Transcript Chapter 6 Energy and Chemical Reactions

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
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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.
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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.
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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
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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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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)
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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.
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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.
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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.
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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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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
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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 …..
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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.
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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
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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
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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.
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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
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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
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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)
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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
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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)
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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
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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
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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
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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
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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.
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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
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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
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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.
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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)
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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
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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
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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
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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).
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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
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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
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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
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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)
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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
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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
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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
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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
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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
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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
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2 CO
2 CO2 + 2 CO + O2
−221.0
ΔH° = −221.0 kJ
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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
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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
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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
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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.
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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.
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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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