Upload
jake
View
94
Download
1
Embed Size (px)
DESCRIPTION
Chemistry: The Molecular Science Moore, Stanitski and Jurs. Chapter 6:Energy and Chemical Reactions. 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. - PowerPoint PPT Presentation
Citation preview
© 2008 Brooks/Cole 1
Chemistry: The Molecular ScienceChemistry: The Molecular ScienceMoore, Stanitski and Jurs
Chapter 6: Energy and Chemical ReactionsChapter 6: Energy and Chemical Reactions
© 2008 Brooks/Cole 2
Energy (Energy (EE)) = the capacity to do work.
All energy is either KineticKinetic or PotentialPotential energy.
Work (w)Work (w) occurs when an object moves against a resisting force:
w = −(resisting force) x (distance traveled)
w = −F d
The Nature of Energy
© 2008 Brooks/Cole 3
Potential energy (Potential energy (EEpp)) – Energy of position. Stored E.
It may arise from: gravity: Ep = m g h (mass x gravity x height).
charges held apart. bond energy.
Kinetic energy (Kinetic energy (EEkk)) - Energy of motion macroscale = mechanical energy random nanoscale = thermal energy periodic nanoscale = acoustic energy
Ek = ½mv2 (m = mass, v = velocity of object)
The Nature of Energy
© 2008 Brooks/Cole 4
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.
joule (J)joule (J) - SI unit (1 J = 1 kg m2s-2)
Energy Units
© 2008 Brooks/Cole 5
calorie (cal)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)Dietary Calorie (Cal) - the “big C” calorie
Used on food products. 1 Cal = 1000 cal
= 1 kcal
Energy Units
© 2008 Brooks/Cole 6
““Energy can neither be created nor destroyed”Energy can neither be created nor destroyed”
E can only change form.
Total E of the universe is constant.
Also called the 11stst Law of Thermodynamics Law of Thermodynamics
Conservation of Energy
© 2008 Brooks/Cole 7
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 8
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 9
Energy and WorkingIn a chemical process, work occurs whenever something expands or contracts.
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.
Expansion pushes back the surrounding air.
© 2008 Brooks/Cole 10
TemperatureTemperature is a measure of the thermal energy of a sample.
Thermal energyThermal energy
• E of motion of atoms, molecules, and ions.
• Atoms of all materials are always in motion.
• Higher T = faster motion.
Energy, Temperature and Heating
© 2008 Brooks/Cole 11
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 12
HeatHeat • Thermal E transfer caused by a T difference.• Heat flows from hotter to cooler objects until
they reach thermal equilibriumthermal equilibrium (have equal T ).
Energy, Temperature and Heating
© 2008 Brooks/Cole 13
Systems, Surroundings & Internal Energy
SurroundingsSurroundings = 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.
UniverseUniverse = System + Surroundings
SystemSystem = the part of the universe under study chemicals in a flask. the coffee in your coffee cup. my textbook.
© 2008 Brooks/Cole 14
Internal energyInternal 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 …..
Systems, Surroundings & Internal Energy
© 2008 Brooks/Cole 15
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.
Systems, Surroundings & Internal Energy
© 2008 Brooks/Cole 16
Einitial
SURROUNDINGS
SYSTEM
Efinal
Efinal
SURROUNDINGS
SYSTEM
Einitial
Calculating Thermodynamic Changes
ΔE > 0
ΔE positive: internal energy increases
Energy change of system = final E – initial E
ΔΔEE = = EEfinalfinal – – E Einitialinitial
A system can gain or lose E
E in ΔE < 0
ΔE negative: internal energy decreases
E out
© 2008 Brooks/Cole 17
• No subscript? Refers to the system: E = Esystem
• E is transferred by heat or by work.• Conservation of energy becomes: ΔE = q + w
Note the same sign convention for q and w
heat work
Calculating Thermodynamic Changes
SURROUNDINGS
SYSTEM
ΔE = q + w
Heat transfer outq < 0
Work transfer inw > 0
Heat transfer inq > 0
Work transfer outw < 0
© 2008 Brooks/Cole 18
Heat Capacity
Heat capacityHeat capacity = E required to raise the T of an object by 1°C. Varies from material to material.
Specific heat capacity (Specific heat capacity (cc))• E needed to heat 1 g1 g of substance by 1°C.
Molar heat capacity (Molar heat capacity (ccmm))
• E needed to heat 1 mole1 mole of substance by 1°C.
© 2008 Brooks/Cole 19
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 20
Heat Capacities
Substance c (J g-1 °C-1) cm (J mol-1 °C-1)
ElementsAl(s) 0.902 24.3C (graphite) 0.720 8.65Fe(s) 0.451 25.1Cu(s) 0.385 24.4Au(s) 0.129 25.4 CompoundsNH3(l) 4.70 80.1H2O(l) 4.184 75.3C2H5OH(l) 2.46 112.(CH2OH)2(l) 2.42 149.H2O(s) 2.06 37.1CCl4(l) 0.861 132.CCl2F2(l) 0.598 72.3Common solidswood 1.76concrete 0.88glass 0.84granite 0.79
© 2008 Brooks/Cole 21
Heat Capacity
Example 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 22
Heat Capacity
ExampleExample
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)
Thus Tfinal = ΔT + Tinital = −107 + 125°C = 18°C
heat is lost, q is negativenegative
ΔT = −24.1 x 103 J /(250. g x 0.902 J g-1 °C-1)
ΔT = Tfinal – Tinital = −107 °C
© 2008 Brooks/Cole 23
Heat CapacityA 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 24
Conservation of Energy and Changes of State
When heat is: AddedAdded to a system q is positive the change is endothermicendothermic
Removed Removed from a system q is negative the change is exothermicexothermic.
Water Boils: H2O(l) H2O(g) endothermic
Steam Condenses: H2O(g) H2O(l) exothermic
Work occurs as the sample expands or contracts. Overall: ΔE = q + w
© 2008 Brooks/Cole 25
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
Conservation of Energy and Changes of State
© 2008 Brooks/Cole 26
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
Conservation of Energy and Changes of State
© 2008 Brooks/Cole 27
Because ΔE = q + w:
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
At Constant V: ΔE = qV • subscript V shows fixed V• work requires motion against an opposing force.• constant V = no motion, so w = 0.
Enthalpy: Heat Transfer at Constant P
© 2008 Brooks/Cole 28
T
empe
ratu
re (
°C)
-50
-25
0
25
50
0 100 200 300 400 500 600 Quantity of energy transferred (J)
Water warms from 0 to 50°C
During freezing (or melting)
Example:Example:
Convert ice at -50°C to water at +50°C
• Substance loses (or gains) E, but…• T remains constant.
Ice is melting. T remains at 0°C
Ice warms from -50 to 0°C
Freezing and Melting (Fusion)
© 2008 Brooks/Cole 29
Change Name value for H2O (J/g)
solid → liq enthalpy of fusion 333
liq → gas enthalpy of vaporization 2260
ΔHfusion = qP = heat to melt a solid.
liq → solid enthalpy of freezing −333
gas → liq enthalpy of condensation −2260
Note: ΔHfusion = − ΔHfreezing etc.
qfusion = −qfreezing
Changes of State
© 2008 Brooks/Cole 30
State functionsState 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
HH E P V T etc.
State Functions and Path Independence
State function changes are path independentpath independent.
ΔH = Hfinal – Hinitial is constant.
© 2008 Brooks/Cole 31
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 = qP can be added to a balanced equation.
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 waterliquid water and release 890 kJ of heat Change a physical state, change ΔH° : H2O(l) vs. H2O(g)
Thermochemical Expressions
© 2008 Brooks/Cole 32
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 33
Where Does the Energy Come From?
Bond Enthalpy (bond energy)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 34
Bond Enthalpies
During a chemical reaction:During a chemical reaction:
Old bonds break: requires E (endothermic)
Both typically occur:
H2(g) + Cl2(g) 2 H(g) + 2 Cl(g) 2 HCl(g)
endothermicΔH= +678 kJ/mol
exothermic ΔH= -862 kJ/mol
New bonds form: releases E (exothermic)
© 2008 Brooks/Cole 35
Bond Enthalpies
Endothermic reactionsEndothermic reactions (ΔH > 0) E is absorbed. New bonds are less stable than the old,
or Fewer bonds are formed than broken
Overall, heat may be absorbed or released:
Exothermic reactionsExothermic reactions (ΔH < 0) E is released. New bonds are more stable than the old,
or More bonds are formed than broken.
reactants
products
ener
gyle
ss s
tabi
lity
reactants
productsener
gyle
ss s
tabi
lity
© 2008 Brooks/Cole 36
Measuring Enthalpy Changes
Heat transfers are measured with a calorimetercalorimeter.
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 37
or
−qreaction = qbomb + qwater
with
qbomb = mcalccalΔT = CcalΔT
Bomb CalorimeterBomb Calorimeter
Measure ΔT of the water. Constant V: qV = ΔE
Conservation of E:
qreaction + qbomb + qwater = 0
Measuring Enthalpy Changes
A constant for a calorimeter
© 2008 Brooks/Cole 38
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
Measuring Enthalpy Changes
© 2008 Brooks/Cole 39
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
Measuring Enthalpy Changes
© 2008 Brooks/Cole 40
Coffee-cup calorimeterCoffee-cup calorimeterNested 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)
Measuring Enthalpy Changes
© 2008 Brooks/Cole 41
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 = 0.03291 mol
nHCl = 0.250 L = 0.100 mol
1 mol24.31 g
0.4 mol1 L
1Mg ≡ 2HCl Mg is limitingMg is limiting
Measuring Enthalpy Changes
© 2008 Brooks/Cole 42
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
So = or ΔH = – 462 kJΔH1 mol Mg
–15.22 kJ0.03291mol
Heat fromfrom the reaction went intointo the solution. So:qsolution = – qreaction
qreaction = –15.22 kJ = ΔH
exothermicexothermic
… 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 = ?
Measuring Enthalpy Changes
© 2008 Brooks/Cole 43
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:Another version:“ΔH° for a reaction is the same whether it takes place in a single step or several steps.”
HH is a state function is a state function
© 2008 Brooks/Cole 44
Hess’s LawMultiply a reaction, multiply ΔH.
Reverse a reaction, change the sign of ΔH.
Then
2 CO2(g) → 2 CO(g) + O2(g) ΔH = –1(–566.0 kJ)
= + 566.0 kJ
4 CO2(g) → 4 CO(g) + 2 O2(g) ΔH = –2(–566.0 kJ)
= +1132.0 kJ
2 CO(g) + O2(g) → 2 CO2 (g) ΔH = −566.0 kJ
© 2008 Brooks/Cole 45
Use Hess’s Law to find ΔH for unmeasured reactions.
ExampleExample
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) CO2(g) ΔH = −393.5 kJ
2 CO(g) + O2(g) 2 CO2(g) ΔH = −566.0 kJ
Hess’s Law
© 2008 Brooks/Cole 46
Rearrange:
or:
+2[C + O2 CO2] +2(−393.5) = −787.0
−1[2 CO + O2 2 CO2] −1(−566.0) = +566.0
Calculate ΔH for the reaction: 2C(graphite) + O2(g) → 2CO(g)
2 C + 2 O2 2 CO2 ΔH° = −787.0 kJ
2 CO2 2 CO + O2 ΔH° = +566.0 kJ
Hess’s Law
Add, then cancel:
2 C + 2 O2 + 2 CO2 2 CO2 + 2 CO + O2 −221.0
2 C + O2 2 CO ΔH° = −221.0 kJ
© 2008 Brooks/Cole 47
Determine ΔH° for the production of coal gas:2 C(s) + 2 H2O(g) CH4(g) + CO2(g)
C(s) + H2O(g) CO(g) + H2(g) ΔH° = 131.3 kJ
CO(g) + H2O(g) CO2(g) + H2(g) ΔH° = −41.2 kJ
CH4(g) + H2O(g) CO(g) + 3 H2(g) ΔH° = 206.1 kJ
AA
BB
CC
Using:
Hess’s Law
© 2008 Brooks/Cole 48
Want: 2 C(s) + 2 H2O(g) CH4(g) + CO2(g)
2 C on left, use 2 x AA
2 C(s) + 2 H2O(g) 2 CO(g) + 2 H2(g) +262.6
1 CO2 on right, so use 1 x BB
CO(g) + H2O(g) CO2(g) + H2(g) −41.2
1 CH4 on right, use −1 x CC
CO(g) + 3 H2(g) CH4(g) + H2O(g) −206.1
Add and cancel:2C + 3H2O + 2CO + 3H2 2CO + 3H2 + CH4 + H2O + CO2 15.3 kJ
change to 2 H2Ochange to 2 H2O 2 C + 2 H2O → CH4 + CO2 ΔH = 15.3 kJ2 C + 2 H2O → CH4 + CO2 ΔH = 15.3 kJ
Hess’s Law
© 2008 Brooks/Cole 49
1
but the formation reaction is:
H2(g) + ½ O2(g) H2O(l) ΔHf° = −285.83 kJ
Hess’s law problems often use a combustion or …
H2 combustion:
2 H2(g) + O2(g) 2 H2O(l) ΔH° = −571.66 kJ
f = formation
Standard Molar Enthalpy of Formation
Formation reactionFormation reaction
Make 1 mol of compound from its elements in their standard states.
© 2008 Brooks/Cole 50
Standard stateStandard 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 at 298 K
ΔHf° (Br2(g) ) ≠ 0 at 298 K
Standard Molar Enthalpy of Formation
© 2008 Brooks/Cole 51
Formula Name Hf°, kJ/mol
Al2O3(s) aluminum oxide −1675.7
CaO(s) calcium oxide −635.09
CH4(g) methane −74.81
C2H2(g) acetylene +226.73
C2H4(g) ethylene +52.26
C2H6(g) ethane −84.68
C2H5OH(l) ethanol −277.69
H2O(g) water vapor −241.818
H2O(l) liquid water −285.830
NaF(s) sodium fluoride −573.647
NaCl(s) sodium chloride −411.153
SO2(g) sulfur dioxide −296.830
SO3(g) sulfur trioxide −395.72
Appendix J
NotesNotes• Most are negative
(formation releases E), but can be positive.
• If the physical state changes, ΔHf° changes.
Standard Enthalpy of Formation (25°C)
© 2008 Brooks/Cole 52
ΔH° ={(nproducts)(ΔHf° products)}
– {(nreactants)(ΔHf° reactants)}
ΔH° = [1ΔHf°(HCN) + 33ΔHf°(H2)] −
[1ΔHf°(NH3) + 1ΔHf°(CH4)]
Standard Molar Enthalpies of Formation
= [+134 + 3(0)] − [− 46.11 + (−74.85)] = 255 kJ/mol
ExampleExample
Calculate ΔH° for:
CH4(g) + NH3(g) HCN(g) + 3 H2(g)
© 2008 Brooks/Cole 53
Hydrazine
N2H4(g) + O2(g) → N2(g) + 2H2O(g)
Chemical Fuels for Home and Industry
Chemical FuelChemical Fuel – reacts exothermically with O2 in air.
A good fuel has weak bonds and/or strong product bonds.
Ebond(kJ/mol) 160(NN), 391(NH) 498(OO) 946(NN) 467(OH)
Emolecule 1724 498 946 1864Ereagents 2222 2814
ΔE -592 kJ/mol
© 2008 Brooks/Cole 54
• Fuel value = (E released) / (mass of fuel in g)
• Energy density = (E released) / Vfuel
U.S. sources of E:
Mostly fossil fuels.
Chemical Fuels for Home and Industry
Good fuels have large:
© 2008 Brooks/Cole 55
Foods: Fuels for Our BodyCarbohydrates, 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 56
Approximate Composition
Food per 100. g Caloric Value
Fat Carbohydrate Protein Cal/g kJ/g
All-purpose flour 0.0 73.3 13.3 3.33 13.95
Apple 0.5 13.0 0.4 0.59 2.47
Brownie with nuts 16.0 64.0 4.0 4.04 16.9
Egg 0.7 10.0 13.0 1.40 5.86
Hamburger 30.0 0.0 22.0 3.60 15.06
Peanuts 50.0 21.4 28.6 5.71 23.91
Rice 1.0 77.6 8.2 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.
Foods: Fuels for Our Body
© 2008 Brooks/Cole 57
Summary ProblemSO2 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 58
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 ΔH° = ΔHf° = −393.509 kJ/mol
ncoal = 7000 tons = 5.292 x 108 mol1 mol
12.011 g
9.08 x 105g
1 ton
Heat released = 5.292x108 mol = −2.08 x 1011 kJ
= −2.08 x 1014 J
–393.509 kJ
1 mol
(b) What % of thermal E becomes electrical E?
Efficiency = (Eout/Ein) x 100% = x100% = 41.5%8.64 x 1013 J
2.08 x 1014 J
© 2008 Brooks/Cole 59
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 60
Summary ProblemSO2 + 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 addedadded to the combustion