Chapter 5 Thermochemistry

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Chapter 5Thermochemistry
Chemistry, The Central Science, 10th
edition Theodore L. Brown H. Eugene LeMay, Jr.
and Bruce E. Bursten

• John D. Bookstaver
• St. Charles Community College
• St. Peters, MO
• ? 2006, Prentice Hall, Inc.

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Energy

• The ability to do work or transfer heat.
• Work Energy used to cause an object that has
  mass to move.
• Heat Energy used to cause the temperature of an
  object to rise.

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Potential Energy

• Energy an object possesses by virtue of its
  position or chemical composition.

PE mgh g gravitational constant 9.8 m/s2
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Kinetic Energy

• Energy an object possesses by virtue of its
  motion.

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Units of Energy

• The SI unit of energy is the joule (J).
• An older, non-SI unit is still in widespread use
  The calorie (cal).
• 1 cal 4.184 J

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System and Surroundings

• The system includes the molecules we want to
  study (here, the hydrogen and oxygen molecules).
• The surroundings are everything else (here, the
  cylinder and piston).

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Work

• Energy used to move an object over some distance.
• w F ? d,
• where w is work, F is the force, and d is the
  distance over which the force is exerted.

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Heat

• Energy can also be transferred as heat.
• Heat flows from warmer objects to cooler objects.

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Transferal of Energy

• The potential energy of this ball of clay is
  increased when it is moved from the ground to the
  top of the wall.

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Transferal of Energy

• The potential energy of this ball of clay is
  increased when it is moved from the ground to the
  top of the wall.
• As the ball falls, its potential energy is
  converted to kinetic energy.

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Transferal of Energy

• The potential energy of this ball of clay is
  increased when it is moved from the ground to the
  top of the wall.
• As the ball falls, its potential energy is
  converted to kinetic energy.
• When it hits the ground, its kinetic energy falls
  to zero (since it is no longer moving) some of
  the energy does work on the ball, the rest is
  dissipated as heat.

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Check Work must be done in part (b) to increase
the potential energy of the ball, which is in
accord with our experiences. The units are
appropriate in both parts (b) and (c). The work
is in units of J and the speed in units of m/s.
In part (c) we have carried an additional digit
in the intermediate calculation involving the
square root, but we report the final value to
only two significant figures, as appropriate.
Comment A speed of 1 m/s is roughly 2 mph, so
the bowling ball has a speed greater than 10 mph
upon impact.
PRACTICE EXERCISE What is the kinetic energy, in
J, of (a) an Ar atom moving with a speed of 650
m/s, (b) a mole of Ar atoms moving with a speed
of 650 m/s? (Hint 1 amu 1.66 ? 10-27kg)
Answers  (a) 1.4 ? 10-20J, (b) 8.4 ? 103J
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First Law of Thermodynamics

• Energy is neither created nor destroyed.
• In other words, the total energy of the universe
  is a constant if the system loses energy, it
  must be gained by the surroundings, and vice
  versa.

Use Fig. 5.5
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Internal Energy

• The internal energy of a system is the sum of
  all kinetic and potential energies of all
  components of the system we call it E.

Use Fig. 5.5
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Internal Energy

• By definition, the change in internal energy,
  ?E, is the final energy of the system minus the
  initial energy of the system
• ?E Efinal - Einitial

Use Fig. 5.5
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Changes in Internal Energy

• If ?E gt 0, Efinal gt Einitial
• Therefore, the system absorbed energy from the
  surroundings.
• This energy change is called endergonic.

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Changes in Internal Energy

• If ?E lt 0, Efinal lt Einitial
• Therefore, the system released energy to the
  surroundings.
• This energy change is called exergonic.

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Changes in Internal Energy

• When energy is exchanged between the system and
  the surroundings, it is exchanged as either heat
  (q) or work (w).
• That is, ?E q w.

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?E, q, w, and Their Signs
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SAMPLE EXERCISE 5.2 Relating Heat and Work to
Changes of Internal Energy
PRACTICE EXERCISE Calculate the change in the
internal energy of the system for a process in
which the system absorbs 140 J of heat from the
surroundings and does 85 J of work on the
surroundings.
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SAMPLE EXERCISE 5.2 Relating Heat and Work to
Changes of Internal Energy
PRACTICE EXERCISE Calculate the change in the
internal energy of the system for a process in
which the system absorbs 140 J of heat from the
surroundings and does 85 J of work on the
surroundings.
Answer 55 J
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Exchange of Heat between System and Surroundings

• When heat is absorbed by the system from the
  surroundings, the process is endothermic.

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Exchange of Heat between System and Surroundings

• When heat is absorbed by the system from the
  surroundings, the process is endothermic.
• When heat is released by the system to the
  surroundings, the process is exothermic.

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State Functions

• Usually we have no way of knowing the internal
  energy of a system finding that value is simply
  too complex a problem.

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State Functions

• However, we do know that the internal energy of a
  system is independent of the path by which the
  system achieved that state.
• In the system below, the water could have reached
  room temperature from either direction.

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State Functions

• Therefore, internal energy is a state function.
• It depends only on the present state of the
  system, not on the path by which the system
  arrived at that state.
• And so, ?E depends only on Einitial and Efinal.

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State Functions

• However, q and w are not state functions.
• Whether the battery is shorted out or is
  discharged by running the fan, its ?E is the
  same.
• But q and w are different in the two cases.

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Work

• When a process occurs in an open container,
  commonly the only work done is a change in volume
  of a gas pushing on the surroundings (or being
  pushed on by the surroundings).

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Work

• We can measure the work done by the gas if the
  reaction is done in a vessel that has been fitted
  with a piston.
• w -P?V

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Enthalpy

• If a process takes place at constant pressure (as
  the majority of processes we study do) and the
  only work done is this pressure-volume work, we
  can account for heat flow during the process by
  measuring the enthalpy of the system.
• Enthalpy is the internal energy plus the product
  of pressure and volume

H E PV
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Enthalpy

• When the system changes at constant pressure, the
  change in enthalpy, ?H, is
• ?H ?(E PV)
• This can be written
• ?H ?E P?V

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Enthalpy

• Since ?E q w and w -P?V, we can substitute
  these into the enthalpy expression
• ?H ?E P?V
• ?H (qw) - w
• ?H q
• So, at constant pressure the change in enthalpy
  is the heat gained or lost.

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Endothermicity and Exothermicity

• A process is endothermic, then, when ?H is
  positive.

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Endothermicity and Exothermicity

• A process is endothermic when ?H is positive.
• A process is exothermic when ?H is negative.

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PRACTICE EXERCISE Suppose we confine 1 g of
butane and sufficient oxygen to completely
combust it in a cylinder like that in Figure
5.13. The cylinder is perfectly insulating, so no
heat can escape to the surroundings. A spark
initiates combustion of the butane, which forms
carbon dioxide and water vapor. If we used this
apparatus to measure the enthalpy change in the
reaction, would the piston rise, fall, or stay
the same?
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Solution   Analyze Our goal is to determine
whether ?H is positive or negative for each
process. Because each process appears to occur at
constant pressure, the enthalpy change of each
one equals the amount of heat absorbed or
evolved, ?E qp. Plan We must predict whether
heat is absorbed or released by the system in
each process. Processes in which heat is absorbed
are endothermic and have a positive sign for ?H
those in which heat is evolved are exothermic and
have a negative sign for ?H. Solve In (a) the
water that makes up the ice cube is the system.
The ice cube absorbs heat from the surroundings
as it melts, so ?H is positive and the process is
endothermic. In (b) the system is the 1 g of
butane and the oxygen required to combust it. The
combustion of butane in oxygen gives off heat, so
?H is negative and the process is exothermic.
PRACTICE EXERCISE Suppose we confine 1 g of
butane and sufficient oxygen to completely
combust it in a cylinder like that in Figure
5.13. The cylinder is perfectly insulating, so no
heat can escape to the surroundings. A spark
initiates combustion of the butane, which forms
carbon dioxide and water vapor. If we used this
apparatus to measure the enthalpy change in the
reaction, would the piston rise, fall, or stay
the same?
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Enthalpies of Reaction

• The change in enthalpy, ?H, is the enthalpy of
  the products minus the enthalpy of the reactants
  
• ?H Hproducts - Hreactants

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Enthalpies of Reaction

• This quantity, ?H, is called the enthalpy of
  reaction, or the heat of reaction.

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The Truth about Enthalpy

• Enthalpy is an extensive property.
• ?H for a reaction in the forward direction is
  equal in size, but opposite in sign, to ?H for
  the reverse reaction.
• ?H for a reaction depends on the state of the
  products and the state of the reactants.

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The negative sign indicates that 250 kJ is
released by the system into the surroundings.
Answer 14.4 kJ
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Calorimetry

• Since we cannot know the exact enthalpy of the
  reactants and products, we measure ?H through
  calorimetry, the measurement of heat flow.

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Heat Capacity and Specific Heat

• The amount of energy required to raise the
  temperature of a substance by 1 K (1?C) is its
  heat capacity.
• We define specific heat capacity (or simply
  specific heat) as the amount of energy required
  to raise the temperature of 1 g of a substance by
  1 K.

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Heat Capacity and Specific Heat

• Specific heat, then, is

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Solution   Analyze In part (a) we must find the
total quantity of heat needed to warm the sample
of water, given the mass of water (m), its
temperature change (?T)? and its specific heat
(s). In part (b) we must calculate the molar heat
capacity (heat capacity per mole) of water from
its specific heat (heat capacity per gram). Plan
(a) Given s, m, and ?T, we can calculate the
quantity of heat, q, using Equation 5.22. (b) We
can use the molar mass of water and dimensional
analysis to convert from heat capacity per gram
to heat capacity per mole. Solve (a) The water
undergoes a temperature change of ?T 98ºC
22ºC 76ºC 76 K Using Equation 5.22, we
have q ?s ??m ? ?T (4.18 J/g-K)(250
g)(76 K) 7.9 ? 104J
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Answers  (a) 4.9 ? 105 J, (b) 11 K 11ºC
decrease
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Solution   Analyze In part (a) we must find the
total quantity of heat needed to warm the sample
of water, given the mass of water (m), its
temperature change (?T)? and its specific heat
(s). In part (b) we must calculate the molar heat
capacity (heat capacity per mole) of water from
its specific heat (heat capacity per gram). Plan
(a) Given s, m, and ?T, we can calculate the
quantity of heat, q, using Equation 5.22. (b) We
can use the molar mass of water and dimensional
analysis to convert from heat capacity per gram
to heat capacity per mole. Solve (a) The water
undergoes a temperature change of ?T 98ºC
22ºC 76ºC 76 K Using Equation 5.22, we
have q ?s ??m ? ?T (4.18 J/g-K)(250
g)(76 K) 7.9 ? 104J
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Answers  (a) 4.9 ? 105 J, (b) 11 K 11ºC
decrease
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Constant Pressure Calorimetry

• By carrying out a reaction in aqueous solution
  in a simple calorimeter such as this one, one can
  indirectly measure the heat change for the system
  by measuring the heat change for the water in the
  calorimeter.

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Constant Pressure Calorimetry

• Because the specific heat for water is well
  known (4.184 J/mol-K), we can measure ?H for the
  reaction with this equation
• q m ? s ? ?T

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To express the enthalpy change on a molar basis,
we use the fact that the number of moles of HCl
and NaOH is given by the product of the
respective solution volumes (50ml 0.050 L) and
concentrations (0.050 L)(1.0 mol/L) 0.050
mol
Thus, the enthalpy change per mole of HCl (or
NaOH) is
Check ?H ?is negative (exothermic), which is
expected for the reaction of an acid with a base.
The molar magnitude of the heat evolved seems
reasonable.
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To express the enthalpy change on a molar basis,
we use the fact that the number of moles of HCl
and NaOH is given by the product of the
respective solution volumes (50ml 0.050 L) and
concentrations (0.050 L)(1.0 mol/L) 0.050
mol
Thus, the enthalpy change per mole of HCl (or
NaOH) is
Check ?H ?is negative (exothermic), which is
expected for the reaction of an acid with a base.
The molar magnitude of the heat evolved seems
reasonable.
Answer 76,000 J/mol 76 kJ/mol 
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Bomb Calorimetry

• Reactions can be carried out in a sealed bomb,
  such as this one, and measure the heat absorbed
  by the water.

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Bomb Calorimetry

• Because the volume in the bomb calorimeter is
  constant, what is measured is really the change
  in internal energy, ?E, not ?H.
• For most reactions, the difference is very small.

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When 4.00 g of methylhydrazine is combusted in a
bomb calorimeter, the temperature of the
calorimeter increases from 25.00C to 39.50C. In
a separate experiment the heat capacity of the
calorimeter is measured to be 7.794 kJ/C. What
is the heat of reaction for the combustion of a
mole of CH6N2 in this calorimeter?
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When 4.00 g of methylhydrazine is combusted in a
bomb calorimeter, the temperature of the
calorimeter increases from 25.00C to 39.50C. In
a separate experiment the heat capacity of the
calorimeter is measured to be 7.794 kJ/C. What
is the heat of reaction for the combustion of a
mole of CH6N2 in this calorimeter?
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When 4.00 g of methylhydrazine is combusted in a
bomb calorimeter, the temperature of the
calorimeter increases from 25.00C to 39.50C. In
a separate experiment the heat capacity of the
calorimeter is measured to be 7.794 kJ/C. What
is the heat of reaction for the combustion of a
mole of CH6N2 in this calorimeter?
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When 4.00 g of methylhydrazine is combusted in a
bomb calorimeter, the temperature of the
calorimeter increases from 25.00C to 39.50C. In
a separate experiment the heat capacity of the
calorimeter is measured to be 7.794 kJ/C. What
is the heat of reaction for the combustion of a
mole of CH6N2 in this calorimeter?
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When 4.00 g of methylhydrazine is combusted in a
bomb calorimeter, the temperature of the
calorimeter increases from 25.00C to 39.50C. In
a separate experiment the heat capacity of the
calorimeter is measured to be 7.794 kJ/C. What
is the heat of reaction for the combustion of a
mole of CH6N2 in this calorimeter?
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PRACTICE EXERCISE A 0.5865-g sample of lactic
acid (HC3H5O3) is burned in a calorimeter whose
heat capacity is 4.812 kJ/C. The temperature
increases from 23.10C to 24.95C. Calculate the
heat of combustion of lactic acid (a) per gram
and (b) per mole.
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Hesss Law

• ?H is well known for many reactions, and it is
  inconvenient to measure ?H for every reaction in
  which we are interested.
• However, we can estimate ?H using ?H values that
  are published and the properties of enthalpy.

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Hesss Law

• Hesss law states that If a reaction is carried
  out in a series of steps, ?H for the overall
  reaction will be equal to the sum of the enthalpy
  changes for the individual steps.

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Hesss Law

• Because ?H is a state function, the total
  enthalpy change depends only on the initial state
  of the reactants and the final state of the
  products.

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Calculation of ?H
C3H8 (g) 5 O2 (g) ?? 3 CO2 (g) 4 H2O (l)

• Imagine this as occurring
• in 3 steps

C3H8 (g) ?? 3 C(graphite) 4 H2 (g) 3
C(graphite) 3 O2 (g) ?? 3 CO2 (g) 4 H2 (g) 2
O2 (g) ?? 4 H2O (l)
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Calculation of ?H
C3H8 (g) 5 O2 (g) ?? 3 CO2 (g) 4 H2O (l)

• Imagine this as occurring
• in 3 steps

C3H8 (g) ?? 3 C(graphite) 4 H2 (g) 3
C(graphite) 3 O2 (g) ?? 3 CO2 (g) 4 H2 (g) 2
O2 (g) ?? 4 H2O (l)
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Calculation of ?H
C3H8 (g) 5 O2 (g) ?? 3 CO2 (g) 4 H2O (l)

• Imagine this as occurring
• in 3 steps

C3H8 (g) ?? 3 C(graphite) 4 H2 (g) 3
C(graphite) 3 O2 (g) ?? 3 CO2 (g) 4 H2 (g) 2
O2 (g) ?? 4 H2O (l)
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Calculation of ?H
C3H8 (g) 5 O2 (g) ?? 3 CO2 (g) 4 H2O (l)

• The sum of these equations is

C3H8 (g) ?? 3 C(graphite) 4 H2 (g) 3
C(graphite) 3 O2 (g) ?? 3 CO2 (g) 4 H2 (g) 2
O2 (g) ?? 4 H2O (l)
C3H8 (g) 5 O2 (g) ?? 3 CO2 (g) 4 H2O (l)
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Calculation of ?H
C3H8 (g) 5 O2 (g) ?? 3 CO2 (g) 4 H2O (l)

• ??????H 3(-393.5 kJ) 4(-285.8 kJ) -
  1(-103.85 kJ) 5(0 kJ)
• (-1180.5 kJ) (-1143.2 kJ) - (-103.85
  kJ) (0 kJ)
• (-2323.7 kJ) - (-103.85 kJ)
• -2219.9 kJ

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SAMPLE EXERCISE 5.8 continued
Answer ?H3 1.9 kJ
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SAMPLE EXERCISE 5.9 continued
Check The procedure must be correct because we
obtained the correct net equation. In cases like
this you should go back over the numerical
manipulations of the ?H values to ensure that you
did not make an inadvertent error with signs.
Answer 304.1kJ
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Calculation of ?H

• We can use Hesss law in this way
• ?H ??n??Hf(products) - ??m??Hf(reactants)
• where n and m are the stoichiometric
  coefficients.

?
?
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Enthalpies of Formation

• An enthalpy of formation, ?Hf, is defined as the
  enthalpy change for the reaction in which a
  compound is made from its constituent elements in
  their elemental forms.

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Standard Enthalpies of Formation
?

• Standard enthalpies of formation, ?Hf, are
  measured under standard conditions (25C and 1.00
  atm pressure).

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PRACTICE EXERCISE Write the equation
corresponding to the standard enthalpy of
formation of liquid carbon tetrachloride (CCl4).
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SAMPLE EXERCISE 5.11 continued
We can calculate ?H for this reaction by using
Equation 5.31 and data in Table 5.3. Remember to
multiply the value for each substance in
the reaction by that substances stoichiometric
coefficient. Recall also that
for any element in its most stable form under
standard conditions, so
Comment Both propane and benzene are
hydrocarbons. As a rule, the energy obtained from
the combustion of a gram of hydrocarbon is
between 40 and 50 kJ.
Answer 1367 kJ  
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Check We expect the enthalpy of formation of a
stable solid such as calcium carbonate to be
negative, as obtained.
Answer 156.1 kJ/mol 
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Energy in Foods

• Most of the fuel in the food we eat comes from
  carbohydrates and fats.

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Solution If cellulose does not provide fuel
value, we must conclude that it is not converted
in the body into CO2 and H2O, as starch is. A
slight, but critical, difference in the
structures of starch and cellulose explains why
only starch is broken down into glucose in the
body. Cellulose passes through without undergoing
significant chemical change. It serves as fiber,
or roughage, in the diet, but provides no caloric
value.
PRACTICE EXERCISE The nutritional label on a
bottle of canola oil indicates that 10 g of the
oil has an energy value of 86 kcal. A similar
label on a bottle of pancake syrup indicates that
60 mL (about 60 g) has an energy value of 200
kcal. Account for the difference.
Answer The oil has a fuel value of 8.6 kcal/g,
whereas the syrup has a fuel value of about 3.3
kcal/g. The higher fuel value for the canola oil
arises because the oil is essentially pure fat,
whereas the syrup is a solution of sugars
(carbohydrates) in water. The oil has a higher
fuel value per gram in addition, the syrup is
diluted by water.
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Fuels

• The vast majority of the energy consumed in this
  country comes from fossil fuels.