Chemistry 101 : Chap' 5

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Chemistry 101 Chap. 5
Thermochemistry

• The Nature of Energy
• (2) The First Law of Thermodynamics
• (3) Enthalpy
• (4) Enthalpy of Reaction
• (5) Calorimetry
• (6) Hesss Law
• (7) Enthalpy of Formation

NOTE We will discuss chapter 19, Chemical
Thermodynamics, after this chapter
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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
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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
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Nature of Energy
? Kinetic Energy The energy an object has due
to its motion
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)
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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.
Ex. (1) Gravitation Energy
(or gravitational potential energy)
Gravitational Force
For the gravitational energy between an object
(m1 m) on earth and earth (m2 mearth)
itself, r radius of earth rearth.
This is the formal definition of weight
Fg, earth m ? (Gmearth/rearth2) mg
Gravitation potential energy (on earth) mgh
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Nature of Energy
(2) Electrostatic Energy
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.
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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
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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?
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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
No exchange of particles between system
and surrounding (closed system)
surrounding
Energy (heat and work) exchange is allowed
between system and surrounding
system
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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
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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.
heat flow
hot
cold
Heat is transferred from
surrounding to system
Heat is transferred from
system to surrounding
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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
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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

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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
Heat (q) transferred from surroundings to system
DE gt 0
Efinal lt Einitial
DE lt 0
Efinal gt Einitial
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The First Law of Thermodynamics
? First Law of Thermodynamics and internal
energy change
?E q w
heat (thermal energy) transferred in
q endothermic
heat (thermal energy) transferred out
-q exothermic
work performed by system on surrounding
-w
work performed by surrounding on system
w
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The First Law of Thermodynamics
w 500 J
work
?E
heat
q -1200J
C (s)
A(g) B(g)
?E q w -1200 J 500 J -700 J
The system lost 700 J of energy to surrounding
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The First Law of Thermodynamics
Most of the work involved in chemical or physical
changes are associated with volume change.
?V gt 0 ? w lt 0 w done by the system
?V lt 0 ? w gt 0 w done on the system
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State Functions
? State Function A property of a system that is
determined by the
systems 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).
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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
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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.
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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
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Enthalpy
? Example What is the enthalpy change (in kJ)
for the sublimation of 15 g
of iodine I2 (s)
I2(g) ?Hsub 62.4kJ
MW254
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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
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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) ?H - 806
kJ
MW44
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Enthalpy of Reactions
2 H2 (g) O2 (g) ? 2 H2O (l)
DH - 483.6 kJ
H
2 H2 (g) O2 (g)
DH lt 0
DH gt 0
2 H2O (l)
DH 483.6 kJ
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Calorimetry
? Calorimetry The measure of heat flow during
physical and chemical
changes
dissolves
How can we measure q ?
heat (q)
? Calorimeter A device for measuring heat flow
during physical and
chemical processes
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Calorimetry
We need to carry out the reaction in a container
that is thermally insulated
simple calorimeter
heat (q)
This is under constant pressure condition
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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
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Calorimetry
? Unit of specific heat capacity
1 g of H2O 11oC
1 g of H2O 12oC
4.184 J of heat
heat
unit of specific heat capacity
mass
temperature change
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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?
CH4(g)
Hg(l) H2O(l) specific heat
2.20 0.14
4.18 capacity (J/g?K)
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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
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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
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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)

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Calorimetry
? Molar Heat Capacity (Cm) The amount of energy
required to raise
the temperature of 1 mol of substance by 1 oC
heat
unit of molar heat capacity
mol
temperature change
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Calorimetry

• Example The specific heat capacity of iron is
  0.451 J/g?oC.
• Determine the molar heat
  capacity of iron.

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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)

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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
)
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Hesss 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
H2O (l)
-6 kJ
H2O (s)
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Hesss Law
Consider the following imaginary reactions.

• A B ? C ?H -125 kJ
• (B) D3 ? C B ?H -41 kJ

How can we calculate the ?H for the following
reaction?
(C) A 2B ? D3
To use Hesss law, we have to rearrange reaction
(A) and (B) so that the overall reaction becomes
(C) A2B ? D3. Be sure that you make appropriate
changes for ?Hs when you rearrange chemical
equations.
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Hesss Law
This is how we can rearrange the reactions.

• Compare (A) and (C)
• (A) A B ? C
• (C) A 2B ? 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
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Hesss Law
(3) Add (A) and (B)
A B ? C
?H -125kJ C B ? D3
?H 41kJ A C
2B ? C D3 ?H -84 kJ
A 2B ? D3 ?H -84 kJ
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Hesss 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)
?H -802 kJ H2O(g) ?
H2O(l) ?H -44
kJ
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Hesss 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 C(diamond) O2(g) ? CO2(g)
?H -395.4 kJ
graphite
diamond
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Hesss 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) ?H
-483.6 kJ 3 O2 (g) ? 2 O3 (g)
?H 284.6 kJ
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Standard Enthalpy of Formation
DH for the formation of one mole of compound from
elements, with all substances in their standard
states 25oC, 1atm
H2 (g) ½ O2 (g) ? H2O
(l) DHof -285.8 kJ
½ N2 (g) ½ O2 (g) ? NO (g) DHof
90.4 kJ
2 C (s) 3 H2 (g) ½ O2 (g) ? C2H5OH (g)
DHof -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
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Standard Enthalpy of Formation
? Example For which of the following reactions
would DH represent a standard
enthalpy of formation?
2 Na (s) ½ O2 (g) ? Na2O (s)
K (l) ½ Cl2 (g) ? KCl
(s)
CO (g) ½ O2 (g) ? CO2 (g)
2 Na (s) Cl2 (g) ? 2 NaCl
(s)
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Standard Enthalpy of Formation
Enthalpies of formations can be used to calculate
enthalpies of reactions (under standard
conditions)
DHorxn sum of all DHof(products) - sum of all
DHof(reactants)
DHorxn S n DHof(products) S m DHof(reactants)
sum
moles of products
moles of reactants
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Standard Enthalpy of Formation
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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, ?HofC2H6O(l) -277.7kJ

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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