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2/9/09
1
Chemistry 123: Physical and Organic Chemistry Topic 2: Thermochemistry http://people.ok.ubc.ca/orcac/chem123out.html
Winter 2009 Page 1
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)
Topic 2: Introduction,
From: http://www.dilbert.com
Chemistry 123: Physical and Organic Chemistry Topic 2: Thermochemistry http://people.ok.ubc.ca/orcac/chem123out.html
Winter 2009 Page 2
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
Topic 2.0: Introduction,
2/9/09
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Chemistry 123: Physical and Organic Chemistry Topic 2: Thermochemistry http://people.ok.ubc.ca/orcac/chem123out.html
Winter 2009 Page 3
Topic 2.0: Introduction, Kinetic Energy
ek = 1 2
mv2 [ek ] = m s
= J kg 2
Work
w = Fd [w ] = kg m s2
= J m
From: http://www.dilbert.com
Chemistry 123: Physical and Organic Chemistry Topic 2: Thermochemistry http://people.ok.ubc.ca/orcac/chem123out.html
Winter 2009 Page 4
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
Topic 2.0: Introduction,
James Joule (1818-1889)
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Chemistry 123: Physical and Organic Chemistry Topic 2: Thermochemistry http://people.ok.ubc.ca/orcac/chem123out.html
Winter 2009 Page 5
Topic 2.0: Introduction,
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 ◦ Mass times c = C specific heat.
◦ Water: c = 4.184 g j-1 °C-1
q = m c ΔT
q = C ΔT
Heat Capacity
Chemistry 123: Physical and Organic Chemistry Topic 2: Thermochemistry http://people.ok.ubc.ca/orcac/chem123out.html
Winter 2009 Page 6
Topic 2.0: Introduction,
2/9/09
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Chemistry 123: Physical and Organic Chemistry Topic 2: Thermochemistry http://people.ok.ubc.ca/orcac/chem123out.html
Winter 2009 Page 7
Conservation of Energy Topic 2.0: Introduction,
From: http://www.dilbert.com
In interactions between a system and its surroundings the total energy remains constant— energy is neither created nor destroyed.
qsystem + qsurroundings = 0
qsystem = -qsurroundings
Chemistry 123: Physical and Organic Chemistry Topic 2: Thermochemistry http://people.ok.ubc.ca/orcac/chem123out.html
Winter 2009 Page 8
Determination of Specific Heat Topic 2.0: Introduction,
From: http://www.dilbert.com
2/9/09
5
Chemistry 123: Physical and Organic Chemistry Topic 2: Thermochemistry http://people.ok.ubc.ca/orcac/chem123out.html
EXAMPLE 7-2
Winter 2009 Page 9
Topic 2.0: Introduction,
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
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…
Chemistry 123: Physical and Organic Chemistry Topic 2: Thermochemistry http://people.ok.ubc.ca/orcac/chem123out.html
Winter 2009 Page 10
Topic 2.0: Introduction,
From: http://www.dilbert.com
♦ 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.
Heats of Reaction and Calorimetry
2/9/09
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Chemistry 123: Physical and Organic Chemistry Topic 2: Thermochemistry http://people.ok.ubc.ca/orcac/chem123out.html
Winter 2009 Page 11
Topic 2.0: Introduction, Heats of Reaction and Calorimetry ♦ Exothermic reactions.
• Produces heat, qrxn < 0. ♦ Endothermic reactions.
• Consumes heat, qrxn > 0.
♦ Calorimeter • A device for measuring quantities
of heat.
Add water
Chemistry 123: Physical and Organic Chemistry Topic 2: Thermochemistry http://people.ok.ubc.ca/orcac/chem123out.html
qrxn < 0 qrxn > 0 Winter 2009 Page 12
Topic 2.0: Introduction,
2/9/09
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Chemistry 123: Physical and Organic Chemistry Topic 2: Thermochemistry http://people.ok.ubc.ca/orcac/chem123out.html
Winter 2009 Page 13
Topic 2.0: Introduction,
qrxn = -qcal
qcal = q bomb + q water + q wires +…
Define the heat capacity of the calorimeter: qcal = ΣmiciΔT = CΔT
all i
heat
C = heat capacity of the calorimeter
Chemistry 123: Physical and Organic Chemistry Topic 2: Thermochemistry http://people.ok.ubc.ca/orcac/chem123out.html
Winter 2009 Page 14
Topic 2.0: Introduction,
From: http://www.dilbert.com
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
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Chemistry 123: Physical and Organic Chemistry Topic 2: Thermochemistry http://people.ok.ubc.ca/orcac/chem123out.html
Winter 2009 Page 15
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
Calculate qrxn: qrxn = -qcal = -16.7 kJ per 1.010 g
qrxn = -qcal = -16.7 kJ
1.010 g
= -16.5 kJ/g
343.3 g
1.00 mol = -16.5 kJ/g
= -5.65 × 103 kJ/mol
qrxn
(a)
Calculate qrxn for one teaspoon:
4.8 g
1 tsp
= (-16.5 kJ/g)( qrxn (b) )( )= -19 kcal/tsp 1.00 cal 4.184 J
Chemistry 123: Physical and Organic Chemistry Topic 2: Thermochemistry http://people.ok.ubc.ca/orcac/chem123out.html
Winter 2009 Page 16
Coffee cup Calorimeter Topic 2.0: Introduction,
A simple calorimeter. • Well insulated and therefore isolated. • Measure temperature change.
qrxn = -qcal
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Chemistry 123: Physical and Organic Chemistry Topic 2: Thermochemistry http://people.ok.ubc.ca/orcac/chem123out.html
Winter 2009 Page 17
Topic 2.0: Introduction,
Chemistry 123: Physical and Organic Chemistry Topic 2: Thermochemistry http://people.ok.ubc.ca/orcac/chem123out.html
Winter 2009 Page 18
Topic 2.0: Introduction,
2/9/09
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Chemistry 123: Physical and Organic Chemistry Topic 2: Thermochemistry http://people.ok.ubc.ca/orcac/chem123out.html
Winter 2009 Page 19
Topic 2.0: Introduction,
From: http://www.dilbert.com
Chemistry 123: Physical and Organic Chemistry Topic 2: Thermochemistry http://people.ok.ubc.ca/orcac/chem123out.html
Winter 2009 Page 20
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.
Pressure- Volume Work of a reaction
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Chemistry 123: Physical and Organic Chemistry Topic 2: Thermochemistry http://people.ok.ubc.ca/orcac/chem123out.html
Winter 2009 Page 21
Topic 2.0: Introduction, Pressure- Volume Work of a reaction
Chemistry 123: Physical and Organic Chemistry Topic 2: Thermochemistry http://people.ok.ubc.ca/orcac/chem123out.html
Winter 2009 Page 22
Topic 2.0: Introduction, Pressure- Volume Work of a reaction
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
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.
ΔV = 2.04 L-1.02 L = 1.02 L
w = -PΔV = -(1.20 atm)(1.02 L)( = -1.24 × 102 J
) -101 J
1 L atm