Homework Problems

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Homework Problems Chapter 5 Homework Problems
1, 12, 16, 18, 20, 28, 30, 36, 38, 40, 48, 50,
60, 62, 66, 87, 96, 108, 122
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CHAPTER 5 Thermochemistry
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Thermochemistry, Work, and Heat Thermochemistry
is the study of energy flow in chemical
systems. Work (w) is (force) .
(displacement) Heat, or thermal energy, (q)
is the energy that moves from a hot object to a
cold object when placed in contact with each
other.
T1 gt T3 gt T2 heat flows from hot to cold block
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Energy Energy is the capacity to do work or
transfer heat. In principal any kind of energy
can be converted into an equivalent amount of
work or heat. We divide energy into two general
types kinetic energy (due to motion in a
particular direction) EK 1/2 mv2 potential
energy (due to position or composition) Examples
- chemical potential energy, gravitational
potential energy. The total amount of all the
different kinds of energy for a system is called
the internal energy (U).
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Units of Energy All forms of energy can be
expressed in the same units. To find the MKS
unit for energy, it is convenient to use the
equation for kinetic energy. EK 1/2mv2 So
units are (kg) (m/s)2 kg.m2 1 Joule 1 J

s2 Since 1 J is a
small amount of energy, we often express energy
in terms of kJ (kilojoule). 1 kJ 1000.
J Other common units for energy include
calorie (cal) Amount of heat needed to raise
the temperature of 1 g of water by 1 ?C 1 cal
4.184 J (exact) 1 kcal 1 food calorie 1000
cal 4184. J (exact)
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Heat and Chemical Reactions Some chemical
reactions release heat into their surroundings.
Other reactions require that the surroundings
provide heat so that the reaction can
proceed. An exothermic reaction is a reaction
that releases heat to the surroundings.
Combustion reactions are one common type of
exother-mic reaction 2 H2(g) O2(g) ? 2
H2O(l) releases 572. kJ of heat per mole
of reaction An endothermic reaction takes up
heat from the surroundings. 2 HgO(s) ? 2
Hg(l) O2(g) takes up 181. kJ of heat per
mole of reaction By convention, q lt 0 for an
exothermic reaction, and q gt 0 for an endothermic
reaction.
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System, Surroundings, and Universe The system
is a part of the universe that we have separated
off for study. The surroundings are everything
not included in the system. The universe is
everything - system surroundings.
While in principle the surroundings are
everything not included in the system, in
practice we can usually focus on that part of the
surroundings in the immediate vicinity of the
system.
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Open, Closed, and Isolated Systems Systems
can be divided into various categories based on
how they interact with their surroundings. Open
system - A system that can exchange both energy
and mass with its surroundings. Closed system -
A system that can exchange energy with its
surroundings, but which cannot exchange mass with
its surroundings. Isolated system - A system
that cannot exchange either energy or mass with
its surroundings.
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State Function A state function is any function
whose change in value depends only on the initial
and final state of the system, but which is
independent of the pathway used to go between
them. The value for something that is not a
state function depends not only on the initial
and final state but also on the pathway used to
travel between them.
Altitude is a state function, distance traveled
is not a state function. In thermodynamics, U is
a state function, q and w are not state functions
(in fact, they are not functions at all!)
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Mechanical Work Mechanical work is the work
associated with the change in the volume of a
system. Consider the expansion of a gas
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Derivation of the Expression For Mechanical
Work w F . d (from the definition of
work) But p F/A, and so F p . A So w p
. A . d However, from the diagram d ?h So w
p . A . ?h But A . d ?V where ?V is the
change in volume. So w p . ?V (absolute
value)
p d
But notice that when a gas expands its volume
change is in a direction opposite to that of the
applied pressure. This intro-duces a negative
sign into the expression for mechanical work.
Therefore w - p . ?V
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First Law of Thermodynamics The first law of
thermodynamics gives a relationship between
internal energy, work, and heat. ?U q w ?U
Uf - Ui change in internal energy q
heat w work For now, we limit ourselves to
mechanical work, where w - p ?V. Note that the
first law is based on experimental observation
and is not derived from some other law or
principle. The first law relates the change in a
state function (?U) to the sum of two things that
are not state functions (q and w).
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Sign Convention for Heat and Work We use the
following sign convention for heat and work.
q gt 0 heat flows from the surroundings into the
system (endothermic) q lt 0 heat flows from the
system to the surroundings (exothermic) w gt 0
work is done on the system (system is
compressed) w lt 0 work is done by the system
(system expands) Note that qsur - qsys, and
wsur - wsys.
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Use of the First Law A gas is confined inside of
a cylinder. The applied pressure is 1.000 atm,
and the initial volume occupied by the gas is
2.000 L. 8000. J of heat is added to the gas
under conditions of constant applied pressure.
The final volume occupied by the gas is 6.000 L.
What are q, w, and ?U for the process?
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A gas is confined inside of a cylinder. The
applied pressure is 1.000 atm, and the initial
volume occupied by the gas is 2.000 L. 8000. J
of heat is added to the gas under conditions of
constant applied pressure. The final volume
occupied by the gas is 6.000 L. What are q, w,
and ?U for the process? From the first law, ?U
q w. Heat is added to the system, so q
8000. J For constant pressure w - p ?V
- (1.00 atm) (6.000 L 2.000 L) - 4.00
L.atm 101.3 J - 405. J 1
L.atm Finally, ?U q w 8000. J (- 405. J)
7595. J Note To find the conversion between
L.atm and J we do the following J 1 L.atm
1 m3 1.013 x 105 N/m2 101.3 N.m 101.3
J 1000 L
1 atm
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Constant Volume Processes Consider some general
process taking place at constant volume. From
the first law ?U q w But for mechanical
work, w - p?V So at constant volume, ?V
0, so w 0 Therefore, for a process carried out
at constant volume ?U qV (V
constant) What does this mean? For a process
carried out at constant volume, q is a state
function, and so no information is needed
concerning path. This makes it far easier to
calculate and keep track of heat flow for these
kinds of processes.
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Enthalpy Most processes in the laboratory are
carried out at constant pressure instead of
constant volume. It would be nice to have a
state function whose change in value was equal to
q for constant pressure processes. Enthalpy is
such a function. We define enthalpy, H, as
follows H U pV Since U, p, and V are
state functions, it follows that enthalpy is also
a state function.
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Constant Pressure Processes and
Enthalpy Consider some general process taking
place at constant pressure. From the definition
of enthalpy H U pV ?H ?U ?(pV) ?U
(pfVf - piVi) Now, if pressure is held constant,
then pf pi p, and so ?H ?U p(Vf - Vi)
?U p?V Now, from the first law ?U q w
If we only have mechanical work ?U q - p ?V
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If we now substitute into the expression for
enthalpy, we get ?H ?U p?V (q - p ?V) p
?V or (finally!) ?H qp (p constant) What
does this mean? For a process carried out at
constant pressure, q is a state function, and so
no information is needed concerning path. This
makes it far easier to calculate and keep track
of heat flow for these kinds of processes. To
summarize ?U qV (constant volume
processes) ?H qp (constant pressure
processes) Since q is something that can be
measured experimentally, we now have a way to
relate this information to changes in state
functions (internal energy or enthalpy).
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Enthalpy of Reaction (?H?rxn) Because of the
importance of enthalpy, we define enthalpy
changes for specific kinds of processes. ?H?rxn
(enthalpy of reaction) - The enthalpy change when
one mole of a specific reaction is carried out at
p 1.00 atm (?) equals qp for the
process. Example CaCO3(s) ? CaO(s)
CO2(g) ?H?rxn 178.3 kJ/mol 2
CaCO3(s) ? 2 CaO(s) 2 CO2(g) ?H?rxn
356.6 kJ/mol CaO(s) CO2(g) ? CaCO3(s)
?H?rxn - 178.3 kJ/mol Notice that
multiplying a reaction by a constant multiplies
the value for ?H?rxn by the same constant.
Changing the direction of a reaction changes the
sign for ?H?rxn .
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We can see the last result as follows
CaCO3(s) ? CaO(s) CO2(g) ?H?rxn
178.3 kJ/mol CaO(s) CO2(g) ? CaCO3(s)
?H?rxn ? ________________________
__________________________________
CaCO3(s) ? CaCO3(s) ?H?total 0.0
kJ/mol Carrying out the first process followed
by the second process means that our final state
is the same as our initial state. There is no
change in the system, and so ?H?total 0.0
kJ/mol. Therefore, the change in enthalpy for
the second process must be the same as the change
in enthalpy for the first process, except for the
sign.
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The Physical State of the Reactants and
Products The value for the enthalpy change for a
process depends not only on the identity and
amounts of the reactants and products but also
their state. Consider the following example
2 H2(g) O2(g) ? 2 H2O(l) ?H?rxn
- 571.6 kJ/mol 2 H2(g) O2(g) ? 2 H2O(g)
?H?rxn - 483.6 kJ/mol The difference
in is due to the fact that heat has to be added
to liquid water to convert it into a gas. In
fact, at T 25.0 C H2O(l) ? H2O(g)
?H?rxn 44.0 kJ/mol
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We can check this as follows. If we add the
first two reactions below we get the third
reaction as our final result. 2 H2(g)
O2(g) ? 2 H2O(l) ?H?rxn - 571.6
kJ/mol 2 H2O(l) ? 2 H2O(g)
?H?rxn 88.0 kJ/mol ___________________________
_______________________________ 2 H2(g)
O2(g) ? 2 H2O(g) ?H?rxn - 483.6
kJ/mol Since enthalpy is a state function (path
independent) the change in enthalpy for the
combination of the first two processes has to be
the same as the change in enthalpy for the third
process. This is a simple example of a general
principle called Hess law.
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Relationship Between ?Hrxn and ?Urxn By
definition, H U pV. For a process carried
out at constant pressure ?H ?U ?(pV) ?U
p ?V If ?V gt 0 then ?H gt ?U ?V 0 then ?H
?U ?V lt 0 then ?H lt ?U For chemical reactions
it is usually true that ?U gtgt p?V and so
?Urxn ? ?Hrxn For precise work we can apply a
correction to correctly convert between ?Urxn and
?Hrxn. (using ?H ?U p?V).
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Heat Capacity Consider adding a known amount of
heat to a substance. One would expect that the
temperature of the substance would increase.
Heat capacity (C) is a measure of how much heat
is required to cause the temperature of a
substance to change by a given amount.
Heat capacity is defined as C q/?T where q
amount of heat added ?T Tf - Ti change
in temperature
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Notes on Heat Capacity 1) Since C q/?T, the
units for C are (energy)/(temperature). The
derived MKS units for C are J/K, though it is
often given in J/?C. 2) Note that the numerical
value for C when expressed in J/K or J/?C is
identical. That is because we are using the
change in temperature. Since the size of a
degree Kelvin and a degree Celsius is the same,
the numerical value for ?T is the same whether
expressed in K or ?C. Example Ti 15.0 ?C
(288.2 K) and Tf 21.5 ?C (294.7 K) ?T Tf
Ti 21.5 ?C 15.0 ?C 6.5 ?C 294.7 K -
288.2 K 6.5 K 3) C is not in general a state
function, since q is not a state function.
However, if we restrict ourselves to processes
occurring under conditions of constant pressure
then C is a state function.
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Specific Heat and Molar Heat Capacity There are
two quantities that are closely related to heat
capacity. Specific heat, (s) - The heat capacity
per gram of substance. Molar heat capacity, (Cm)
- The heat capacity per mole of substance. The
relationships between C, s, and Cm (heat
capacity, specific heat, and molar heat capacity)
are as follows s C/m where m mass of
substance Cm C/n where n moles of
substance So s has units of J/g?C and Cm has
units of J/mol?C. Note that C is an extensive
property, but s and Cm are intensive properties.
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Heat Capacity Problem The initial temperature
of a metal block of mass 38.44 g is Ti 18.3 C.
After the addition of 132. J of heat the
temperature of the metal increases to a final
value Tf 25.7 C. What are C and s for the
metal block?
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The initial temperature of a metal block of mass
38.44 g is Ti 18.3 C. After the addition of
132. J of heat the temperature of the metal
increases to a final value Tf 25.7 C. What
are C and s for the metal block? From the
definition of heat capacity C heat
absorbed q 132. J
18. J/C change in
temperature ?T (25.7 C - 18.3 C) s
C 18. J/C 0.46 J/g?C m
38.44 g Note that C is an extensive property
but s is an intensive property of the metal.
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Specific Heat For Common Substances
Substance s (J/gC) Al(s)
0.900 Au(s) 0.129 C(graphite)
0.720 C(diamond) 0.502 Cu(s)
0.385 Fe(s) 0.444 Hg(l) 0.139
H2O(l) 4.184 Pb(s) 0.128
Sn(s) 0.227 Zn(s) 0.389
The above table is at T 25. ?C. Note that
water has an unusual-ly large value for heat
capacity, which acts to moderate temperatures for
cities surrounded by large bodies of water.
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Calorimetry Calorimetry is the experimental
method used to measure the heat produced or taken
up in a chemical reaction or physical process.
The device used in these measurements is called a
calorimeter. If the heat capacity and
temperature change for the calorimeter are known,
we can find the value for q for the process If
the process is carried out at constant pressure,
then qsys ?H If the process is carried out at
constant volume, then qsys ?U
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Constant Pressure Calorimetry
In a coffee cup calorimeter a chemical reaction
or other process is carried out under conditions
of constant pressure.
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Example A metal block of mass 14.486 g at an
initial temperature of 74.25 C is placed inside
a coffee cup calorimeter containing 100.00 g of
water at a temperature of 17.42 C. After
equilibrium is reached the final temperature of
the water is 20.84 C. What is s, the specific
heat, of the metal? (Note s 4.184 J/gC for
water.)
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A metal block of mass 14.486 g at an initial
temperature of 74.25 C is placed inside a coffee
cup calorimeter containing 100.00 g of water at a
temperature of 17.42 C. After equilibrium is
reached the final temperature of the water is
20.84 C. What is s, the specific heat, of the
metal? (Note s 4.184 J/gC for
water.) qwater - qmetal swater mwater
?Twater - smetal mmetal ?Tmetal smetal -
swater (mwater/mmetal)( ?Twater/?Tmetal
) So smetal - (4.184 J/gC) 100.00 g
(20.84 C - 17.42 C)
14.486 g (20.84 C - 74.25 C)
1.85 J/gC Note we did not take into
account the heat capacity of the calorimeter
itself, but assumed it was small. For more
precise work the heat capacity of the calorimeter
would need to be included in the calculations.
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Bomb Calorimetry In bomb calorimetry a measured
mass of a chemical substance reacts with excess
oxygen to form combustion products. The reaction
taking place is thus a combustion reaction.
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Bomb Calorimetry Calculations If the heat
capacity of the calorimeter apparatus is known
(and this can be determined experimentally)
then q - C ?T C heat capacity of
calorimeter ?T Tf - Ti change in
temperature The negative sign in the above
equation occurs because we are measuring the
value of q for the surroundings, and qsys -
qsur. If we know the energy of combustion for a
compound, in units of kJ/g, then we can say q
m ?Ucom m mass of compound burned ?Ucom
energy of combustion (in kJ/g) Note that by
burning a known amount of a compound whose value
for ?Ucom is known, and measuring ?T, we can find
C, the heat capacity of the calorimeter.
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Bomb Calorimetry Example Example 1.412 g of
carbon (M 12.01 g/mol) are burned in a bomb
calorimeter (C 10325. J/?C). The temperature
of the calorimeter increases by 4.48 ?C. What
are ?U (change in internal energy) and ?Um
(change in internal energy per mole of carbon)
for the process.
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Example 1.412 g of carbon (M 12.01 g/mol) are
burned in a bomb calorimeter (C 10325. J/?C).
The temperature of the calorimeter changes by
4.48 ?C. What are ?U (change in internal energy)
and ?Um (change in internal energy per mole of
carbon) for the process. q - C ?T - (10325.
J/?C) (4.48 ?C) - 46300. J For a process
carried out at constant volume q ?U, so ?U
- 46300. J Finally, the change in internal
energy per mole of carbon is ?Um ?U/n n
1.412 g 1 mol 0.1176 mol

12.01 g ?Um - 46300. J - 393000
J/mol - 393. kJ/mol
0.1176 mol Note that ?Hm ? ?Um - 393.
kJ/mol, as previously noted.
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Standard Conditions and Standard State It is
convenient to report thermodynamic data for a
particular set of conditions. These conditions
are called the thermodynamic standard conditions,
and are usually (but not always) taken to be p
1.00 atm, T 25.0 ?C 298.2 K. The standard
state for an element is the most stable form of
the element for standard conditions. carbon
C(s) nitrogen N2(g) bromine Br2(l) oxygen
O2(g) iron Fe(s) argon Ar(g) mercury Hg
(l) chlorine Cl2(g) sulfur S(s) ________________
__________________________________________
Technically, standard pressure is now taken to be
1.00 bar 0.987 atm. This makes only a minor
difference in values for thermodynamic quantities.
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Enthalpy of Formation (?H?f) The formation
reaction for a substance is defined as the
reaction that produces one mole of a single
product out of elements in their standard state.
Because of the way we have defined the formation
reaction, we may have to use fractional
stoichiometric coefficients for some or all of
the reactants. The enthalpy change for this
reaction is defined as the enthalpy of formation
for the substance. H2(g) 1/2 O2(g) ?
H2O(l) ?H?f(H2O(l)) - 285.8 kJ/mol
H2(g) 1/2 O2(g) ? H2O(g) ?H?f(H2O(g)) -
241.8 kJ/mol C(s) O2(g) ? CO2(g)
?H?f(CO2(g)) - 393.5 kJ/mol Pb(s)
C(s) 3/2 O2(g) ? PbCO3(s) ?H?f(PbCO3(s)) -
699.1 kJ/mol 3/2 O2(g) ? O3(g)
?H?f(O3(g)) 143.0 kJ/mol Data for formation
enthalpies are given in Appendix 2 of Burdge
(with separate tables for inorganic and organic
substances).
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Note that based on the above definition it
follows that the enthalpy of formation of an
element in its standard state is 0.0 kJ/mol. We
may see this as follows The formation reaction
for N2(g), by definition (formation of one mole
of a single product out of elements in their
standard state), is N2(g) ? N2(g) But
nothing happens in the above process, and so it
follows that ?H?rxn ?H?f(N2(g) ) 0.0
kJ/mol This will be true for any element in its
standard (thermodynamically most stable) state.
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Example Write the formation reaction for
carbon tetrachloride (CCl4(l)), acetone
(CH3COCH3(l)), and nitrous oxide (N2O(g)).
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Example Write the formation reaction for carbon
tetrachloride (CCl4(l)), acetone (CH3COCH3(l)),
and nitrous oxide (N2O(g)). C(s) 2 Cl2(g) ?
CCl4(l) 3 C(s) 3 H2(g) ½ O2(g) ?
CH3COCH3(l) N2(g) ½ O2(g) ? N2O(g) Note
that the enthalpy change when 1 mole of any of
the above reactions is carried out corresponds to
the enthalpy of formation for the substance.
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Enthalpy of Combustion (?H?com) The combustion
reaction for a substance is defined as the
reaction of one mole of a single substance with
O2(g) to form combustion products. Because of
the way in which we have defined the combustion
reaction we may have to use fractional
coefficients for some of the reactants and
products. The enthalpy change for this reaction
is defined as the enthalpy of combustion for the
substance. Combustion products for common
elements are C ? CO2(g) H ? H2O(l) N ? N2(g) S
? SO2(g) CH4(g) 2 O2(g) ? CO2(g) 2 H2O(l)
?H?com(CH4(g)) - 890.3
kJ/mol C6H6(?) 15/2 O2(g) ? 6 CO2(g) 3
H2O(l) ?H?com(C6H6(?)) - 3267.4
kJ/mol
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Writing Formation and Combustion Reactions We
may use the definition of formation reaction and
combustion reaction to write own the balanced
chemical equations corresponding to these
reactions. Example Hexane (C6H14(l)) is a
hydrocarbon often used as a solvent in organic
reactions. Write the formation reaction and the
combustion reaction for hexane.
CH3CH2CH2CH2CH2CH3
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Example Hexane (C6H14(l)) is a hydrocarbon
often used as a solvent in organic reactions.
Write the formation reaction and the combustion
reaction for hexane. Formation - One mole of a
single product out of elements in their standard
state. ? ? C6H14(l) 6 C(s) 7 H2(g) ?
C6H14(l) (balanced) Combustion - One mole of
a single reactant, plus oxygen, to form
combustion products. C6H14(l) ? O2(g) ?
? C6H14(l) ? O2(g) ? 6 CO2(g) 7
H2O(l) C6H14(l) 19/2 O2(g) ? 6 CO2(g) 7
H2O(l) (balanced)
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Enthalpy Change For Phase Transitions There are
three phase transitions that can occur by adding
heat to a substance. s ? l fusion (melting)
?H?fus l ? g vaporization ?H?vap s ? g
sublimation ?H?sub The enthalpy change when one
mole of a substance undergoes the transition is
defined as the enthalpy change for the phase
transition. It represents the amount of heat
required to convert one mole of the substance
from the initial phase to the final phase.
H2O(l) ? H2O(g) ?H?vap(H2O) 40.7 kJ/mol at T
100.0 ?C Unlike other processes, the
temperature for a phase transition is usually
taken to be the temperature for which the two
phases exist at equilibrium when p 1.0 atm,
called the normal transition point.
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Hess Law We previously noted that the change
in the value for a state function depends only on
initial and final state and is independent of the
path used to travel between the two states. We
may put this in a more formal manner in terms of
Hess law. Hess law The change in value for
any state function will be the same for any
process or combination of processes that have the
same initial and final state. We are
particularly interested in applying Hess law to
chemical reactions.
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Hess Law For a Chemical Reaction Let us use
Hess law to find the value for (?H?rxn) for the
following chemical reaction CaCO3(s) ? CaO(s)
CO2(g) ?H?rxn
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Hess Law For a Chemical Reaction Let us use
Hess law to find the value for (?H?rxn) for the
following chemical reaction CaCO3(s) ? CaO(s)
CO2(g) ?H?rxn We may obtain this same
reaction as follows Ca(s) ½ O2(g) ?
CaO(s) ?H?f(CaO(s)) C(s) O2(g) ?
CO2(g) ?H?f(CO2(g)) CaCO3(s) ? Ca(s) C(s)
3/2 O2(g) - ?H?f(CaCO3(s)) CaCO3(s) ?
CaO(s) CO2(g) ?H?rxn So by Hess law,
?H?rxn ?H?f(CaO(s)) ?H?f(CO2(g)) -
?H?f(CaCO3(s)) We can get ?H?rxn using only
data on enthalpies of formation!
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General Method For Finding ?H?rxn Based on the
same procedure used in the previous example the
following general relationship can be derived
?H?rxn ? ?H?f(products) - ?
?H?f(reactants) Notice what this means. If
we have a table for formation enthalpies we can
find the value for ?H?rxn for any chemical
reaction. In fact, the same general procedure
can be used to find the values for the change in
any state function. Also note that when we use
the superscript ? this indicates not only
standard conditions but also standard
concentrations for reactants and
products. gases p 1.0 atm solutes M 1.0
mol/L solids, liquids, solvents must be present
in the system
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Example Find ?H?rxn for the conversion of
acetylene into dichloroethane C2H2(g) 2
HCl(g) ? CH2ClCH2Cl(l)
53
Example Find ?H?rxn for the conversion of
acetylene into dichloroethane C2H2(g) 2
HCl(g) ? CH2ClCH2Cl(l) ?H?rxn
?H?f(CH2ClCH2Cl(l)) - ?H?f(C2H2(g)) 2
?H?f(HCl(g)) (- 165.2 kJ/mol) - (226.7
kJ/mol) 2 ( - 92.3 kJ/mol) - 207.3
kJ/mol Thermodynamic data are given in the
Appendix 2 of Burdge, with separate tables for
inorganic and organic substances.
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Other Uses of Hess Law Hess law can be used to
find ?H?rxn for any process that can be written
as a combination of other chemical reactions
whose values for ?H?rxn are known. Example
Consider the formation reaction 5 C(s) 6 H2(g)
? C5H12(l) ?H?rxn ? Using the following
information find ?H?rxn for this reaction (1)
C5H12(l) 8 O2(g) ? 5 CO2(g) 6 H2O(g) ?H?rxn
- 3505.8 kJ/mol (2) C(s) O2(g) ? CO2(g)
?H?rxn - 393.5 kJ/mol (3) 2 H2(g) O2(g) ? 2
H2O(g) ?H?rxn - 483.5 kJ/mol Note all
of the above are combustion reactions, which are
particularly easy to carry out experimentally.
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Example Consider the formation reaction 5 C(s)
6 H2(g) ? C5H12(l) ?H?rxn ? Using the
following information find ?H?rxn for this
reaction (1) C5H12(l) 8 O2(g) ? 5 CO2(g) 6
H2O(g) ?H?rxn - 3505.8 kJ/mol (2) C(s) O2(g)
? CO2(g) ?H?rxn - 393.5 kJ/mol (3) 2
H2(g) O2(g) ? 2 H2O(g) ?H?rxn - 483.5
kJ/mol reverse 1st reaction 5 CO2(g) 6 H2O(g)
? C5H12(l) 8 O2(g) ?H?rxn 3505.8 kJ/mol 5
times the 2nd reaction 5 C(s) 5 O2(g) ? 5
CO2(g) ?H?rxn - 1967.5 kJ/mol 3 times 3rd
reaction 6 H2(g) 3 O2(g) ? 6 H2O(g)
?H?rxn - 1450.5 kJ/mol 5 C(s) 6 H2(g) ?
C5H12(l) ?H?rxn 87.8 kJ/mol
56
End of Chapter 5
In this house we obey the laws of
thermodynamics! - Homer Simpson ...the
Dutch physicist Heike Kamerlingh Onnes gave H the
name enthalpy, from the Greek ?? (in) and ??????
(heat), or from the single Greek word ????????
(enthalpos), to warm within. K. J. Laidler, The
World of Physical Chemistry Thermodynamics
is the only physical theory of universal content
which, within the framework of the applicability
of its basic concepts, I am convinced will never
be overthrown. Albert Einstein