Sec 6'1 The Nature of Energy pg' 242246

1

• Sec 6.1 The Nature of Energy pg. 242-246
• Energy- the capacity to do ________ or to produce
  heat.
• Law of conservation of energy Energy can be
  converted from one form to another but can be
  neither created nor ____________ (the energy in
  the universe is constant).
• Potential energy (PE) is energy due to
  ___________ or composition. Example Water
  behind a dam has potential energy and if allowed
  to fall can do work to turn turbines to produce
  electricity. Molecules have potential energy due
  to the attractive and repulsive forces between
  atoms, in a chemical reaction that occurs with no
  added energy, the atoms are rearranged to form
  molecules of lower PE and energy is then released
  as heat.
• Kinetic energy (KE) is the energy due to
  _________. (KE ½ mv2 doubling the mass at
  constant velocity will increase the KE by a
  factor of ___while doubling the velocity at
  constant mass will increase the KE by a factor of
  ___ ).
• Work is force acting over a distance (W F. d).
• Difference of heat and temperature
• Temperature is a property that depends upon the
  motion (avg. ______) of the particles.
• Heat is the ___________ of energy between two
  objects as a result of a temperature difference.
• See figure 6.1 and read text pertaining to the
  figure.
• Energy can be transferred in 2 ways heat or
  work. For a process beginning at some energy and
  ending at some energy, the energy change will
  always be the same (it is independent of pathway
  making it a ________ function). However, the
  manner in which the energy is transferred can be
  divided differently between the heat and work
  (the specific pathway).
• Chemical energy
• Consider the reaction CH4(g) 2 O2(g) ? CO2(g)
  2 H2O (g) energy (heat)
• In order to discuss the reaction, the universe is
  divided into 2 parts the ____________ (the part
  of the universe that we are considering) and the
  _______________ (everything else in the universe
  other than the system). For the reaction above,
  the reactants and products are the ____________.
  The surroundings would be the reaction vessel,
  the room and anything else.

work
destroyed
position
motion
2
4
KE
transfer
state
system
surroundings
system
2
out

• In an exothermic reaction, energy flows _____ of
  the system (the heat released comes from the
  difference between the PE of the products and the
  reactants. Here, the products would have a
  _________ PE than the reactants). In an
  exothermic process the energy gained by the
  surroundings is ________ to the energy lost by
  the system (See fig 6.2 pg. 244). In an
  endothermic process energy flows _______ the
  system from the surroundings to increase PE of
  the system.
• N2 (g) O2 (g) energy (heat) ? 2NO (see
  fig 6.3 pg. 245)

lower
equal
into
3

• Thermodynamics is the study of energy and its
  interconversions.
• First Law of Thermodynamics - the energy of the
  universe is _____________ (cannot be created or
  destroyed).
• Internal Energy (E) of a system is the sum of the
  KE and PE of all particles in the system. The
  internal energy of a system can be changed by a
  flow of heat and/or work.
• ?E q w (q is heat, w is work)
• In an exothermic process ? E lt 0 for an
  endothermic process ? E gt 0
• If the system does work on the surroundings, then
  w is ___________ if the surroundings do work on
  the system, then w is ____________. If the system
  loses heat to the surroundings, then q is
  ___________ if the surroundings give heat to the
  system, then q is ____________ (here the systems
  point of view is considered).
• Sample exercise 6.1 pg. 246. (L atm 101.325
  J)
• ?E q w
• Work performed by gases are commonly associated
  with chemical processes (In an automobile,
  combustion of gases (the system) performs work on
  pistons (the surroundings). Here, the sign of
  the work would be ____________.
• See pg. 244 for derivation of the following
  equation Work - P ? V
• (When the gas is expanding ? V is positive so w
  is _________, when compressed ? V is negative so
  w is __________.
• See sample exercise 6.2 and 6.3

conserved
negative
positive
positive
negative
?E 15.6 kJ (1.4 kJ)
?E 17.0 kJ
negative
negative
positive
w -P?V
?E q w
101.325 J Latm
w -(1.0 atm)(0.50.106L) .
w -P?V
?E 13.107 J - 5.06625.107 J
w -(15 atm)(64L 46 L)
w -5.06625.107 J
w -(15 atm)(18 L)
?E 8.107 J
w -270 Latm
Homework pg. 280-281 9 17 19 21 23 25 27
4
Sec 6.2 Enthalpy and Calorimetry 1. Change in
Enthalpy (?H) of a system is a measure of the
____________ flow as heat. ?H Hproducts -
Hreactants (?H qp q at constant pressure).. See
pg. 249. Enthalpy and heat of reaction are
interchangeable at constant pressure where no
change In volume is present. 2. If ?H is ___
than an exothermic change has occurred if ____
then an endothermic change has occurred. Sample
exercise 6.4 pg 249 and 33a
energy

-
6.4 ?kJ 5.8 g CH4 (mol / 16.043 g)
(-890 kJ /mol CH4) -320 kJ or (320 kJ of
heat released)
(-1652 kJ /4 mol Fe)
33a. ?kJ 4.00 mol Fe
-1650 KJ
Calorimetry the science of measuring heat based
upon observing the______________ change of a body
when it absorbs or discharges heat. Heat Capacity
(C) the heat absorbed per increase in
temperature. Specific heat capacity (s commonly
expressed as Cp) heat capacity per _____. Molar
heat capacity heat capacity per ______.
temperature
gram
mole

• To determine the energy change involving
  reactions in ___________, a constant-pressure
  calorimeter
• can be used (see fig. 6.5 pg. 251).
• When using dilute aqueous solutions approximate s
  to be 4.2 J/goC instead of 4.18 J/goC
• 2. To obtain the heat involved for a temperature
  change in a substance use the following
• q s m ?T (try problem 42 pg. 282) ?T tf -
  ti

solution
5

• qlost by metals - qgained by water
• (ms?t)Al (ms?t)Fe - (ms?t)water
• (ms(tf - ti))Al (ms(tf - ti))Fe - (ms(tf
  ti))water
• (ms(ti- tf))Al (ms(ti- tf))Fe (ms(tf
  ti))water
• (It is OK to start here where all ?ts are
  positive )
• msti,Al msAl tf msti,Fe msFe tf msw tf -
  msti,w
• (msw msAl msFe ) tf msti,Al msti,Fe
  msti,w
• tf ((msAl msFe) ti,Fe,Al msti,w)
• (msw msAl msFe ))
• tf((5.00 g)(0.89 J/goC)(10.00
  g)(0.45 J/goC)) 100.0oC) (97.3g)(4.18J/goC)(22.0o
  C)
• (97.3 g)(4.18 J/goC) (5.00 g)(0.89 J/goC)
  (10.00 g)(0.45 J/goC)
• tf (445.0 450.0 8847.71 ) J

for large equations with many variables working
w/o variable isolation is OK (show your algebra).
6

• qlost by metals - qgained by water
• (ms?t)Al (ms?t)Fe - (ms?t)water
• (ms(tf - ti))Al (ms(tf - ti))Fe - (ms(tf
  ti))water
• (ms(ti- tf))Al (ms(ti- tf))Fe (ms(tf
  ti))water
• ((5.00 g)(0.89 J/goC)(100.0oC- tf))(10.00
  g)(0.45 J/goC)(100.0oC- tf) (97.3g)(4.18J/goC)(tf
  - 22.0oC)
• 445.0 J - 4.450 J/oC tf 450.0 J 4.500 J/oC
  tf406.71 J/oC tf 8947.7 J
• 415.66 J/oC tf 9842.7 J
• tf 9842. J/ 415.66 J / oC
• tf 24oC

7
Heat of a reaction is an _____________ property
(depends upon the amounts of reactants). Sample
Exercise 6.5 pg. 252
extensive
Ba(NO3)2 (aq) Na2SO4 (aq) ? BaSO4 (s) 2 NaNO3
(aq)
Initial 1.0000 mol 1.0000 mol 0 mol
0 mol
Final 0 mol 0 mol 1.0000
mol 2.0000 mol
?H q / mol
q m s ?t
? H - 2.604 .104 J / 1.0000 mol
q (2.00 L (1000 ml / L)(1.0 g/ml) (4.2 J/goC)
(3.1 oC)
? H - 26 kJ /mol
q - 2.604 .104 J (heat released to water)
bomb
Constant-Volume Calorimetry is done through the
use of a _______ calorimeter (see fig. 6.6 pg.
251). No work (w -P?V) done due to no changes
in volume. Read the problem on pg.
253-254. Sample exercise 6.6 pg. 255
?kJ 14.3oC ( 11.3 kJ / oC) g
11.5 g
141 kJ / g for H2
?kJ 7.3oC ( 11.3 kJ / oC) g 1.5
g
55 kJ / g for CH4
Homework Day 1 pg. 281-282 s 29 31
33b,d 37 39 41 (heat lost - heat gained)
43 45 Day 2 pg.
281-285 s 24 26 30 32 36 44 48 49 79
8

• Sec 6.3 Hesss Law
• Change in enthalpy ( ) is a state
  function so it is independent of _____________.
  Therefore, the change in energy that transpires
  between some initial state and some final state
  of a chemical reaction will be the __________
  whether this change occurs in one step or the sum
  of a series of steps.
• Consider the following N2 (g) 2 O2 (g) ?
  2NO2 (g) ?H 68 kJ
• The reaction shown above may occur in 2 steps
• Step 1 N2 (g) O2 (g) ? 2NO ?H 180 kJ
• Step 2 O2 (g) 2 NO (g) ? 2NO2 (g) ?H -112
  kJ
• The sum of steps 1 and 2 yield the reaction
  originally shown (the NO is neither a reactant
  nor a product and is referred to as an
  intermediate).

?H
pathway
same
Consider the formation of diamond from graphite
(an endothermic process so heat is a reactant) C
graphite ? C diamond ?H 2 kJ or C graphite
2 kJ ? C diamond The reverse of this process
is an exothermic process (exothermic processes
tend to occur more spontaneously-not necessarily
quickly (see pg. 496)) C diamond ? C graphite
?H - 2 kJ or C diamond ? C graphite 2 kJ
Both compounds have a heat of combustion as shown
below C graphite (s) O2 (g) ? CO2 (g) ?H
- 394 kJ C diamond (s) O2 (g) ? CO2 (g)
?H - 396 kJ
If the diamond reaction is inverted and then sum
of the two is taken, the following is obtained. C
graphite (s) O2 (g) ? CO2 (g) ?H -394 kJ CO2
(g) ? C diamond (s) O2 (g) ?H 396 kJ C (s)
graphite ? C (s) diamond ?H 2 kJ
9

• Try Sample Exercise 6.8 pg. 259

2B (s) 3H2 (g)? B2H6 (g)
2B (s) 3/2 O2 (g)? B2O3 (g)
?H -1273 kJ
B2O3 (s) 3 H2O (g)? B2H6 (g) 3 O2 (g)
?H 2035 kJ
?H 3 ( -286 kJ)
H2(g) ½ O2 (g) ? H2O (l)
3( )

3 H2(g) 3/2 O2 (g) ? 3 H2O (l)
H2O (l) ? H2O (g)
3 3
?H 3 (44 kJ)
?H 36 kJ
2B (s) 3H2 (g)? B2H6 (g)
Homework pg. 280-281 s 14 47 51 55 (and
additional 1 and 2 below) Additional 1. S
(s) 3/2 O2 (g) ? SO3 (g) ?H -395.2 kJ
2. 2 NH3(g) 3 N2O(g) ? 4 N2(g) 3 H2O(l)
?H -1010. kJ 2 SO2 (g) O2 (g) ? 2SO3 (g)
?H -198.2 kJ N2O (g) 3 H2 (g) ?
N2H4 (g) H2O (l) ?H -317 kJ Determine ?H
for the following 2 NH3 (g) ½ O2
(g) ? N2H4 (g) H2O (l) ?H -143 kJ S (s)
O2 (g) ? SO2 (g) ?H ?
H2 (g) ½ O2 (g) ? H2O (l) ?H -286
kJ Determine ?H for the following N2H4
(g) O2 (g) ? N2 (g) 2 H2O (l) ?H ?
10

• Sec 6.4 Standard Enthalpies of Formation
• 1. Often it is impossible to determine the ?H of
  a reaction through the use of calorimetry because
  the
• process may be too _______ under normal
  conditions. Such as the conversion of graphite
  to diamond
• Compounds are formed from ____________. The ?H
  of formation of compounds from their
• elements can be used to determine the ?H of
  reactions.
• Standard enthalpy of formation ( ) is
  the enthalpy that accompanies the formation of 1
  mole of
• a substance from its elements under
  ______________ state conditions. The degree
  symbol is used to
• indicate standard state condition. See pg. 260
  for standard states.
• When writing chemical equations for standard
  enthalpy of formation, all substances are to be
  in
• their standard state and the equation should be
  balanced so as to form ___ mole of compound.
• (see formation of nitrogen dioxide and methanol
  pg. 261).
• See table 6.2 pg. 261 and back of book (A21-A23)
  for standard enthalpy of formation values.

slow
elements
?Hof
standard
1
?Horeaction S(n ?Hoproducts) - S(n
?Horeactants)
0
5. The enthalpy of formation of an element is ___.
Lets look at an example using Hesss Law
CH4 (s) 2 02(g) ? CO2 (g) 2 H2O (g)
?Hfo -75 kJ
C (s) 2 H2(g) ? CH4 (g)
?Hfo 75 kJ
CH4 (g) ? C (s) 2 H2(g)
?Hfo -286 kJ
H2 (g) ½ O2(g) ? H2O (g)
2 H2 (g) O2(g) ? 2 H2O (g)
?Hfo 2(-286 kJ)
?Hfo -394 kJ
C (s) O2(g) ? CO2 (g)
C (s) O2(g) ? CO2 (g)
?Hfo -394 kJ
?Horeaction (1(-394 kJ) 2 (-286 kJ)) 1 (75
kJ)
-891 kJ
11
Sample exercises 6.9-6.11 pg. 263-267
Homework Day1 pg. 280-286 s 13 15
50 59 61 65, 67 Day 2 pg. 280-286 s 54
56 62a 66 68 (have water as a liquid product
here unlike the combustion reactions of 61 and
62a) 86 (not required but for more Hesss law
practice 63)
12

• The End is Near (at least the end of Chapter 6)
• Sec 6.6 New Energy Sources
• Most of the energy we now use comes from coal,
  hydro and nuclear power and fossil fuels (see
  figure 6.11 pg. 267). As we look to the future,
  there are a few energy sources to consider the
  sun (________ energy), ________ (biomass) and
  fission and fusion reactions (nuclear energy
  which will be explored in Chapter 21).
• Coal Conversion
• 1. Transportation of solid coal is costly so
  more energy-efficient fuels are being developed
  from coal such as a ____________ fuel.
• To convert coal to a gas, its structure must be
  broken apart. This can be done by treating the
  coal with oxygen and steam at _____ temperature
  to break apart many of the C-C bonds and replace
  them with
• C-H and C-O bonds. This produces much smaller
  molecules (CH4(g), H2O (g), CO2(g), CO (g),
  H2(g)). This process is referred to as coal
  ______________ (see figure 6.14 pg. 272).
• 3. The desired products of coal gasification is
  a mixture of carbon monoxide ( ) and
  hydrogen ( ) (called syngas) and methane (
  ). These products are combustible and
  release energy upon burning (?H is negative due
  to the exothermic process).
• 4. Most of the reactions to make these products
  from coal are exothermic, one is endothermic (pg.
  272). Therefore, in producing syngas and methane
  from coal conditions such as feed rate of coal,
  air, steam must be controlled so that the
  reactions energy can ___________ the process
  with no external source of energy.
• Syngas can be used as a fuel itself or may be
  converted to another form of fuel,
  _________(CH3OH). Methanol can be converted to
  gasoline (half of South Africas gasoline supply
  comes from methanol produced from syngas).
• Hydrogen as a fuel
• Combustion of hydrogen releases large amounts of
  energy (-286 kJ/mol which is the ?Hof of water).
  On advantage of combustion of hydrogen is that is
  produces no ______ (a greenhouse gas) only water.
• Problems with using hydrogen as a fuel.

solar
plants
gaseous
high
gasification
CO
H2
CH4
sustain
methanol
CO2
13

• 1. Cost of production.
• Hydrogen can be produced by treating methane
  (natural gas) with ________. However, this
  process required 206 kJ per 3 mole of hydrogen
  formed (?H is _______).See page 273.
• Another method of producing hydrogen is the
  decomposition of water (the worlds largest
  resource) . This required _______ kJ per mole of
  hydrogen produced. This decomposition can be
  done electrolytically (fig. 17.20 pg. 857) or by
  thermochemical decomposition or through a
  modification of ____________________(read pg.
  275).
• 2. Transportation and storage
• Hydrogen gas makes metal structures ________ .
• Hydrogen does not give much energy per _________
  of hydrogen gas although it yields much per gram.
  It yields about one third the energy per volume
  as __________.
• Sample Exercise 6.12-6.13 pg. 275-276.
• Hydrogen would need to be stored as a _________
  (it has a boiling point of 20 K and a critical
  temperature (temperature above which, no matter
  how great the pressure the gas cannot be
  liquified) of 33 K. Therefore, storage of
  hydrogen would require super-insulated containers
  that can handle very high pressures.
• A better alternative is the use of metal
  __________ (read pg. 277).
• Other Sources of energy
• 1. Oil shale contains kerogen (a fuel).
  Difficult to recover because it cant be pumped
  so the rock is ___________ (250 oC) to remove the
  kerogen.
• 2. Ethanol is a supplement to gasoline. Made
  from fermentation of _________ (mostly from corn
  but can come from other fruits and vegetables) by
  the action of yeast.
• The burning of ______ ethanol is not possible at
  low temperatures because it has a low vapor
  pressure so it can only be used in ________
  climates such as Brazil.
• 3. Oil can be squeezed from seeds. Farmers in
  North Dakota, South Africa and Australia are now
  using sunflower oil to replace _________ fuel.
  The oil is a combustible hydrocarbon so it will
  release heat when burned.

steam

286
photosynthesis
brittle
mole
methane
liquid
hydrides
heated
sugar
pure
warm
diesel