CH 201 Chapter 9-notes - Linn–Benton Community...
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Transcript of CH 201 Chapter 9-notes - Linn–Benton Community...
2/28/2014
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Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Chapter 9
Energy Flow & Chemical Change
The First Law of Thermodynamics (Law of
Conservation of Energy): The total energy of the universe is constant. Energy can be
neither destroyed nor created.
∆Euniverse = 0
Kinetic Energy: Energy due to motion.
E = ½ mv2
Potential Energy: Stored energy (energy due to position)
Chemical Energy: Energy related to bonds breaking and bonds forming during
chemical reactions
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Examples of Energy Relationships
Units of Energy
• Joule (J)
• calorie (cal)
• Calorie (Cal)
Conversions to know
1,000 J = 1 kJ
1 cal = 4.184 J
1 Cal = 1,000 cal = 1 kcal
Problem Solving
During a snow day, Ron eats 75% of a bag
of Tostitos chips. The bag holds 13.5 oz. One serving is considered to be 6 chips.
The 6 chips is 1 oz. and contains 120. Cal.
How much energy, in kJ, will Ron have to go snowboarding?
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More about Thermodynamics
The system and the surroundings.
The system: The part of the universe whose
change is being observed.
The surroundings: Everything around the system relevant to the change being
observed.
Energy Changes
∆E = Efinal - Einitial or,
∆E = Eproducts – Ereactants
Flow of Energy
Energy can flow into the system from the surroundings
Endothermic process
∆Esystem > 0 ; positive
∆Esurroundings < 0 ; negative
Income Expense
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Or …,
Energy can flow out of the system and into the surroundings
Exothermic process
∆Esystem < 0 ; negative
∆Esurroundings > 0 ; positive
Income Expense
Delta E
The magnitude of ∆E is the same for the
system and the surroundings. However, the sign is opposite.
∆Esystem = -∆Esurroundings or,
-∆Esystem = ∆Esurroundings
Initialstate
Finalstate
Finalstate
Initialstate
Efinal Einitial
En
erg
y, E
En
erg
y, E
Einitial Efinal
A E of system decreases B E of system increases
Efinal < Einitial
∆E < 0Efinal > Einitial
∆E > 0Energy lost tosurroundings
Energy gainedfrom surroundings
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∆E > 0
Einitial Efinal
Tsys > Tsurr
Tsurr
Tsys = Tsurr
∆E < 0
Tsurr
Efinal
Tsys = Tsurr
Tsurr
Tsys < Tsurr
Tsurr
Einitial
En
erg
y, E
En
erg
y, E
Hot H2OTsys
Room tempH2OTsys
B E gained as heatA E lost as heat
Ice H2OTsys
Room tempH2OTsys
Heat (q) lostto surroundings(q < 0)
Heat (q) gainedfrom surroundings(q > 0)
Energy Transfer
Energy can be transferred as heat and/or work.
∆E = heat + work
∆E = q + w
However, in THIS chemistry class: ∆E = q
because systems are open and therefore no work is done.
Enthalpy (H)
Enthalpy (H): Heats of reactions in chemical change.
H = E + PV
However, PV goes to 0 in open systems.
Thus,
∆H = ∆E or,
∆H = qp also,
∆H = ∆Hf - ∆Hi
∆H = ∆Hproducts - ∆Hreactants
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Delta H for Reactions
Assume 0 energy
Example
2 SO2(g) + O2(g) � 2 SO3(g) ∆Hrxn = -198.4 kJ
What does the sign of delta H indicate?
What does this mean in terms of heat
transfer with respect to the system and the surroundings?
Initialstate
Finalstate
Finalstate
Initialstate
Efinal Einitial
En
erg
y, E
En
erg
y, E
Einitial Efinal
A E of system decreases B E of system increases
Efinal < Einitial
∆E < 0Efinal > Einitial
∆E > 0Energy lost tosurroundings
Energy gainedfrom surroundings
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Specific Heat Capacity (c)Specific heat capacity – The quantity of heat
required to change the temperature of 1.00 gram of substance by 1 Kelvin.
The relationship between specific heat, mass, and heat energy is:
q = mc∆Τ
m = mass
c = specific heat capacity
∆T = Tfinal - Tinitial
Note: If a substance absorbs heat ∆T is positive
If a substance releases heat ∆T is negative
Examples of Specific Heats
Al(s) 0.900 J/g-K
Au(s) 0.129 J/g-K
H2O(l) 4.184 J/g-K * Know this one!
CCl4(l) 0.862 J/g-K
Molar Heat Capacity (C)
Molar heat capacity (C) – Amount of heat
required to change the temperature of 1.00 mole of a substance by 1 K.
Problem Solving: What is the molar heat capacity of gold?
Gold's atomic weight = 196.97 g/mol.
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More Examples
1. How much heat is required to increase
the temperature of 3.00 g of gold from 25oC to 55oC?
2. If 9.05 grams of aluminum at 30. oC lost 407.25 J of heat, what is the final
temperature?
Two More Examples
Calcium chloride dissolving in water.
Ammonium nitrate dissolves in water.
Calorimetry
Calorimetry: Lab measurements of ∆Hrxn
- Coffee Cup Calorimeter
- Bomb Calorimeter
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Enthalpy (H)
Enthalpy (H): Heats of reactions in chemical change.
H = E + PV
However, PV goes to 0 in open systems.
Thus,
∆H = ∆E or,
∆H = qp also,
∆H = ∆Hf - ∆Hi
∆H = ∆Hproducts - ∆Hreactants
Coffee Cup Calorimeter
Bomb Calorimeter
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Calorimeters
Recall that
∆Euniverse = ∆Esystem + ∆Esurroundings
And 0 = qsystem + qsurroundings
Then, -qsystem = q surroundings
Reactions in Aqueous SolutionsWhen reactions occur we consider them to be the
system.
Let’s consider the acid/base reaction below in an aqueous solution in a coffee cup calorimeter.
HCl(aq) + NaOH(aq) � H2O(l) + NaCl(aq)
Net Ionic: H+ + OH- � H2O(l)
What is the system? The surroundings? The heat exchange? Explain.
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Measuring Heats of Rxns
Using a coffee cup calorimeter:
System = Rxn
Surroundings = Solution in which the rxn occurs
qsystem = - qsurroundings
= - mc∆T
Bomb Calorimetry
There are TWO ways of looking at bomb
calorimetry.
#1: -qrxn = qcal where qcal = qbomb + qwater
#2: -qrxn = qbomb where the water is already included in the heat capacity of the
bomb calorimeter.
Bomb Calorimeter
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Bomb Calorimetry Problem Solving
Suppose a 0.562 g sample of graphite is
placed in a bomb calorimeter with an excess of oxygen gas at 25.00oC and 1.00
atm. The calorimeter temperature rises to
25.89oC. The heat capacity of the calorimeter is 20.7 kJ/oC. What is the
∆Hrxn per mole of Cgraphite?
Cgraphite + O2(g) � CO2(g)
Chemical Rxns and Bonds
During chemical reactions bonds break and
bonds form.
Bonds breaking is an endothermic process.
Bonds forming is an exothermic process.
Enthalpy and Rxn Stoichiometry
When 2 moles of H2(g) react with 1 mole of
O2(g), water is produced along with 572 kJ of heat. Write a “thermochemical
equation.”
What unit factors can be derived from this
thermochemical equation?
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Problem Solving
How much heat evolves if 52.568 grams of
hydrogen gas react with oxygen gas to produce gaseous water vapor?
Problem Solving
Consider the rxn of solid calcium oxide reacting with solid carbon to form calcium carbide and carbon monoxide. For this reaction ∆H = -464.8 kJ.
(a) Is heat released or absorbed?
(b) Is the reaction exothermic or endothermic?
(c) If 15.00 grams of carbon react with excess calcium oxide, how much energy is produced?
(d) If 986.21 kJ of heat are produced, how many grams of carbon monoxide are produced?
Standard Heat of Reaction (∆Horxn)
For standard heats of reaction solutions are
1 Molar, gases are at 1 atm, temperature is 25oC. The pure substances (elements
and compounds) must be in their most
stable form.
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Formation Reactions (∆Hof)
A formation reaction is one that forms 1 mole of a compound from the most stable form of its elements.
Note: ∆Hof = 0 for the most stable form of an
element. See appendix
Examples: Write the formation reaction for ammonium nitrate.
Write the formation reaction for sodium sulfite.
Delta H for Reactions
Assume 0 energy
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Using ∆Hof to Determine ∆Ho
rxn
∆Horxn = Σn ∆Ho
f(products) - Σn ∆Hof(reactants)
Example: Determine the heat of reaction for the following unbalanced equation:
NH3(g) + O2(g) � NO(g) + H2O(g)
Hess’s Law
The enthalpy change, ∆H, of an overall process is the sum of the individual enthalpy changes of
each step.
Example: Calculate ∆Hrxn for . . .
2H2(g) + O2(g) � 2 H2O(l)
given the following:
H2(g) + ½ O2(g) � H2O(g) ∆H = -241.8 kJ
H2O(g) � H2O(l) ∆H = -44.0 kJ
Example: Calculate ∆Hrxn for . . .
2H2(g) + O2(g) � 2 H2O(l)
given the following:
H2(g) + ½ O2(g) � H2O(g) ∆H = -241.8 kJ
H2O(g) � H2O(l) ∆H = -44.0 kJ
2H2(g) +O2(g) � 2H2O(g) ∆Hrxn = -482.6 kJ
2H2O(g) � 2H2O(l) ∆Hrxn = -88.0 kJ
2H2(g) +O2(g) + 2H2O(g) � 2H2O(g) + 2H2O(l) ∆Hrxn = -562.6 kJ
2H2(g) +O2(g) � 2H2O(l) ∆Hrxn = -562.6 kJ
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Hess’s Law
E
H2O(l)
H2O(g)
H2, O2
ΔHrxn
ΔHrxn
ΔHrxn
ΔHrxn = ΔHrxn + ΔHrxn
Algebra Review
2X + 1 = 3Y + 4
6Y + 8 = 9
4X + 2 = 6Y + 8
6Y + 8 = 9
4X + 2 + 6Y + 8 = 6Y + 8 + 94X + 2 = 9
Another Example
Determine ∆Ho for the rxn . . .
C2H6(g) � C2H4(g) + H2(g)
Given,2 C2H6(g) + 7O2(g) ���� 4 CO2(g) + 6 H2O(l) ∆∆∆∆Ho = -3119.4 kJ
C2H4(g) + 3 O2(g) ���� 2 CO2(g) + 2 H2O(l) ∆∆∆∆Ho = -1410.9 kJ
2 H2(g) + O2(g) ���� 2 H2O(l) ∆∆∆∆Ho = -571.66 kJ
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Review: Phase Changes - Changes in Physical States
Solid � Liquid Melting (Fusion)
Liquid � Solid Freezing
Liquid � Gas Vaporization
Gas � Liquid Condensation
Solid � Gas Sublimation
Gas � Solid Deposition
Examples of Physical Properties of
Three States of Matter
Which of the physical changes are endothermic
or exothermic?
Endothermic Exothermic
Melting Freezing
Sublimation Deposition
Vaporization Condensation
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∆H of a Phase Change: The amount of
energy required to produce a phase change
for one mole of a substance (kJ/mol).
Note: The VALUE of ∆Hvap. = ∆Hcond.
However, the sign is different
depending on exothermic or
endothermic processes.
Phase Changes of pure substances require a
specific amount of energy per mole (∆H)
Examples
H2O(l) � H2O(g) ∆Hvap = 40.7 kJ/mol
H2O(g) � H2O(l) ∆Hvap = - 40.7 kJ/mol
H2O(s) � H2O(l) ∆Hfus = 6.02 kJ/mol
H2O(l) � H2O(s) ∆Hfus = - 6.02 kJ/mol
Enthalpy of Vaporization & Fusion for Various Substances
HgMP = -38oC
BP = 357oC
ArMP = -189oC
BP = -186oC
C6H6
MP = 6 oC
BP = 80 oC
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Quantitative Aspects of Phase Changes
Calculating the amount of energy for a substance to undergo phase changes.
If you had 25.00 grams of water at 130.0oC, how much energy would be released when the water is cooled to
-40.0oC.
Some important facts to know:
Specific Heat of H2O(g): 33.1 J/mol-oCSpecific Heat of H2O(l) : 75.4 J/mol-oC (4.184 J/go C)
Specific Heat of H2O(s): 37.6 J/mol-oC
Stage 1 Stage 5
GAS – LIQUID
LIQUID
∆H 0vap
LIQUID–SOLID
Stage 4Stage 3Stage 2
Tem
pera
ture
(ºC)
SOLID
GAS
100
0
130
– 40
∆H 0fus
Heat removed
Tem
pera
ture
(ºC)
GAS
100
0
130
– 40Heat removed
Stage 1
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Calculating the amount of heat absorbed or released when a substance undergoes a temperature change:
q = mc∆T or q = nc∆T
n = (25.00gH2O)(1mol H2O)
(18.02 g H2O)
n = 1.387 mol H2O
q = (1.387 mol)(33.1J/moloC)(100.0oC -130.0oC)
q = -1377.291 J
GAS – LIQUID
∆H 0vap
Tem
pera
ture
(ºC)
GAS
100
0
130
– 40Heat removed
Stage 1 Stage 2
Calculating the amount of heat gained or
lost during a phase change.
q = n∆H?
q = 1.387 mol (- 40.7 kJ/mol)
q = -56.4509 kJ
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GAS – LIQUID
LIQUID
∆H 0vap
Tem
pera
ture
(ºC)
GAS
100
0
130
– 40Heat removed
Stage 1 Stage 2 Stage 3
q = nc∆T
q = 1.387mol (75.4 J/moloC)(0.0oC – 100.0oC)
q = -10457.98 J
GAS – LIQUID
LIQUID
∆H 0vap
LIQUID–SOLID
Tem
pera
ture
(ºC)
GAS
∆H 0fus
100
0
130
– 40Heat removed
Stage 1 Stage 4Stage 3Stage 2
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q = n∆Hfus
q = 1.387mol(-6.02 kJ/mol)
q = -8.34974 kJ
GAS – LIQUID
LIQUID
∆H 0vap
LIQUID–SOLID
Heat removed
Tem
pera
ture
(ºC)
SOLID
GAS
∆H 0fus
100
0
130
– 40
Stage 1 Stage 5Stage 4Stage 3Stage 2
q = nc∆T
q = 1.387mol (37.6 J/moloC)(-40.0oC – 0.0oC)
q = -2086.048 J
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Total Heat Change
q = -1377.291 J + -56.4509 kJ +
-10457.98 J + -8.34974 kJ + -2086.048 J
However, you can’t add kJ with J!
1,000 J = 1 kJ
-1.377291 kJ
-56.4509 kJ
-10.45798 kJ
-8.34974 kJ
-2.086048 kJ
-78.721959 kJ
Final Answer: - 78.7 kJ of heat are released
GAS – LIQUID
LIQUID
∆H 0vap
LIQUID–SOLID
Heat removed
Tem
pera
ture
(ºC)
SOLID
GAS
∆H 0fus
100
0
130
– 40
Stage 1 Stage 5Stage 4Stage 3Stage 2
-40°C to 130°C:
130°C to -40°C: - 78.7 kJ
+78.7 kJ
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Possible Exam Question(s)
You should be able to calculate the amount
of heat released or absorbed for any substance given ∆H’s, C’s, and temps.
ΔEuniverse = 0
ΔHrxn ≈ ΔE ≈ qsys
qsys = –qsurr
q = mcΔT or ncΔT
q = nΔHvap or fus
ΔH°rxnA = ΔH°rxnB + ΔH°rxnC (Hess’s Law)
ΔH°rxn = ΣmΔHf°prdt – ΣnΔHf°react. (Hess’s Law)