Download Energy, Enthalpy, and Thermochemistry

CHEM 1310: Update
Final Exam: Mon, Dec 10 (2:50 - 5:40 pm)
Exam 1
Chapters
1-5
Exam 3
Fri., Nov 16
Week 12
Exam 2
Exam 3
Chapters
12, 13, 16, 17
Chapters
6-9
Chapters
10, 15
 Chemical Equilibrium
 Acids and Bases
 Applications of Aqueous Equilibria
 Energy, Enthalpy, and Thermochemistry
CHEM 1310 - Sections L and M
1
Energy, Enthalpy, & Thermochemistry
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9.1
9.2
9.3
9.4
9.5
9.6
9.7
9.8
Week 12
The Nature of Energy
Enthalpy
Thermodynamics of Ideal Gases
Calorimetry
Hess’s Law
Standard Enthalpies of Formation
Present Sources of Energy
New Energy Sources
CHEM 1310 - Sections L and M
2
What is Energy?

Electromagnetic Energy: inversely related to wavelength
∆E = (h x c)/ λ
Week 12
From Ch. 12 - Quantum Mechanics
CHEM 1310 - Sections L and M
3
1
What is Energy?

Chemical (Mechanical) Energy: the capacity to do work
or to produce heat
ΔE = q + w
where E is the internal energy of a system
• q = Heat Absorbed by a system
 If q > 0 , heat is absorbed
 If q < 0 , heat is given off
• w is the work done on the “body”
In Ch. 9,
focus on the
transfer of E
via heat
Recall
 w=Fxd
 w = ΔEKinetic= Δ (½ mv2)
 w = ΔE potential= mgΔh
Week 12
CHEM 1310 - Sections L and M
4
Units of Energy
 Joule = 1 N m = kg x m2 x s-2
 SI Unit
 Calories
 Food energy
 1 calorie = 4.186 J
 British Thermal Units (BTU)
James Prescott Joule
(1818-1889)
 Heating & Air, Power, etc.
Week 12
CHEM 1310 - Sections L and M
5
Nature of Energy
 Law of Conservation of Energy
 Energy can be converted from one form to
another, but energy cannot be created or
destroyed.
 Kinetic: energy of an object due to motion
 Potential: “stored” energy; energy an object
has the capability of using if it were in
motion
Week 12
CHEM 1310 - Sections L and M
6
2
Kinetic vs. Potential Energy
 Potential energy (e.g., ΔEpotential= mgΔh)
 Kinetic energy (e.g., ΔEKinetic= Δ (1/2mv2)
 Energy transfer
 Through heat (frictional heating)
 Through work
Work = Force over a distance
Week 12
CHEM 1310 - Sections L and M
7
Endothermic Reactions
 Reactants + Heat
Week 12
Products
CHEM 1310 - Sections L and M
8
Exothermic Reactions
 Reactants
Products + Heat
2 Al (s) + Fe2O3 (s) → Al2O 3 (s) + 2Fe (s)
Thermite Reaction
Week 12
CHEM 1310 - Sections L and M
9
3
Internal Energy Flow
An Exothermic Process
q=–x
Week 12
An Endothermic Process
q=+x
CHEM 1310 - Sections L and M
10
Energy as Work
Work = F x d
Work = F x Δh
Work = P x A x Δh
Since ΔV = A Δh
Work = P x ΔV
Week 12
CHEM 1310 - Sections L and M
11
Energy as Work
 Force exerted by piston.
 Expansion vs. Compression
Week 12
CHEM 1310 - Sections L and M
w = −Pext ΔV
12
4
Energy as Work
w = – Pext ΔV
 Expansion
 ΔV > 0 therefore w < 0
 The system does work on the surroundings
 Compression
 ΔV < 0 therefore w > 0
 The surroundings have done work on the system
Week 12
CHEM 1310 - Sections L and M
13
Enthalpy
Enthalpy:
Heat content
of a system
H = U + PV
Enthalpy
Internal Energy
of System
Normally measure change in enthalpy, ΔH
If ΔH positive then q > 0
Heat is absorbed Endothermic
qpressure = ΔH
Week 12
If ΔH negative then q < 0
Heat is given off  Exothermic
CHEM 1310 - Sections L and M
14
Enthalpy is a State Function
 A state function is a property that depends on its
present state, not on any path leading to that state.
Analogy
Week 12
CHEM 1310 - Sections L and M
15
5
Enthalpy of a Reaction
ΔHrxn = Hproducts – Hreactants
CO (g) + ½ O2 (g) → CO2 (g) ΔH= -283 kJ
Exothermic!
If molar ratios are doubled, then enthalpy doubles…
2 CO (g) + 1 O2 (g) → 2 CO2 (g) ΔH= -566 kJ
If the rxn is reversed, then enthalpy sign changes…
CO2 (g) → CO (g) + ½ O2 (g) ΔH= +283 kJ
Week 12
Endothermic!
CHEM 1310 - Sections L and M
16
Example Problem
How much heat energy is needed to decompose 9.74g of
HBr (MW = 80.9 g/mol) into its elements?
H2 + Br2 → 2HBr (g)
ΔH = -72.8 kJ mol-1
Note: decomposition = reverse reaction
2 HBr (g) → H2 + Br2
Week 12
ΔH = +72.8 kJ mol-1
CHEM 1310 - Sections L and M
17
Example Problem
How much heat energy is needed to decompose
9.74g of HBr (M =80.9 g/mol) into its elements?
Solution
2HBr (g) → H2 + Br2
ΔH = +72.8 kJ mol-1
Ê 72.8 kJ ˆÊ 1 mol HBr ˆ
q = DH = Á
˜Á
˜(9.74 g HBr)
Ë 2 moles HBr ¯Ë 80.91g HBr ¯
= 4.38 kJ
Week 12
CHEM 1310 - Sections L and M
18
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