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เคมีเชิงฟสิกส I
Physical Chemistry
The study of underlying physical principles
that governs the properties and behaviors of
chemical systems
Physical Chemistry I
01403242
สาขาวิชาเคมี คณะศิลปศาสตรและวิทยาศาสตร ม.ก. กพส.
Sc.4-309 : [email protected]
Physical chemistry is the branch of chemistry
establishes and develops the principles of
subject. It concepts are used to explain
interpret observations on the physical
chemical properties of matter.
that
the
and
and
Physical chemistry is also essential for
developing and interpreting the modern
techniques used to determine the structure and
properties of matter, such as new synthetic
materials and biological macromolecules.
Part 2: Structure
1.
Quantum theory
2.
Molecular structure
3.
Spectroscopy
Part 3: Change
1.
Molecular in Motion (g
(gases, liquids)
q
)
2.
The rates of chemical reactions
3.
The kinetics of complex reactions
4.
Processes at solid surfaces
Physical chemistry provides the basic framework for all other
branches of Chemistry-for inorganic chemistry, organic
chemistry, biochemistry, geochemistry and engineering.
Physical Chemistry
Part 1: Equilibrium
1.
The properties of gases
The perfect gas and real gases
2 Thermodynamics
2.
Th
d
i
First, Second and Third Law
3. Phase diagram
One of the roles of physical chemistry is to establish
the link between the properties of bulk matter and the
behaviour of the particles-atoms, ions or molecules-of
which it is composed.
Bulk matter
G iis a fluid
Gas
fl id state off matter that
h fills
fill the
h container
i
it
i
occupies.
Liquid is a fluid state of matter that possesses a welldefined surface
Solid retains its shape independent of the shape of
the container it occupies.
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Physical chemist thinks about the three states of matter:
A liquid consists of particles that are in contact with each
other, but are able to move past each other in a restricted
manner. The particles are in a continuous state of motion, but
travel only a fraction of a diameter before bumping into a
neighbour. The overriding image is one of movement, but with
molecules jostling one another.
A gas is composed of particles in continuous rapid, chaotic
motion. A particle travels several diameters before another
particle, and for most of time the particles are so far apart that
th interact
they
i t
t with
ith eachh other
th only
l very weakly.
kl The
Th overriding
idi
image is one of busy chaos.
Perfect Gases They move rapidly, and collide randomly
and chaotically with one another.
A solid consists of particles that are in contact with one
another and unable to move past one another. Although the
particles oscillate around an average location, they are
essentially trapped in their initial positions. The overriding
image is one of almost stationary, regularly arrayed particles.
The physical properties of an ideal gas are completely
described by four parameters which their respective SI
units are:
the amount of substance of which it is composed, n, in moles;
the temperature of the gas, T, in Kelvin
the pressure of the gas, P, in Pascal;
The essential difference between the three states of matter
comes down to the freedom of the particles to move past one
another.
The volume occupied by the gas, V, in m3
The kinetic theory of gases : a model of molecular nature of gases
The perfect gas equations
Boyle ‘s law; P . V = constant at constant temperature
The kinetic theory of gases is an attempt to describe the
macroscopic properties of a gas in terms of molecular behavior.
Charles ‘s law; V  T
at constant pressure
1.
Avogadro ‘s principle; V n
at constant pressure and
t
temperature
t
2
2.
There are no attractive forces between the molecules;
3.
The molecules travel in straight lines, except during the
course of collisions. Molecules undergo perfectly elastic
collisions; i.e. the kinetic energy of the molecules is
conversed in all collisions, but may be transferred
between them.
The perfect gas equation
PV = nRT
Or ideal gas law was empirically developed from experimental
observations.
The size of the molecules which make up the gas is
negligible compared to the distance between them;
These are considered to be true of real systems at low pressure
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The speed of molecules in gases
Real gases
Total energy = kinetic energy + potential energy
Maxwell distribution of speeds
1/ 2
 2RT 


 M 
Most probable speed =
The molecular origin of pressure
p
Perfect gas is a gas in which the only contribution to the energy
comes from the kinetic energy
In fact all molecules do interact with one another provided they are
close enough together, so the ‘kinetic energy only’ model is only
an approximation.
nMc 
2
3V
At low pressure : The intermolecular forces play no significant role.
Intermolecular interaction
At moderate pressure :
Intermolecular attraction (long-range interaction) is responsible
for the condensation of gases into liquids at low temperatures.
Intermolecular repulsion (short-range interaction) is
responsible for the fact that liquids and solids have a
ddefinite
fi it bbulk
lk and
d ddo nott collapse
ll
tto an iinfinitesimal
fi it i l
point.
The van der Waals equation of state
The attractive forces dominate the repulsive forces
In this case, the gas can be expected to be more compressible than
a perfect gas
At high pressure :
p
forces
f
dominate and the gas
g can be expected
p
to be
The repulsive
less compressible because now the forces help to drive the
molecules apart.
At low temperature:
2

 n  
p  a   V  nb  nRT
V

  
Intermolecular forces may also be important . Molecules travel
with such low mean speeds that they can be captured by one
another
The van der Waals Parameters
The compression factor, Z
-29
-28
3
3
b – roughly the volume of a molecule, (3.5·10
– 1.7 ·10 ) m ~(few Å)
Z
pVm
RT
Vm = the molar volume
-51
-48
3
a – varies a lot [~ (8·10
– 3 ·10 ) J · m ] depending on the intermolecular interactions
(strongest – between polar molecules, weakest – for inert gases).
Substance
. 3a’
2
(J m /mol )
-5
b’ 3 (x10
m /mol)
Pc
(MPa)
For a perfect gas, Z = 1 under all conditions, deviation of Z
from 1 is a measure of departure from perfect behaviour.
behaviour
At high pressures, all the gases have Z > 1, signifying that they
are more difficult to compress than a perfect gas.
At intermediate pressures, most gases have Z  1, indicating that
the attractive forces are dominant and favour compression.
Tc
(K)
Air
.1358
3.64
3.77
133 K
Carbon Dioxide (CO2)
.3643
4.27
7.39
304.2 K
Nitrogen (N2)
.1361
3.85
3.39
126.2 K
Hydrogen (H2)
.0247
2.65
1.30
.5507
3.04
Ammonia (NH3)
.4233
Helium (He)
.00341
Freon (CCl2F2)
1.078
When can
Water (H2O)
be reduced to
33.2 K
22.09
647.3 K
3.73
11.28
406 K
2.34
0.23
5.2 K
9.98
4.12
385 K
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The Critical Point
The van der Waals Isotherms
P
Nk BT
aN 2
 2
V  Nb  V
Nk BT  2 aN
abN

V 3   Nb 
V
0
V 
P 
P
P

2
3
The critical point is the unique point where both (dP/dV)T = 0 and
2
2
(d P/dV )T = 0
Critical parameters:
VC  3Nb PC 
1 a
8 a
k BTC 
27 b2
27 b
unstable
0
N·b
substance
H2
He
N2
CO2
H20
TC (K)
33.2
5.2
126
304
647
PC (MPa)
1.3
0.23
3.4
7.4
22.1
Outlines : Thermodynamics
• Definitions
• 1st, 2nd, 3rd laws of thermodynamics
• Helmholtz and Gibbs free energies
• Partial molar properties, chemical potential
andd fugacity
f
i
•Thermodynamics of solutions
•Equilibrium
The branch of physical chemistry know as thermodynamics is
concerned with the study of the transformation of energy, and in
particular the transformation of energy from heat into work
and vice versa.
Th
Thermodynamics
d
i iis concerned
d with
ith equilibrium
ilib i
states off
matter and has nothing to do with time.
Thermodynamics was originally formulated by physicists and
engineers interested in the efficiency of steam engines.
Definitions
The system is the part of the world in which we have
a special interest.
Heat
Adiabatic system
The surroundings are where we make our observations.
Universe = surrounding + system
Open
Closed
Isolated
Types of system
An open system is a system that can exchange matter
with its surroundings.
An closed system is a system that can exchange
energy with its surroundings.
An isolated system is a system that can exchange
neither matter nor energy with its surroundings.
Biochemical cell
Water in a vacuum
Stoppered flask
The energy of a closed system can be changed: by transferring
energy as work and by transferring energy as heat.
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The measurement of work
Work is a transfer of energy that can be used to change the height of
a weight somewhere in the surrounding
In chemistry, a very important type of work is expasion work,
the work done when a system expands against an opposing
pressure.
Work done by system = pexV
Heat is a transfer
f off energy
gy as a result off a temperature
p
difference
ff
between the system and the surroundings.
Explanation of work and heat in the
way of motion of atom ????
We were interested in the maximum work (Isothermal)
(
)
The maximum work (Isothermal) is obtained when the external
pressure is only infinitestimally less than the pressure of the gas
in the system
(b) Free expansion
The expansion work
Expasion against zero opposing force
(a) The general expression for work
pex = 0
The calculation of expansion work starts
from the definition used in Physics.
dw = -Fdz
w sys    0dV
0
Force opposing expansion is F = pexA
dw = -pex Adz
(c) Expansion against constant pressure
w sys    p ex dV
dw = -pex dV
 p ex V
w sys    p ex dV
 p ex V
Chongrak_Thermo 28-78
(d) Reversible expansion
(b) Isothermal reversible expansion
A reversible change in thermodynamics is a change that can
be reversed by an infinitesimal modification of a variable.
w sys    Pext dV    PdV
nRT
dV
dV   nRT 
V
V
V
Vf
  nRT ln V V   nRT ln f
i
Vi
 
A reversible change
Equilibrium
Reversible expansion : pex = p at each state of expansion
W can summarize
We
i by
b the
h following
f ll i remarks
k :
w sys    p ex dV    pdV
1.
A system does maximum expansion work when the external
pressure is equal to that of the system (pex = p)
Note : In every day life “reversible” means a process that can
be reversed; in thermodynamics it has a stronger meaning-it
means that a process can be reversed by an infinitesimal
modification in some property (such as pressure).
2.
A system does maximum expansion work when it is in
mechanical equilibrium with its surroundings.
3.
A system does maximum expansion work when it is changing
reversibly.
Chongrak_Thermo 29-78
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Mechanical equilibrium
Thermal equilibrium
The condition of equality of pressure on either side of a
movable wall (a piston) is a state of mechanical equilibrium
High
Low
pressure
pressure
High temp.
Low temp.
Equal temp.
Low temp.
High temp.
Diathermic
wall
Equal pressure
Phase equilibrium
Low
High
pressure
pressure
Processes (Changes)
Properties of matters
Isothermal process : T const
Isobaric process :
Isochoric process:
Chemical equilibrium
Intensive property : P, T, d, ...
Extensive property : V, m, S , ...
P const
V const
extensive
t i properties
ti
 intensive properties
extensive properties
Isenthalpic process: H const
Adiabatic process :
T
An important classification of thermodynamic properties is
whether they are state functions or path functions
State functions and exact differential
Such properties that are independent of how a sample is prepared
are called state functions, such as pressure and temperature and
internal energy that define the current state of the system.
f
U   dU
i
The value of U depends on the initial and final states of the
system but is independent of the path between them. This pathindependent of the integral is expressed by saying that dU is an
exact differential.
Path functions and inexact differential
Properties that relate to the preparation of state are called
path functions
f
q 
 dq
i , path
When a system is heated, the total energy transferred as heat
is the sum of all individual contributions at each point of the
path. Work is also path function.
This path-dependence is expressed by saying that dq is
inexact differential.
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When a system releases 10kJ of energy to the surrounding as work,
the internal energy of the system decrease by 10kJ
Internal energy, U
A fundamental parameter in thermodynamics is the internal energy .
This is the total amount of energy in a system.
U = -10kJ
Internal energy is the sum total of all kinetic and potential energy
within the system.
If we supply 20kJ of energy as heat, , then the internal energy
increases by 20kJ
U is a state function,
function as a specific system has a specific value at any
given temperature and pressure.
U = -20kJ
Thermodynamics deal with changes in U, denoted as U.
U has negative (-) and positive (+) value.
U > O
work
U > O
heat
The First Law of thermodynamics
U = P.E + K.E
“The total energy of an isolated thermodynamic system is constant”
“The conservation of energy”
(isothermal expansion process)
Real Gas
P.E
P
E0
U  0
“energy can not be created and destroyed”
In other words, energy may be lost from a system in only two ways,
K.E = 0
Perfect Gas
Either as work and as heat.
K.E = 0
P.E = 0
 U = 0
U = q + w
Example:
Adiabatic bomb calorimeter????
1). Nutritionists are interested in the use of energy by the human
body, and we can consider our own body as a thermodynamic
“system”. Calorimeters have been constructed that can
accommodate a person to measure their net energy output. Suppose
in the course of an experiment someone does 622 kJ of work on an
exercise bicycle and loses 82 kJ of energy as heat. What is the
change in internal energy of the person?
Heat Capacity
The internal energy of a substance increases when its temperature
is raised.
 U 
Cv  

 T  V
a ‘partial derivative’
U = q + w = (-82) + (-622) = -704 kJ
We see that the person ‘s internal energy falls by 704 kJ.
2). An electric battery is charged by supplying 250 kJ of energy to it
as electrical work, but in the process it loses 25 kJ of energy as heat
to the surroundings. What is the change in internal energy of the
battery? (+225 kJ)
Chongrak_Thermo 41-78
Heat capacities are extensive properties ; J K-1.
Molar heat capacity at constant volume is an intensive
property ; J K-1 mol-1.
Specific heat capacity ( specific heat) ; J K-1 g-1
Chongrak_Thermo 42-78
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The heat capacity can be used to relate a change in internal
energy to a change in temperature of a constant-volume system.
The measurement of U
dU = CvdT
U = q at constant volume
If the heat capacity is independent of temperature over the
range of temperatures of interest.
U = CvT
Because a change in internal energy can be identified with the
heat supplied at constant volume.
The heat supplied to or absorbed by a system that
cannot change its volume
The apparatus used in practice is called
qv = CvT
a bomb calorimeter
This relation provides a simple way of measuring the heat
capacity of a sample; a measured quantity of heat is supplied to
the sample (electrically, for example), and the resulting increase
in temperature is monitored.
Chongrak_Thermo 43-78
One procedure is to heat the calorimeter electrically by
passing a know current for a measured time through a heater,
and record the increase in temperature. The heat supplied by
the heater is given by
heat supplied to system = Ivt
I is the current in amperes (A),
V is the potential of the supply in volts (V)
If we observe that the temperature of the calorimeter rises by
4.47C, then the calorimeter constant, C, the ratio of the heat
supplied to the increase in temperature, is
C = ?
If in an experiment on an unknown reaction we measure a
temperature rise of 3.22 C in the same calorimeter, then we
could conclude that the heat output of the reaction is
heat output = C x increase in temperature
T is the time in second for which the current flows.
heat supplied to system =(1.23 A) x (12.0 V) x (123 s) = 1.82 kJ
(1 Avs = 1 J)
Exercise
Calculating work, heat and internal energy. Let the initial
state be T, Vi and the final state be T, Vf. The change of state are;
Path 1, in which there is free expansion against zero external
pressure; Path 2, in which there is reversible, isothermal
expansion.
=
x 3.22 C = 1.31 kJ
Similar procedures apply when reaction releases heat or
absorbs heat: the temperature of a calorimeter rises if heat is
released and it falls if heat is absorbed.
Enthalpy
The change in internal energy is not equal to the heat supplied when
system is free to change its volume. Under these circumstances some
of the energy supplied as heat to the system is returned to the
surroundings as expansion work, so dU is less than dq
Then,, the enthalpy
py is defined as
Homework : Compare the quantities of heat required to carry out
the isothermal expansion of 1.00 mol of an ideal gas, initially
occupying 28.0 L at pressure of 0.880 atm and a temperature of 300
K, to a volume of 40.0 L (a) reversible, and (b) irreversible against a
pressure equal to the final pressure of the gas.
(b) 7.49 x 102 J
H = U + pV
State function
H = U + pV
H = qp
Only work done by the system is pV work
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The enthalpy of a perfect gas is related to its internal energy by
using pV = nRT
H = U + ngRT
ng is the change in the amount of gas molecules in the reaction.
Example : Calculating a change in enthalpy
Water is heated to boiling under a pressure of 1.0 atm. When an
electric current of 0.50 A from 12 V supply is passed for 300 s
through a resistance in thermal contact with it, it is found that 0.798 g
of water is vaporized. Calculate the molar internal energy and
enthalpy changes at the boiling point (373.15K)
q = IVt = (0.5 A) x (12 V) x (300 s) = + 1.8 kJ
2H2O(l) ng = -3
3 mol
2H2 (g) + O2(g)
Hm = +1.8
+1 8 kJ/ 0.0443
0 0443 mol = + 41 kJ mol-11
H2O (l)
H - U = +ngRT
H - U = (-3 mol) RT = -7.5 kJ
The variation of enthalpy with temperature
Um  Hm = RT
The molar enthalpy of vaporization of benzene at its boiling point
(353.25 K) is 30.8 kJ mol-1. What is the molar internal energy
change? For how long would the same 12 V source need to supply a
0.50 A current in order to vaporize a 10 g sample?
(+27.9 kJ mol-1, 6.6 x 102 s)
A convenient approximate empirical expression of Cp is
C p , m  a  bT 
The enthalpy of a substance increases as its temperature is raised.
The slope of a graph of enthalpy against temperature at constant
pressure is called the heat capacity at constant pressure, Cp
 H 
Cp  

 T  p
H = CpT
H2O (g)
Um = + 38 kJ mol-1
c
T2
The empirical parameters a, b, and c are independent of temperature.
Excercise :
What is the change
g in molar enthalpy
py of N2 when it is heated from 25C
to 100C? The heat capacity information of N2 is:
a = 28.58
qp = CpT
b = 3.77 x 10-3 K-1
The variation of heat capacity with temperature can sometimes
be ignored if the temperature range in small. This
approximation is highly accurate for a monatomic perfect gas
(one of the noble gas)
c = -0.5 x 105 K2
The Joule-Thomson effect
The analysis of the Joule-Thomson coefficient is central to the
technological problems associated with the liquefaction of gases.
 T 
   
 p H
Joule-Thomson coefficient
They let a gas expand through a porous barrier from one constant
pressure to another,
another and monitored the difference of temperature that
arose from the expansion
The change of internal energy of the gas as it move from
one side of the throttle to the other is
Uf – Ui = w = piVi - pfVf
Uf + pfVf = Ui + piVi or Hf = Hi
Isenthalpic process
The thermodynamic quantity measured is µ ,
 T 
 
 p  H
The physical interpretation of µ is that it is the ratio of the
change in temperature to the change in pressure when a gas
expands under adiabatic conditions
Real g
gases have nonzero Joule – Thomson coefficients and,,
depending on the identity of the gas, the pressure, the relative
magnitudes of the attractive and repulsive intermolecular forces,
and the temperature, the sign of the coefficient may be either
positive and negative
The very useful meaning is µ is positive, then T decreases when P
decreases, or the gas cools upon expansion.
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Thermochemistry
Standard enthalpy changes
The study of the heat produced or required by chemical
reactions is called thermochemistry
Thermochemistry is the branch of the thermodynamics because a
reaction vessel and its contents form a system, and chemical
reactions result in the exchange of energy between the system and
the surroundings.
We can use calorimetry to measure the heat produced or absorbed
by reaction, and can identify q with a change in U and H
H will be discussed in the detail for Thermochemistry
(a) Enthalpies of physical change
The standard enthalpy change that accompanies a change of
physical state is called the standard enthalpy of transition
H2O (g) Hsub
H2O (s)
Overall:
H2O (s)
H2O (l)
Hfus
H2O (l)
H2O (g)
Hvap
H2O (g) Hfus +
H2O (s)
The standard state of a substance at a specified temperature is its
pure form at 1 bar
For example;
The standard state of liquid ethanol at 298K is pure liquid
ethanol at 298K and 1 bar.
The standard state of solid iron at 500 K is pure at 500K
and 1 bar.
Standard enthalpies may be reported for any temperature.
However, the conventional temperature for reporting
thermodynamic data is 298.15 K
Enthalpy is a state property, the enthalpy change in a reverse
process is the negative of the enthalpy change in the forward
process between the same two states:
H (reverse) = -H (forward)
The enthalpy of vaporization of water is +44 kJ mol-1 at 298 K, its
enthalpy of condensation at that temperature is -44 kJ mol-1
(b) Enthalpies of chemical change
H
vap
Because enthalpy is a state function, the enthalpy change in a
direct route between two states is equal to the sum of the enthalpy
changes for a sequence of changes between the same two states:
The standard reaction enthalpy, Hr , is the change in enthalpy
when reactants in their standard states change to products in their
standard states
CH4(g) + 2O2(g)
H (indirect route) = H (direct route)
For general:
2A + B
3C + D
Hr = {3Hm(C) + Hm(D) } – {2 Hm(A) + Hm(B) }
Molar enthalpy
(c) Hess ‘s law
The standard enthalpies of individual of an overall reaction is the
sum of the standard enthalpies of the individual reactions into
which a reaction may be divided.
CO2(g) + 2H2O(l) Hr = -890 kJ mol-1
Thermochemical
equation
Hess ‘s law
1.
The reaction must be balanced.
2.
Every substances in the reaction have to be defined a
physical state (s, l, g)
3.
When the direction of reaction is changed, the sign of the
enthalpy
h l have
h
to be
b changed.
h
d
4.
If any number is multiplied in the reaction, the enthalpy
of that reaction have to multiplied by that number.
The importance of Hess ‘s law is that information about a
reaction of interest, which may be difficult to determine directly,
can be assembled from information on other reactions.
C(graphite) + ½O2(g)
CO(g)
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Chongrak_kinetics
Example: Using the Hess ‘s law
Exercise : Calculate the standard enthalpy of formation of CH4
 H2(g) + O2(g)  H2O(l) H1 = -285.8 kJ
 H2(g) + O2(g)  H2O(g) H2 = -241.8 kJ
H  of H2O(g)  H2O(l) = ?
C(s) + 2H2(g)  CH4(g)
 C(s) + O2(g)  CO2(g) H1 = -393 kJ
 H2(g) + O2(g)  H2O(l) H2 = -285 kJ
 CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)
H3 = -890 kJ
The Born - Haber cycle is a specific application of the
first law of thermodynamics using Hess ‘s law.
The cycle allows indirect determination of the lattice
enthalpy of an ionic solid.
nMm+(g) + mXn-
MnXm(s)
Because direct measurement of this process is generally
impractical, and indirect path is created.
The Born – Haber cycle ??? And How to calculate the
standard of formation of ionic lattice.
Standard enthalpies of formation, Hf
The standard enthalpy of formation of a substance is the standard
reaction enthalpy for the formation of the compound from its
elements in their reference states.
For example, at 298 K
The reference state of nitrogen is a gas of N2 molecules.
The reference state of mercury is liquid mercury
The reference state of carbon is graphite
The standard enthalpy of formation of liquid benzene at 298 K
refers to the reaction
6C(s, graphite) + 3H2(g)
C6H6 (l) Hf = +49.0 kJ mol-1
Chongrak_Thermo 64-78
Element
Reference state
Arsenic
grey arsenic
Bromine
liquid
Carbon
graphite
Hydrogen
gas
Iodine
solid
Mercury
liquid
Nitrogen
gas
Oxygen
gas
Phosphorus
white phosphorus
Sulfur
rhombic sulfur
Tin
white tin
The reaction enthalpy in terms of enthalpies of formation
Stoichiometric coefficients
Hr =  ν Hf +  ν Hf
products
Reactants
The standard enthalpy of combustion, Hcom
Th standard
The
d d enthalpy
h l off combustion
b i is
i the
h standard
d d change
h
in
i
enthalpy per mole of combustible substance.
CH4(g) + 2O2(g)
CO2(g) + 2H2O(l) Hcom = -890 kJ
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Chongrak_kinetics
Exercise: Using Hess ‘s law
The temperature dependence of reaction enthalpies
Given the thermochemical equations
C3H8(g) H = -124 kJ
C3H6(g) + H2(g)
C3H8(g) + 5O2(g)
3CO2(g) + 4H2O(l) H = -2220 kJ
Calculate the standard enthalpy of combustion of propene
The standard enthalpies of many important reactions have been
measured at different temperatures. The standard reaction
enthalpy at different temperatures may be estimated from heat
capacities and the reaction enthalpy at some other temperature.
T2
H r (T2 )  H r (T1 )   C P, r dT
T1
Kirchhoff ‘s law
Cp,r =
 ν Cp,r +
products
 ν Cp,r
Reactants
Example : Using Kirchhoff ‘s law
The standard enthalpy of formation of gaseous water at 25 C is
-241.82 kJ mol-1. Estimate its value at 100 C
T2
H r (T2 )  H r (T1 )   C P,r dT
T1
-241 82 kJ mol-1
-241.82
The molar heat capacities at constant pressure :
H2O(g) = 33.58 J K-1mol-1
H2(g) = 28.84 J K-1mol-1
O2(g) = 29.37 J K-1mol-1 : Assume that the heat capacities
are independent of temperature.
Hr (373K) = -242.6 kJ mol-1
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