Transcript Thermochemistry

Thermochemistry
The study of heat change in chemical reactions
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Thermochemistry is part of a broader
subject called thermodynamics, which
is the scientific study of the
interconversion of heat and other
kinds of energy.
The first law of thermodynamics, which
is based on the law of conservation of
energy, states that energy can be
converted from one form to another,
but cannot be created or destroyed.
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Types of Energy
• Thermal energy - The energy
associated with the random motion of
atoms and molecules.
– Temperature - the average kinetic
energy of the particles in a sample of
matter
– Heat (q)- the total kinetic energy of the
particles in a sample of matter
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The internal energy of a system has two components:
kinetic and potential energy.
Kinetic energy - the energy produced by a moving
object (various types of molecular motion and the
movement of electrons within molecules).
Potential energy - energy available by virtue of an
object’s position (determined by the attractive
and repulsive forces at the atomic level) .
It is impossible to measure all these contributions
accurately, so we cannot calculate the total energy
of a system with any certainty. Changes in energy,
on the other hand, can be determined
experimentally.
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• Chemical energy - The energy stored
within the structural units of
chemical substances; its quantity is
determined by the type and
arrangement of constituent atoms.
– When substances participate in chemical
reactions, chemical energy is released, stored,
or converted to other forms of energy.
– Chemical energy can be considered a form of
potential energy because it is associated with
the relative positions and arrangements of
atoms within a given substance.
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Energy can assume many different
forms that are interconvertible.
When one form of energy disappears,
some other form of energy (of equal
magnitude) must appear, and vice
versa.
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Energy Changes in Chemical Reactions
Almost all chemical reactions absorb or
produce (release) energy, generally in
the form of heat.
In order to analyze energy changes
associated with chemical reactions we
must first define the terms system
and surroundings.
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• System - the specific part of the
universe that is of interest to us.
For chemists, systems usually include
substances involved in chemical and
physical changes.
•Surroundings - the rest of the universe
outside the system.
For example, in an acid-base neutralization
experiment, the system would be the contents of the
beaker, the 50.0 mL of HCl to which 50.0 mL of NaOH
are added.
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2H2(g) + O2(g)
2H2O(l) + energy
The combustion of hydrogen gas is one
of many chemical reactions that
release considerable quantities of
energy.
System - ?
Surroundings - ?
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Since energy cannot be created or
destroyed, any energy lost by the
system must be gained by the
surroundings. Thus the heat
generated by the combustion process
is transferred from the system to its
surroundings. This is an example of
an exothermic process, which is any
process which gives off heat - that is,
transfers thermal energy to the
surroundings.
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heat + 2HgO(s)
2Hg(l) + O2(g)
The decomposition of mercury (II)
oxide is an endothermic process, in
which heat has to be supplied to the
system by the surroundings.
System - ?
Surroundings - ?
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Calorimetry
Our discussion of calorimetry, the
measurement of heat changes, will
depend on an understanding of
specific heat and heat capacity.
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Specific Heat and Heat Capacity
Specific heat (s) of a substance is the
amount of heat required to raise the
temperature of one gram of the
substance by one degree Celsius.
• Specific heat is an intensive property,
its magnitude doesn’t depend on the
amount of the substance present
• Specific heat has the units
J/g.oC or kJ/g.oC
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• If we know the specific heat and the
amount (mass in g) of a substance (m),
then the change in the sample’s
temperature (DT = Tfinal - Tinitial) will
tell us the amount of heat (q) that
has been absorbed or released in a
particular process.
q = msDT
The sign convention for q is positive for endothermic
processes and negative for exothermic processes.
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Example: A 466g sample of water is
heated from 8.50 oC to 74.60 oC.
Calculate the amount of heat
absorbed by the water in kJ.
q = msDT
q = (466g)(4.184 J/g.oC)(74.60 oC - 8.50 oC)
q = 129000 J
q = 129 kJ
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The heat capacity (C) of a substance is the
amount of heat required to raise the
temperature of a given quantity of the
substance by one degree Celsius.
• Heat capacity is an extensive property (it
varies dependent on the amount of substance
present).
• Heat capacity has the units J/oC or kJ/oC
• The relationship between the heat capacity
and specific heat of a substance is
C = ms
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Determine the heat capacity for 60.0g
of water.
C = ms
(60.0g)(4.184 J/g.oC) = 251 J/oC
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Calorimeters
In a laboratory, heat changes in physical
and chemical processes are measured
with a calorimeter, a closed container
designed specifically for this purpose.
• The special design of a calorimeter
allows us to assume that no heat is lost
to the surroundings during the time it
takes to make measurements.
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• If we know the heat capacity,then
the change in the sample’s
temperature (DT = Tfinal - Tinitial) will
tell us the amount of heat (q) that
has been absorbed or released in a
particular process.
q = CDT
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A certain calorimeter has a heat capacity
of 1.55 kJ/oC. Calculate the heat
absorbed by a calorimeter in kJ that
increases in temperature from 23.60 oC
to 25.87 oC.
q = CDT
q = (1.55 kJ/oC)(25.87 oC - 23.60 oC)
q = 3.52 kJ
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Heat of combustion, the amount of energy given off when a
certain amount of a substance is “burned” in the presence of
oxygen, is usually measured by placing a known mass of a
compound in a steel container called a constant-volume bomb
calorimeter. (With this being a steel container, its volume
obviously doesn’t change, but the pressure certainly would.)
This calorimeter is filled with
oxygen before the sample is
placed in the cup. The sample is
ignited electrically, and the
heat produced by the reaction
can be accurately determined
by measuring the temperature
increase in the known amount of
surrounding water.
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Because no heat enters or leaves the
system throughout the process, we
can write
qrxn = -(qwater + qcalorimeter)
where qwater is obtained by
q = msDT
and qcalorimeter is determined by
q = CDT
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A quantity of 1.435g of naphthalene (C10H8), a pungent-smelling substance used in moth
repellents, was burned in a constant-volume bomb calorimeter. Consequently, the
temperature of the water rose from 20.17 oC to 25.84 oC. If the mass of water
surrounding the calorimeter was exactly 2000g and the heat capacity of the bomb
calorimeter was 1.80 kJ/ oC, calculate the heat of combustion of naphthalene on a
molar basis; that is, find the molar heat of combustion.
q = -(qwater + qcalorimeter)
qwater
= msDT
= (2000.g)(4.184J/g.oC)(25.84 oC - 20.17 oC)
= 47400 J or 47.4 kJ
qcalorimeter = CDT
= (1.80 kJ/oC)(25.84 oC - 20.17 oC)
= 10.2 kJ
q = -(47.4 kJ + 10.2 kJ) = -57.6 kJ
The molar mass of naphthalene is 128.173g/mol, so the heat of combustion of 1 mol
of naphthalene is
-57.6 kJ
=
-5140 kJ__
1.435 g C10H8
128.173 g C10H8
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A simpler device than the
constant-volume calorimeter is
the constant-pressure
calorimeter used to determine
the heat changes for
noncombustion reactions. An
operable constant-pressure
calorimeter can be constructed
from two Styrofoam coffee
cups.
This device measures the heat
effects of a variety of
reactions, such as acid-base
neutralization, as well as the
heat of solution and heat of
dilution.
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As the following example show, we can study the heat of
acid-base reactions using a constant-pressure calorimeter.
1.00 x 102 mL of 0.500 M HCl is mixed with 1.00 x 102 mL of 0.500 M NaOH in a constant-pressure
calorimeter that has a heat capacity of 335 J/oC. The initial temperature of the HCl and NaOH solution
is the same, 22.50 oC, and the final temperature of the mixed solution is 24.90 oC. Calculate the heat of
neutralization for the following reaction
NaOH(aq) + HCl(aq)
NaCl(aq) + H2O(l)
(Assume the densities and specific heats of the solutions are the same as for water.)
Assuming that no heat is lost to the surroundings, we write
qsoln
qrxn = -(qsoln + qcalorimeter)
= msDT
= (1.00 x 102g + 1.00 x 102g)(4.184 J/g.oC)(24.90 oC - 22.50 oC)
= 2010 J or 2.01 kJ
qcalorimeter= CDT
= (335 J/oC )(24.90 oC - 22.50 oC)
= 804 J or .804 kJ
qrxn
= -(2.01 kJ + .804 kJ) = -2.81 kJ
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2.81 kJ of heat are given off by the neutralization of 1.00 x 102 mL of 0.500 M HCl
and 1.00 x 102 mL of 0.500 M NaOH. To calculate the heat of neutralization for
this reaction, you need to know the number of kJ released when 1 mol of HCl reacts
with 1 mol of NaOH. (Remember, the coefficients in a thermochemical equation
represent moles.)
NaOH(aq) + HCl(aq)
NaCl(aq) + H2O(l)
0.500 M HCl = 0.500 mol HCl = 0.0500 mol HCl
1L
.1 L
0.500 M NaOH = 0.500 mol NaOH = 0.0500 mol NaOH
1L
.1 L
yield .0500 mol
H2O
-2.81 kJ = -56.2 kJ
0.0500 mol
1 mol
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Enthalpy
To quantify the heat flow into or out of
a system, chemists use a property
called enthalpy, represented by the
symbol H. Enthalpy is an extensive
property; its magnitude depends on
the amount of the substance present.
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Enthalpy of a Reaction
It is the change in enthalpy, DH, that
we measure in a chemical reaction.
The change of enthalpy of a reaction,
DHrxn commonly represented as just
DH, is the difference between the
enthalpies of the products and the
enthalpies of the reactants:
DH = H(products) - H(reactants)
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In other words, DH represents the heat
given off or absorbed during a
reaction.
• The enthalpy of a reaction can be
positive or negative, depending on the
process.
• For an endothermic process DH is
positive. For an exothermic process
DH is negative.
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Thermochemical equations
Thermochemical equations show the
enthalpy changes as well as the mass
relationships in a thermochemical
reaction.
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In a thermochemical equation...
• the stoichiometric coefficients always refer to the
number of moles of a substance.
• When we reverse an equation, we change the roles of
reactants and products. Consequently, the magnitude of
DH for the equation remains the same, but its sign
changes.
• If we multiply both sides of a thermochemical equation
by a factor n, then DH must also change by the same
factor.
• When writing thermochemical equations, we must always
specify the physical states of all reactants and
products, because they help determine the actual
enthalpy changes.
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Examples of thermochemical equations
H2O(s)
H2O(l)
DH =
6.01 kJ
H2O(l)
H2O(s)
DH =
-6.01 kJ
2H2O(l)
2H2O(s)
DH = -12.02 kJ
Thermochemical equations show the enthalpy
changes that occur in a physical or chemical process.
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How much energy is absorbed or released when
25.0 g of water is frozen?
H2O(l)
H2O(s)
DH =
-6.01 kJ
25.0 g H2O x 1 mol H2O
= 1.39 mol H2O
18.02 g H2O
1 mol H2O
-6.01 kJ
=
1.39 mol H2O
-8.35 kJ
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It is impossible to determine the absolute
enthalpy of a substance. This problem is
similar to the one geographers face in
expressing the elevations of specific
mountains or valleys. Rather than trying to
devise some type of “absolute” elevation
scale (perhaps based on distance from the
center of the Earth?), by common
agreement all geographic heights and
depths are expressed relative to sea level,
an arbitrary reference with a defined
elevation of “zero” meters or feet.
Similarly, chemists have agreed on an
arbitrary reference point for enthalpy.
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Enthalpy of Formation
The “sea level” reference point for all enthalpy
expressions is called the standard enthalpy of
formation, (DHof), which is defined as the heat
change that results when one mole of a compound
is formed from its elements at a pressure of 1 atm.
Elements are said to be in the standard state at 1
atm, hence the term “standard enthalpy”. The
superscript “o” represents standard-state
conditions (1 atm), and the subscript “f” stands for
formation. Although the standard state does not
specify a temperature, we will always use DHof
values measured at 25 oC.
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By convention, the standard enthalpy
of formation of any element in its
most stable form is zero.
Using your reference packet, state
the most stable (allotrope) form for
molecular oxygen
O2 - because the DHof for O2 = 0, and
the DHof for O3 = 0
State the most stable allotrope for carbon
C(graphite) is the most stable allotrope
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You can usually get DHf values for
compounds from your
thermochemistry reference packet.
DHf that aren’t listed in the reference
packet can be determined in one of
two ways, the direct method or the
indirect method.
• Both of these methods require an
understanding of thermochemical
equations.
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The Direct Method for Determining DHf
• Works best for compounds that can be readily
synthesized from their elements. Suppose we want to
know the enthalpy of formation of carbon dioxide. We
must measure the enthalpy of the reaction when carbon
(graphite) and molecular oxygen in their standard
states are converted to carbon dioxide in its standard
state:
C(graphite) + O2(g)
CO2(g) DHorxn = -393.5 kJ
DHorxn = SnDHof(products) - SnDHof(reactants)
-393.5 kJ = 1 mol(x) - [1 mol(0 kJ/mol) + 1 mol(0 kJ/mol)]
-393.5 kJ = 1(x)
x must be equal to -393.5 kJ/mol, therefore, DHof for
CO2 is equal to -393.5 kJ/mol.
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The Indirect Method for Determining DHf
Many compounds cannot be directly
synthesized from their elements. In
some cases, the reaction proceeds too
slowly, or side reactions produce
substances other than the desired
compound. In these cases DHof can be
determined by an indirect approach,
which is based on the law of heat
summation (or simply Hess’s law).
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• Hess’s law can be stated as follows:
When reactants are converted to
products, the change in enthalpy is
the same whether the reaction takes
place in one step or in a series of
steps. In other words, if we break
down the reaction of interest into a
series of reactions for which DHorxn
can be measured, we can calculate
DHorxn for the overall reaction.
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What is DHof for methane?
We would represent the synthesis of CH4
from its elements as
C(graphite) + 2H2(g)
CH4(g)
However, this reaction does not take place
as written, so we cannot measure the
enthalpy change directly. We must use
Hess’s law.
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The following reactions involving
C(graphite), H2(g) and CH4(g) with
O2(g) have all been studied and the
DHorxn values are accurately known
(a) C(graphite) + O2(g)
CO2(g)
(b) 2H2(g) + O2(g)
2H2O(l)
(c) CH4(g) + 2O2(g)
CO2(g) + 2H2O(l)
DHorxn = -393.5 kJ
DHorxn = -571.6 kJ
DHorxn = -890.4 kJ
Remember, we are trying to determine the DHof for
C(graphite) + 2H2(g)
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CH4(g)
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Since we want to obtain one equation containing only C and H2 as
reactants and CH4 as a product, we must reverse (c) to get
(d) CO2(g) + 2H2O(l)
CH4(g) + 2O2(g)
DHorxn = 890.4 kJ
The next step is to add equations (a), (b), and (d)
(a) C(graphite) + O2(g)
CO2(g)
(b) 2H2(g) + O2(g)
2H2O(l)
(d) CO2(g) + 2H2O(l)
CH4(g) + 2O2(g)
C(graphite) + 2H2(g)
CH4(g)
DHorxn = -393.5 kJ
DHorxn = -571.6 kJ
DHorxn = 890.4 kJ
DHorxn = -74.7 kJ
As you can see, all the nonessential species (O2, CO2, and H2O) cancel in
this operation. Because the above equation represents the synthesis of 1
mol of CH4 from its elements, we have DHof(CH4) = -74.7 kJ/mol.
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The general rule in applying Hess’s law
is that we should arrange a series of
chemical equations (corresponding to
a series of steps) in such a way that,
when added together, all species will
cancel except for the reactants and
products that appear in the overall
reaction.
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Once we know the standard enthalpies of
formation, we can calculate the standard
enthalpy of reaction, DHorxn, defined as the
enthalpy of a reaction carried out at 1 atm.
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Consider this hypothetical reaction
aA + bB
cC + dD
where a, b, c, and d are stoichiometric
coefficients.
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For this reaction DHorxn is given by
DHorxn = [cDHof(C) + dDHof(D)] - [aDHof(A) + bDHof(B)]
We can generalize this equation as:
DHorxn = SnDHof(products) - SnDHof(reactants)
where n denotes the stoichiometric coefficients
for the reactants and products, and S (sigma)
means “the sum of”.
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Calculate the kJ of heat released
when 1 mol of liquid pentaborane,
B5H9, is exposed to gaseous
oxygen producing solid diboron
trioxide and liquid water.
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(*Note - DHf values vary slightly depending on which
reference book you choose, these values are from
Raymond Chang Chemistry, 6th Edition)
Substance
DHf
(kJ/mol)
O2(g)
0
B2O3(s)
-1263.6
H2O(l)
-285.8
B5O9(l)
73.2
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Remember,
these are kJ
released or
absorbed per
one mole
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B5H9(l) + 6O2(g)  2.5 B2O3(s) + 4.5 H2O(l)
Notice, this was
balanced in such
a way that you
only started with
1 mol of B5H9
aA + bB
 cC + dD
If it had been written
with 2B5H9, you would
have to divide the
final kJ by 2 to get
the amount of heat
released when 1 mol of
pentaborane reacted
with oxygen.
DH = H(products) - H(reactants)
DHorxn = [cDHof(C) + dDHof(D)] - [aDHof(A) + bDHof(B)]
DHorxn = [2.5(-1263.6) + 4.5(-285.8)] - [1(73.2) +6(0)]
DHorxn = -4518.3 kJ
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Hess’s Law can also be used to calculate the
change in enthalpy for a reaction. The following
slide shows the steps involved in forming sodium
chloride from elemental sodium and molecular
chlorine.
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Although we have focused so far on the
thermal energy effects resulting
from chemical reactions, many
physical processes, such as the
melting of ice or the condensation of
a vapor, also involve the absorption or
release of heat.
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Change of State Processes for Water
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Thermal energy
increases the
kinetic energy of
molecules, allowing
them to break free
of intermolecular
forces. With
enough added
thermal energy,
molecules in a solid
enter the liquid
phase and molecules
in a liquid enter the
gas phase.
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Enthalpy changes dramatically during a change of state.
The added thermal energy
not only causes a change
of state but also
increases the molar
enthalpy of water.
Water’s freezing
point
oC
_____
0
273 K
_____
Water’s boiling point
_____
100 oC
_____
373 K
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• The molar enthalpy change that
occurs during melting is called the
molar enthalpy of fusion, or heat of
fusion (DHfus). It is the difference
between the enthalpy of 1 mol of a
substance in its liquid and solid
states.
Water’s DHfus equals 6.01 kJ/mol
Since DH is a positive value, this is an endothermic
process, as expected for an energy-absorbing change
of melting ice.
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Calculate the heat absorbed when 25.0 g of H2O(s) is
converted to H2O(l).
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Example: Given the thermochemical equation:
H2O(l)
H2O(s)
DH =
-6.01 kJ
You need to reverse the equation to represent
melting…
H2O(s)
H2O(l)
DH =
Then determine the mol of H2O you are melting
6.01 kJ
Note the sign
change
25.0 g H2O x 1 mol H2O = 1.39 mol H2O
18.02 g
6.01 kJ
1 mol H2O
=
8.35 kJ
1.39 mol H2O
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Use a ratio to
determine the
kJ of heat
absorbed
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• The molar enthalpy change that
occurs during evaporation is called
the molar enthalpy of vaporization, or
heat of vaporization (DHvap). It is the
difference between the enthalpy of 1
mol of a substance in its liquid and
gaseous states.
Water’s DHvap equals 40.79 kJ/mol
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Heat of Solution and Dilution
Enthalpy changes occur when a solute
dissolves in a solvent or when a
solution is diluted.
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Heat of Solution
In the vast majority of cases, dissolving a solute in a
solvent produces measurable heat change. At
constant pressure, (coffee cup calorimeter) the
heat change is equal to the enthalpy change. The
heat change of solution , or enthalpy of solution,
DHsoln, is the heat generated or absorbed when a
certain amount of solute dissolves in a certain
amount of solvent. The quantity DHsoln represents
the difference between the enthalpy of the final
solution and the enthalpies of its original
components (solute and solvent) before they are
mixed.
• Like other enthalpy changes, DHsoln is positive for
endothermic processes and negative for
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exothermic processes.
Consider the heat of solution of a process in which an ionic
compound is the solute and water is the solvent. For
example, what happens when solid NaCl dissolves in water?
In solid NaCl, the Na+ and the Cl- ions are held together by
strong positive-negative (electrostatic) forces, but when a
small crystal of NaCl dissolves in water, the three
dimensional network of ions breaks into its individual units.
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The energy required to completely
separate one mole of a solid ionic
compound into gaseous ions is called
lattice energy (U). The lattice energy
of NaCl is 788 kJ/mol. In other
words, we would need to supply 788
kJ of energy to break 1 mole of solid
NaCl into 1 mole of Na+ and Cl- ions.
NaCl(s)
Na+(g) + Cl-(g)
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U = 788 kJ
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Next the separated Na+ and Cl- ions are
stabilized in solution by their
interaction with water molecules.
These ions are said to be hydrated.
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The enthalpy change associated with
the hydration process is called the
heat of hydration, DHhydr
• Heat of hydration is a negative
quantity for cations and anions.
Na+(g)
+
Cl-(g)
H2O
Na+(aq) + Cl-(aq)
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DHhydr = -784 kJ
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Applying Hess’s law, it is possible to
consider DHsoln as the sum of two
related quantities, lattice energy (U)
and heat of hydration (DHhydr):
NaCl(s)
NaCl(s)
H2O
Na+(g) + Cl-(g)
Na+(g) + Cl-(g)
NaCl(s)
Na+(aq) + Cl-(aq)
H2O
H2O
Na+(aq) + Cl-(aq)
Na+(aq) + Cl-(aq)
DHsoln = ?
U = 788 kJ
DHhydr = -784 kJ
DHsoln =
4 kJ
Therefore, when 1 mol of NaCl dissolves in water, 4 kJ
of heat will be absorbed from the surroundings
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The solution process for NaCl.
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Heat of Dilution
When a previously prepared solution is
diluted, that is, when more solvent is
added to lower the overall
concentration of the solute,
additional heat is usually given off or
absorbed. The heat of dilution is the
heat change associated with the
dilution process.
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Therefore, always be cautious when working on a dilution
procedure in the laboratory. Because of its highly
exothermic heat of dilution, concentrated sulfuric acid
(H2SO4) poses a particularly hazardous problem if its
concentration must be reduced by mixing it with additional
water. The process is so exothermic that you must NEVER
attempt to dilute the concentrated acid by adding water to
it. The heat generated could cause the acid solution to boil
and splatter.
• The recommended procedure is to add the concentrated acid
slowly to the water (while constantly stirring).
“Do as you oughter, add acid to water.”
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THE END
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