Advanced physical chemistry notes

Advanced Physical Chemistry
CHEM540
Dr. Fateh Eltaboni
Assistant Professor of Physical Chemistry at the University of Benghazi
Omer Al Mokhtar University
Faculty of Science
Chemistry Department
CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 1

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CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni)
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Physical
Chemistry
Equilibrium
Thermodynamics
Quantum and
Spectroscopy
Statistical
Thermodynamics
Special Topics
Chemical kinetics

Gas Laws
❶

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CHEM540 Course Syllabus
1. Kinetic molecular theory of gases
2. Thermochemistry
3. Thermodynamics
4. Chemical Kinetics
5. Nuclear chemistry and Radio chemistry
6. Electrochemistry
7. Symmetry and group theory
8. Quantum mechanics and atomic structure
9. Spectroscopy and Photochemistry
Course General Information

The properties of gases
• The simplest state of matter is a gas, a form of matter that fills any container it
occupies.
• The perfect (ideal) gas:
• Ideal gas molecules are widely separated from one another and move in paths that
are largely unaffected by intermolecular forces.
• The states of gases:
• The physical state of a sample of a substance, its physical condition, is defined by
its physical properties.
• Two samples of a substance that have the same physical properties are in the
same state.
• The state of a pure gas, for example, is specified by giving its
• volume, V, amount of substance (number of moles), n pressure, p, and
temperature, T.
7 CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni)

The properties of gases
• Experimentally, is sufficient to specify only three of these variables, for then
the fourth variable is fixed. That is, it is an experimental fact that each
substance is described by an equation of state, an equation that interrelates
these four variables.
• The general form of an equation of state is:
• This equation tells us that, if we know the values of n, T, and V for a particular
substance, then the pressure has a fixed value.
• One very important example is the equation of state of a ‘perfect gas’, which
has the form p = nRT/V, where R is a constant.

The states of gases:
• (a) Pressure
• Pressure, p, is defined as force, F, divided by the area, A, to which the force
is applied:
• That is, the greater the force acting on a given area, the greater the
pressure.
• The origin of the force exerted by a gas is the continuous collisions of the
molecules on the walls of its container.
• The SI unit of pressure, the pascal (Pa, 1 Pa = 1 N m−2) . But there are
several other units are still widely used (Table 1.1).

Pressure
• Fig. 1.1 When a region of high pressure is separated
from a region of low pressure by a movable wall
(piston), the wall will be pushed into one region or
the other, as in (a) and (c). However, if the two
pressures are identical, he wall will not move (b).
The latter condition is one of mechanical
equilibrium between the two regions.
• This condition of equality of pressure on either side
of a movable wall is a state of mechanical
equilibrium between the two gases.

Pressure units

The measurement of pressure:
• The pressure exerted by the atmosphere is
measured with a barometer.
• The original version of a barometer (which was
invented by Torricelli, a student of Galileo) was an
inverted tube of mercury sealed at the upper end.
• When the column of mercury is in mechanical
equilibrium with the atmosphere, the pressure at its
base is equal to that exerted by the atmosphere.
• It follows that the height of the mercury column is
proportional to the external pressure.

• Example 1.1 Calculating the pressure exerted by a column of liquid
• Derive an equation for the pressure at the base of a column of liquid of mass
density ρ (rho) and height h at the surface of the Earth. The pressure exerted
by a column of liquid is commonly called the ‘hydrostatic pressure’.
• Solution:
where and
The pressure at the base of the column is therefore:
‘hydrostatic pressure’.
• Hydrostatic pressure is independent of the shape and cross-sectional area of
the column. The mass of the column of a given height increases as the area
(F= mg), so the two cancel.

14
1. The pressure of a sample of gas inside a container is measured by using a pressure
gauge, which is a device with electrical properties that depend on the pressure.
 For instance, a Bayard–Alpert pressure gauge is based on the ionization of the molecules
present in the gas and the resulting current of ions is interpreted in terms of the pressure.
2. In a capacitance manometer, the deflection of a diaphragm relative to a fixed electrode
is monitored through its effect on the capacitance of the arrangement.
3. Certain semiconductors also respond to pressure and are used as transducers in solid
state pressure gauges.

(c) Temperature
• The concept of temperature springs from the observation that a change in
physical state (for example, a change of volume) can occur when two objects
are in contact with one another, as when a red-hot metal is plunged into
water.
• The temperature, T, is the property that indicates the direction of the flow of
energy through a thermally conducting, rigid wall.
• If energy flows from A to B when they are in contact, then we say that A has
a higher temperature than B (Fig. 1.2).

(c) Temperature
16
Fig. 1.2 Energy flows as heat from a region at a
higher temperature to one at a lower temperature if
the two are in contact through a diathermic wall
(Boundary), as in (a) and (c). However, if the two
regions have identical temperatures, there is no
net transfer of energy as heat even though the two
regions are separated by a diathermic wall (b). The
latter condition corresponds to the two regions
being at thermal equilibrium.
• Two types of boundary:
1. Diathermic (thermally conducting; ‘dia’ is from the Greek word for ‘through’) if a
change of state is observed when two objects at different temperatures are
brought into contact. A metal container has diathermic walls.
2. Adiabatic (thermally insulating) if no change occurs even though the two objects
have different temperatures. A vacuum flask is an approximation to an adiabatic
container. CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni)

(c) Temperature
• The temperature is a property that indicates whether two objects would be
in ‘thermal equilibrium’ if they were in contact through a diathermic
boundary.
• Zeroth Law of thermodynamics:
17
Fig. 1.3 The experience
summarized by the Zeroth Law of
thermodynamics:
if an object A is in thermal
equilibrium with B and B is in
thermal equilibrium with C, then C is
in thermal equilibrium with A.CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni)

thermometer
• The Zeroth Law justifies the concept of temperature and the use of a
thermometer, a device for measuring the temperature.
• Suppose that B is a glass capillary containing a liquid, such as mercury, that
expands as the temperature increases.
• When A is in contact with B. According to the Zeroth Law, if the mercury
column in B has the same length when it is placed in thermal contact with
another object C, then we can predict that no change of state of A and C will
occur when they are in thermal contact.
• Moreover, we can use the length of the mercury column as a measure of
the temperatures of A and C.

Thermometry
• In the thermometry: temperatures are related to the length of a column of
liquid, and the difference in lengths shown when the thermometer was first
in contact with melting ice and then with boiling water was divided into 100
steps called ‘degrees’, the lower point being labelled 0.
• This procedure led to the Celsius scale of temperature. The Celsius scale are
denoted θ (theta) and expressed in degrees Celsius (°C).
• Because different liquids expand to different extents, thermometers
constructed from different materials showed different numerical values of
the temperature between their fixed points.

Thermodynamic temperature scale
• The pressure of a gas can be used to construct a perfect-gas
temperature scale that is independent of the identity of the gas.
•
• Thermodynamic temperature scale: in which temperatures are
denoted T and are normally reported in kelvins (K; not °K).
• Thermodynamic and Celsius temperatures are related:

(c) Temperature
• Example:
• Express 25.00°C as a temperature in kelvins.
• Answer:
T/K = (25.00°C)/°C + 273.15 = 25.00 + 273.15
= 298.15
• Note how the units (in this case, °C) are cancelled like
numbers. This is the procedure called ‘quantity calculus’
in which a physical quantity (such as the temperature) is
the product of a numerical value (25.00) and a unit (1°C).
• Multiplication of both sides by the unit K then gives T =
298.15 K.

The gas laws
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(a) The perfect gas laws:
A. Avogadro’s principale: V α n at constant p, T
B. Boyle’s law: p α 1/V at constant n, T
C. Charles’s law: V α T at constant n, p
p α T at constant n, V
Avogadro’s principle is a principle rather than a law because
it depends on the validity of a model, in this case the
existence of molecules.

The gas laws
• Boyle’s and Charles’s laws are examples of a
limiting law, a law that is strictly true only in a
certain limit, in this case p → 0.
• The empirical observations summarized by (A,B,
and C) can be combined into a single expression
• pV = constant × nT
• Perfect gas law:

The gas laws
24
Linear relationship

The gas laws
25
 The perfect gas law (or perfect gas equation of
state). It is the approximate equation of state
of any gas, and becomes increasingly exact as
the pressure of the gas approaches zero.
 A gas that obeys perfect gas law exactly under
all conditions is called a perfect gas (or ideal
gas).
 A real gas, an actual gas, behaves more like a
perfect gas at lower the pressure.
 The gas constant R can be determined by
evaluating R = pV/nT for a gas.
 More accurate value for R can be obtained by
measuring the speed of sound in a low-
pressure gas (argon is used in practice) and
extrapolating its value to zero pressure.

The gas laws

The gas laws
• Example 1.2 Using the perfect gas law
• In an industrial process, nitrogen is heated to 500 K in a vessel of constant
volume. If it enters the vessel at 100 atm and 300 K, what pressure would it
exert at the working temperature if it behaved as a perfect gas?
• Method: We expect the pressure to be greater on account of the increase
in temperature.
• The perfect gas law in the form pV/nT = R
Combined gas law:

Perfect gas law
• The perfect gas law is important in physical chemistry because it is used to
derive a wide range of relations that are used in thermodynamics.
• It is also used to calculate the properties of a gas under a variety of
conditions. For instance, the molar volume, Vm =V/n, of a perfect gas under
the conditions called standard ambient temperature and pressure.
• (SATP), which means 298.15 K and 1 bar (that is, exactly 105 Pa), is easily
calculated from Vm = RT/p to be 24.789 dm3 mol−1
• Standard temperature and pressure (STP), was 0°C and 1 atm; at STP, the
molar volume of a perfect gas is 22.414 dm3 mol−1.

The kinetic model of gases
• A gas may be pictured as a collection
of particles in ceaseless, random
motion (Fig. 1.6).
• The ‘kinetic model’ (or the ‘kinetic
molecular theory’, KMT) of gases: is a
technique proposes a qualitative
model and then expressing that model
mathematically.
• The perfect gas law is quantitative
model very simple, and experimentally
confirmable.

KMT
• The kinetic model of gases is based on three assumptions:
1. A gas consists of molecules in ceaseless random motion.
2. The size of the molecules is negligible in the sense that their diameters
are much smaller than the average distance travelled between
collisions.
3. The molecules do not interact, except during collisions.
• Assumption 3 suggests that the potential energy of the molecules (their
energy due to their position) is independent of their separation and may
be set equal to zero.
• The total energy of a sample of gas is therefore the sum of the kinetic
energies (the energy due to motion) of all the molecules present in it.
• It follows that the faster the molecules travel (and hence the greater
their kinetic energy), the greater the total energy of the gas.

The pressure of a gas according to the kinetic model
• The kinetic model accounts for the steady pressure exerted by a gas in terms
of the collisions the molecules make with the walls of the container.
• The pressure exerted by a gas of mass m in a volume V is:
32
Compare it to pV = nRT. This
conclusion is a major success of the
kinetic model, for the model implies an
experimentally verified result.

KMT: Root mean square speed
• Here c (or μ)is the root-mean-square speed (rms speed) of the molecules.
This quantity is defined as the square root of the mean value of the squares
of the speeds, v, of the molecules. That is, for a sample consisting of N
molecules with speeds v1, v2, . . . , vN,
• The rms speed might at first sight seem to be a rather peculiar measure of
the mean speeds of the molecules, but its significance becomes clear when
we make use of the fact that the kinetic energy of a molecule of mass m
travelling at a speed v is Ek = ½ mv2, which implies that the mean kinetic
energy, 〈Ek〉, is the average of this quantity, or mc2. It follows from the
relation ½ mc2 = 〈Ek〉 that

KMT: rms
• rms its a measure of the mean kinetic energy of the molecules of the gas.
The rms speed is a measure of molecular speed.
• The mean speed, ć, of the molecules:
• For samples consisting of large numbers of molecules, the mean speed is
slightly smaller than the rms speed.
• The precise relation is:

rms speed of gas molecules
•
• Which gives:
• The ns now cancel to give:
• Rearrange: rms =
• Substitution of the molar mass of O2 (32.0 g mol−1) and a temperature
corresponding to 25°C (that is, 298 K) gives an rms speed for these
molecules of 482 m s−1.
• The same calculation for nitrogen molecules gives 515 m s−1. Both these
values are not far off the speed of sound in air (346 m s−1 at 25°C).

Speeds of gas molecules
• That similarity is reasonable, because sound is a
wave of pressure variation transmitted by the
movement of molecules, so the speed of
propagation of a wave should be approximately
the same as the speed at which molecules can
adjust their locations.
• The rms and mean speeds of molecules in a gas
is proportional to the square root of the
temperature.
• Therefore, doubling the thermodynamic
temperature (that is, doubling the temperature
on the Kelvin scale) increases the mean and the
rms speed of molecules by a factor of 21/2 =
1.414. . . .

Maxwell distribution of speeds
• Not all molecules travel at the same speed: some move more slowly than
the average (until they collide, and get accelerated to a high speed, like the
impact of a bat on a ball), and others move at much higher speeds than the
average, but be brought to a sudden stop when they collide.
• There is a ceaseless redistribution of speeds among molecules as they
undergo collisions. Each molecule collides once every nanosecond (1 ns =
10−9 s) or so in a gas under normal conditions.
• The mathematical expression that tells us the fraction of molecules that
have a particular speed at any instant is called the distribution of molecular
speeds.

The Maxwell distribution of speeds
• The distribution tell us that at 20°C 19 out of 1000 O2 molecules have a
speed in the range between 300 and 310 m s−1, that 21 out of 1000 have a
speed in the range 400 to 410 m s−1, and so on.
• The precise form of the distribution was worked out by James Clerk Maxwell
towards the end of the 19th century, and his expression is known as the
Maxwell distribution of speeds.
• According to Maxwell, the fraction f of molecules that have a speed in a
narrow range between s and s + Δs (for example, between 300 m s−1 and
310 m s−1, corresponding to s = 300 m s−1 and Δs = 10 m s−1) is

Maxwell equation
• One of the skills to develop in physical chemistry is the ability to interpret
the message carried by equations.
• Equations carry information, and it is more important to be able to read
that information than simply to remember the equation.
• Let’s read the information in Maxwell equation piece by piece. It will be
useful to know the shape of exponential functions. Here, we deal with two
types, e−ax and e−ax2
.
• An exponential function, a function of the form e−ax, starts off at 1 when x =
0 and decays toward zero, which it reaches as x →∞ (Fig. 1.7). This function
approaches zero more rapidly as a increases.

Maxwell equation
40
• A Gaussian function, a
function of the form e-ax2
, also
starts off at 1 when x = 0 and
decays to zero as x increases,
however, its decay is initially
slower but then plunges down to
zero more rapidly than an
exponential function (Fig. 1.7).
• The illustration also shows the
behaviour of the two functions
for negative values of x. The
exponential function e-ax rises
rapidly to infinity, but the
Gaussian function falls back to
zero and traces out a bell-CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni)

Maxwell equation
• Maxwell equation includes a decaying exponential function, the term:
• Its presence implies that the fraction of molecules with very high speeds will
be very small because e−s2
becomes very small when s2 is large.
• The factor Ms2/2RT in the exponent, is large when the molar mass, M, is
large, so the exponential factor goes most rapidly towards zero when M is
large.
• That tells us that heavy molecules are unlikely to be found with very high
speeds.

Maxwell equation
• The opposite is true when the temperature, T, is high: then the factor
M/2RT in the exponent is small, so the exponential factor falls towards zero
relatively slowly as s increases.
• This tells us that at high temperatures, a greater fraction of the molecules
can be expected to have high speeds than at low temperatures.
• A factor s2 goes to zero as s goes to zero, so the
fraction of molecules with very low speeds will also be very small.
• The remaining factors (the term ) simply ensure that
when we add together the fractions over the entire range of speeds from
zero to infinity, then we get 1.

Maxwell equation: T & M

Maxwell Equation: Experiment
• The Maxwell distribution has been verified experimentally by:

Diffusion and effusion
• Diffusion is the process by which two or more molecules of different substances
mix with each other.
• The atoms of two solids diffuse into each other when the two solids are in
contact, but the process is very slow.
• The diffusion of a solid through a liquid solvent is much faster but mixing
normally needs to be encouraged by stirring or shaking (the process is then no
longer pure diffusion).
• Gaseous diffusion is much faster. It accounts for the largely uniform composition
of the atmosphere, for if a gas is produced by a localized source (such as CO2from
the breathing of animals, O2 from photosynthesis by green plants, and pollutants
(oxides) from vehicles and industrial sources), then the molecules of gas will
diffuse from the source and will be distributed throughout the atmosphere
(CO2+O2+Oxides).

Diffusion and effusion
• The process of effusion is the escape of a gas through a small hole, as in a
puncture in an inflated balloon or tyre (Fig. 1.10).

Diffusion
• The rates of diffusion and effusion of gases increase with increasing
temperature, as both processes depend on the motion of molecules, and
molecular speeds increase with temperature (Maxwell).
• The rates also decrease with increasing molar mass, as molecular speeds
decrease with increasing molar mass (Maxwell). The dependence on molar
mass, however, is simple only in the case of effusion.
• The experimental observations on the dependence of the rate of effusion of a
gas on its molar mass are summarized by Graham’s law of effusion, proposed
by Thomas Graham in 1833:

Diffusion
• At a given pressure and temperature, the
rate of effusion of a gas is inversely
proportional to the square root of its molar
mass:
• Rate in this context means the number (or
number of moles) of molecules that escape
per second.
• The rate of effusion of gases was used to
determine molar mass by comparison of the
rate of effusion of a gas or vapour with that
of a gas of known molar mass.

Molecular collisions
• The average distance that a molecule travels between collisions is called its
mean free path, λ (lambda).
• The mean free path in a liquid is less than the diameter of the molecules,
because a molecule in a liquid meets a neighbour even if it moves only a
fraction of a diameter.
• In gases, the mean free paths of molecules can be several hundred molecular
diameters.
• If we think of a molecule as the size of a tennis ball, then the mean free path
in a typical gas would be about the length of a tennis court.

Collision
• The collision frequency, z, is the average rate of collisions made
by one molecule. Specifically, z is the average number of
collisions one molecule makes in a given time interval divided by
the length of the interval.
• The inverse of the collision frequency, 1/z, is the time of flight,
the average time that a molecule spends in flight between two
collisions:
• For instance, if there are 10 collisions per second, so the
collision frequency is 10 s−1, then the average time between
collisions is 0.1 of a second and 0.1 the time of flight is s).
• The collision frequency in a typical gas is about 109 s−1 at 1 atm
and room temperature, so the time of flight in a gas is typically
1 ns.

Collision
• The rms speed c:
51
• To find expressions for λ and z we need a
decorative version of the kinetic model.
• The basic kinetic model supposes that the
molecules are point-like; however, to obtain
collisions, we need to assume that two ‘points’
score a hit whenever they come within a certain
range d of each other, where d can be thought of
as the diameter of the molecules (Fig. 1.11).
• The collision cross-section, σ (sigma), the
target area presented by one molecule to
another, is therefore the area of a circle of radius
d, so σ = πd2 . When this quantity is built into the
kinetic model, it is possible to show that:

Collision

Thermodynamics
❷

Thermodynamics: First Law
• The basic concepts
• Work, heat, and energy
• The internal energy
• Expansion work
• Heat transactions
• Enthalpy
• Adiabatic changes
• Thermochemistry
• Standard enthalpy changes
• Standard enthalpies of formation
• The temperature dependence of
reaction enthalpies
• The Joule–Thomson effect

System and Surrounding
• The basic concepts:
• The universe is divided into two parts, the system and its surroundings.
1. System is the part of the world exists under study
• Examples: a reaction vessel, an engine, a biological cell, ……
2. Surroundings is the region outside the system and are where we make our
measurements.
• Types of System:
• 1.Open 2. Closed 3.Isolated

Types of Systems
A. Open system: exchange matter and energy with its
surroundings.
B. Closed system: exchange energy with its surroundings
but it cannot exchange matter.
C. Isolated system: exchange neither energy nor matter
with its surroundings.

Work, heat, and energy
A. Work: is done to achieve motion against an opposing force.
B. Energy: is the capacity to do work.
• Types of thermodynamic processes:
1. An exothermic process: is a process that releases energy as heat
into its surroundings (combustion reactions).
• 2. An endothermic process: is a process that absorbs heat from its
surroundings.
(a)When an
endothermic
process occurs in an
adiabatic system,
the temperature
falls;
(a)if the process is
exothermic, the
temperature rises.
(c) When an endothermic
process occurs in a
diathermic container, the
system remains at the
same temperature.
(d) If the process is
exothermic, the process is
isothermal.

The molecular interpretation of heat and work
Transfer heat to
surroundings
stimulates
random motion
Transfer work to
surroundings
stimulates
orderly motion

The internal energy
• Total energy of a system is called its internal energy, U.
• The internal energy is the total kinetic and potential energy of the molecules in
the system.
• The change in internal energy (ΔU): when a system changes from an initial state i
with internal energy Ui to a final state f of internal energy Uf:
• ΔU = Uf − Ui
• The internal energy is a state function, its value depends only on the final and
initial states of the system.
• The internal energy is an extensive property of a system (depends on amount)
and is measured in joules (1 J = 1 kg m2 s−2).
• The molar internal energy(Um) is the internal energy divided by the amount of
substance in a system
• Um = U/n: it is an intensive property (independents on amount) its unit (kJ mol−1).

(a) Molecular interpretation of internal energy
• Modes of motion of a molecule:
1. The ability to translate (the motion of its centre of mass through space)
2. Rotate around its centre of mass
3. Vibrate (as its bond lengths and angles change, centre of mass unmoved).
• Many physical and chemical properties depend on the energy associated with each
of these modes of motion.
• For example, a chemical bond might break if a lot of energy becomes concentrated in
it, for instance as strong vibration.
• Total energy of a monatomic perfect gas free to move in three dimensions is 3
2 𝑛𝑅𝑇
where Um(0) is the molar internal energy at T = 0,

Modes of motions
• When the gas consists of molecules, we need to take into account the effect
of rotation and vibration:
(a) A linear molecule can
rotate about two axes
perpendicular to the line of the
atoms.
(b) A nonlinear molecule can rotate
about three perpendicular axes.

First Law of thermodynamics
• First Law of Thermodynamics: The internal energy of an isolated system is
constant.
• Mathematical Statement of the First Law: ΔU = q + w (q: heat & w: work)
• w and q are positive if energy is transferred to the system as work or heat
and are negative if energy is lost from the system.

Expansion work
• Expansion work: is the work arising from a change in volume.
• (a) The general expression for work:
• dw = -F x dz (opposing force x distance)
• F = pexA
• dw = -pexA x dz (A x dz = dV)
• dw = -pex dV (Expansion work)
• To obtain the total work done when the volume
changes from an initial value Vi to a final value Vf we
integrate this expression between the initial and final
volumes:

Types of Work

• (b) Free expansion: It occurs when pex = 0. w = 0 J
• (c) Expansion against constant pressure:

• (d) Reversible expansion: To achieve reversible expansion we set pex
equal to p (gas pressure).
• (e) Isothermal reversible expansion( T is constant):

• Example 2.1 Calculating the work of gas production
• Calculate the work done when 50 g of iron reacts with hydrochloric acid to
produce FeCl2(aq) and hydrogen in (a) a closed vessel of fixed volume, (b)
an open beaker at 25°C.
• Answer:
• (a) Fixed volume ∆V = 0, so no expansion work is done and w = 0.
• (b) Gas drives back the atmosphere and therefore w = −pexΔV.
• Neglect Vi because Vf (after the production of gas) is so much larger
• ΔV = Vf − Vi ≈ Vf = nRT/pex, where n is the amount of H2 produced.
Therefore,
n can be taken as the amount of Fe
atoms that react. Because the
molar mass of Fe is 55.85 g mol−1

Heat transfer
• The heat transfer as at constant volume is equal to
the change in internal energy of the system.
∆U = qv
• Calorimetry is the measurement of heat transfer.
• Calorimeter constant (C): q = CΔT
• At constant volume C is the slope of the internal
energy with respect to temperature.
• Heat capacity at constant V:

• Heat capacities are extensive properties: 100 g of water, for instance,
has 100 times the heat capacity of 1 g of water.
• The molar heat capacity at constant volume, CV,m = CV/n (intensive
property)
• Specific heat capacity: CV,s = CV/m
Cp − CV = nR
• Enthalpy: H = U + pV
• Energy transferred as heat at constant pressure is equal to the change
in enthalpy of a system: ΔH = qp

• Justification: The relation ΔH = qp
• For a general infinitesimal change :
• H changes from U + pV to
• H + dH = (U + dU) + (p + dp)(V + dV)
• H + dH = U + dU + pV + pdV + Vdp + dpdV
(dpdV is small and is neglected)
• After recognizing U + pV = H
• H + dH = H + dU + pdV + Vdp
and hence that
• dH = dU + pdV + Vdp
• If dU = dq + dw
• dH = dq + dw + pdV + Vdp
• If dw = −pdV
• dH = dq + Vdp
• If the heating occurs at
constant pressure (dp = 0).
• dH = dq

The measurement of an enthalpy change
• An enthalpy change measured by monitoring the temperature change that
accompanies a physical or chemical change occurring at constant pressure.
1. Isobaric calorimeter: used for studying processes at constant pressure.
2. Adiabatic flame calorimeter: used for a combustion reaction to measure ΔT.
3. Bomb calorimeter: used to measure ΔU and convert it to ΔH.
4. Differential scanning calorimeter (DSC)
5. Noncalorimetric technique
• ΔH = ΔU for solids and liquids
• Because solids and liquids have small molar volumes:
• Hm = Um + pVm ≈ Um

• Example 2.2 Relating ΔH and ΔU
• The change in molar internal energy when CaCO3(s) as calcite
converts to another form, aragonite, is +0.21 kJ mol−1. Calculate the
difference between the molar enthalpy and internal energy changes
when the pressure is 1.0 bar given that the densities of the
polymorphs are 2.71 g cm−3 and 2.93 g cm−3, respectively.
• Answer:
• ΔHm = Hm(aragonite) − Hm(calcite)
= {Um(a) + pVm(a)} − {Um(c) + pVm(c)}
= ΔUm + p{Vm(a) − Vm(c)}
• by substituting Vm =M/ρ

• Substitution of the data, using M= 100 g mol−1, gives
= −2.8 × 105 Pa cm3 mol−1 = −0.28 Pa m3 mol−1
• Because 1 Pa m3 = 1 J, ΔHm − ΔUm = −0.28 J mol−1

The enthalpy of a perfect gas (PV=nRT)
• H = U + pV = U + nRT
• ΔH = ΔU + ΔngRT (Δng is the change in moles of gas)
• Example: In the reaction 2 H2(g) + O2(g)→2 H2O(l)
• Δng = (0- 3) = −3 mol.
• Therefore, at 298 K, when RT = 2.48 kJ mol−1, the enthalpy and internal
energy changes taking place in the system are related by:
• ΔHm − ΔUm = (−3 mol) × RT ≈ −7.4 kJ mol−1

• Example 2.3 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 a 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
(373.15 K).
• Answer The enthalpy change is
• ΔH = qp = I ΔΦt = (0.50A) × (12 V) × (300 s) = 0.50 × 12 × 300 J
• In the process H2O(l) → H2O(g): Δng = +1 mol, so
• ΔUm = ΔHm − RT = (+41 kJmol-1) – (8.314x10-3 kJmol-1K-1 x 373.15K)
• = +38 kJ mol−1

The variation of enthalpy with temperature
• Heat capacity at constant pressure:
• ΔH = CpΔT (at constant pressure)
• qp = CpΔT (ΔH = qp)
• Cp,m/(J K−1 mol−1) = a + bT + c/T2 (a,b,c constants)
•
•

Answer: 2.20 kJ mol-1Method: ∆H

Adiabatic changes
• For the reversible adiabatic expansion of a perfect gas, pressure and volume are
related by an expression that depends on the ratio of heat capacities.
• ΔU = CV (Tf − Ti) = CVΔT
• Because the expansion is adiabatic, q = 0
• And because ΔU = q + w,
• So work done during an adiabatic expansion
• wad = CVΔT
• Ti and Tf for reversible adiabatic expansion:
c = CV,m/R

Adiabatic Changes
• Vi and Vf for reversible adiabatic expansion:

Thermochemistry
• Thermochemistry: is a branch of thermodynamics that study of the heat
transferred during the chemical reactions.
• If we know ΔU or ΔH for a reaction, we can predict the heat the reaction can
produce
• Exothermic process: ΔH < 0 (-) Endothermic process: ΔH > 0 (+)
• Standard enthalpy change, ΔHo, the change in enthalpy for a process in which the
initial and final substances are in their standard states:
• Standard state of a substance at a specified temperature is its pure form at 1 bar (1
atm).
• Standard enthalpy of vaporization (ΔHo
vap):

Enthalpies of
physical
changes

Enthalpy is a state function:
• Enthalpy of sublimation:
• single step:
• Multi steps
• Inverse enthalpy:
For example, because 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.

Enthalpies of chemical change
• Thermochemical equation, a combination of a chemical equation and the
corresponding change in standard enthalpy:
• Standard reaction enthalpy, ΔrHo:
• For a reaction of the form 2 A + B→3 C + D
• Example, slandered enthalpy of combustion:

Hess’s law:
• The standard enthalpy of an overall reaction is the sum of the standard enthalpies of the individual reactions into which
a reaction may be divided.
• Example 2.5 Using Hess’s law
• From the following data:
• CH2=CHCH3(g) + H2(g)→CH3CH2CH3(g) ∆H1= −124 kJ mol−1
• CH3CH2CH3(g) + 5 O2(g)→3 CO2(g) + 4 H2O(l) ∆H2 = −2220 kJ mol−1
• H2O(l)→H2(g) + 1
2 O2(g) ∆H3 = +286 kJ mol−1
Calculate the standard enthalpy of combustion of propene.
Answer: combustion reaction we require is
CH2=CHCH3(g) + 9
2O2(g)→ 3CO2(g) + 3H2O(l) (balance CHO)
----------------------------------------------------------------------------------------------------------------------------------------------------------
CH2=CHCH3(g) + H2(g)→CH3CH2CH3(g) ∆H1= −124 kJ mol−1
CH3CH2CH3(g) + 5 O2(g)→3 CO2(g) + 4 H2O(l) ∆H2 = −2220 kJ mol−1
H2O(l)→H2(g) + 1
2 O2(g) ∆H3 = +286 kJ mol−1
--------------------------------------------------------------------------------------------------------------------------------------------------------
CH2=CHCH3(g) + 4.5 O2(g)→ 3CO2(g) + 3H2O(l)
∆H = ∆H1 + ∆H1+ ∆H1 = ( −124 – 2220 +286) kJ mol−1 = -2058 kJ mol−1

Standard enthalpies of formation:
• The standard enthalpy of formation, ΔfHo,
of a substance is the standard reaction
enthalpy for the formation of the compound
from its elements in their reference states:
• The reference state of an element is its most
stable state at the specified temperature and
1 bar.
•

• The Joule–Thomson effect:
• The Joule–Thomson effect is the change in temperature of a gas when it
undergoes isenthalpic expansion.
•

Fig. 2.30 The sign of the Joule–Thomson coefficient, μ,
depends on the conditions. Inside the boundary, the
blue area, it is positive and outside it is negative. The
temperature corresponding to the boundary at a given
pressure is the inversion temperature’ of the gas at that
pressure. For a given pressure, the temperature must
be below a certain value if cooling is required but, if it
becomes too low, the boundary is crossed again and
heating occurs. Reduction of pressure under adiabatic
conditions moves the system along one of the
isenthalps, or curves of constant enthalpy. The inversion
temperature curve runs through the points of the
isenthalps where their slope changes from negative to
positive.
Fig. 2.32 The principle of the Linde refrigerator is shown
in this diagram. The gas is recirculated, and, so long as it
is beneath its inversion temperature, it cools on
expansion through the throttle. The cooled gas cools the
high-pressure gas, which cools still further as it expands.
Eventually liquefied gas drips from the throttle.

Thermodynamics: Second Law
• The direction of spontaneous change
• The dispersal of energy
• Entropy
• Entropy changes accompanying specific
processes
• The Third Law of thermodynamics
• Concentrating on the system
• The Helmholtz and Gibbs energies
• Standard molar Gibbs energies
• Combining the First and Second Laws
• The fundamental equation
• Properties of the internal energy
• Properties of the Gibbs energy

The dispersal of energy
• During a spontaneous change in an
isolated system the total energy is
dispersed into random thermal motion
of the particles in the system.
• Spontaneous: some things happen
naturally.
• The energy of the system that tends
towards a minimum
Fig. 3.2 The direction of spontaneous
change for a ball (system)bouncing on a
floor (surrounding). On each bounce some
of its energy is degraded into the thermal
motion of the atoms of the floor, and that
energy disperses. The reverse process is
nonspontaneous.

Entropy
• Entropy: is a measure of the energy dispersed in a process.
• The Second Law of thermodynamics (entropy):
• The entropy of an isolated system increases in the course of a spontaneous change:
ΔStot > 0 (where Stot is the total entropy of the system and its surroundings)
• Thermodynamically irreversible processes like:
1. Cooling to the temperature of the surroundings
2. free expansion of gases
are spontaneous processes, and hence must be accompanied by an increase in total
entropy.

The thermodynamic definition of entropy
•
• dS is change in entropy, qrev is the heat supplied reversibly and T is
temperature in kelvin.
• The units of entropy are (J K−1). Entropy is an extensive property.
• The units of molar entropy are (J K−1 mol−1). Entropy is an intensive property.
• Calculating the entropy change for the isothermal expansion of a perfect
gas:

• If ΔU = q + w and ΔU = 0 for Isothermal process
• Therefore that qrev= −wrev
• (Isothermal expansion work)
•

Entropy change of the surroundings
• Tsur is constant:
• Psur is constant:
• For an adiabatic change:
qsur = 0
ΔSsur = 0
∆Hsur

The statistical view of
entropy:
• Boltzmann formula for the entropy:
• S = k lnW
• where k = 1.381 × 10−23 J K−1 and W
is the number of microstates
• When W = 1, S =0
• Increasing W leads to increase S

Carnot cycle:

Carnot Cycle:
PA VA Th
PB VB Th
PC VC TC
PD VD TC
• To prove entropy is a state
function we need to show that
the integral of dS is independent
of path.
• To do so, it is sufficient to prove
that the integral of dS around an
arbitrary cycle is zero, for that
guarantees that the entropy is
the same at the initial and final
states of the system regardless
of the path taken between them

Carnot Cycle:
1. Step (1): Reversible isothermal expansion from A to B at Th; dS = +qh/Th,
qh is the energy supplied to the system as heat from the hot source.
2. Step (2): Reversible adiabatic expansion from B to C. No energy leaves the system
as heat, dS = 0 (Th decreases to Tc)
3. Step (3): Reversible isothermal compression from C to D at Tc. Energy is released
as heat to the cold sink; dS = -qh/Th
4. Step (4): Reversible adiabatic expansion from B to C. No energy leaves the system
as heat, dS = 0 (Tc increases to Th)
• The total change in entropy around the cycle:

Heat during the two isothermal stages are:
• Step (1): Step (3):
• From the relation between T and V for reversible adiabatic processes:
• Multiplication of the above gives:
• which, on cancellation of the temperatures, simplifies to:
• With this relation we can write:
• And therefore:
• Since and
• So =
−𝑞𝑐 𝑇ℎ
𝑇ℎ 𝑇 𝑐
+
𝑞𝑐
𝑇𝑐
= 0

Efficiency,η (eta), of a heat engine:
•
• Carnot Efficiency:
• The Second Law of thermodynamics suggests that all reversible engines have
the same efficiency regardless of their construction!!.
• Coefficient of performance:

• The greater the coefficient of performance
and the more efficient is the refrigerator:

Entropy changes accompanying specific processes:
• Entropy change for the isothermal expansion of a perfect gas:
• This change is the negative of the change in the system, so we can conclude
that ΔStot = 0, which is what we should expect for a reversible process.
• On the other hand the isothermal expansion occurs freely (w = 0), then q = 0
(because ΔU = 0). Consequently, ΔSsur = 0, and the total entropy change is
given by:
• In this case, ΔStot > 0, as we expect for an irreversible process.

Entropy of phase transition:
•
• If the phase transition is exothermic (ΔtrsH< 0, as in freezing or condensing)
• If the phase transition is endothermic (ΔtrsH> 0, as in melting or vaporization)
∆trs S = 85 JK-1mol-1taken from table

Entropy of heating
•
•
•

The measurement of entropy
• The entropy of a system at a temperature T is
related to its entropy at T = 0
• At T= 0 Cp,m = aT3

The Third Law of thermodynamics
• Nernst heat theorem: ΔS→0 as T→0
• The Third Law: entropy of all perfect crystalline
substances is zero at T = 0.

The Helmholtz and Gibbs energies
• Helmholtz energy, A, which is defined as: dA = dU − TdS
• Gibbs energy, G , which is defined as: dG = dH − TdS
• The criteria of spontaneous change as: (a) dAT,V ≤ 0 (b) dGT, p ≤ 0
• The criterion of equilibrium, when neither the forward nor reverse process has
a tendency to occur, is (a) dAT,V = 0 (b) dGT, p = 0
• The tendency of a system to move to lower A is due to its tendency to move
towards states of lower internal energy and higher entropy.

Gibbs energy:
• The Gibbs energy (the ‘free energy’) is more common in chemistry than the
Helmholtz energy because, at least in laboratory chemistry, we are usually
more interested in changes occurring at constant pressure than at constant
volume.
• Standard molar Gibbs energies:

The fundamental equation:
• The fundamental equation, a combination of the First and Second Laws, is an
expression for the change in internal energy that accompanies changes in the
volume and entropy of a system.
The equilibrium constant for a reaction is related to its standard reaction
Gibbs energy by: ΔrGo = −RTln K

Chemical
kinetics
❸

The Rates of Chemical Reactions
 chemical kinetics, the study of reaction rates.
 mechanism of reaction, the sequence of elementary steps involved in a reaction.
CHEMICAL KINETICS
Experimental techniques
 real-time analysis, a procedure in which the composition of a system is analysed while
the reaction is in progress.
(1)flow method, a procedure in which the composition of a system is analysed as the
reactants flow into a mixing chamber.
(2)stopped-flow technique, a procedure in which the reagents are mixed very quickly in a
small chamber fitted with a syringe instead of an outlet tube.

(3) flash photolysis, a procedure in which the reaction is initiated by a brief flash of light.
 quenching methods, techniques based on stopping the reaction after it has been allowed to
proceed for a certain time.
(1)chemical quench flow method, a technique in which the reactants are mixed as in the flow
method but the reaction is quenched by another reagent.
(2)freeze quench method, a technique in which the reaction is quenched by cooling the
mixture.

The rates of reactions
(a) The definition of rate
 rate of consumption of a reactant R, –d[R]/dt.
 rate of formation of a product P, d[P]/dt.
 rate of reaction, v = (1/V)dξ/dt where ξ is the extent of reaction.
 rate of homogeneous reaction, v = (1/vJ)d[J]/dt.
 rate of heterogeneous reaction, v = (1/vJ)dσJ/dt.
(b) Rate laws and rate constants
 rate law, the rate as a function of concentration, v = f([A],[B], ...).
 rate constant, the constant k in a rate law.
 hydrogen–bromine reaction: the observed rate law is d[HBr]/dt = kr[H2][Br2]3/2/([Br2] + kr[HBr]).
(c) Reaction order
 reaction order, the power to which the concentration of a species is raised in a rate law of the form
v = [A]a[B]b... .
 first-order reaction, a reaction with a rate law of the form v = kr[A].
 second-order reaction, a reaction with a rate law of the form v = kr[A]2.
 overall order, the sum of the orders a + b +..., in a rate law of the form v = kr[A]a[B]b....
 zero-order rate law, a rate law of the form v = kr.

(d) The determination of the rate law
 isolation method, a procedure in which the concentrations of all the reactants except one
are in large excess.
 Pseudo first-order rate law, v = kr[A] with kr = kr[B]0 by maintaining B in large excess.
S. W. Han et al., Chem. Lett., 2007, 36, 1350.
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 method of initial rates, a procedure in which the rate is measured at the beginning
of the reaction for several different initial concentrations of reactants; v0 = kr [A]0
a
log v0 = log kr + a log [A]0.
Example 21.2

Integrated rate laws
integrated rate law, the integrated form of a rate law for concentration as a function of time.
(a) First-order reactions
 first-order integrated rate law, -d[A]/dt= kr[A] ln([A]/[A]0) = –krt, [A] = [A]0e–krt.
 half life, t1/2 = (ln 2)/kr.
 time constant, the time required for the concentration of a reactant to fall to 1/e of its initial
value,τ = 1/kr.
Example 21.3

(c) Second-order reactions
 second-order integrated rate law, -d[A]/dt= kr[A]2  1/[A] – 1/[A]0 = krt
[A] = [A]0/(1 + krt[A]0).
 half life, t1/2 = 1/kr[A]0.
 half life for nth-order reaction (n>1), t1/2 = 2n-1-1/(n-1)kr[A]0
n-1.
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The temperature dependence of reaction rates
 Arrhenius equation, ln k = ln A – Ea/RT.
 pre-exponential factor (frequency factor), the parameter A in the Arrhenius equation.
 activation energy, the parameter Ea in the Arrhenius equation; the minimum kinetic
energy for reaction during a molecular encounter.
 Arrhenius parameters, the parameters A and Ea.
generalized activation energy, Ea = RT2(d ln k/dT).
 activated complex, the cluster of atoms that corresponds to the region close to the
maximum potential energy along the reaction coordinate.
 transition state, a configuration of atoms in the activated complex which, if attained,
leads to products.

ACCOUNTING FOR THE RATE LAWS
21.6 Elementary reactions
 elementary reaction, a single step in a reaction mechanism.
H + Br2  HBr + Br
 molecularity, the number of molecules coming together to react in an elementary
reaction.
 reaction order, the power to which the concentration of a species is raised in a rate
law of the form v = [A]a[B]b... ; an empirical quantity, and obtained from the
experimental rate law.
 unimolecular reaction, an elementary reaction involving a single reactant molecule.
 bimolecular reaction, an elementary reaction involving the encounter of two
reactant molecules.
CH3I(alc) + CH3CH2O-(alc)  CH3OCH2CH3(alc) + I-(alc)
Mechanism: CH3I + CH3CH2O-  CH3OCH2CH3 + I-, a single elementary step
Rate law: v=kr[CH3I][CH3CH2O-]

Chapter 21: The Rates of Chemical
Reactions21.7 Consecutive elementary reactions
 consecutive first-order reactions, a sequence of first-order reactions.
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Reactions
 rate-determining step, the step in a mechanism that controls the overall rate of the
reaction; commonly but not necessarily the slowest step.

Reactions pre-equilibrium, a state in which an intermediate is in equilibrium with the reactants
and which arises when the rates of formation of the intermediate and its decay back
into reactants are much faster than its rate of formation of products.
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ReactionsExamples of reaction mechanisms
21.8 Unimolecular reactions
 Lindemann–Hinshelwood mechanism, a theory of ‘unimolecular’ reactions.
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Reactions21.9 POLYMERIZATION KINETICS
 stepwise polymerization, a polymerization reaction in which any two monomers
present in the reaction mixture can link together at any time and the growth of the
polymer is not confined to chains that are already forming.
 chain polymerization, a polymerization reaction in which an activated monomer attacks
another monomer, links to it, then that unit attacks another monomer, and so on.
stepwise
polymerization
chain
polymerization

Reactions21.9(a) Stepwise polymerization
 degree of polymerization, the average number of monomer residues per polymer
molecule, n = 1/(1 – p), where p is the average number of monomers per polymer
molecule; n = 1 + krt[A]0.
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Reactions21.9(b) Chain polymerization
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rate of chain
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Reactions
 kinetic chain length, v, the ratio of the number of monomer units consumed per
activated centre produced in the initiation step; v = k[M][I]–½.
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Reactions21.10 PHOTOCHEMISTRY
 primary process, a process in which products are formed directly from the excited
state of a reactant.
 secondary process, a process in which products originate from intermediates formed
directly from the excited state of a reactant.

Reactions21.10(a) The primary quantum yield
 primary quantum yield, ϕ, the number of photophysical or photochemical events
that lead to primary products divided by the number of photons absorbed by the
molecule in the same interval, ϕ = v/Iabs.
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Reactions21.10(b) Mechanism of decay of excited singlet states
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Reactions21.10(c) Quenching
 quenching, shortening of the lifetime of an excited state.
 Stern–Volmer equation, φf,0/φf = 1 + τ0kQ[Q].
equationrSternVolme];[1
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Reactions Modified Stern–Volmer equation
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Example 21.9
Q: Fe(OH2)6
3+

Reactions Three common mechanism of quenching
+--+


+++
++
++
QSorQSQS
QSQS
QSQS
:ransferElectron t
:nsferenergy traResonance
:ondeactivatilCollisiona
21.10(d) Resonance energy transfer
 Förster theory, a theory of resonance energy transfer,
- efficiency (ηT =1-φf,0/φf); ηT  1/R6 [ηT = R0
6/(R0
6 + R6)].
- hdonor  hacceptor

Reactions
 fluorescence resonance energy transfer (FRET), a technique used to measure distances
in biological systems.
ηT = R0
6/(R0
6 + R6)
7.9 nm
Protein rhodopsin
ηT =1-φf,0/φf

Quantum
theory

Quantum theory: introduction and principles
• The origins of quantum mechanics
• Energy quantization
• Wave–particle duality
• The dynamics of microscopic systems
• The Schrödinger equation
• The Born interpretation of the wavefunction
• Quantum mechanical principles
• The information in a wavefunction
• The uncertainty principle
• The postulates of quantum mechanics

The origins of quantum mechanics
• Classical Mechanics: the laws of atoms and subatomic particles motion introduced in
the 17th century by Isaac Newton,
• Quantum Mechanics: towards the end of the 19th century, experimental evidence
accumulated showing that classical mechanics failed when it was applied to particles as
small as electrons, and it took until the 1920s to discover the appropriate concepts and
equations for describing them.
• In classical physics, light is described as electromagnetic radiation: an oscillating electric
and magnetic disturbance that spreads as a harmonic wave through the vacuum.
• Wave displacements that can be expressed as sine or cosine functions.
• The speed of light, c, The constant speed of wave traveling (c ≈ 3 × 108 m s−1)

• Wavelength,λ (lambda), the distance
between the neighbouring peaks of
the wave.
• Frequency, ν (nu), the number of
times per second at which its
displacement at a fixed point returns
to its original value.
• The wavelength is usually measured in
nm, where 1 nm = 10−9 m.
• The frequency is measured in hertz,
where 1 Hz = 1 s−1.
• Wavenumbers (ῡ)are normally reported in reciprocal
centimetres (cm-1).

The electromagnetic spectrum and the
classification of the spectral regions:

Energy quantization:
• Energy can be transferred only in
separate amounts.
• Black-body radiation (19 th-century):
• Black body: is an ideal emitter object
capable of emitting and absorbing all
wavelengths of radiation uniformly.
• A good approximation to a black body is
a pinhole in an empty container
maintained at a constant temperature,
Energy distribution in a blackbody
at several temperatures.
The total energy density (the area
under the curve) increases as the
temperature is increased (as T4).

• The energy density (dE, J m−3) is proportional to the width, dλ
• ρ (rho), the constant of proportionality, is called the density of states (J m−4).
• The total energy density in a region is the integral over all wavelengths:

The electromagnetic vacuum can be
regarded as able to support oscillations
of the electromagnetic field. When a
high frequency, short wavelength
oscillator (a) is excited, that frequency
of radiation is present. The presence of
low frequency, long wavelength
radiation (b) signifies that an oscillator
of the corresponding frequency has
been excited.
The Rayleigh–Jeans law predicts an infinite energy
density at short wavelengths. This approach to
infinity is called the ultraviolet catastrophe.

• Quantization of energy: Planck (1900) found that the permitted energies of
an electromagnetic oscillator of frequency ν are integer multiples of hν :
• E = nhν n = 0, 1, 2, . . .
• h is Plank’s constant = 6.626 × 10−34 J s.
The Einstein temperature, θE = hν/k

Atomic and molecular spectra: Bohr theory
Spectrum of radiation emitted
by excited iron atoms: consists
of radiation at a series of
discrete wavelengths (or
frequencies).
When a molecule changes its state, it does so
by absorbing radiation at definite frequencies.
This spectrum is part of that due to the
electronic, vibrational, and rotational excitation
of SO2 molecules. This observation suggests
that molecules can possess only discrete
energies, not an arbitrary energy.
Spectroscopic transitions, such as those
shown above, can be accounted for if we
assume that a molecule emits a photon
as it changes between discrete energy
levels. Note that high-frequency radiation
is emitted when the energy change is
large.

Wave–particle duality
• The particle character of electromagnetic radiation:
• The photoelectric effect establishes the view that electromagnetic radiation,
regarded in classical physics as wave-like, consists of particles (photons).
• Each particle having an energy hν, if two the energy is 2hν, and so on.
• E = Nhν
• Number of photons (N): P is power in watts.

• Wave function (Φ): the energy required to remove
an electron from the metal.
• The photoelectric effect: is supposed that the
incident radiation is composed of photons that
have energy proportional to the frequency of the
radiation.
• (a) The energy of the photon is insufficient to drive
an electron out of the metal.
• (b) The energy of the photon is more than enough
to eject an electron, and the excess energy is
carried away as the kinetic energy of the
photoelectron (the ejected electron).

The wave character of particles
• Louis de Broglie when, in 1924,
• linear momentum p =mv (with m the mass and v the speed of the particle)
• Δφ= potential difference, m = mass of electron
• e = charge of electron

The Schrödinger equation, 1926
• The mathematical representation of the wave that in quantum mechanics replaces
the classical concept of trajectory is called a wavefunction, ψ (psi).
• The Schrödinger equation is a second-order differential equation used to calculate
the wavefunction of a system.
• The factor V(x) is the potential energy of the particle at the point x
• ħ = h/2π (which is read h-cross or h-bar) is a modification of Planck’s constant with
the value 1.055 × 10−34 J s.

• The Schrödinger equation
1. For one-dimensional systems:
• V(x) is potential energy of the particle and E is its
total energy.
2. For three-dimensional systems:
where V may depend on position and ∇2 (‘del squared’) is:
In systems with spherical symmetry is:
where

• In the general case the Schrodinger equation is written:
• where Ĥ is the hamiltonian operator for the system:
• For the evolution of a system with time, it is necessary to solve the
time-dependent Schrödinger equation:
• Using the Schrödinger equation to develop the de Broglie relation:
• For a freely moving particle in a region where its potential energy V is
constant.
• After writing V(x) = V, we can rearrange into:

•
• cos kx is a wave of wavelength λ = 2π/k, as can be seen by comparing cos
kx with the standard form of a harmonic wave, cos(2πx/λ).
• The quantity E − V is equal to the kinetic energy of the particle, Ek, so k =
(2mEk/ħ2)1/2, which implies that Ek = k2ħ2/2m. Because Ek = p2/2m, it
follows that p = kħ.
• Therefore, the linear momentum is related to the wavelength of the
wavefunction by:
which is the de Broglie relation.

The Born interpretation of the
wavefunction:
• If the wavefunction of a particle has the value ψ at
some point x, then the probability of finding the
particle between x and x + dx is proportional to
|ψ |2dx.
• |ψ |2 is the probability density
For a one-dimensional
system
The probability of finding the particle in the volume
element dτ = dxdydz at some location r is
proportional to the product of dτ and the value of
|ψ |2 at that location.
For three-dimensional
space

The sign of a wavefunction has no direct
physical significance: the positive and
negative regions of this wavefunction
both correspond to the same probability
distribution (as given by the square
modulus of ψ and depicted by the
density of shading).
a0a constant and r the distance from the nucleus of H
atom.
Relative probabilities of finding the electron
inside a region of volume δV:
P ∝ e−2r/a0 δV

Example (Interpreting a wavefunction)Calculate the relative probabilities of finding the
electron inside a region of volume δV = 1.0 pm3, which is small even on the scale of the
atom, located at (a) the nucleus, (b) a distance a0 from the nucleus.
Answer: P ∝ e−2r/a0 δV
(a) At the nucleus, r = 0, so
P ∝ e0 × (1.0 pm3) = (1.0) × (1.0 pm3) = 1
(b) At a distance r = a0 in an arbitrary direction,
P ∝ e−2* a0 /a0 × (1.0 pm3) = (0.14) × (1.0 pm3) = 0.14
Comment: the ratio of probabilities is 1.0/0.14 = 7.1. Note that it is more probable (by a factor
of 7) that the electron will be found at the nucleus. The negatively charged electron is
attracted to the positively charged nucleus, and is likely to be found close to it.

Normalization:
• This freedom to vary the wavefunction by a constant factor means that it is
always possible to find a normalization constant, N, such that the
proportionality of the Born interpretation becomes an equality.
In one dimension, the wavefunction is normalized if:
In three dimensions, the wavefunction is normalized if:

x = r sin θ cos φ, y = r sin θ sin φ, z = r cos θ
r, the radius, ranges from 0 to ∞
θ, the colatitude, ranges from 0 to π
φ, the azimuth, ranges from 0 to 2π
Then
dτ = r2 sin θ drdθ dφ
Spherical polar coordinates

• Example: (Normalizing a wavefunction) Normalize the wavefunction
• Answer:

Quantization of Ψ
• An acceptable wavefunction cannot be
zero everywhere, because the particle
it describes must be somewhere. Since
the energy of a particle is quantized.
• ψ must be:
1. continuous
2. have a continuous slope
3. be single-valued
4. be square-integrable
(d) Unacceptable because it
is infinite over a finite
region
(c) Unacceptable
because it is not
single-valued
(b) Unacceptable
because its slope is
discontinuous
(a) Unacceptable
because it is not
continuous

Quantum mechanical principles
• The information in a wavefunction:
• The Schrödinger equation for a particle of mass m free to move parallel to
the x-axis with zero potential energy is obtained from
by setting V = 0, and is
The solutions of this equation have the form and
A and B are constants

The probability density
Suppose that B = 0
Where is the particle? To find
out, we calculate the probability
density:
• This probability density is independent of x
so, wherever we look along the x-axis, there
is an equal probability of finding the particle
• The same would be true if A = 0; then the
probability density would be |B|2, a
constant.
The square modulus of a wavefunction
corresponding to a definite state of
linear momentum is a constant; so it
corresponds to a uniform probability of
finding the particle anywhere.

Now suppose that the wavefunction A = B. Then
The probability density now has the form
• The probability density periodically
varies between 0 and 4|A|2.
• The locations where the probability
density is zero correspond to nodes in
the wavefunction.
• Node is a point where a wavefunction
passes through zero.

Operators, eigenvalues, and eigenfunctions
with (in one dimension)
Eigenvalue
equation:
(E)Energy is eigenvalue

The construction of operators
• The importance of eigenvalue equations to measure an observable
• Observables is measurable properties of a system, such as the momentum or
the electric dipole moment.
operator (^) for
location along the
x-axis is
multiplication (of
the wavefunction)
by x
operator for linear
momentum parallel to
the x-axis is proportional
to taking the derivative
(of the wavefunction)
with respect to x.

Construct the operator for kinetic energy
Operator for the total energy, the hamiltonian operator:

Hermitian operators

• To say that two different functions ψ i and ψ j are orthogonal means
that the integral (over all space) of their product is zero:

Example: The wavefunctions sin x and sin 2x are eigenfunctions of the
hermitian operator d2/dx2, with eigenvalues −1 and −4. Verify that the two
wavefunctions are mutually orthogonal.
Answer: we integrate the product (sin x)(sin 2x) over all space, which we
may take to span from x = 0 to x = 2π,

The uncertainty principle
Δp is the ‘uncertainty’ in the linear momentum parallel to the
axis q, and Δq is the uncertainty in position along that axis.

Advanced physical chemistry notes

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Advanced physical chemistry notes