Chemistry: The Molecular Science
Moore, Stanitski and Jurs
Chapter 6: Energy and Chemical Reactions
© 2008 Brooks/Cole
1
The Nature of Energy
Energy (E) = the capacity to do work.
Work (w) occurs when an object moves against a
resisting force:
w = −(resisting force) x (distance traveled)
w = −F d
All energy is either Kinetic or Potential energy.
© 2008 Brooks/Cole
2
The Nature of Energy
Kinetic energy (Ek) - Energy of motion
 macroscale = mechanical energy
 random nanoscale = thermal energy
 periodic nanoscale = acoustic energy
Ek = ½mv2
(m = mass, v = velocity of object)
Potential energy (Ep) – Energy of position. Stored E.
It may arise from:
 gravity: Ep = m g h (mass x gravity x height).
 charges held apart.
 bond energy.
© 2008 Brooks/Cole
3
Energy Units
joule (J) - SI unit (1 J = 1 kg m2s-2)
2.0 kg mass moving at 1.0 m/s (~2 mph):
Ek
= ½ mv2 = ½ (2.0 kg)(1.0 m/s)2
= 1.0 kg m2 s-2
= 1.0 J
1 J is a relatively small amount of energy.
1 kJ (1000 J) is more common in chemical problems.
© 2008 Brooks/Cole
4
Energy Units
calorie (cal)
Originally:
“The energy needed to heat of 1g of water from 14.5
to 15.5 °C.”
Now:
1 cal = 4.184 J (exactly)
Dietary Calorie (Cal) - the “big C” calorie
Used on food products.
1 Cal = 1000 cal
= 1 kcal
© 2008 Brooks/Cole
5
Conservation of Energy
“Energy can neither be created nor destroyed”
E can only change form.
Total E of the universe is constant.
Also called the 1st Law of Thermodynamics
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6
Conservation of Energy
A diver:
• Has Ep due to macroscale position.
• Converts Ep to macroscale Ek.
• Converts Ek,macroscale to Ek,nanoscale (motion of water, heat)
© 2008 Brooks/Cole
7
Energy and Working
If an object moves against a force, work is done.
• Lift a book
 you do work against gravity. The book’s Ep increases.
• Drop the book:
 Ep converts into Ek
 The book does work pushing the air aside.
• The book hits the floor
 no work is done on the floor (it does not move).
 Ek converts to a sound wave and T of the book and floor
increase (Ek converts to heat).
© 2008 Brooks/Cole
8
Energy and Working
In a chemical process, work occurs whenever
something expands or contracts.
Expansion pushes back the surrounding air.
Imagine the gas inside a balloon
 heat the gas.
 the gas expands and the balloon grows.
 the gas does work pushing back the rubber and
the air outside it.
© 2008 Brooks/Cole
9
Energy, Temperature and Heating
Temperature is a measure of the
thermal energy of a sample.
Thermal energy
• E of motion of atoms, molecules, and ions.
• Atoms of all materials are always in motion.
• Higher T = faster motion.
© 2008 Brooks/Cole
10
Energy, Temperature and Heating
Consider a thermometer. As T increases:
 Atoms move faster, and on average get farther apart.
 V of the material increases.
 Length of liquid column increases.
© 2008 Brooks/Cole
11
Energy, Temperature and Heating
Heat
• Thermal E transfer caused by a T difference.
• Heat flows from hotter to cooler objects until
they reach thermal equilibrium (have equal T ).
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12
Systems, Surroundings & Internal Energy
System = the part of the universe under study
 chemicals in a flask.
 the coffee in your coffee cup.
 my textbook.
Surroundings = rest of the universe (or as much as
needed…)
 the flask.
 perhaps the flask and this classroom.
 perhaps the flask and all of the building, etc.
Universe = System + Surroundings
© 2008 Brooks/Cole
13
Systems, Surroundings & Internal Energy
Internal energy = E within the system because of
nanoscale position or motion
Einternal= sum of all nanoscale Ek and Ep
 nanoscale Ek = thermal energy
 nanoscale Ep
• ion/ion attraction or repulsion
• nucleus/electron attraction
• proton/proton repulsion …..
© 2008 Brooks/Cole
14
Systems, Surroundings & Internal Energy
Internal energy depends on
• Temperature
 higher T = larger Ek for the nanoscale particles.
• Type of material
 nanoscale Ek depends upon the particle mass.
 nanoscale Ep depends upon the type(s) of particle.
• Amount of material
 number of particles.
 double sample size, double Einternal, etc.
© 2008 Brooks/Cole
15
Calculating Thermodynamic Changes
Energy change of system = final E – initial E
ΔE = Efinal – Einitial
A system can gain or lose E
SURROUNDINGS
SYSTEM
SURROUNDINGS
SYSTEM
Efinal
ΔE > 0
E in
Einitial
ΔE positive: internal energy increases
© 2008 Brooks/Cole
Einitial
ΔE < 0
E out
Efinal
ΔE negative: internal energy decreases
16
Calculating Thermodynamic Changes
• No subscript? Refers to the system: E = Esystem
• E is transferred by heat or by work.
• Conservation of energy becomes: ΔE = q + w
heat
work
SURROUNDINGS
SYSTEM
Heat transfer out
q<0
Heat transfer in
q>0
ΔE = q + w
Work transfer in
w>0
Work transfer out
w<0
Note the same sign convention for q and w
© 2008 Brooks/Cole
17
Heat Capacity
Heat capacity = E required to raise the T of an object
by 1°C. Varies from material to material.
Specific heat capacity (c)
• E needed to heat 1 g of substance by 1°C.
Molar heat capacity (cm)
• E needed to heat 1 mole of substance by 1°C.
© 2008 Brooks/Cole
18
Heat Capacity
E required to change the T of an object is:
Heat required = mass x specific heat x ΔT
q = m c ΔT
or…
Heat required = moles x molar heat capacity x ΔT
q = n cm ΔT
© 2008 Brooks/Cole
19
Substance
Heat
Capacities
© 2008 Brooks/Cole
Elements
Al(s)
C (graphite)
Fe(s)
Cu(s)
Au(s)
Compounds
NH3(l)
H2O(l)
C2H5OH(l)
(CH2OH)2(l)
H2O(s)
CCl4(l)
CCl2F2(l)
Common solids
wood
concrete
glass
granite
c (J g-1 °C-1) cm (J mol-1 °C-1)
0.902
0.720
0.451
0.385
0.129
24.3
8.65
25.1
24.4
25.4
4.70
4.184
2.46
2.42
2.06
0.861
0.598
80.1
75.3
112.
149.
37.1
132.
72.3
1.76
0.88
0.84
0.79
20
Heat Capacity
Example
How much energy will be used to heat 500.0 g of iron
from 22°C to 55°C? cFe = 0.451 J g-1 °C-1.
Heat required = mass x specific heat x ΔT
q = m c ΔT
q = 500.0 g (0.451 J g-1 °C-1)(55−22)°C
q = 7442 J = +7.4 kJ
+ sign, E added to the system (the iron)
© 2008 Brooks/Cole
21
Heat Capacity
Example
24.1 kJ of energy is lost by a 250. g aluminum block.
If the block is initially at 125.0°C what will be its final
temperature? (cAl = 0.902 J g-1 °C-1)
q = m c ΔT
ΔT = q / (m c)
heat is lost, q is negative
ΔT = −24.1 x 103 J /(250. g x 0.902 J g-1 °C-1)
ΔT = Tfinal – Tinital = −107 °C
Thus Tfinal = ΔT + Tinital = −107 + 125°C = 18°C
© 2008 Brooks/Cole
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Heat Capacity
A 200. g block of Cu at 500.°C is plunged into 1000. g of water
(T = 23.4 °C) in an insulated container. What will be the final
equilibrium T of water and Cu? (cCu = 0.385 J g-1 °C-1)
Cu cools (−q); water heats (+q);
q = m c ΔT
Heat lost by Cu = −(200. g)(0.385 J g-1 °C-1)(Tfinal− 500)
Heat gained by H2O = +(1000. g)(4.184 J g-1 °C-1)(Tfinal− 23.4)
So: −77.0(Tfinal – 500) = 4184(Tfinal – 23.4)
(4184 + 77.0)Tfinal = 38500 + 97906
Tfinal = 32.0°C
(Note: Tfinal must be between Thot and Tcold)
© 2008 Brooks/Cole
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Conservation of Energy and Changes of State
When heat is: Added to a system
 q is positive
 the change is endothermic
Removed from a system
 q is negative
 the change is exothermic.
H2O(l)  H2O(g)
endothermic
Steam Condenses: H2O(g)  H2O(l)
exothermic
Water Boils:
Work occurs as the sample expands or contracts.
Overall:
ΔE = q + w
© 2008 Brooks/Cole
24
Conservation of Energy and Changes of State
A liquid cools from 45°C to 30°C, transferring 911 J to
the surroundings. No work is done on or by the
liquid. What is ΔEliquid?
ΔEliquid = qliquid + wliquid
here wliquid = 0
Heat transfers from the liquid to the surroundings:
qliquid = -911 J
(qsurroundings = +911 J)
ΔEliquid = -911J
© 2008 Brooks/Cole
25
Conservation of Energy and Changes of State
A system does 50.2 J of work on its surroundings and
there is a simultaneous 90.1 J heat transfer from the
surroundings to the system. What is ΔEsystem?
Work done on the surroundings by the system
Heat transfers from the surroundings to the system
wsystem = -50.2 J
qsystem = +90.1 J
ΔEsystem = qsystem + wsystem
ΔEsystem = -50.2 J +90.1 J = +39.9 J
© 2008 Brooks/Cole
26
Enthalpy: Heat Transfer at Constant P
Because ΔE = q + w:
 At Constant V:
ΔE = qV
• subscript V shows fixed V
• work requires motion against an opposing force.
• constant V = no motion, so w = 0.
 At Constant P:
ΔE = qP + watm= ΔH + watm
• Subscript P shows fixed P.
• watm = work done to push back the atmosphere
• H = enthalpy. ΔH = qp
© 2008 Brooks/Cole
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Freezing and Melting (Fusion)
During freezing (or melting)
-50
Example:
Convert ice at
-50°C to water
at +50°C
Temperature (°C)
25
0
-25
50
• Substance loses (or gains) E, but…
• T remains constant.
© 2008 Brooks/Cole
Ice is melting. T
remains at 0°C
Water warms
from 0 to 50°C
Ice warms from -50
to 0°C
0
100
200
300
400
500
Quantity of energy transferred (J)
600
28
Changes of State
ΔHfusion = qP = heat to melt a solid.
Change
Name
value for H2O (J/g)
solid → liq
enthalpy of fusion
liq → gas
enthalpy of vaporization
2260
liq → solid
enthalpy of freezing
−333
gas → liq
enthalpy of condensation
Note:
ΔHfusion = − ΔHfreezing
333
−2260
etc.
qfusion = −qfreezing
© 2008 Brooks/Cole
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State Functions and Path Independence
State functions
Always have the same value whenever
the system is in the same state.
Two equal mass samples of water produced by:
1. Heating one from 20°C to 50°C.
2. Cooling the other from 100°C to 50°C.
have identical final H (and V, P, E…).
State
functions
H
P
T
E
V
etc.
State function changes are path independent.
ΔH = Hfinal – Hinitial is constant.
© 2008 Brooks/Cole
30
Thermochemical Expressions
ΔH = qP can be added to a balanced equation.
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g) ΔH° = −803.05 kJ
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)
ΔH° = −890.36 kJ
ΔH° is the standard enthalpy change
 P = 1 bar.
 T must be stated (if it isn’t, assume 25°C).
 ΔH° is a molar value. Burn 1 mol of CH4 with 2 mol O2 to
form 2 mol of liquid water and release 890 kJ of heat
 Change a physical state, change ΔH° : H2O(l) vs. H2O(g)
© 2008 Brooks/Cole
31
Enthalpy Changes for Chemical Reactions
Calculate the heat generated when 500. g of propane
burns in excess O2.
C3H8(l) + 5 O2(g)
3 CO2(g) + 4 H2O(l)
ΔH° = – 2220. kJ
Molar mass of C3H8 = 44.097 g/mol.
nC3H8 = (500. g) / (44.097 g/mol)
= 11.34 mol C3H8
Since ΔH° = qp = –2220. kJ/(1 mol C3H8)
q = (11.34 mol C3H8) (–2220. kJ / mol C3H8 )
= –2.52 x 104 kJ
© 2008 Brooks/Cole
32
Where Does the Energy Come From?
Bond Enthalpy (bond energy)
• Equals the strength of 1 mole of bonds
• Always positive
 It takes E to break a bond
 Separated parts are less stable than the molecule.
 Less stable = higher E
• E is always released when a bond forms
 Product is more stable than the separated parts.
 More stable = lower E
© 2008 Brooks/Cole
33
Bond Enthalpies
During a chemical reaction:
Old bonds break: requires E (endothermic)
New bonds form: releases E (exothermic)
Both typically occur:
H2(g) + Cl2(g)
2 H(g) + 2 Cl(g)
endothermic
ΔH= +678 kJ/mol
© 2008 Brooks/Cole
2 HCl(g)
exothermic
ΔH= -862 kJ/mol
34
Bond Enthalpies
Overall, heat may be absorbed or released:
 E is released.
 New bonds are more stable than the old,
or
 More bonds are formed than broken.
energy
less stability
Exothermic reactions (ΔH < 0)
reactants
products
 E is absorbed.
 New bonds are less stable than the old,
or
 Fewer bonds are formed than broken
© 2008 Brooks/Cole
energy
less stability
Endothermic reactions (ΔH > 0)
products
reactants
35
Measuring Enthalpy Changes
Heat transfers are measured with a calorimeter.
Common types:
• Bomb calorimeter.
 rigid steel container.
 filled with O2(g) and a small sample to be burnt.
 constant V, so qV = ΔE
• Flame calorimeter.
 samples burnt in an open flame.
 constant P, so qp = ΔH
• Coffee-cup calorimeter in lab (constant P).
© 2008 Brooks/Cole
36
Measuring Enthalpy Changes
Bomb Calorimeter
Measure ΔT of the water. Constant V:
qV = ΔE
Conservation of E:
qreaction + qbomb + qwater = 0
or
−qreaction = qbomb + qwater
with
qbomb = mcalccalΔT = CcalΔT
A constant for a calorimeter
© 2008 Brooks/Cole
37
Measuring Enthalpy Changes
Octane (0.600 g) was burned in a bomb calorimeter containing
751 g of water. T increased from 22.15°C to 29.12°C. Calculate
the heat evolved per mole of octane burned. Ccal = 895 J°C-1.
2 C8H18(l) + 25 O2(g)
16 CO2(g) + 18 H2O(l)
−qreaction = qbomb + qwater
qbomb = CcalΔT = 895 J°C-1 (29.12 – 22.15)°C = +6238 J
qwater = m c ΔT = 751 g (4.184 J g-1 °C-1)(29.12 – 22.15)°C
= +2.190 x 104 J
So −qreaction = +6238 + 2.190 x 104 J = 2.81 x 104 J = 28.1 kJ
qreaction = −28.1 kJ
© 2008 Brooks/Cole
38
Measuring Enthalpy Changes
Octane (0.600 g) was burned in a bomb calorimeter… Calculate the heat
evolved per mole of octane burned
Molar mass of C8H18 = 114.23 g/mol.
nC8H18 = (0.600 g) / (114.23 g/mol)
= 0.00525 mol C8H18
Heat evolved /mol octane =
−28.1 kJ
0.00525 mol
= −5.35 x 103 kJ/mol
= −5.35 MJ/mol
© 2008 Brooks/Cole
39
Measuring Enthalpy Changes
Coffee-cup calorimeter
Nested styrofoam cups prevent heat transfer
with the surroundings.
Constant P. ΔT measured.
q = qp = ΔH
Assume the cups do not absorb heat.
(Ccal = 0 and qcal = 0)
© 2008 Brooks/Cole
40
Measuring Enthalpy Changes
0.800g of Mg was added to 250. mL of 0.40 M HCl in
a coffee-cup calorimeter at 1 bar. Tsolution increased
from 23.4 to 37.9°C. Assume csolution = 4.184 J g-1°C-1
and complete the equation:
Mg(s) + 2 HCl(aq)
H2(g) + MgCl2(aq) ΔH = ?
nMg = 0.800 g 1 mol = 0.03291 mol
24.31 g
nHCl = 0.250 L 0.4 mol = 0.100 mol
1L
1Mg ≡ 2HCl
© 2008 Brooks/Cole
Mg is limiting
41
Measuring Enthalpy Changes
… 0.800g of Mg was added to 250. mL of 0.40 M HCl. T increased from 23.4 to
37.9°C. Assume the solution has c = 4.184 J g-1°C-1.
ΔH = ?
qsolution = msolution c ΔT
(msolution = macid + mMg )
= 250.8 g (4.184 J g-1 °C-1)(37.9 − 23.4)°C
= 15,220 J
Heat from the reaction went into the solution. So:
qsolution = – qreaction
qreaction = –15.22 kJ = ΔH
So
ΔH = –15.22 kJ
0.03291mol
1 mol Mg
© 2008 Brooks/Cole
or
ΔH = – 462 kJ
exothermic
42
Hess’s Law
“If the equation for a reaction is the sum of the
equations for two or more other reactions, then ΔH°
for the 1st reaction must be the sum of the ΔH°
values of the other reactions.”
Another version:
“ΔH° for a reaction is the same whether it takes place
in a single step or several steps.”
H is a state function
© 2008 Brooks/Cole
43
Hess’s Law
Multiply a reaction, multiply ΔH.
Reverse a reaction, change the sign of ΔH.
2 CO(g) + O2(g) → 2 CO2 (g)
Then
2 CO2(g) → 2 CO(g) + O2(g)
4 CO2(g) → 4 CO(g) + 2 O2(g)
© 2008 Brooks/Cole
ΔH = −566.0 kJ
ΔH = –1(–566.0 kJ)
= + 566.0 kJ
ΔH = –2(–566.0 kJ)
= +1132.0 kJ
44
Hess’s Law
Use Hess’s Law to find ΔH for unmeasured reactions.
Example
It is difficult to measure ΔH for:
2 C(graphite) + O2(g)
2 CO(g)
Some CO2 always forms. Calculate ΔH given:
C(graphite) + O2(g)
2 CO(g) + O2(g)
© 2008 Brooks/Cole
CO2(g)
2 CO2(g)
ΔH = −393.5 kJ
ΔH = −566.0 kJ
45
Hess’s Law
Calculate ΔH for the reaction:
Rearrange:
+2[C + O2
−1[2 CO + O2
2C(graphite) + O2(g) → 2CO(g)
CO2]
2 CO2]
or:
2 C + 2 O2
2 CO2
2 CO2
2 CO + O2
+2(−393.5) = −787.0
−1(−566.0) = +566.0
ΔH° = −787.0 kJ
ΔH° = +566.0 kJ
Add, then cancel:
2 C + 2 O2 + 2 CO2
2 C + O2
© 2008 Brooks/Cole
2 CO
2 CO2 + 2 CO + O2
−221.0
ΔH° = −221.0 kJ
46
Hess’s Law
Determine ΔH° for the production of coal gas:
2 C(s) + 2 H2O(g)
CH4(g) + CO2(g)
Using:
C(s) + H2O(g)
CO(g) + H2(g)
ΔH° = 131.3 kJ
A
CO(g) + H2O(g)
CO2(g) + H2(g)
ΔH° = −41.2 kJ
B
CH4(g) + H2O(g)
CO(g) + 3 H2(g) ΔH° = 206.1 kJ
C
© 2008 Brooks/Cole
47
Hess’s Law
Want:
2 C(s) + 2 H2O(g)
CH4(g) + CO2(g)
2 C on left, use 2 x A
2 C(s) + 2 H2O(g)
2 CO(g) + 2 H2(g)
+262.6
1 CH4 on right, use −1 x C
CO(g) + 3 H2(g)
CH4(g) + H2O(g)
−206.1
1 CO2 on right, so use 1 x B
CO(g) + H2O(g)
CO2(g) + H2(g)
−41.2
Add and cancel:
2C + 3H2O + 2CO + 3H2
15.3 kJ
change to 2 H2O
© 2008 Brooks/Cole
2CO + 3H2 + CH4 + H2O + CO2
2 C + 2 H2O → CH4 + CO2 ΔH = 15.3 kJ
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Standard Molar Enthalpy of Formation
Hess’s law problems often use a combustion or …
Formation reaction
Make 1 mol of compound from its elements in their
standard states.
H2 combustion:
2 H2(g) + O2(g)
2 H2O(l)
ΔH° = −571.66 kJ
but the formation reaction is:
H2(g) + ½ O2(g)
1 H2O(l)
ΔHf° = −285.83 kJ
f = formation
© 2008 Brooks/Cole
49
Standard Molar Enthalpy of Formation
Standard state = most stable form of the pure
element at P = 1 bar.
 e.g. C standard state = graphite (not diamond)
ΔHf° for any element in its standard state is zero.
(take 1 mol of the element and make… 1 mol of element)
ΔHf° (Br2(l) ) = 0
ΔHf° (Br2(g) ) ≠ 0
© 2008 Brooks/Cole
at 298 K
at 298 K
50
Standard Enthalpy of Formation (25°C)
Appendix J
Notes
• Most are negative
(formation releases E),
but can be positive.
• If the physical state
changes, ΔHf° changes.
© 2008 Brooks/Cole
Formula
Al2O3(s)
CaO(s)
CH4(g)
C2H2(g)
C2H4(g)
C2H6(g)
C2H5OH(l)
H2O(g)
H2O(l)
NaF(s)
NaCl(s)
SO2(g)
SO3(g)
Name
DHf°, kJ/mol
aluminum oxide −1675.7
calcium oxide
−635.09
methane
−74.81
acetylene
+226.73
ethylene
+52.26
ethane
−84.68
ethanol
−277.69
water vapor
−241.818
liquid water
−285.830
sodium fluoride −573.647
sodium chloride −411.153
sulfur dioxide
−296.830
sulfur trioxide
−395.72
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Standard Molar Enthalpies of Formation
ΔH° ={(nproducts)(ΔHf° products)}
– {(nreactants)(ΔHf° reactants)}
Example
Calculate ΔH° for:
CH4(g) + NH3(g)  HCN(g) + 3 H2(g)
ΔH° = [1ΔHf°(HCN) + 3ΔHf°(H2)] −
[1ΔHf°(NH3) + 1ΔHf°(CH4)]
= [+134 + 3(0)] − [− 46.11 + (−74.85)] = 255 kJ/mol
© 2008 Brooks/Cole
52
Chemical Fuels for Home and Industry
Chemical Fuel – reacts exothermically with O2 in air.
A good fuel has weak bonds and/or strong product
bonds.
Hydrazine
+ O2(g) → N2(g) + 2H2O(g)
N2H4(g)
Ebond(kJ/mol) 160(NN), 391(NH)
Emolecule
1724
Ereagents
ΔE
© 2008 Brooks/Cole
498(OO)
498
946(NN)
946
2222
467(OH)
1864
2814
-592 kJ/mol
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Chemical Fuels for Home and Industry
Good fuels have large:
• Fuel value = (E released) / (mass of fuel in g)
• Energy density = (E released) / Vfuel
U.S. sources of E:
Mostly fossil fuels.
© 2008 Brooks/Cole
54
Foods: Fuels for Our Body
Carbohydrates, Cx(H2O)y, are converted to glucose, C6H12O6
Glucose is the body’s fuel
C6H12O6 + 6 O2 → 6 CO2 + 6 H2O
ΔH° = −2801.6 kJ
(average carbohydrate =17 kJ/g or 4 Cal/g)
Excess glucose  fat. Fat is metabolized when needed:
2 C57H110O6 + 163 O2 → 114 CO2 + 110 H2O ΔH° = −75,520 kJ
(average fat = 38 kJ/g or 9 Cal/g)
Metabolism of dietary protein releases 17 kJ/g or 4 Cal/g.
© 2008 Brooks/Cole
55
Foods: Fuels for Our Body
Food
All-purpose flour
Apple
Brownie with nuts
Egg
Hamburger
Peanuts
Rice
Approximate Composition
per 100. g
Fat
Carbohydrate Protein
0.0
73.3
13.3
0.5
13.0
0.4
16.0
64.0
4.0
0.7
10.0
13.0
30.0
0.0
22.0
50.0
21.4
28.6
1.0
77.6
8.2
Caloric Value
Cal/g
kJ/g
3.33
13.95
0.59
2.47
4.04
16.9
1.40
5.86
3.60
15.06
5.71
23.91
3.47
14.52
Basal metabolic rate (BMR) = minimum energy to maintain a body at rest.
≈ 1 Cal kg-1 hr-1
Metabolizing food requires energy and this adds ≈ 10% to the BMR.
© 2008 Brooks/Cole
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Summary Problem
SO2 is a major pollutant emitted by coal-fired electric power
plants. A large plant can produce 8.64 x 1013 J of electricity
each day by burning 7000 tons of coal (1 ton = 9.08 x 105 g).
(a) Assume coal ≈ graphite. Calculate the energy transferred
per day to the surroundings by coal combustion.
(b) What is the efficiency of the plant; what % of thermal
energy becomes electrical energy?
(c) SO2 can be trapped by MgO in the smokestack to form
MgSO4:
SO2 + MgO + ½ O2 → MgSO4
If 140 tons of SO2 is emitted each year, how much MgO is
needed? How much MgSO4 is produced?
(d) How much heat does the reaction above add/remove?
© 2008 Brooks/Cole
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Summary Problem
…8.64 x 1013 J/day from 7000 tons of coal (1 ton = 9.08 x 105 g).
(a) Calculate the E transferred/day to the surroundings.
C + O2 → CO2
ncoal = 7000 tons
ΔH° = ΔHf° = −393.509 kJ/mol
9.08 x 105g 1 mol = 5.292 x 108 mol
1 ton
12.011 g
Heat released = 5.292x108 mol –393.509 kJ = −2.08 x 1011 kJ
1 mol
= −2.08 x 1014 J
(b) What % of thermal E becomes electrical E?
13 J
8.64
x
10
Efficiency = (Eout/Ein) x 100% =
x100% = 41.5%
14
2.08 x 10 J
© 2008 Brooks/Cole
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Summary Problem
(c) SO2 can be trapped: SO2 + MgO + ½O2 → MgSO4
140 tons of SO2 is emitted/year. MgO needed? MgSO4 produced?
140 tons = 140 x 9.08 x 105g = 1.271 x 108g
nSO2 = 1.271 x 108g /64.07 g mol-1 = 1.984 x 106 mol
1 MgO ≡ 1 SO2
So 1.984 x 106 mol MgO required
= 1.984 x 106 mol x (40.30 g/mol) = 7.996 x 107 g
= 88.1 tons of MgO
1 MgO ≡ 1MgSO4 So 1.984 x 106 mol MgSO4 produced
= 1.984 x 106 mol x (120.4 g/mol) = 2.39 x 108 g
= 263 tons of MgSO4
© 2008 Brooks/Cole
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Summary Problem
SO2 + MgO + ½O2 → MgSO4
ΔH° = ?
(d) How much heat does the reaction add/remove?
nSO2 = 1.984 x 106 mol
1 ΔH° ≡ 1 SO2
ΔH°? Use ΔHf° values
ΔH° = ΔHf°(MgSO4) − ΔHf°(SO2 ) − ΔHf°(MgO) − ½ΔHf°(O2 )
= −1284.9 −(−296.83) − (−601.70) − ½(0) = −386.4 kJ
Heat produced = 1.984 x 106 x (-386.4) kJ
= −7.67 x 108 kJ
7.67 x 1011 J of heat are added to the combustion
© 2008 Brooks/Cole
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