Download Unit 5 1 Thermodynamics Intro / Thermochemistry/ Entropy

NAME ____________________________ UNIT 5 (1): INTRO TO THERMODYNAMICS & ENTROPY
Big Idea 5: The laws of thermodynamics describe the essential role of
energy & explain & predict the direction of changes in matter.
Enduring understanding 5.A: Two Essential knowledge 5.A.1: Temperature is a measure of the average kinetic energy of atoms and
systems with different temperatures molecules.
that are in thermal contact will
Essential knowledge 5.A.2: The process of kinetic energy transfer at the particulate scale is referred to in
exchange energy. The quantity of this course as heat transfer, and the spontaneous direction of the transfer is always from a hot to a cold body.
thermal energy transferred from one
system to another is called heat.
Enduring understanding 5.B:
Energy is neither created nor
destroyed, but only transformed from
one form to another.
Essential knowledge 5.B.1: Energy is transferred between systems either through heat transfer or through
one system doing work on the other system.
Essential knowledge 5.B.2: When two systems are in contact with each other and are otherwise isolated,
the energy that comes out of one system is equal to the energy that goes into the other system. The combined
energy of the two systems remains fixed. Energy transfer can occur through either heat exchange or work.
Essential knowledge 5.B.3: Chemical systems undergo three main processes that change their energy:
heating/ cooling, phase transitions, and chemical reactions.
Essential knowledge 5.B.4: Calorimetry is an experimental technique that is used to determine the heat
exchanged/transferred in a chemical system.
Enduring understanding 5.C:
Essential knowledge 5.C.1: Potential energy is associated with a particular geometric arrangement of
Breaking bonds requires energy, and atoms or ions and the electrostatic interactions between them.
making bonds releases energy.
Essential knowledge 5.C.2: The net energy change during a reaction is the sum of the energy required
to break the bonds in the reactant molecules and the energy released in forming the bonds of the product
molecules. The net change in energy may be positive for endothermic reactions where energy is required, or
negative for exothermic reactions where energy is released.
Enduring understanding 5.D:
Essential knowledge 5.D.1: Potential energy is associated with the interaction of molecules; as molecules
Electrostatic forces exist between draw near each other, they experience an attractive force.
molecules as well as between atoms
Essential knowledge 5.D.2: At the particulate scale, chemical processes can be distinguished from physical
or ions, and breaking the resultant
processes because chemical bonds can be distinguished from intermolecular interactions.
intermolecular interactions requires
Essential knowledge 5.D.3: Noncovalent and intermolecular interactions play important roles in many
energy.
biological and polymer systems.
Enduring understanding 5.E:
Essential knowledge 5.E.1: Entropy is a measure of the dispersal of matter and energy.
Chemical or physical processes are
driven by a decrease in enthalpy or Essential knowledge 5.E.2: Some physical or chemical processes involve both a decrease in the internal
an increase in entropy, or both.
energy of the components (ΔH ° < 0) under consideration and an increase in the entropy of those components
(ΔS ° > 0). These processes are necessarily “thermodynamically favored” (ΔG ° < 0).
Essential knowledge 5.E.3: If a chemical or physical process is not driven by both entropy and enthalpy
changes, then the Gibbs free energy change can be used to determine whether the process is
thermodynamically favored.
Essential knowledge 5.E.4: External sources of energy can be used to drive change in cases where the
Gibbs free energy change is positive.
Essential knowledge 5.E.5: A thermodynamically favored process may not occur due to kinetic constraints
(kinetic vs. thermodynamic control).
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Learning objective 5.1 The student is able to create or use graphical representations in order to connect the dependence of potential energy to the
distance between atoms and factors, such as bond order (for covalent interactions) and polarity (for intermolecular interactions), which influence the
interaction strength. [See SP 1.1, 1.4, 7.2, connects to Big Idea 2; Essential knowledge components of 5.A–5.E]
Learning objective 5.2 The student is able to relate temperature to the motions of particles, either via particulate representations, such as
drawings of particles with arrows indicating velocities, and/or via representations of average kinetic energy and distribution of kinetic energies of
the particles, such as plots of the Maxwell-Boltzmann distribution. [See SP 1.1, 1.4, 7.1; Essential knowledge 5.A.1]
Learning objective 5.3 The student can generate explanations or make predictions about the transfer of thermal energy between systems based
on this transfer being due to a kinetic energy transfer between systems arising from molecular collisions. [See SP 7.1; Essential knowledge 5.A.2]
Learning objective 5.4 The student is able to use conservation of energy to relate the magnitudes of the energy changes occurring in two or more
interacting systems, including identification of the systems, the type (heat versus work), or the direction of energy flow. [See SP 1.4, 2.2, connects to
Essential knowledge 5.B.1, 5.B.2]
Learning objective 5.5 The student is able to use conservation of energy to relate the magnitudes of the energy changes when two nonreacting
substances are mixed or brought into contact with one another. [See SP 2.2, connects to Essential knowledge 5.B.1, 5.B.2]
Learning objective 5.6 The student is able to use calculations or estimations to relate energy changes associated with heating/cooling a substance to
the heat capacity, relate energy changes associated with a phase transition to the enthalpy of fusion/vaporization, relate energy changes associated
with a chemical reaction to the enthalpy of the reaction, and relate energy changes to PΔV work. [See SP 2.2, 2.3; Essential knowledge 5.B.3]
Learning objective 5.7 The student is able to design and/or interpret the results of an experiment in which calorimetry is used to determine the change
in enthalpy of a chemical process (heating/cooling, phase transition, or chemical reaction) at constant pressure. [See SP 4.2, 5.1, 6.4; Essential
knowledge 5.B.4]
Learning objective 5.8 The student is able to draw qualitative and quantitative connections between the reaction enthalpy and the energies involved in
the breaking and formation of chemical bonds. [See SP 2.3, 7.1, 7.2; Essential knowledge 5.C.2]
Learning objective 5.9 The student is able to make claims and/or predictions regarding relative magnitudes of the forces acting within collections of
interacting molecules based on the distribution of electrons within the molecules and the types of intermolecular forces through which the molecules
interact. [See SP 6.4; Essential knowledge 5.D.1]
Learning objective 5.10 The student can support the claim about whether a process is a chemical or physical change (or may be classified as both)
based on whether the process involves changes in intramolecular versus intermolecular interactions. [See SP 5.1; Essential kn owledge 5.D.2]
Learning objective 5.11 The student is able to identify the noncovalent interactions within and between large molecules, and/or connect the shape and
function of the large molecule to the presence and magnitude of these interactions. [See SP 7.2; Essential knowledge 5.D.3]
Learning objective 5.12 The student is able to use representations and models to predict the sign and relative magnitude of the entropy change
associated with chemical or physical processes. [See SP 1.4; Essential knowledge 5.E.1]
Learning objective 5.13 The student is able to predict whether or not a physical or chemical process is thermodynamically favored by determination of
(either quantitatively or qualitatively) the signs of both ΔH° and ΔS°, and calculation or estimation of ΔG° when needed. [See SP 2.2, 2.3, 6.4;
Essential knowledge 5.E.2, connects to 5.E.3]
Learning objective 5.14 The student is able to determine whether a chemical or physical process is thermodynamically favorable by calculating the
change in standard Gibbs free energy. [See SP 2.2; Essential knowledge 5.E.3, connects to 5.E.2]
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Learning objective 5.15 The student is able to explain how the application of external energy sources or the coupling of favorable with unfavorable
reactions can be used to cause processes that are not thermodynamically favorable to become favorable. [See SP 6.2; Essential knowledge 5.E.4]
Learning objective 5.16 The student can use Le Chatelier’s principle to make qualitative predictions for systems in which cou pled reactions that share
a common intermediate drive formation of a product. [See SP 6.4; Essential knowledge 5.E.4, connects to 6.B.1]
Learning objective 5.17 The student can make quantitative predictions for systems involving coupled reactions that share a common intermediate,
based on the equilibrium constant for the combined reaction. [See SP 6.4; Essential knowledge 5.E.4, connects to 6.A.2]
Learning objective 5.18 The student can explain why a thermodynamically favored chemical reaction may not produce large amoun ts of product
(based on consideration of both initial conditions and kinetic effects), or why a thermodynamically unfavored chemical reaction ca n produce large
amounts of product for certain sets of initial conditions. [See SP 1.3, 7.2; Essential knowledge 5.E.5, connects to 6.D.1]
There are two huge divisions in this unit, per the AP curriculum. The following ideas are not limited to one
division or the other. In fact, many ideas overlap and interconnect. I list some of these ideas, only to share with
you some of the topics with which we are about to concern ourselves:
Thermodynamics
Thermochemistry: deals with energy absorbed or
released in chemical reactions and/or physical
changes
Conditions Involved in
Predicting Thermodynamically
Favored Reactions
concerned with
(rather arbitrarily)
concerned with
(rather arbitrarily)
 Energy / Law of the
Conservation of Energy
 Heat (q)
 heating / cooling curves
 phase changes
 average kinetic energy
 potential energy
 enthalpy of reaction,
combustion, formation,
fusion, vaporization






Hess’s Law
State Function(s)
Enthalpy
Entropy
Free Energy (Gibb’s)
Equilibrium / Acid-Base
Theory
 Kinetics
The SI unit for energy is a derived unit: the joule (J) … The joule
corresponds to the amount of kinetic energy possessed by a 2
kilogram object moving at a speed of 1 meter per second. Using
the equation for kinetic energy: K.E. = ½ mv2 …..
1 J = ½ (2 kg) 1 m
1s
2
… 1 J = 1 kg m2s-2
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To begin our work, we must re-familiarize ourselves with energy … which is the ability to do work or to create a
change….
Internal Energy is the sum of P.E. and K.E.
can be divided
The kinetic energy of chemicals
is expressed in the temperature,
and motion of the chemicals
into 2 really big categories
Potential Energy
can be converted to
defined as
Kinetic Energy
defined as
energy of position of
between species relative
to an assumed standard
K.E. = ½ mv2 with mass in kilograms
energy of motion (possessed by
an object in motion)
examples of kinetic energy
examples of potential energy
Chemical
Bond Energy
P.E. is associated with the chemical
energy of bonds.
Electromagnetic Spectrum [Light, Thermal, Microwave];
& then there is/are: Sound energy, motion,
Between bonded species, there is a
bond length or distance and thus the
energy possessed by a chemical is
associated closely with potential
energy … in this case, it’s called,
chemical energy
There are 3 important terms over which we must be most concerned, with respect to energy:
1) Entropy
2) Enthalpy
3) Free Energy (or Gibbs Free Energy)
Very often, we are most concerned with the *change in entropy (ΔS), change in enthalpy (ΔH) and
change in Gibbs Free Energy (ΔG).
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Working Vocabulary: These will be refined over the course of our work …and are simplified, here.
Heat: (a.k.a. thermal energy) the transfer of energy between objects having different temperatures. The
transfer continues until the objects are at the same temperature, and in thermal equilibrium.
Work: force acting over distance …. w = F x d
or w = -P∆V where gases are involved.
Temperature: proportional to the average kinetic energy of the molecule (KEavg)
Heat ‘em up, speed ‘em up.
Enthalpy (H): *flow of energy (heat exchange) at constant pressure, when 2 systems are in contact with each
other.
Enthalpy of reaction: ∆Hrxn amount of heat released (negative value) or absorbed (positive value) by a
chemical reaction at constant pressure (kJ/molrxn or kJ …[an older unit])
Enthalpy of combustion:
the heat absorbed or released by burning (usually with O2) in kJ/molrxn.
Combustion reactions produce oxides of that which is combusted (fuel)
Enthalpy of formation: ∆Hf heat absorbed or released (transferred) when 1 mol of compound is
formed from its elements in their standard states in kJ/molrxn
Enthalpy of fusion ∆Hfus
heat absorbed to melt (overcome IMFs) 1 mol of solid to liquid at a
constant temperature (the melting point) ( kJ/molrxn )
Enthalpy of vaporization:
Entropy (S): *the measure of the dispersal of matter and energy … the change may be a +ΔS or a -ΔS
System: area of the universe on which we are focusing (e.g. the experiment)
Surroundings: everything outside of the system
Endothermic: +ΔH in kJ/molrxn
Exothermic:
-ΔH in kJ/molrxn
Gibbs Free Energy: criteria for determining the thermodynamic favorability (ΔG = ΔH -TΔS)
State Function: a property of matter, INDEPENDENT of pathway (e.g. internal energy, mass, volume,
entropy, ΔH ….)
Standard Conditions: Lab conditions of 1 atm, 25°C and 1 M (when solutions are involved). All of this is known by
adding the symbol ° to G, H or S. Thus, given ∆H° you automatically know the
pressure and temperature conditions which apply to the measurement. The symbol
° is pronounced as naught
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TRY THIS DIAGNOSTIC! Recall: ΔH = mcΔT, ΔH = mHfusion, ΔH = mHvap
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I) Thermodynamics is all about energy… it is technically the study of the transformations of energy from
one form to another…...And as Dr. Peter Atkins writes; “Energy is the basis of civilization”.
The laws of thermodynamics govern chemistry and life. They explain why reactions take place and let us
predict how much heat reactions release and how much work they can do. Thermodynamics plays a role in
every part of our lives. For example, the energy released as heat can be used to compare fuels, and the
energy resources of food are used to assess its nutritional value. (Chemical Principles Quest for Insight p. 235)
A) *Heat and work
are equivalent means of transferring energy between a system and its
surroundings. The total energy of an isolated system is constant. The enthalpy change (ΔH)
for a process is equal to the heat released at constant pressure.
B) There are, 4 Laws of Thermodynamics
Zeroth law of thermodynamics – When two thermodynamic systems are each in thermal equilibrium
with a third, then they are in thermal equilibrium with each other. …

Its big implication is that: * Energy flows from high to low
First law of thermodynamics – Energy can neither be created nor destroyed, but it can change forms.
In any process, the total energy of the universe remains the same. For a thermodynamic cycle the net
heat supplied to the system equals the net work done by the system.

The First law is all about the Law of the Conservation of Energy … and it is essentially our
“bookkeeper” (Hey! the work bookkeeper has 3 sets of double letters in a row!)

It addresses the familiar pieces of chemistry – as covered in the first year course …
How much energy is involved in a change? (±∆H… change in enthalpy)
Does energy flow into or out of the system? (endothermic / exothermic)
What form does the energy finally assume? (EMS  chemical or PE  KE)

Its big implication is that: *The energy of the universe is constant
Second law of thermodynamics – The entropy of an isolated system not in equilibrium will tend to
increase over time, approaching a maximum value at equilibrium …

Its big implication is that: *The entropy of the universe is increasing
Third law of thermodynamics – The entropy of a perfect crystal at 0 Kelvin is zero

It is insanely rare to find a perfect crystal …so entropy values (even for elements) are rarely 0

Its big implication is that: It tells us that the absolute entropy of a substance can be
determined at any temperature, higher than zero K …and this is the defense as to why Gibb’s
Free Energy (G) and the Enthalpy (H) of an element = 0…. but entropy (S) does NOT
equal zero!!!!
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SUMMARY:
Law
Zeroth
4 Laws of Thermodynamics
Interpretation
Energy flows from zones of high
temperature to zones of lower temperature
Implication(s)
When a hot metal is placed in contact with a cooler metal,
energy will flow from the hotter to the cooler, until a
thermal equilibrium is established and both metals are at
the same temperature.
Temperature measures average kinetic energy of atoms and
molecules.
The process of kinetic energy transfer at the molecular level
is referred to as heat transfer.
First
Energy can neither be created nor
destroyed. It can only do 2 things …It can
be transferred, or be converted to another
form.
The energy of the universe is constant.
e.g. Calculate ΔE for a system
undergoing an endothermic process in
which 15.6 kJ of heat flows and where
1.4 kJ of work is done on the system.
ΔE =
ans. 17.0 kJ
Second
The entropy of the universe is increasing
For 2 isolated systems in contact with each other, the energy
that comes out of one system is equal to the energy that goes
into the other system. The combined energy of the two
systems remains fixed.
q
+ w
energy heat transfer
work
Energy is transferred between systems whether through heat
transfer or through one system doing work on the other
system.
Chemical systems undergo 3 main processes that change
their energy: heating/cooling, phase changes, and chemical
reactions.
The universe is constantly increasing the dispersal of matter
and energy …. This is an expanding universe.
There is no such thing as 100% conversion of chemical
energy into useable work.
Third
The entropy of a perfect crystal = 0
We can deduce entropy at any temperature above 0 K.
Check out:
1) Crash Course #17: heat and work: https://www.youtube.com/watch?v=GqtUWyDR1fg
2) State Functions https://www.youtube.com/watch?v=lL2CXpVti34
3) State Functions (more) https://www.youtube.com/watch?v=SEapOFaYDlA
4) Crash Course #18: enthalpy: https://www.youtube.com/watch?v=SV7U4yAXL5I
5) Heat and work (alt) https://www.youtube.com/watch?v=SfHHZxGpwr8
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C) Systems, Work and Heat
1) System: the region in which we are interested: such as a flask of gas, a beaker of acid, a
reaction mixture, or a muscle fiber (Atkins. p. 236). Everything else, such as a water
bath, in which a reaction is immersed (in another beaker) is referred to as the
surroundings.
The surroundings are where we make observations of the energy transferred into our
out of the system. The system and the surroundings jointly make up the universe …
but we understand that only a very localized portion of the universe is affected
significantly.
a) Open system: allows for the exchange of matter and energy between the chemicals
and the surroundings …. for example, automobile engines and our bodies!
b) Closed system: … possesses a fixed amount of matter, but it can exchange energy
with the surroundings. For example, a closed system could be a chemical cold pack
used for treating athletic injuries/inflammation.
c) Isolated system: … has no contact with its surroundings, so that there is no exchange
of matter or energy with the surroundings …. e.g.) a rigid, thermally insulated vessel
holding the system …. such as a hot liquid inside a sealed vacuum flask
2) Work… The process of achieving motion against an opposing force… all forms of work can
be thought of as equivalent to the work of raising a weight against the pull of gravity. … e.g
pushing an electrical current through a circuit. W = Force opposing x distance
or work = -P∆V where gases are involved
a) unit = Joule where 1 J = 1 kg∙m2∙s-2
(1 J = 1 N∙m in physics)
3) Energy is the capacity to do work … or to create a change …OR to supply heat(!)
a) Internal Energy (E or U) … the total store of energy in a system …
b) The absolute value of internal energy cannot be measured (we would need to know
the energies of every atom from their electrons to their nuclei)… hence we tend to
measure the change in internal energy …
4) When work is done ON a system, (assuming no other changes), the internal energy is
increased
5) A system can do 2 types of work: expansion and non-expansion work
a) expansion work involves a change in volume of the system
e.g. a gas expanding in a cylinder, moving a piston … hence w= -P∆V
b) non-expansion work does not involve a change in volume of the system.
e.g. the chemical reaction in a battery.
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6) Heat: The second law is a bit more involved… As written by Patrick Coffey in
Cathedrals of Science: …. The second law implies that
“While work can always be converted to heat, only a portion of the heat of a system,
even under ideal circumstances, can be converted into work in a cyclic process – like a
steam engine. The first law is sometimes stated as “You can’t get something for
nothing” and the second law as “You can’t even break even”
a) Heat and temperature are very different and to grasp the 2nd Law … we need
to understand a bit more about “heat”.
i) temperature is a property that reflects the random motions of the species of a
substance.
ii) HEAT is the energy transfer (frictional heating) between two objects due to
a difference in temperature. While we often talk about an object being
contained by an object in reality. It is not.
We call the transfer, q … as in q = mc∆T
yeah, “q” = heat
iii) Heat is the sibling of work, and heat is often referred to as a form of
energy (but “heat” per se, is not on the electrochemical spectrum [infrared
energy is … but not, heat] Heat is equivalent to the amount of energy that
is transferred between the reacting chemicals and surroundings.
We limit the work produced during a chemical reaction, by measuring the
progression of the reaction in rigid, non-flexible containers (a calorimeter). This
limits the work by limiting the movement of the container atoms, over a distance.
So, in our measurements (calorimetry) we tend to refer to the energy exchange
as heat, (which represents the change in internal energy) …and this is at the heart
of the confusion.
Intrinsically we recognize that “heat” is a transfer (a verb) not a specific type of
energy. e.g. We need to "heat up" a cup of coffee. It’s really heating up out
here”.
And, most experts don’t even like talking about it as a verb. See: http://hyperphysics.phy
astr.gsu.edu/hbase/thermo/heat.html …. According to the late Dr. Mark Zemansky:
Don't refer to the "heat in a body", or say "this object has twice as much heat as that
body". The First Law of Thermodynamics identifies both heat and work as methods
of energy transfer which can bring about a change in the internal energy of a system.
After that, neither the words work or heat have any usefulness in describing the final
state of the system - we can speak only of the internal energy of the system.
b) Essentially the 2nd Law states that there is no such thing as 100% conversion of
chemical energy into useable work (energy)
c) Thus, one big implication of the 2nd Law is that …. While the energy of the
universe is constant … not all of it remains useable… as the entropy of
universe is increasing. Hence: Energy is constant…Entropy is NOT
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II) Reading: Edited from Peter Atkin’s Chemical Principles: A Quest for Insight (p 287)
The first law of thermodynamics tells us that, if a reaction takes place, then the total energy of the
universe (the reaction system and its surroundings) remains unchanged. But the first law does not
address the questions that lie behind the word, “IF”.
Why do some reactions have a tendency to occur, whereas others do not? That is, why are some
reactions thermodynamically favorable while others are not?
Why does anything happen at all? To answer these deeply important questions about the world
around us, we need to take a further step into thermodynamics and learn more about energy,
beyond the fact that it is conserved…
Hence, the second law of thermodynamics is the key to understanding why one chemical reaction
has a natural tendency to occur but another one does not. We apply the second law using the
very important concepts of entropy and Gibbs free energy. The third law (which does not play as
large a role in AP Chemistry) is the basis of the numerical values of these two quantities. The
second and third laws jointly provide a way to predict the effects of changes in temperature and
pressure on physical and chemical processes. They also lay the thermodynamic foundations for
discussing chemical equilibrium (Big Idea 6).
We will from this point on be working to blend …to draw together concepts related to the first law
(particularly enthalpy) and work. We need to be aware that molecules can occupy only a series
of discrete energy levels and so possess only certain energies.
A) Thermodynamically Favored Processes (spontaneous processes) and Entropy
Thermodynamically Favored Processes
definition
those processes which occur
without outside intervention
may be slow or fast
may be physical changes OR
chemical reactions
e.g cooling a block of hot metal
the expansion of gas into a vacuum
all about direction of procession not speed
e.g it is thermodynamically favored
for a diamond to revert to graphite
(sp3 to sp2), but it is a very slow
process …. Kinetics and Thermodynamics are both required for a
full explanation
some require to be initiated (activation energy)
“favored” is concerned with the tendency to occur.
Whether the process is realized depends on its rates
(kinetics)
e.g. it is thermodynamically favored
for syrup to flow out of a bottle, but
this can be very slow, at exceedingly
low temperatures
A process is spontaneous (thermodynamically favored) when it has a tendency to occur without being driven
by an external influence (excluding initiation); spontaneous changes need not be fast.
B) A classic list of thermodynamically favored changes might include:

a ball rolls down a hill – but not “spontaneously” up the hill (recall… spontaneously is an
older term for thermodynamically favored … and indicates the lack of outside
intervention)
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
a gas fills a vessel uniformly – it never spontaneously collects at one end of a container

an iron nail exposed to moisture and oxygen rusts – but iron(III) oxide does not
spontaneously revert back to iron and dioxygen

heat transfer occurs from an object of greater temperature (hot) to one of lesser
temperature (cooler). It does not flow from cool to hot spontaneously (if it did we would
not need to have motors on our refrigerators!)

wood burns exothermically and spontaneously to produce carbon dioxide and water ….
but carbon dioxide and water, when heated together, do not produce wood.

At standard pressure and temperatures below 0°C water freezes (spontaneously), in a
thermodynamically favored process, and at temperatures superior to 0°C, ice melts
to water.
As we focus upon the thermodynamically favored reaction we must develop a more fluid grasp of
enthalpy and entropy … we will then tie the two together with Gibbs Free Energy
III) Entropy: A state function which is a measure of disorder of the matter and energy… It captures how the
energy is dispersed throughout the microstates of the matter.
A) The entropy of an isolated system increases in any spontaneous (thermodynamically favored)
process.
1) The common denominator to each of the spontaneous scenarios in IIB) turns out
to be *entropy … which is the dispersion of energy (and very, very often,
linked to the matter) throughout the matter of the system. Many see entropy as measure
of the molecular randomness or disorder … but that focuses only upon the matter
A slightly more sophisticated view is the disorder of the energy….(not just the matter …)
B) Entropy is closely linked to the second law of thermodynamics …it is another way of interpreting
the law … While it is a measure of the energy dispersion, it may be seen also in a roundabout way
*indicating the energy NOT available for work …
Remember this as Gibb’s Free Energy accounts for this inaccessible energy …when determining the “Free Energy”
to do work…
1) symbol for entropy (S) and symbol for the change in entropy (±∆S)
2) unit: J∙K-1 or J/molrxn•K …. NOTE…Enthalpy is often given in kJ … beware!!!!
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3) entropy is a thermodynamic function that describes the number of arrangements (positions
&/or energy levels) that are available to a system existing in a given state. These probable
positions and/or energy levels are called microstates
a)
a microstate is a single possible arrangement of the positions and kinetic energies of
the molecules when the molecules are in a specific thermodynamic state.
i) e.g. A mixture of gases will exhibit a greater entropy than a similar volume
of a single gas … because the mixture has a greater “randomness” or ability
to exist in a greater number of variations of molecular order or energy level
4) According to Zumdahl “The key concept is that the more ways a particular state can be
achieved, the greater is the likelihood (probability) of finding that state. In other words,….
Nature spontaneously proceeds toward the states that have the highest probabilities of
existing” (p. 790) … Keep thinking then, that the driving force is …. well …. probability!
a) Predicting the entropy change of a system is based on physical evidence
(NMSI Entropy and Free Energy)

The greater the dispersal of matter and/or energy in a system, the larger the
entropy

The entropy of a substance always increases as it changes from solid to liquid to
gas.

When a pure solid or liquid dissolves in a solvent, the entropy of the substance
increase (there are exceptions …e.g. carbonates …in water they bring more order,
and decrease entropy …due to the extensive water of hydration)

When a gas molecule escapes from a solvent, the entropy increases

Entropy generally increases with increasing molecular complexity (species with
more electrons and bonding have greater complexity) e.g KCl < CaCl2

Reactions increasing the number of moles often increase entropy … and those in
which the number of moles of product is less than the number of reactants,
entropy decreases.
So remember … The greater the number of arrangements, the greater the entropy of the system.
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b) Many at this level argue positional probability as a driving force (the reason)
but I urge you to consider it from the point of view of the energy.
For instance: Compare a solid to its liquid phase and to its gaseous phase …What can
we say about the entropy? *solid < liquid < gas
As some would argue, there is a greater positional probability in the gas phase, when
compared to the solid phase (and they would be correct … but I think that’s limited)
Consider this … The solid essentially is a single state … We do not even speak about
molecules when considering solid para-dichlorobenzene (for instance), because discrete
entities do not really exist … hence the energy of the system is essentially dispersed and
shared by the entire (single micro) state. When converted to a gas, the energy is
dispersed to individualized molecules, comprising millions of microstates … Hence the
energy has been dispersed across a greater level of disorder, and is at a greater
entropy.
When it comes to the formation of solutions, positional entropy works … but again,
I urge you to consider it really from the point of view of the dispersion of energy.
The tendency to mix is due to the increased volume available to the particles of each
component of the mixture. Granted: when two liquids are mixed, the molecules of
each liquid has more available volume and thus more available positions.
Perhaps so, but the energy of the system is allowed to be dispersed across a greater
number of microstates as those solutions are mixed. Using the argument of energy
is a powerful means of visualizing entropy, since entropy is all about energy levels.
Fascinating: A little reading on fusion and microstates from 1885 …The second chapter is entitled “Equilibrium
in Dissociation,” and attempts to dispel the older belief, based on the caloric theory, that heat is a repulsive
force which acts in opposition to chemical affinity. The reason compounds eventually dissociate upon heating
is not because the repulsion of the added heat finally overwhelms the attractions of the internal bonds, nor
because, in keeping with the newer mechanical theory of heat, the increasing violence of the intramolecular
vibrations finally break the internal bonds. Rather it is because the net increase in the number of
independently moving species formed upon dissociation is more able to effectively dissipate the system’s
internal kinetic energy. George Downing Liveing Chemical Equilibrium The result of the Dissipation of Energy 1885 Deighton, Bell and Co.
London. As cited in Bulletin for the History of Chemistry Volume 38 No. 1 2013 p. 41-2
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5) A gas has more energetically equivalent configurations than a solid, because the gas
has more ways to distribute its energy than a solid.
a) A solid’s energy is essentially found in its vibrational motion between its molecules
….it is essentially a single microstate …few options of re-arranging the energy….
b) A gas may exhibit more “places” to put the energy … there is rotational energy or
translational energy (straight-line motion)
c) Ludwig Boltzmann came up with its expression as: S = k lnW where:
k = Boltzmann constant (The universal gas constant / Avogadro’s number)
thus k = R/Na = 1.38 x 10-23 J/K
W = the number of possible individual microstates that can result in an overall
arrangement.
d) A chemical system proceeds in a direction that increases the entropy of the universe –
it proceeds in a direction that has the largest number of energetically equivalent ways
to arrange its components (Tro 820)… its matter, but most importantly … its energy.
http://www.wikipremed.com/image.php?img=010304_68zzzz131150_30902_68.jpg&image_id=131150
6) Recall: The average kinetic energy of the molecules of an ideal gas is directly proportional
to the absolute (Kelvin) temperature of the gas.
a) Consequence: The higher the temperature, the faster the molecules move and the
greater the kinetic energy they possess (K.E. = ½ mv2) …with mass
in kilograms.
Additionally, as we saw in the Maxwell-Boltzmann distribution
graph, systems at higher temperatures have a broader distribution
of molecular speeds.
481
b) Real gases can exhibit 3 kinds of complex motion …
Vibrational
Rotational
Translational (a molecule moving in one direction)
http://catalog.flatworldknowledge.com/bookhub/4309?e=averill_1.0-ch18_s04
i) solids have the least, rotational and translational motion.
http://education-portal.com/academy/lesson/properties-of-water.html
gases can exhibit the highest form of all three:
http://wateronthemove.wikispaces.com/Forms+of+Water
ii) Consequence: These more complex expressions of kinetic energy, are the
means by which molecules can store more energy, as
temperature increases.
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TRY THIS: Right from your Honors Notes … Identify the change in positional entropy as + or a –
Basic entropy interpretations:
kJ + H2O(l)  H2O(g)
S = + l<g
H2O(g)  H2O(s) + kJ
S = - g>s
kJ + CH3OH(l)  CH2OH(g)
C6H6(l)  C6H6(s) + kJ
Slightly more:
complex
2 Fe2O3(s) + 3 C(s) 
4 Fe(s) + 3 CO2(g)
*S
= +
S = -
S = +
2 AgI(s) + F2(g)  I2(s) + 2 AgF(s)
S = -
4 Li(s) + O2(g) 
S = -
2 H2O2(l) 
2 Li2O(s)
S = +
2 H2O(l) + O2(g)
2 KClO3(s)  2 KCl(s) + 3 O2(g)
S = +
4 Cr(s) + 4 K(s) + 7 O2(g)  2K2Cr2O7(s)
S = -
Advanced: Darn It
2 N2O3(g)  2 N2(g) + 3 O2(g)
There's no phase change
N2(g) + 3 H2(g)  2NH3(g)
S = +
2 SO2(g) + O2(g)  2SO3(g)
S = +
2 PaI5(s)  2 Pa(s) + 5 I2(s)
S = +
S = -
TRY THIS: Right from your Honors Notes …Again! The answers to the multiple choice are after question
#7 of this TRY THIS section
1) Given:
C6H4Cl2(s) + kJ → C6H4Cl2(g)
Which statement describes this change?
(1) It is endothermic, and entropy decreases.
(2) It is endothermic, and entropy increases.
(3) It is exothermic, and entropy decreases.
(4) It is exothermic, and entropy increases.
2) Which process is accompanied by a decrease in entropy?
(1) Deposition of carbon dioxide
(2) Evaporation of water
(3) Allowing a gas to expand into a larger volume
(4) Heating a balloon filled with a gas
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3) At a pressure of 101.3 kilopascals and a temperature of 373 K, energy is lost from a sample of water vapor
causing the sample to change from the gaseous phase to the liquid phase. This phase change is represented
by the equation: H2O(g) → H2O(l) + kJ
Explain, in terms of particle arrangement, why entropy decreases during this phase change.
Be sure to use
entropy’s definition … and what is going on in the phase change (per the number of microstates) in your answer….
*Entropy is associated with the dispersion of matter and the energy of that matter. As water
vapor condenses the molecules decrease their relative position (a loss of positional entropy)
to each other, and in the process, some energy is lost from the system to the surroundings. In the
phase change, the number of possible microstates (or possible ways the matter and energy may exist)
is more limited, than in the gas phase, the entropy is decreasing
4) Systems in nature tend to undergo changes
(1) lower energy and less disorder
(2) lower energy and more disorder
(3) higher energy and less disorder
(4) higher energy and more disorder
5) Which of these biological reactions represents one that decreases the entropy of a cell?
(1) digestion
(2) hydrolysis
(3) catabolism
(4) dehydration reactions
6) Even though the process is endothermic, snow can sublime. Which tendency in nature accounts for this
phase change?
(1) a tendency toward greater entropy
(2) a tendency toward greater energy
(3) a tendency toward less entropy
(4) a tendency toward less energy
water
7)
Study the dissolving of potassium nitrate in water:
KNO3(s) + 34.89 kJ
K1+(aq) + NO3-1 (aq)
Given all you know, thus far identify which best describe the process
(1)
(1)
(3)
(4)
It is endothermic,
It is endothermic,
It is exothermic,
It is exothermic,
physical
chemical
chemical,
physical
and the
and the
and the
and the
entropy increases
entropy decreases
entropy decreases
entropy increases
TRY THIS answers 1) 2 3 see online 2) 1 4) 2
5) 4 6) 1
7) 1
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Before going on … what do you know (about)…..?
___entropy is about the dispersion of energy amidst the matter
___entropy has the symbol S
___the change in entropy at standard conditions has the symbol ΔS°
___the greater the number of molecular arrangements, the greater the entropy
___entropy changes with phase
___a substance in its solid phase has a lower entropy than in its gas phase
___ the particles of a gas have a greater entropy because of greater vibrational,
rotational, and translational options “putting” the energy in more “places”
___when moles decrease, concerning reactants to products, entropy decreases
___a crystal of AlCl3 has a greater entropy than a crystal of NaCl, due to
the greater number of electrons
___the units of entropy are often different from the units of enthalpy
___how to determine a change in entropy given a balanced equation?
___the first three laws of thermodynamics?
C) Calculating Standard Molar Entropy (∆S°) at Standard Conditions
1) When calculating molar entropy change: For AP Chemistry the granddaddy of
all entropy equations is:
∆S°rxn = ∑∆S°products - ∑∆S°reactants
a) Entropy is a state function (it is not pathway-dependent). We can calculate the
entropy change for a given process, by taking the difference between the
standard molar entropy values (S°) of products and reactants. It is an extensive
property … so it depends upon the number of moles
2) Absolute values for entropy can be assigned …. And hence, into our story, enters
the third law of thermodynamics.
a) The entropy of a perfect crystal of an element at 0 K (absolute zero) represents the
lowest possible entropy … the internal arrangement of a perfect crystal is absolutely
regular at absolute zero … for molecular motion virtually ceases.
b) Essentially the above is a re-working of the third law of thermodynamics.
i) Caveat: “A perfect crystal at 0 K is an unattainable ideal, taken as a standard
but never actually observed” (Zumdahl 803)
3) As the temperature of a perfect crystal is increased, the random molecular motion increases
increasing the “disorder” … in the distribution of the energy among slightly differing
microstates, increases (due to the more complex motions) Thus entropy increases.
a) S for a perfect crystal is 0 at 0 K … the entropy value for a substance at a particular
temperature can be calculated by knowing the temperature dependence of entropy.
485
b) Study the table from your text and complete the following observations:
Standard Molar
Entropies at 298 K
Substance S° (J/K∙mol)
H2(g)
130.6
N2(g)
191.5
O2(g)
205.0
H2O(g)
188.8
NH3(g)
192.5
CH3OH(g)
237.6
C6H6(g)
269.2
69.9
H2O(ℓ)
126.8
CH3OH(ℓ)
172.8
C6H6(ℓ)
Li(s)
29.1
Na(s)
51.4
K(s)
64.7
Fe(s)
27.23
FeCl3(s)
142.3
NaCl(s)
72.3
Observations of the Data:
1) The enthalpy of formation = 0, but standard molar entropies of
elements at the reference temperature are NOT zero
proof / examples _____________________________
2) The standard molar entropies of gases are greater than or less
than those of liquids and solids.
proof: compare _____________________________
3) Standard molar entropies generally increase or decrease with
increasing molar mass
proof: compare _____________________________
4) Standard molar entropies generally * increase
with
increasing number of atoms (& bonds) in the formula of a
substance
(it * increases
with molecular complexity).
proof: compare _____________________________
TRY THIS! Given all of the information below, predict a numerical range or value for the S° (standard
molar entropy) for butane. Then using one or more of the observations from the table,
explain your reasoning.
Substance Name
Formula
methane
ethane
propane
butane
CH4(g)
C2H6(g)
C3H8(g)
C4H10(g)
Value for S°
J/K∙mol
186.3
229.6
270.3
????
* The standard molar enthalpy for butane should be somewhere very close to 310 J/K ∙mol. Based upon
the data there appears to be an increase of 40 J/K for the increase in 1 carbon and 2 hydrogen atoms
(which references the general formula for the alkanes of CnH2n+2). This prediction should be borne out
by the observations from the table, because observation 3 & 4 reference in increasing value of S° for
more massive molecules and those molecules exhibiting greater complexity. Butane’s greater mass
is relatively unassailable. I submit that it also has a greater complexity due to the possible rotational
possibilities given the availability of C – C free rotation. This rotation increases the possible number of
microstates, relative to the other examples and thus should exhibit a greater S°
486
TRY THIS: This uses that Granddaddy of an equation… ∆S°rxn = ∑∆S°products - ∑∆S°reactants
Calculate the entropy change at 25°C, in J/molrxn•K) from the following standard values, for the reaction:
Entropy (J/mol•K)
SO2(g)
248.1
O2(g)
205.3
SO3(g)
256.6
2 SO2(g) + O2(g) → 2 SO3(g)
Hint 1: * This sort of problem is a
repeating theme … You need to
calculate the entropy (using the
coefficients of the balanced equation)
Hint 2:* Multiply each mole value by
the entropy value for each substance.
Get the total for the product and sum
total for the reactants.
ans: -188.3 J/ mol•K
TRY THIS: Calculate ∆S° at 298 K for the reaction: N2(g) + 3 H2(g) → 2 NH3(g)
(B&L p. 829)
Table of Standard Entropy Values
Substance
NH3(g)
N2(g)
H2(g)
S° (J/K∙mol)
192.5
191.5
130.6
*∆S°rxn = ∑S°products - ∑S°reactants
∆S°rxn = 2(192.5 J/ K∙mol) - [(191.5 J/K∙mol) + 3 (130.6 J/K∙mol)]
385 - (191.5 + 391.8) = -198.3 J/K∙mol
ans: -198.3 J/K∙mol
*Note that given our qualitative work the negative value of ∆S is right in line with this response … As we can see that there
is a decrease in the number of moles of gas … or rather, a decrease in the number of microstates …. All we have done in the
above example is put a number to that decrease in entropy.
TRY THIS! Calculate the standard entropy change, ∆S°, for the following reaction at 298 K.
Al2O3(s) + 3 H2(g) → 2 Al(s) + 3 H2O(g)
Table of Standard Entropy
Values
Substance
S° (J/K∙mol)
H2O (g)
188.8
Al(s)
28.32
Al2O3(s)
51.00
H2(g)
130.6
* ∆S°rxn = 3(188.8 J/K mol) + 2 (28.32 J/K mol) – (51.00 J/K mol) + 3(130.6 J/K mol)
566.4
+ 56.64
623.04
-
- 442.8
442.8
= 180.24 J/K mol
ans: 180.24 J/K mol
487
D) Entropy changes as they apply to phase changes: ΔSsurroundings = qrev
T
or rather … ΔSsurr = -ΔHsys
Tsurr
1) this assumes: a constant temperature (as in a phase change), where
 q is the energy transferred as heat
 T is the Kelvin temperature at which the transfer occurs.
 The “rev” signifies that the energy must be transferred reversibly in the sense that any a
reversible process is one that can take place in either direction …like a phase change!
The equation tells us that when a lot of energy is transferred as heat (a large qrev), a lot of
disorder is stirred up in the system and we expect a correspondingly big increase in entropy.
For a given transfer of energy, we expect a greater change in disorder when the temperature
is low than when it is high. The arriving energy stirs up the molecules of a cool system, which
the molecules are already moving vigorously. (Think of a sneeze attracting attention in a
library …but very little on a noisy street) Atkins p. 289
2) For instance: Calculate the ∆Ssurr for melting 1 mol of ice at 273 K. The constant for
melting water is ∆Hfusion = 6.01 x 103 J/mol. (It’s positive because melting is
an endothermic change … also since entropy is extensive, it depends upon the # of mols)
∆Ssurr = qrev which = -∆Hfusion = -(1 mol)( 6.01 x 103 J/mol) = -22.0 J/K
T
T
273
3) This sort of problem blends enthalpy and entropy … and there is an issue with units!!
You want to have common units … it would be advisable to change the enthalpy value
from kJ/mol to J/mol
a) Also, be aware that ∆Ssurr is * temperature dependent.
b) Because of this dependency, when you are calculating the change in entropy of the
surroundings … when work is done on the surroundings (as in an exothermic reaction,
from the point of view of the chemicals) … the change in entropy (for the surroundings)
is positive.
When the enthalpy indicates that the reaction is endothermic, that means that the energy
transfer was from the surroundings to the chemical system …. causing a greater order in
the surroundings or a negative entropy value (for the surroundings)
488
TRY THIS! When a process is exothermic, the entropy of the surroundings …
a) always increases
b) always decreases
c) increases or decreases depending upon the process
Defend your answer:* a) ∆Ssurr is temperature dependent, and in an exothermic reaction the system
transfers energy to the surroundings, causing and increasing in temperature, and thus an increase
in the disorder (the number of microstates) of the matter and energy distribution of the surroundings.
Entropy must increase. OR: ∆Ssurr= ∆H/T and in an exothermic process, ∆H must be negative for the
system BUT positive for the surroundings, hence hence the value for ∆Ssurr must be positive.
TRY THIS: Calculate ΔSsurr for the following reaction at 25°C and 1 atm
(NMSI: Entropy and Free Energy)
In the metallurgy of antimony, the pure metal is recovered via different reactions, depending on the composition
of the ore. For example, iron is used to reduce antimony in sulfide ores.
Sb2S3(s) + 3 Fe(s) → 2 Sb(s) + 3 FeS(s)
ΔH = -125 kJ/mol
*ΔH in joules = 125,000 J/mol
* T in K = 298 K
*ΔSsurr = ΔH/T = 125,000/298
Why is the ΔSsurr a positive value?
Hint 1: *While this is not a phase change,
you are asked for ΔSsurr so use the
equation ΔSsurr = ΔH/T
Hint 2:* Enthalpy is given in kJ …but
entropy is in J … convert that enthalpic
value to Joules… and check that
temperature … is it in Kelvin?
ans: 419 J/mol∙K
* The reaction is exothermic (based upon the change in enthalpy of -125 kJ/mol. This means that the
chemicals lost energy while the surroundings gained it. Thus work was done on the surrounding,
causing an increase in the entropy …hence ΔSsurr is positive.
489
TRY THIS: Calculate ΔSsurr for the following reaction at 25°C and 1 atm
Given the same scenario as above in the first “TRY THIS”… the antimony ore can be reduced
using carbon … according to:
Sb4O6(s) + 6 C(s) → 4 Sb(s) + 6CO(g) ΔH = 778 kJ/mol
ans: -2.61 x 103 J/mol∙K
*ΔSsurr = -ΔHsys
Tsurr
*= -778,000 J/mol
298 K
Why is the ΔSsurr a negative value? *The reaction is endothermic, according to the positive value
for the change in enthalpy. This means that the reaction chemicals absorbed energy from the
surroundings. Hence the surroundings experience a decrease in entropy …signified by a negative
value.
Assignment Give it a whirl … Get to Trivedi 18.9
Summary:
Hold On! Take a minute out and look at what this packet has covered:
Aside from the summary a few pages back … take a whirl at asking:
1) Do you know your basic definitions for such important terms as: Enthalpy, Entropy, and System?
2) Do you know how to work with and when to work with: q = mcΔT
3) Do you know that q = mcΔT is most closely associated with enthalpy?
4) Do you know the condition(s) under which q = ΔH?
5) Do you know how to calculate ΔH, using the stoichiometry of a balanced reaction?
6) Do you know the 3 types of motion: vibrational, rotational, and translational & how they mesh with
entropy?
7) Do you know how to calculate standard molar entropy (∆S°) using a balanced equation and data?
8) Do you know how to calculate the change in entropy when give ΔH?
9) Have you taken a good hard look at the learning objectives at the front of this packet?
490