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Intersection 6: Thermochemistry
10/10/05
6.4 p228-232; 6.7 p240-242; 6.10-6.11 p 248-256
11.3-11.4 p494-507
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A Review of Thermochemistry (from
Studio)
• Thermochemical Equations
C(graphite) + O2(g)→ CO2(g) ΔH = -393.509 kJ/mol
• Endothermic and exothermic
• Heats of reactions are additive (Hess’s Law)
• Calorimetry
– How heats of reaction are measured
– Heat capacity q = mCDT
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Problem 1
Suppose you burn 0.300 g C(graphite) in an excess
of O2(g) to give CO2(g). The temperature of
the calorimeter which contains 775 g of water
increases from 25.00oC to 27.38oC.
What is the quantity of thermal energy evolved
per mole of graphite?
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Question 1
100 mL of water at 25°C and 100 mL of alcohol at 25°C
are both heated at the same rate under identical
conditions. After 3 minutes the temperature of the
alcohol is 50°C. Two minutes later the temperature
of the water is 50°C.
Which liquid received more heat as it warmed to 50°C?
a. The water.
b.The alcohol.
c. Both received the same amount of heat.
d.It is impossible to tell from the information given.
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Do all substances have the same
heat capacity?
It takes 145 J to raise the temperature of 1 mole of
air by 5oC.
Is the heat capacity of air higher, lower, or the same
as water?
Explain why at the beginning of the summer, the air
is really warm, but the water at the beach is not.
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Thermochemistry (in IS)
• Heat and bonds
– Breaking bonds takes energy
– Making bonds releases energy
– Trends in bond enthalpy
• Calculate Heats of Reactions
– Heats of reaction can be calculated using bond enthalpy
– Heat of reactions can be calculated using heats of
formation
• Change of state and heat energy
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Bonds
Why do bonds form?
Does it take energy or give off energy to
break a bond?
Picture from: www.nios.ac.in/ sc10/ch5sc10.htm
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The energy given off to make H-H bonds is shown
below.
H(g) + H(g) → H2(g) Δ H = -435 kJ/mol
The opposite process, breaking H-H bonds requires
the input of the same amount of energy that was
given off in making the bond
H2(g) → H(g) + H(g)
Δ H = +435 kJ/mol
Which process is endothermic? Exothermic?
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Bond Enthalpy (kJ/mol)
(average energy needed to break bonds)
H-F
H-Cl
H-Br
H-I
C-Si
C-P
C-S
C-Cl
566
431
366
299
301
264
272
327
C=C
C=N
C=O
C-C
C-N
C-O
C-F
598
C-O
418
C=O
695-803 C≡O
356
285
336
486
C-C
C=C
C≡C
336
695
1073
356
598
813
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Energy terms
Heat energy: enthalpy, DH, q
Enthalpy (H) is the quantity of thermal energy
transferred into a system at constant pressure.
Δ E = Δ H + wexpansion
"ΔH accounts for all the energy transferred except
the quantity that does the work of pushing back
the atmosphere, which is usually relatively
small. Whenever heat transfer is measured at
constant pressure, it is ΔH that is determined."
(Moore, p232)
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Holy Contradiction!
It takes energy to break
bonds….and yet we get
energy from fats (and
carbohydrates and
alcohol) by breaking
them down ???
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How do we get energy from
food?
Food is broken down by adding O2 and
converting all of the carbon and hydrogen in
the food molecules to CO2 and H2O.
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Energy from Ethanol
CH3CH2OH + 3 O2 → 2 CO2 + 3H2O
H
H
H
C
C
H
H
O
moles
H
+
3 O
Bonds
broken
2 O
O
Enthalpies
kJ/mol
moles
C
O
+
O
3 H
Bonds Enthalpies
made kJ/mol
5
C-H
416
4
C=O
803
1
C-C
356
6
O-H
467
1
C-O
336
1
O-H
467
3
O=O
498
Total
4733 kJ
(-) 6014 kJ
H
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A Picture of Breaking Down Ethanol
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Total D H
H
H
H
C
C
H
H
O
H
+
4733 kJ into reactants
3 O
O
2 O
C
O
+
O
3 H
6014 kJ out of products
Δ Hreaction = Σ Hreactants - Σ Hproducts
= -1281 kJ
Watch your signs!!!!!!!!!!….!!!!!
H
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Breaking Down Glucose
H OH
C CH2
H
O
6 2
C6H12O6 + ___O
C
HO
HO
C
C
H
H
OH
C
OH
6
6
____CO
2 + ___H2O
H
moles
7
bonds kJ/mol
moles bonds kJ/mol
C-H
416
12
C=O
803
5
C-C
356
12
O-H
467
7
C-O
336
5
O-H
467
6
O=O
498
Total
12217 kJ
Δ Hreaction = Σ Hreactants - Σ Hproducts = -3023 kJ
15240 kJ
Inflaming Nitrocellulose
O
N
H OH
C CH2
H
HO
C
C
C
H
H
O
N
H
O
C
O
O
OH
H
C CH2
HNO3/H2SO4
O
C
O
O
H O
C
C
C
H
H
O
O
O
H
O
N
O
n
O
n
Cellulose Oxidation
C6O5H10 + 6 O2
6 CO2 + 5 H2O
Nitrocelluose Oxidation
4 C6O11N3H7 + 9 O2
24 CO2 + 14 H2O + 6 N2
DHreaction Comparison
Nitrocellulose Combustion
Cellulose Combustion
Reactants
Reactants
Products
Products
Moles
Bonds
kJ/mol
Moles
Bonds
kJ/mol
Moles
Bonds
kJ/mol
Moles
Bonds
kJ/mol
7
C-H
416
12
C=O
803
28
C-H
416
48
C=O
803
5
C-C
356
10
O-H
467
20
C-C
356
28
O-H
467
7
C-O
336
28
C-O
336
6
N≡N
946
3
O-H
467
-
O-H
467
6
O=O
498
9
O=O
498
24
N-O
201
12
N=O
607
Total
11,433
14306
Δ Hreaction = -2870 kJ/mol
Total
44,776
57,296
Δ Hreaction = -12,520 kJ/mol
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Conversion to food Calories
• 1 Calorie = 1 kilocalorie = 1000 calories
• 1 calorie = 4.184 J
• Convert the enthalpies for glucose and ethanol to Cal and
Cal/g:
For glucose
3023 kJ
mol
1000J
1 kJ
722.5 Cal
1 cal
4.184 J
1Cal
1000 cal
1 mol glucose
= 4.0 Cal/g
mol glucose
180 g
= 722.5 Cal/mol
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Comparing Energy Sources
Bond
Energy
Nutritional
Guidelines*
Ethanol
6.7 Cal/g
7 Cal/g
Glucose
4.0 Cal/g
4 Cal/g
Stearic Acid 8.9 Cal/g
9 Cal/g
C18H36O2 + 26 O2
18 CO2 + 18 H2O
DH = -10622 kJ
Stearic acid
*http://www.unlimited-health.com/nutrition/
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Evolution due to fat storage…..
"Animals, though they store small amounts of glycogen in
the muscles and liver, and rely on glucose in the
bloodstream for immediate energy needs, bank most of
their energy reserves in fat. Economy in weight is
important for an organism that must work not only to
move, but simply to stand up against the force of
gravity. The body of a typical woman is about 25% fat
by weight (a man's body is closer to 15%). This means
that if her fat supply were converted to its energy
equivalent in carbohydrates, a 120-pound woman
would weigh 150 pounds. In a world where swiftness
and agility increase an animal's chances for survival, fat
is clearly the preferable means of energy storage.
(Harold McGee, On Food and Cooking. Scribners,
New York: 1984, p. 597.)
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Who wants to count up bonds every time?
Δ Hof standard enthalpy of formation is the energy
given off or needed to form a substance from its
elements in their standard states [ie, O2(g), H2(g),
N2(g), Br2(l), Na(s)..etc].
The o stands for standard conditions of 1 atm and
298K.
The formation reaction of gaseous water would be written:
H2(g) + 1/2 O2(g) → H2O(g)
Δ Hrxn (1 atm, 298K) = Δ Hof H2O(g) = -241.83 kJ/mol
Δ Hof found in Appendix J of Moore or back of coursepack
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Table 6-2, p.250
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Standard Enthalpy of Formation
H2(g) + 1/2 O2(g) → H2O(g)
Δ Hof H2O(g) = -241.83
kJ/mol
Do you get a different answer from bond enthalpies?
moles
Total
bonds kJ/mol
H-H
436
O=O
498
moles
bonds
kJ/mol
O-H
467
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DHof : Signs and Multiplicity
C(graphite) + O2(g)→ CO2(g)
ΔH = -393.509 kJ/mol
CO2(g) → C(graphite) + O2(g) ΔH = +393.509 kJ/mol
2C(graphite) +2 O2(g)→ 2CO2(g)
ΔH = -787.018 kJ/mol
4CO2(g) → 4C(graphite) + 4O2(g) ΔH = +1574.04 kJ/mol
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Calculate the DHrxn using heats of formation
CH3CH2CH3 (g) + 5 O2 (g) → 3 CO2 (g) + 4 H2O (g)
3C(s,graphite)+ 4H2(g) → CH3CH2CH3 (g) DH = - 103.82 kJ/mol
O2 (g)
C(s,graphite)+ O2 (g) → CO2 (g)
H2(g) + ½ O2(g) → H2O(g)
DH = 0 kJ/mol
DH = - 393.5 kJ/mol
DH = - 241.82 kJ/mol
CH3CH2CH3(g) → 3C(s,graphite) + 4H2(g)
3C(s,graphite)+ 3O2 (g) → 3CO2 (g)
4H2(g) + 2 O2(g) → 4H2O(g)
DH = + 103.82 kJ/mol
3(DH = - 393.5 kJ/mol)
4(DH = - 241.82 kJ/mol)
-2043.96 kJ/mol
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Calculate the DHrxn using heats of formation
CH3CH2CH3 (g) + 5 O2 (g) → 3 CO2 (g) + 4 H2O (g)
Δ Hrxn = Σ nHof products - Σ nHof reactants
Form the products (keep sign)
Unform the reactants (change sign)
Σ nΔ Hof products = -393.5 kJ/mol (3 mol CO2) + -241.83 kJ/mol (4 mol H2O)
= - 2147.82 kJ
Σ nΔ Hof reactants = -103.82 kJ/mol (1 mol propane) + 0 kJ/mol (5 mol O2)
= -103.82 kJ
Δ Hrxn = Σ nHof products - Σ nHof reactants
= -2147.82 kJ - (-103.82 kJ)
= -2044 kJ
Iron Oxidation
4 Fe + 3 O2
2 Fe2O3
DHf for Fe2O3 = -824.2 kJ/mol
What is DHf for Fe and O2??
Δ Hrxn = Σ nHof products - Σ nHof reactants
Δ Hrxn = Σ nHof products - Σ nHof reactants
= 2(-824.2 kJ/mol) - 0
= -1648 kJ/mol
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Problem 2
2 C8H6(l) + 25 O2(g) 16 CO2(g) + 18 H2O(g) DH = -10,992 kJ
Write a thermochemical equation for:
a) The production of 4.00 mol CO2(g)
b) The combustion of 100. mol C8H6
c) How much heat energy is produced when 50 g of
C8H6 is burned?
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Problem 3
The standard molar enthalpy of formation of
AgCl(s) is –127.1 kJ/mol. Write a balanced
thermochemical equation for which the
enthalpy of reaction is –127.2 kJ
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Problem 4
The Romans used CaO as mortar in stone
structures. The CaO was mixed with water
to give Ca(OH)2, which slowly reacted with
CO2 in the air to give limestone.
Ca(OH)2(s) + CO2(g)  CaCO3(s) + H2O(g)
Calculate the enthalpy change for this reaction
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Question 2
Heat is given off when hydrogen burns in air
according to the equation 2H2 + O2 2H2O
Which of the following is responsible for the heat?
a. Breaking hydrogen bonds gives off energy.
b. Breaking oxygen bonds gives off energy.
c. Forming hydrogen-oxygen bonds gives off energy.
d. Both (a) and (b) are responsible.
e. (a), (b), and (c) are responsible.
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Debate: _________is the safest and
best artificial sweetener
Friday, October 13th in studio
–
–
–
–
Splenda (sucralose)
Aspartame
Saccharin
Acesulfame K
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Opening
Rebuttal
Question 1
Response 1
Question 2
Response 2
Question 3
Response 3
Question 4
Response 4
Closing
Chips Points
3
9
3
5
1
2
1
3
1
2
1
3
1
2
1
3
1
2
1
3
3
6
Time
6 min
3 min
30 sec
1 min
30 sec
1 min
30 sec
1 min
30 sec
1 min
3 min