction
at standard condition?
C3H8 (g) + 5 O2 (g) → 3 CO2 (g) + 4 H2O (l)
Hof [CO2 (g)] = -394kJ, Hof [C3H8(g)]= -104kJ, Hof [H2O(l)] = -286kJ