6-Ch6

Report
1
All Bold Numbered Problems
2
Chapter 6 Outline
•
•
•
•
•
Energy -vs- Heat
Specific Heat
First Law of Thermodynamics
q (Heat)
Hess’s Law
THERMOCHEMISTRY
or
Thermodynamics
3
4
Energy & Chemistry
• Burning peanuts
supply sufficient
energy to boil a cup
of water.
• Burning sugar
(sugar reacts with
KClO3, a strong
oxidizing agent)
5
Energy and Chemistry
2 H2(g) + O2(g) --> 2 H2O(g) + heat and light
This can be set up to provide
ELECTRIC ENERGY in a fuel
cell.
Oxidation:
2 H2 ---> 4 H+ + 4 e-
Reduction:
4 e- + O2 + 2 H2O ---> 4 OH-
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Energy and Chemistry
ENERGY is the capacity to do work
or transfer heat.
HEAT is the form of energy that flows
between 2 samples because of their
difference in temperature.
Other types of energy
• light
• electrical
• kinetic and potential
7
Kinetic and Potential Energy
Potential energy:
energy stored in
chemical due to
its structure.
The energy of a
motionless body
due to its position.
8
Kinetic and Potential Energy
Kinetic energy:
energy of motion.
• Translational
• Rotational
• Vibrational
(IR)
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Thermodynamics
Thermodynamics is the science
of heat (energy) transfer.
Heat energy is associated
with molecular motions.
10
Energy and Chemistry
All of thermodynamics
depends on the law of
THE CONSERVATION
OF ENERGY.
• The total energy of a
system is constant.
11
UNITS OF ENERGY
1 calorie = heat required to raise
temp. of 1.00 g of H2O by 1.0 oC.
1000 cal = 1 kilocalorie = 1 kcal
1 kcal = 1 Calorie (a food “calorie”)
The S.I sytem uses
the unit called the
JOULE
1 cal = 4.18 Joules
James Joule
1818-1889
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Specific Heat Capacity
Thermochemistry is the science of
heat (energy) flow. A difference in
temperature leads to energy transfer.
The heat “lost” or “gained” is related to
a)
sample mass
b)
change in T
c)
specific heat capacity
Specific heat capacity
=
heat lost or gained by substance (J)
(mass, g)(T change, K)
13
Specific Heat Capacity
Substance
H2O
Al
glass
Spec. Heat (J/g•K)
4.18
0.902
0.84
Aluminum
14
CHEMICAL REACTIVITY
What drives chemical reactions?
How do they occur?
The first is answered by THERMODYNAMICS
and the second by KINETICS.
Have already seen a number of “driving forces”
for reactions that are PRODUCT-FAVORED.
• formation of a precipitate
• gas formation
• H2O formation (acid-base reaction)
• electron transfer in a battery
CHEMICAL REACTIVITY
15
But ENERGY TRANSFER also allows us to
predict reactivity.
In general, reactions that transfer energy to
their surroundings are product-favored.
So, let us consider heat transfer in chemical
processes.
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Heat Energy Transfer in
Chemical Processes
CO2 (s, -78 oC) ---> CO2 (g, -78 oC)
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FIRST LAW OF
THERMODYNAMICS
heat energy transferred (to the system)
DE = q + w
energy
change
work done
(by the
surroundings)
Energy is conserved!
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The First Law of Thermodynamics is
the law of conservation of energy.
DE = q + w
DE is the change in Energy, q is heat, w is work
• Enthalpy (H), defined H = E + PV.
DH = DE + PDV at constant pressure (only V and T change).
• If w = - PDV (because V does not change in a typical lab experiment,
only T changes) and
DH = q + w + PDV
DH = qp (for chemist in a lab which means
only T changes)
ENTHALPY
Most chemical reactions occur at
constant P.
qp =
DH
DH = change in heat content of the system
DH = Hfinal - Hinitial
How do we measure q in the lab?
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20
Specific Heat Capacity
If 25.0 g of Al cool
from 310 oC to 37 oC,
how many joules of
heat energy are lost
by the Al?
Substance
(J/g•K)
H 2O
Al
glass
Specific heat capacity =
heat lost or gained by substance (J)
(mass, g)(T change, K)
q  (C )(m)(DT )
C = Specific Heat, units of J/(gK)
Spec. Heat
4.18
0.902
0.84
or
21
Specific Heat Capacity
If 25.0 g of Al cool from 310. oC to 37
oC, how many joules of heat energy are
lost by the Al?
heat gain/lost = q = (C)(mass)(DT)
where DT = Tfinal - Tinitial
q = (0.902 J/g•K)(25.0 g)(37 - 310.)K
q = - 6160 J
Heat Energy Transfer in
Chemical Processes
Heat flows FROM the SYSTEM into the
SURROUNDINGS is an EXOTHERMIC
process.
q is “-”
The reaction
feels hot.
Energy given
Off!
22
Heat Energy Transfer in
Chemical Processes
Heat flows INTO the SYSTEM from the
SURROUNDINGS is an ENDOTHERMIC
process.
q is “+”
The reaction
feels cold.
Energy Added!
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24
Specific Heat Capacity
If 25.0 g of Al cool from 310 oC to 37 oC, how
many joules of heat energy are lost by the Al?
q = - 6160 J
q = - 6.16 kJ
Notice that the negative sign on q
signals heat “lost by” or transferred
out of Al. Is this Exo or Endothermic?
25
Specific Heat Capacity
A 500.0 g piece of metal is heated to 75.0 0C. It
is placed into 300. mL of water at 22.0 0C and
the final temperature is 31.0 0C. Calculate the
specific heat of the metal.
qwater +
qmetal = 0
qwater = -qmetal
(300. g)(4.18 J/gK)(9.0oC) = -(500.0 g)(c)(-44.0oC)
c = 0.51 J/gK
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Specific Heat Capacity
To 150 g of water at 25oC is added 45 g of Al at
115oC, what will be the final temperature?
Al 0.902
Heat Transfer and
Changes of State
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Changes of state involve energy
Ice -----> Water
333 J/g (heat of fusion)
+ energy
Exo or Endo? Sign of q?
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Heat Transfer and
Changes of State
Liquid ---> Vapor
Requires energy (heat).
This is the reason
a) you cool down after
swimming
b) you use water to put
out a fire.
+ energy
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29
Heat and Changes of State
What quantity of heat is required to change
10.0 g of ice at -50.0 oC to steam at 200.0 oC?
Endo or Exo? Sign of q?
Heat of fusion of ice = 333 J/g
Heat of vaporization = 2260 J/g
+333 J/g
+2260 J/g
30
Heating/Cooling Curve for Water
4
3
2
1
5
Heat and Changes of State
1.
2.
3.
4.
5.
To heat ice
q = (10.0 g)(2.09 J/g•K)(50.0K) = 1050J
To melt ice
q = (10.0 g)(333 J/g) = 3330J
To heat water
q = (10.0 g)(4.18 J/g•K)(100.K) = 4180J
To evaporate water
q = (10.0 g)(2260 J/g) = 22600J
To heat steam
q = (10.0 g)(2.03 J/g•K)(100.0K) = 2030J
Total heat energy = 33190.J = 33.190kJ
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32
Heat and Changes of State
Calculate the amount of heat energy necessary
to change 25.0 g of copper solid at 925o C to
liquid at 1083o C. Melting point = 1083o C,
csolid = 0.382 J/g.K, Heat of fusion = 205 J/g.
q1 = (25.0 g)(.382 J/gK)(1083oC-925oC)= 1510 J
q2 = (25.0 g)(205 J/g) = 5120 J
qT = q1 + q2 = 6630 J
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Endo- and Exothermic
qsystem > 0
T(system) goes up
ENDOTHERMIC
Heat goes in the system
qsystem < 0
T(system) goes down
EXOTHERMIC
Heat leaves
the system
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USING ENTHALPY
Consider the decomposition of water
H2O(g) + 242 kJ ---> H2(g) + 1/2 O2(g)
Endo or Exo?
Endothermic reaction
Heat is a “reactant”
DH = + 242 kJ
USING ENTHALPY
Making H2 from H2O involves two steps.
Each step requires energy.
Liquid H2O
H2 + O2 gas
H2O vapor
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USING ENTHALPY
36
Making H2 from H2O involves two steps.
H2O(liq) + 44 kJ ---> H2O(g)
H2O(g) + 242 kJ ---> H2(g) + 1/2 O2(g)
-----------------------------------------------------------------------
H2O(liq) + 286 kJ --> H2(g) + 1/2 O2(g)
Example of HESS’S LAW
If a rxn. is the sum of 2 or more
others, the net DH is the sum of
the DH’s of the other rxns.
USING ENTHALPY
Calc. DH for:
37
S(s) + 3/2 O2(g) --> SO3(g)
Given:
S(s) + O2(g) --> SO2(g)
SO2(g) + 1/2 O2(g) --> SO3(g)
DH1 = -320.5 kJ
DH2 = -75.2 kJ
The two equations add up to give the
desired equation, so -
DHnet = DH1 + DH2 = -395.7 kJ
ENERGY
38
S solid
direct path
+ 3/2 O2
DH =
-395.7 kJ
SO3 gas
+O2
DH 1 =
-320.5 kJ
SO2 gas
+ 1/2 O2
DH 2 = -75.2 kJ
 DH along one path =
 DH along another path
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DH along one path = DH along another path
• This equation is valid
because DH is a STATE
FUNCTION
• State functions depend only
on the state of the system and
not how it got there.
• Examples: V, T, P, energy —
and your bank account!
• Unlike V, T, and P, one cannot
measure absolute H. We can
only measure DH.
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41
Standard Enthalpy Values
Most DH values are labeled ΔHo
Measured under standard conditions
P = 1 atmosphere or 1 bar,
(105Pascals)
Concentration = 1 mol/L
T = 25 oC
with all species in standard states
e.g., C = graphite and O2 = gas, etc.
42
Standard Enthalpy Values
NIST (National Institute for Standards and
Technology) gives values of
DHof = standard molar enthalpy of
formation
This is the enthalpy change when 1 mole of
compound is formed from elements under
standard conditions.
See Table 6.2 and Appendix L
DHof, standard molar
enthalpy of formation
H2(g) + 1/2 O2(g) --> H2O(g)
DHof = -241.8 kJ/mol
By definition, DHof = 0
for elements in their
standard states.
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44
Using Standard Enthalpy Values
Use DHf’s to calculate enthalpy change for:
H2O(g) + C(graphite) --> H2(g) + CO(g)
(product is called “water gas”)
45
Using Standard Enthalpy Values
H2O(g) + C(graphite) --> H2(g) + CO(g)
From reference books we find:
H2(g) + 1/2 O2(g) --> H2O(g)
DHf of H2O vapor = - 241.8 kJ/mol
C(s) + 1/2 O2(g) --> CO(g)
DH f of CO = - 110.5 kJ/mol
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Using Standard Enthalpy Values
H2O(g) --> H2(g) + 1/2 O2(g) DHo = +241.8kJ
C(s) + 1/2 O2(g) --> CO(g) DHo = -110.5 kJ
--------------------------------------------------------------------------------
H2O(g) + C(graphite) --> H2(g) + CO(g)
DHonet = +131.3 kJ
To convert 1 mol of water to 1 mol each of
H2 and CO requires 131.3 kJ of energy.
The “water gas” reaction is ENDOthermic.
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Using Standard Enthalpy Values
Calculate DH of reaction?
In general, when ALL enthalpies
of formation are known,
DHorxn =
 DHof (products) -  DHof (reactants)
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Using Standard Enthalpy Values
Calculate the heat of combustion of
methanol, i.e., DHorxn for:
CH3OH(g) + 3/2 O2(g) --> CO2(g) + 2 H2O(g)
DHorxn =  DHof (prod) -  DHof (react)
Use Appendix L
Using Standard Enthalpy Values
CH3OH(g) + 3/2 O2(g) --> CO2(g) + 2 H2O(g)
DHorxn =  DHof (prod) -  DHof (react)
DHorxn = {DHof (CO2) + 2 DHof (H2O)}
- { 3/2 DHof (O2) + DHof (CH3OH) }
= {(-393.5 kJ) + 2 (-241.8 kJ)}
- { 0 + (-200.7 kJ) }
DHorxn = - 676.4 kJ per mol of methanol
49
Using Standard Enthalpy Values
OF2(g) + H2O(g) --> O2(g) + 2 HF(g)
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DHorxn =- 318.0kJ
DHof (H2O) = - 241.8 kJ and DHof (HF) = - 271.1 kJ
Calculate DHof (OF2)
DHorxn = {DHof (O2) + 2 DHof (HF) }
- { DHof (OF2) + DHof (H2O) }
-318.0 kJ = {( 0 ) + 2 (-271.1 kJ) }
- {DHof (OF2) + (-241.8 kJ) }
DHof (OF2) = 17.6 kJ /mole
Measuring Heats of Reaction with a
BOMB CALORIMETRY
51
Coffee Cup
Calorimeter
52
Measuring Heats of Reaction
CALORIMETRY
Calculate heat of combustion of octane.
C8H18 + 25/2 O2 --> 8 CO2 + 9 H2O
• Burn 1.00 g of octane
• Temp rises from 25.00 to 33.20 oC
• Calorimeter contains 1300. g water
• Heat capacity of bomb = 427 J/K
CALORIMETRY
53
Step 1 Calc. heat transferred from reaction to
water.
q = (4.18 J/g•K)(1300. g)(8.20 K) = 44,600 J
Step 2 Calc. heat transferred from reaction to
bomb.
q = (bomb heat capacity)(DT)
= (427 J/K)(8.20 K) = 3500 J
Step 3 Total heat evolved =
44,600 J + 3500 J = 48,100 J
Heat of combustion of 1.00 g of octane =
- 48.1 kJ
54
Measuring Heats of Reaction
CALORIMETRY
Calculate heat of combustion of ethanol.
C2H5OH + 3 O2 --> 2 CO2 + 3 H2O
• Burn 0.888 g of ethanol
• Temp rises from 25.00 oC to 35.00 oC
• Calorimeter contains 210. g water
• Heat capacity of bomb = 0.950 kJ/K
CALORIMETRY
55
Step 1 Calc. heat transferred from reaction to
water.
q = (4.18 J/g•K)(210. g)(10.00 K) = 8.80 kJ
Step 2 Calc. heat transferred from reaction to
bomb.
q = (bomb heat capacity)(DT)
= (0.950 kJ/K)(10.00 K) = 9.50 kJ
Step 3 Total heat evolved =
8.80 kJ + 9.50 kJ = 18.30 kJ
(-18.30 kJ/0.888g)(46.0 g/mole) = - 948 kJ/mole
56
Sample Problem
A 0.105 g sample of C2H4 is burned in
a calorimeter with a heat capacity of
0.47 kJ/oC and containing 2,000. g of
water. The temperature increased from
25.00oC to 27.14oC. Calculate the heat
of combustion of C2H4 in kJ/mole.
CALORIMETRY
57
Step 1 Calc. heat transferred from reaction to
water.
q = (4.18 J/g•K)(2000. g)(2.14 K) = 17,900 J
Step 2 Calc. heat transferred from reaction to
bomb.
q = (bomb heat capacity)(DT)
= (0.47 kJ/K)(2.14 K) = 1.0 kJ
Step 3 Total heat evolved =
17.9 kJ + 1.0 kJ = 18.9 kJ
(-18.9 kJ/0.105g)(28.0 g/mole) = - 5040 kJ/mole
58
Practice Problems
1. How many kcal is 75.0 kJ?
2. Convert 88.6 kcal/mole to kJ/mole.
3. Calculate the kinetic energy of a 2000. lb car
traveling at 50. miles per hour.
4. Is the kinetic energy of a 500. g object moving
at 50. km/hr greater, less than, or the same as
that of a 1.00 kg object moving at 25 km/hr?
5. How much heat is required to raise the
temperature of 204 grams of lead from 22.8oC
to 64.9oC? (C= 0.038 cal/gK)
59
Practice Problems
6. If the specific heat of silver is 0.0573 cal/gK,
how much heat would it take to raise the
temperature of 1 kg of silver from -50.oC to
150.oC?
7. To what temperature will one pound of nickel
be raised, beginning at 25.0oC, if 750 calories
are absorbed by it? (c = 0.106 cal/gK)
8. A 50. g chunk of unknown metal X is heated to
98.5oC and then dropped into 450 g of water
initially at 25.00oC. The water temperature is
observed to rise to 26.47oC. Calculate the
specific heat of X.
60
Practice Problems
9. 2 H2O (l) --> 2 H2 (g) + O2 (g) DH = 273.2 kcal
How many mLs of water will be decomposed to
hydrogen and oxygen by 85 kcal?
10. CaCO3 (s) --> CaO (s) + CO2 (g) DH = 42.5 kcal
How much heat is required to react 186 g CaCO3?
11. C6H12O6 (s) + 6 O2 (g) --> 6 H2O (g) + 6 CO2 (g)
DH = -673 kcal. How much heat is evolved in the
oxidation of 240 g of C6H12O6?
12. MgCO3(s)+2 HCl(g)->MgCl2(s)+CO2(g) + H2O(g)
DH = -5.6 kcal. How many grams of MgCl2 will be
produced if 80.0 kilocalories are released?
61
Practice Problems
13. 3 Cu2S(s) + 16 HNO3(l) --> 6 Cu(NO3)2(s) +
3 S(s) + 4 NO(g) + 8 H2O(l) DH = -543 kcal/mole
How much energy will be released by the reaction
of 325 grams of Cu2S with HNO3.
14. 2 B5H9 (l) + 12 O2 (g) --> 9 H2O (l) + 5 B2O3 (s)
DH = -4188 kcal/mole. How much B5H9 must be
burned to produce 4500 kcal of energy?
15. Upon dissolving 8.63 g of ammonium
dihydrogen phosphate in 100. mL of water, it was
found that DT= - 2.7oC. Calculate the value of the
molar heat of reaction.
62
Practice Problems
16. A 0.25 mole sample of ammonium nitrate was
dissolved in 500. mL of water at 21.0oC. As the
solute dissolved the temperature of the solution
dropped to a minimum value of 18.0oC. Calculate
the heat of reaction in kcal/mole.
17. Given the following data:
DH(kcal)
MnO2 + CO --> MnO + CO2
-36.0
Mn3O4 + CO --> 3 MnO + CO2 -13.0
3 Mn2O3 + CO --> 2 Mn3O4 + CO2 -34.0
Find DH in kcal for the following reaction:
2 MnO2 + CO --> Mn2O3 + CO2
63
Practice Problems
18. Given the following two reactions:
DH(kcal)
2 Na(s) + 2HCl(g) --> 2 NaCl(s) + H2(g)
-152.34
H2(g) + Cl2(g) --> 2 HCl(g)
-44.12
What is DHfo,in kcal/mole for NaCl(s)?
19. Calculate the DHfo of glycerol,
(CH2OH)2CHOH, knowing that its heat of
combustion is -397.0 kJ/mole.
20. Find the DH for the reaction:
2 SO2(g) + 2 H2O(g) + O2(g) --> 2 H2SO4 (l)
64
Practice Problems
21. Determine the enthalpy of formation in
kcal/mole for chloroethane, C2H5Cl, from the
following data:
DH(kcal)
H2 + 1/2 O2 --> H2O
-68.3
C + O2 --> CO2
-94.1
C2H5Cl --> C2H4 + HCl
17.2
2 CO2 + 2 H2O --> C2H4 + 3 O2
337.2
H2 + Cl2 --> 2 HCl
-44.2
65
Practice Problems
22. The combustion of 0.450 g of ethene (C2H4)
causes a temperature rise of 2.00oC in a bomb
calorimeter that has a specific heat capacity
of 600. cal/oC and containing 500. g of water.
What is the heat of combustion in kJ/mole for
ethene?
66
Practice Problems
23. A bomb calorimeter has a heat capacity of
385 J/oC and containing 200. g of water. A
temperature rise of 3.577oC is observed when
the calorimeter is used in the combustion of a
0.7600 g sample of powdered tantalum in excess
oxygen to product Ta2O5. Find DHfo of Ta2O5.
24. How many calories would be required to
change the temperature of 750 g of water from
15.0 oC to 90.0 oC?
25. How many joules of heat would be liberated if
the temperature of 300. g of iron were changed
from 75 oC to 17 oC?
67
Practice Problems
26. The heat of formation of water is
-68.4 kcal/mole. How many joules would
be liberated by burning 12 g of hydrogen?
27.
4 Fe + 3 O2 --> 2 Fe2O3 + 398 kcal
4 Al + 3 O2 --> 2 Al2O3 + 798 kcal
Calculate DH for the reaction
Fe2O3 + 2 Al --> Al2O3 + 2 Fe
68
Practice Problems
28. A 0.757 g sample of C was placed in a
calorimeter with a specific heat capacity of 1550
cal/oC and containing 1,000. G of water and
burned. The temperature rose from
22.54 oC
to 24.87 oC. Calculate the heat of combustion in
kcal/mole?
69
Practice Problems Answers
1. 17.9 kcal
3. 2 x 105 J
5. 330 cal
7. 41oC
9. 11 mL
11. 9.0 x 102 kcal
13. 370. Kcal
15. 15,000 J/mole
17. -52.0 kJ
2. 370. kJ/mole
4. Greater
6. 10,000 cal
8. .8 J/gK
10. 79.0 kcal
12. 1400 g
14. 140 g
16. 6.0 kcal/mole
18. -98.23kcal/mole
70
Practice Problems Answers
19. -1750.8 kJ
21. -26.9 kJ
23. -2080 kJ/mole
25. 7800 J
27. -200 kcal
20. -550.682 kJ
22. -572 kJ/mole
24. 56000 cal
26. 1.7 x 106 J
28. -94.2 kcal/mole
71
Specific Heat Capacity
How much heat is required to raise the
temperature of 57 g of water from 25.6oC to
66.5oC?
q = (57 g)(4.18 J/g K)(40.9oC)
q = 9700 J
72
Energy and Reactions
Endothermic
C+D
DH
H
A+B
Reaction-->
A + B + energy -->C + D
Exothermic
W+X
H DH
Y+Z
Reaction-->
W + X -->Y + Z + energy
73
Energy and Reactions
1. The reaction of 14 g of hydrogen
will produce how much energy?
3 H2 + N2 --> 2 NH3 + 92 kJ
14 g
? kJ
14 g H2 mole H2 92 kJ
2.0 g H2 3 mole H2 = 210 kJ
74
Energy and Reactions
2. If 126 kJ are produced how many
grams of ammonia are produced?
3 H2 + N2 --> 2 NH3 + 92 kJ
?g
126 kJ
126 kJ 2 mole NH3 17.0 g NH3
92 kJ
mole NH3
= 47g
NH3
75
Hess’s Law
N2 (g) + O2 (g) --> 2 NO (g)
DH = 180.8 kJ
N2 (g) + 2 O2 (g) --> 2 NO2 (g) DH = 67.7 kJ
Find DH for the following equation:
2 NO (g) + O2 (g) --> 2 NO2 (g)
2 NO (g) --> N2 (g) + O2 (g) DH = -180.8 kJ
1
N2 (g) + 2 O2 (g) --> 2 NO2 (g)
DH = 67.7 kJ
2 NO (g) + O2 (g) --> 2 NO2 (g)
DH = -113.1 kJ
76
Hess’s Law
2 C2H4 + 2 H2 + 7 O2 --> 4 CO2 + 6 H2O DH = -3394 kJ
2 C2H6 + 7 O2 --> 4 CO2 + 6 H2O
DH = -3120. kJ
Find DH for the following equation:
C2H4 + H2 --> C2H6
1
1
7/2
2
3
-1697 kJ
2 C2H4 + 2 H2 + 7 O2 --> 4 CO2 + 6 H2O DH = -3394 kJ
2
3
1
7/2
4 CO2 + 6 H2O --> 2 C2H6 + 7 O2
C2H4 + H2 --> C2H6
+1560. kJ
DH = +3120. kJ
DH = -137 kJ
77
Hess’s Law
1/2 H2 + N2 + 5/2 O2 --> HNO3 + NO2 DH = -101 kJ
2 NH + H2 --> N2H4
DH = -567 kJ
2 N2O5 --> 2 N2 + 5 O2
DH = 23 kJ
Find DH for the following equation:
N2H4 + 2 N2O5 --> 2 HNO3 + 2 NO2 + 2 NH
1
2
5
2
2
-202 kJ
N2H4 --> 2 NH + H2
DH = 567 kJ
1/2 H2 + N2 + 5/2 O2 --> HNO3 + NO2 DH = -101 kJ
2 N2O5 --> 2 N2 + 5 O2
DH = 23 kJ
N2H4 + 2 N2O5 --> 2 HNO3 + 2 NO2 + 2 NH
DH = 388 kJ
Using Standard Enthalpy Values
Calculate the DH for the following reaction:
Al2O3 (s) + 3 CO (g) --> 3 CO2 (g) + 2 Al (s)
DHorxn = { 3 DHof (CO2) + 2 DHof (Al) }
- { DHof (Al2O3) + 3 DHof (CO) }
= { 3 (-393.5 kJ) + 2 ( 0 )}
- { -1675.7 kJ + 3 (-110.5 kJ) }
DHorxn = 826.7 kJ
78
Using Standard Enthalpy Values
79
C2H4 + 3 O2 --> 2 CO2 + 2 H2O (g) DH = -1323.0 kJ
2 C + 2 H2 --> C2H4
DH = 52.3 kJ
H2 + 1/2 O2 --> H2O (g)
DH = -241.8 kJ
Calculate DHf for CO2.
1/2
3/2 1
1
1
-661.5 kJ
C2H4 + 3 O2 --> 2 CO2 + 2 H2O (g) DH = -1323.0 kJ
1
1
1/2
2 C + 2 H2 --> C2H4
H2O (g) --> H2 + 1/2 O2
DHf =
26.0 kJ
52 kJ
DHf = 241.8 kJ
C + O2 --> CO2
DHf = -393.7 kJ
Using Standard Enthalpy Values
80
C2H4 + 3 O2 --> 2 CO2 + 2 H2O (g) DH = -1323.0 kJ
2 C + 2 H2 --> C2H4
DHf = 52.3 kJ
H2 + 1/2 O2 --> H2O (g)
DH = -241.8 kJ
Calculate DHf for CO2.
f
DHorxn = { 2 DHof (CO2) + 2 DHof (H2O) }
- {DHof (C2H4) + 3 DHof (O2) }
-1323.0 kJ = { 2 DHof (CO2) + 2 ( -241.8 kJ )}
- { 52.3 kJ + 3 ( 0 ) }
DHof (CO2) = -393.6 kJ

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