2/9/09
Chemistry 123: Physical and Organic Chemistry
Topic 2: Thermochemistry
http://people.ok.ubc.ca/orcac/chem123out.html
Topic 2: Introduction,
Topic 2: Thermochemistry
Text: Chapter 7 and 19 (~ 3 weeks)
2.0 Introduction, terminology and scope
2.1 Enthalapy and Energy Change in a chemical process; 1st law of
Thermodynamics
2.2 Enthalapy of Formation, Calorimetry, Heat capacity, Bond Energies,
and Standard State
2.3 Entropy; 2nd and 3rd laws of thermodynamics
2.4 Free Energy; Spontaneity and Equilibrium.
2.5 Relationship between dH°,dS°, dG° and K
2.6 Change of dG with Temperature adn Concentration; Effect of
Temperature on K (van't Hoff)
Winter 2009
From: http://www.dilbert.com
Page 1
Chemistry 123: Physical and Organic Chemistry
Topic 2: Thermochemistry
http://people.ok.ubc.ca/orcac/chem123out.html
Topic 2.0: Introduction,
2.0 Introduction, terminology and scope
• A System is the part of the universe under
study.
•  The Surroundings is anything outside of the
System.
•  An open system allows for exchange of both
material and energy
•  A closed system can exchange only energy
• An isolated system, does not permit the
transfer of mass or Energy.
•  Energy is the ability to do work
•  Work occurs when a force acts through a
distance.
•  Kinetic Energy is the energy of a moving
object.
•  Thermal Energy is the kinetic energy
associated with random molecular motion
Winter 2009
Page 2
1
2/9/09
Chemistry 123: Physical and Organic Chemistry
Topic 2: Thermochemistry
http://people.ok.ubc.ca/orcac/chem123out.html
Topic 2.0: Introduction,
Kinetic Energy
ek =
1
2
2
mv2
m
= J
[ek ] = kg
s
Work
w = Fd
Winter 2009
[w ] =
kg m m
= J
s2
From: http://www.dilbert.com
Page 3
Chemistry 123: Physical and Organic Chemistry
Topic 2: Thermochemistry
http://people.ok.ubc.ca/orcac/chem123out.html
Topic 2.0: Introduction,
Understanding
Calorie (cal): The quantity of heat required to change the
temperature of one gram of water by one degree Celsius.
Joule (J): International System of Units (SI) unit for heat
1 cal = 4.184 J
James Joule (1818-1889)
Winter 2009
Page 4
2
2/9/09
Chemistry 123: Physical and Organic Chemistry
Topic 2: Thermochemistry
http://people.ok.ubc.ca/orcac/chem123out.html
Topic 2.0: Introduction,
Heat Capacity
The quantity of heat (q) required to change the
temperature of a system by one degree.
•  Molar heat capacity.
◦  System is one mole of substance.
•  Specific heat capacity, c.
◦  System is one gram of substance
•  Heat capacity
q
◦  Mass times c = C specific heat.
◦  Water: c = 4.184 g
j-1
°C-1
= m c ΔT
q = C ΔT
Winter 2009
Page 5
Chemistry 123: Physical and Organic Chemistry
Topic 2: Thermochemistry
http://people.ok.ubc.ca/orcac/chem123out.html
Topic 2.0: Introduction,
Winter 2009
Page 6
3
2/9/09
Chemistry 123: Physical and Organic Chemistry
Topic 2: Thermochemistry
http://people.ok.ubc.ca/orcac/chem123out.html
Topic 2.0: Introduction,
Conservation of Energy
In interactions between a system and its surroundings the total
energy remains constant— energy is neither created nor
destroyed.
qsystem + qsurroundings = 0
qsystem = -qsurroundings
Winter 2009
From: http://www.dilbert.com
Page 7
Chemistry 123: Physical and Organic Chemistry
Topic 2: Thermochemistry
http://people.ok.ubc.ca/orcac/chem123out.html
Topic 2.0: Introduction,
Determination of Specific Heat
Winter 2009
From: http://www.dilbert.com
Page 8
4
2/9/09
Chemistry 123: Physical and Organic Chemistry
Topic 2: Thermochemistry
http://people.ok.ubc.ca/orcac/chem123out.html
Topic 2.0: Introduction,
EXAMPLE 7-2
Determining Specific Heat from Experimental Data. Using the data
presented on the last slide to calculate the specific heat of lead. Water is the
surroundings…
qlead = -qwater
qwater = mcΔT = (50.0 g)(4.184 J/g °C)(28.8 - 22.0)°C
qwater = 1.4×103 J
qlead = -1.4×103 J = mcΔT = (150.0 g)(clead)(28.8 - 100.0)°C
clead = 0.13 Jg-1°C-1
Winter 2009
Page 9
Chemistry 123: Physical and Organic Chemistry
Topic 2: Thermochemistry
http://people.ok.ubc.ca/orcac/chem123out.html
Topic 2.0: Introduction,
Heats of Reaction and Calorimetry
♦ Chemical energy.
•  Contributes to the internal energy of a
system.
♦ Heat of reaction, qrxn.
•  The quantity of heat exchanged between a
system and its surroundings when a
chemical reaction occurs within the system,
at constant temperature.
Winter 2009
From: http://www.dilbert.com
Page 10
5
2/9/09
Chemistry 123: Physical and Organic Chemistry
Topic 2: Thermochemistry
http://people.ok.ubc.ca/orcac/chem123out.html
Topic 2.0: Introduction,
Heats of Reaction and Calorimetry
♦  Exothermic reactions.
•  Produces heat, qrxn < 0.
♦  Endothermic reactions.
•  Consumes heat, qrxn > 0.
Add water
♦  Calorimeter
•  A device for measuring quantities
of heat.
Winter 2009
Page 11
Chemistry 123: Physical and Organic Chemistry
Topic 2: Thermochemistry
http://people.ok.ubc.ca/orcac/chem123out.html
Topic 2.0: Introduction,
Winter 2009
qrxn < 0
qrxn > 0
Page 12
6
2/9/09
Chemistry 123: Physical and Organic Chemistry
Topic 2: Thermochemistry
http://people.ok.ubc.ca/orcac/chem123out.html
Topic 2.0: Introduction,
qrxn = -qcal
qcal = q bomb + q water + q wires +…
Define the heat capacity of the
calorimeter:
qcal = ΣmiciΔT = CΔT
heat
all i
C = heat capacity of the calorimeter
Winter 2009
Page 13
Chemistry 123: Physical and Organic Chemistry
Topic 2: Thermochemistry
http://people.ok.ubc.ca/orcac/chem123out.html
Topic 2.0: Introduction,
Using Bomb Calorimetry Data to Determine a Heat of Reaction. The
combustion of 1.010 g sucrose, in a bomb calorimeter, causes the temperature
to rise from 24.92 to 28.33°C. The heat capacity of the calorimeter assembly
is 4.90 kJ/°C. (a) What is the heat of combustion of sucrose, expressed in kJ/
mol C12H22O11? (b) Verify the claim of sugar producers that one teaspoon of
sugar (about 4.8 g) contains only 19 calories.
1 food Calorie ( 1 Calorie with a capital C) is actually 1000 cal or Kcal
Winter 2009
From: http://www.dilbert.com
Page 14
7
2/9/09
Chemistry 123: Physical and Organic Chemistry
Topic 2: Thermochemistry
http://people.ok.ubc.ca/orcac/chem123out.html
Topic 2.0: Introduction,
Calculate qcalorimeter:
qcal = CΔT = (4.90 kJ/°C)(28.33-24.92)°C = (4.90)(3.41) kJ
= 16.7 kJ
per 1.010 g
Calculate qrxn:
qrxn = -qcal = -16.7 kJ
-16.7 kJ
qrxn = -qcal =
= -16.5 kJ/g
1.010 g
343.3 g
qrxn = -16.5 kJ/g
1.00 mol
(a)
= -5.65 × 103 kJ/mol
Calculate qrxn for one teaspoon:
qrxn = (-16.5 kJ/g)(
4.8 g
1 tsp
)(
1.00 cal
4.184 J
)= -19 kcal/tsp
(b)
Winter 2009
Page 15
Chemistry 123: Physical and Organic Chemistry
Topic 2: Thermochemistry
http://people.ok.ubc.ca/orcac/chem123out.html
Topic 2.0: Introduction,
Coffee cup Calorimeter
A simple calorimeter.
•  Well insulated and therefore isolated.
•  Measure temperature change.
qrxn = -qcal
Winter 2009
Page 16
8
2/9/09
Chemistry 123: Physical and Organic Chemistry
Topic 2: Thermochemistry
http://people.ok.ubc.ca/orcac/chem123out.html
Topic 2.0: Introduction,
Winter 2009
Page 17
Chemistry 123: Physical and Organic Chemistry
Topic 2: Thermochemistry
http://people.ok.ubc.ca/orcac/chem123out.html
Topic 2.0: Introduction,
Winter 2009
Page 18
9
2/9/09
Chemistry 123: Physical and Organic Chemistry
Topic 2: Thermochemistry
http://people.ok.ubc.ca/orcac/chem123out.html
Topic 2.0: Introduction,
From: http://www.dilbert.com
Winter 2009
Page 19
Chemistry 123: Physical and Organic Chemistry
Topic 2: Thermochemistry
http://people.ok.ubc.ca/orcac/chem123out.html
Pressure- Volume Work of a reaction
Topic 2.0: Introduction,
In addition to heat effects chemical reactions
may also do work.
Gas formed pushes against
the atmosphere.
Volume changes.
Pressure-volume work.
Winter 2009
Page 20
10
2/9/09
Chemistry 123: Physical and Organic Chemistry
Topic 2: Thermochemistry
http://people.ok.ubc.ca/orcac/chem123out.html
Pressure- Volume Work of a reaction
Topic 2.0: Introduction,
Winter 2009
Page 21
Chemistry 123: Physical and Organic Chemistry
Topic 2: Thermochemistry
http://people.ok.ubc.ca/orcac/chem123out.html
Pressure- Volume Work of a reaction
Topic 2.0: Introduction,
Calculating Pressure-Volume Work. Suppose the gas in the
previous figure is 0.100 mol He at 298 K and the each mass in the
figure corresponds to an external pressure of 1.20 atm. How much
work, in Joules, is associated with its expansion at constant pressure.
Assume an ideal gas and calculate the volume change:
Vi = nRT/P
= (0.100 mol)(0.08201 L atm mol-1 K-1)(298K)/(2.40 atm)
= 1.02 L
Vf = 2.04 L
ΔV = 2.04 L-1.02 L = 1.02 L
w = -PΔV
-101 J )
= -(1.20 atm)(1.02 L)(
1 L atm
= -1.24 × 102 J
Winter 2009
Page 22
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