Chemistry and Biochemistry Questions

1A rigorous treatment of classical thermodynamics requires multivariate calculus and
is beyond the scope of this course. Take Chem 110A, if you’re interested!
(Biochemistry majors are required to take 110A)
The discipline of thermodynamics was developed in the 19th century, partly as an
outgrowth of efforts to maximize the efficiency of steam engines.
Classical thermodynamics involves some pretty abstract concepts, like heat, work,
energy, entropy, and free energy. A grasp of these concepts is important although not
easy to attain. Don’t worry if you find some of the ideas confusing. You’re in good
company!
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Albert Einstein (author), Paul Arthur, Schilpp (editor). Autobiographical Notes. A
Centennial Edition. Open Court Publishing Company. 1979. p. 31 [As quoted by Don
Howard, John Stachel. Einstein: The Formative Years, 1879-1909 (Einstein Studies,
vol. 8). Birkhäuser Boston. 2000. p. 1]
https://en.wikiquote.org/wiki/Thermodynamics
https://secondlawoflife.wordpress.com/2009/07/12/what-einstein-thought-aboutthermodynamics/
Q: What do we mean by classical thermodynamics? What other kind of
thermodynamics is there?
Einstein’s non-scientific legacy:
•
Came to the U.S. as a a refugee from Nazi Germany in 1933.
•
Joined the NAACP and campaigned for African American civil rights
•
Considered racism to be America’s “worst disease”
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The original painting is 9 feet tall by 6 feet wide!
She was only 13 when they married.
From Wikipedia article on Antoine Lavoisier: “She was to play an important part in
Lavoisier’s scientific career—notably, she translated English documents for him,
including Richard Kirwan’s Essay on Phlogiston and Joseph Priestley’s research. In
addition, she assisted him in the laboratory and created many sketches and carved
engravings of the laboratory instruments used by Lavoisier and his colleagues for
their scientific works. Madame Lavoisier edited and published Antoine’s memoirs
(whether any English translations of those memoirs have survived is unknown as of
today) and hosted parties at which eminent scientists discussed ideas and problems
related to chemistry.”
Madame Lavoisier is believed to have been a student of David and her artistic skills
show in the sketches she made of her husband’s 200 scientific instruments. The
purpose of several of the instruments portrayed in the painting can be found in this
article: https://www.sciencehistory.org/distillations/revolutionary-instrumentslavoisiers-tools-as-objets-dart
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The glass balloon on the floor was used to determine the masses of various gases. It
would be evacuated and weighed and then filled with a gas and weighed again. Then
after correcting for temperature and pressure, the mass of the gas could be
determined. This was critical for Lavoiser’s experiments to prove that mass was
always conserved in chemical reactions.
The glass cylinder is filled with mercury and then inverted in a basin of mercury just
like Torricelli’s barometer. Then Lavoisier added water and iron which floated to the
top. As the iron reacted with water and became oxidized, a gas was released which
he sampled and characterized. He named this gas hydrogen (which means born of
water). It was proof that water was made of two gases: hydrogen and oxygen.
Here’s a very readable report of a paper read by Lavoisier to the Royal Academy of
Sciences in 1783 (https://web.lemoyne.edu/~giunta/ea/LAVEAUan). It describes
Lavoisier’s experiments to demonstrate that water is not an element, but a
combination of two gases. Interestingly, despite the fact that he is the primary
disprover of the of the phlogiston theory, he nonetheless refers to oxygen as
dephlogisticated air!
The instrument immediately to Lavoisier’s left is a device for capturing oxygen
produced during the reduction of mercury oxide or lead oxide. This oxygen was
critical for his experiments demonstrating the nature of combustion, in which
Lavoisier disproved the phlogiston theory of combustion and showed that
combustion consumed oxygen and produced water and carbon dioxide.
Lavoisier came up with the theory that animal respiration was a slow form of
combustion and showed that human beings consumed oxygen and generated carbon
dioxide.
It’s probably safe to say that these discoveries would not have been made without
the efforts of Madame Lavoisier. Her artistic ability and her knowledge of language
were essential to his efforts.
He was a tax collector and supervised gunpowder manufacturing for Louis XVI and
was guillotined in the French revolution.
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Reinstallation of David’s portrait in The Met’s European Paintings galleries in 2020,
following conservation treatment and technical analysis. Photo credit: Eddie Knox ©
Oxford Films, 2020
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From Wikipedia: “The phlogiston theory is a superseded scientific theory that
postulated the existence of a fire-like element called phlogiston contained within
combustible bodies and released during combustion. The name comes from
the Ancient Greek φλογιστόν phlogistón (burning up), from φλόξ phlóx (flame). The
idea was first proposed in 1667 by Johann Joachim Becher and later put together
more formally by Georg Ernst Stahl. Phlogiston theory attempted to explain processes
such as combustion and rusting, now collectively known as oxidation, and was
abandoned before the end of the 18th century following experiments by Antoine
Lavoisier and others. Phlogiston theory led to experiments which ultimately
concluded with the discovery of oxygen.”
Combustion of paraffin: 𝐶!” 𝐻#$ 𝑠 + 47𝑂% (𝑔) → 31𝐶𝑂% (𝑔) + 32𝐻% 𝑂(𝑙)
In contrast to the phlogiston theory of combustion, in which a substance is released,
our modern understanding is that a substance (oxygen) is consumed. How does the
experiment shown in the video provide support for the idea that something is
consumed?
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From Wikipedia: “Lavoisier was a powerful member of a number of
aristocratic councils, and an administrator of the Ferme générale. The Ferme
générale was one of the most hated components of the Ancien Régime because of
the profits it took at the expense of the state, the secrecy of the terms of its
contracts, and the violence of its armed agents. All of these political and economic
activities enabled him to fund his scientific research. At the height of the French
Revolution, he was charged with tax fraud and selling adulterated tobacco, and was
guillotined.”
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The whole subject of thermodynamics is built upon these three laws.
They are postulates, i.e., we accept them as true without proof.
The predictions to which they lead are always correct, thus validating these laws.
We’ll spend the first half the course learning about these laws. Most of the second
half of the course is about learning to apply these laws.
Before you can understand these laws, you need to know what we mean by a
thermodynamic “system”.
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Types of systems:
•
Open system: The boundaries of the system permit exchange of matter and
energy with the surroundings.
•
Closed system: The boundaries of the system permit the exchange of energy,
but not matter.
•
Isolated system: The boundaries of the system prevent the exchange of both
energy and matter. The universe as a whole is an isolated system.
Systems can also be categorized as homogeneous (all one phase) or
heterogeneous (a mixture of two or more phases).
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The only two ways to transfer energy into or out of a system are through work and
heat. As you’ll be seeing, this statement is a form of the first law of thermodynamics.
Transfer of energy as heat occurs when there is a temperature difference between
the system and the surroundings.
Transfer of energy as mechanical work occurs when there is a pressure difference
between the system and the surroundings.
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Clicker 1A.1
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Clicker 1A.2
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Thermodynamic properties are also sometimes called macroscopic variables
because they describe the state of a system on “human scale”. They’re determined
collectively by the behavior of all the atoms and molecules in the system. In
contrast, a microscopic variable is a property of a single atom, ion, or molecule
such as its position, velocity, or charge.
We’ll see that for a system consisting of a fixed mass of a single pure substance,
there are only two independent state functions, meaning if we know the values of
any two state functions the system is fully defined.
Broadest definition of equilibrium: “Thermodynamic properties of the system are
not changing with time”. This does mean that things are not happening at the
microscopic level. For example, chemical reactions are still occurring in a system
at chemical equilibrium. But the forward and reverse reactions are occurring at
equal rates.
Types of equilibrium include:
Mechanical equilibrium: Pressure is everywhere the same and unchanging.
Thermal equilibrium: Temperature is everywhere the same and unchanging.
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Phase equilibrium: Amount of substance in each phase is unchanging.
Chemical equilibrium: Chemical composition is everywhere the same and
unchanging.
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Changes in state due to thermodynamic processes can be:
• Physical: includes changes in phase, volume, temperature, and pressure
• Chemical: changes resulting from chemical reactions
The change in a thermodynamic property X can also be computed by summing up a
series of infinitesimal changes in X, in others words, by determining the definite
integral of dX from the initial state to the final state.
𝒇𝒊𝒏𝒂𝒍
𝒔𝒕𝒂𝒕𝒆
∆𝑿 =
d 𝒅𝑿 = 𝑿𝒇 − 𝑿𝒊
𝒊𝒏𝒊𝒕𝒊𝒂𝒍
𝒔𝒕𝒂𝒕𝒆
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For a cyclic process, the change in the value of a state function is 0. So one way to
prove that a function is a state function is to show that its change is 0 for any
cyclic path.
We can only assign a value to a path function for a thermodynamic process, and
only if we know the exact path by which the process occurs.
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Clicker 1A.4
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Clicker 1A.5
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Because of the simplicity of ideal gases, we’ll frequently use ideal gases to
illustrate important concepts in thermodynamics.
While no gas perfectly satisfies these criteria, at sufficiently low pressure and
high temperature all gases exhibit nearly ideal behavior.
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An equation of state is an equation that relates two or more thermodynamic
properties to one another.
The ideal gas law is an example of a “limiting law” meaning it becomes more and
more valid as certain limits are approached. In this case, the limits are 0 pressure and
infinite temperature. At low enough pressure and high enough temperature, every
gas comes close to obeying this law. “Low enough pressure” and ”high enough
temperature” vary depending on the gas. But for many gases, near ideal behavior is
observed when the temperature is 0 ˚C or above and the pressure is 5 atm or below.
The ideal gas law was originally derived empirically (from experimental observation),
and is based on the following three empirical laws.
”
Boyle’s law: 𝑉 ∝ . (𝑎𝑡 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑇 & 𝑛)
”
Charles’ law: 𝑉 ∝ . (𝑎𝑡 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑇 & 𝑛)
Avogadro’s hypothesis: 𝑉 ∝ 𝑛 (𝑎𝑡 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑇 & 𝑃)
/0
Combining these gives:𝑉 ∝ .
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But it can also be proved through theory, starting with the criteria listed on the
previous slide.
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The torr was originally defined to be the same as one millimeter of mercury at 0 ˚C.
But due to subsequent redefinition, it is now slightly less than a millimeter or
mercury.
Evangelista Torricelli (18th century Italian physicist)
1 torr = 133.322 (368 421 052 631 578 947)∞ Pa This is an infinitely long decimal in
which the last 18 digits repeat forever.
1 mmHg = 133.322 387 415 Pa (exactly)
They differ by less than 1 part in seven million. So it’s almost always safe to use them
interchangeably.
Why mercury?
• It’s dense (13.56 g/cm3) so a shorter column of liquid is adequate. A water
barometer would have to be ~10 m long.
• Low vapor pressure (0.0017 Torr at 25 ˚C). So the space above the liquid mercury is
close to a perfect vaccum at all reasonable temperatures. In contrast, water has
significant vapor pressure, and this vapor pressure varies significantly with
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temperature (e.g., 23.8 torr at 25 ˚C, 31.8 torr at 30 ˚C). So for a given air pressure,
the height of a column of water would vary significantly with temperature.
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Clicker 1A.6
L kPa = (10–3 m3)(103 Pa) = Pa m3 = (N/m2)m3 = Nm = J
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Clicker 1A.7
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Clicker 1A.8
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Clicker 1A.9
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Energy is a very abstract concept. Energy is not something that we can see or touch.
Even Feynman struggled with this concept. The definition that he came up with
seems to be “it’s something that is conserved”. In other words, it’s a mathematical
quantity that does not change as the universe changes. There are many different
forms of energy and for each form, we can come up with a different equation to
calculate the amount of energy in that form. The thing that makes all these things
energy is that the sum of these amounts never changes.
Feynman 1965 Nobel Prize in Physics for work in quantum electrodynamics
I = moment of inertia = sum of the mass of each particle times the square of its
distance from the axis of rotation (kg · m2)
𝜔 = angular velocity (radians/sec)
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http://www.feynmanlectures.caltech.edu/I_04.html
Feynman could not really define energy except to say that it is something that is conserved
– so the definition of energy is inextricably linked to the conservation of energy.
Feyman’s definition of energy: There is a fact, or if you wish, a law, governing all natural
phenomena that are known to date. There is no known exception to this law—it is exact so
far as we know. The law is called the conservation of energy. It states that there is a certain
quantity, which we call energy, that does not change in the manifold changes which nature
undergoes. That is a most abstract idea, because it is a mathematical principle; it says that
there is a numerical quantity which does not change when something happens. It is not a
description of a mechanism, or anything concrete; it is just a strange fact that we can
calculate some number and when we finish watching nature go through her tricks and
calculate the number again, it is the same. (Something like the bishop on a red square, and
after a number of moves—details unknown—it is still on some red square. It is a law of this
nature.)
• When the amount of one type of energy decreases, the amount of another type of
energy must increase by the same amount.
• Any apparent violation of the conservation of energy is a consequence of our inability
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to identify all the different types of energy that are being interconverted.
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By particles, I mean atoms, ions, molecules, and subatomic particles.
The internal energy is the energy required to create the system.
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Clicker 1B.1
Gravitational potential energy = mgh
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Often, in physics, the opposite sign convention is used.
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Expansion work is a kind of mechanical work.
Any work that doesn’t involve a change of volume is called “isochoric”. Examples of
isochoric work:
1) Stirring
2) Electrical
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Expansion work is a kind of mechanical work.
Any work that doesn’t involve a change of volume is called “isochoric”. Examples of
isochoric work:
1) Stirring
2) Electrical
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•
In the next few slides, you’ll explore reversibility by comparing reversible
and irreversible paths for the expansion of a gas.
•
Later, we’ll come back to to the concept of reversibility when we learn about
entropy, because the definition of entropy invokes reversibility.
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Numerical integration by the rectangle rule
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Clicker 1B.3
What kind of walls surrounded the system in experiments 1 and 2?
How would you determine the work in experiment 1 in units of J?
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Thermal motion has translational, rotational, and vibrational components.
Thermal energy is part of the internal energy.
Heat is NOT the same as temperature!! You can have a temperature change
without heat transfer and vice versa.
Adiabatic compression or expansion of a gas leads to a temperature change without
heat transfer. Isothermal compression or expansion of a gas leads to heat transfer
without temperature change.
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Clicker 1B.4
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Examples
Ø
Heat capacity of a system consisting of 100 g of air at 298 K ≅ 100. J∙K–1
Ø
Heat capacity of a system consisting of 100 g of liquid water at 298 K ≅ 418
J∙K–1
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Clicker 1B.5
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Ways to deliver a known amount of heat to the calorimeter:
1) Carry out a chemical reaction (such as a combustion reaction) in the calorimeter
with a known heat of reaction. In this case, qreaction = -qcal
2) Perform a known amount of electrical work on the calorimeter by passing a
known amount of current at a known voltage for a specific period of time through
a resister in the calorimeter: electrical work = (V)(I)t. As we’ll be seeing soon,
work and heat are equivalent and so the amount of electrical work is the same as
the amount of heat delivered to the calorimeter.
Why the minus sign in the equation relating q to Ccal?
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Clicker 1B.6
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In the 18th and early 19th centuries, it was commonly believed that heat was a selfrepellent fluid or gas called caloric. Because it was self-repellent it flowed from
warmer bodies, which contain more caloric, to cooler bodies, which contain less
caloric. Caloric was assumed to be a substance that could neither be created not
destroyed implying that heat is conserved. Experiments such as that shown here, in
which it was discovered that:
1) Thermodynamic systems could be warmed without the transfer of heat and
2) The change in temperature when performing work on a system surrounded by
adiabatic walls was proportional to the amount of work and
3) That the proportionality constant was always the same regardless of the type of
work,
led to the conclusion that heat was not conserved but that energy was conserved.
Thus, the caloric theory was ultimately replaced by the modern notion that heat and
work are ways of transferring energy. Once a system gains energy through work or
heat, there is no way to know if that energy was transferred into the system by work
or by heat.
This also provides a nice demonstration that work and heat are path functions, but
that the sum of the work and the heat equal the change in a state function.
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We can’t prove the first law any more than Euclid could prove that there is a straight
line connecting any two points.
We can’t prove the first law from a more fundamental postulate.
•
It’s one of the premises upon which thermodynamics is built.
•
The success of thermodynamics as a field demonstrates the truth of the first
law.
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We can’t prove that internal energy is a state function. But the first law depends on
this. If internal energy were not a state function, then it should be possible to find
cyclic paths that result in an increase in internal energy. This would indicate that
perpetual motion machines were possible. But we know from vast experience that
this is impossible. As Leonardo Da Vinci said “Oh, ye seekers after perpetual motion,
how many vain chimeras have you pursued? Go and take your place with the
alchemists!”
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Clicker 1B.7
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Clicker 1B.8
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Clicker 1C.1
3
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Clicker 1C.2
Before you make your choice, consider the following: Why is a system at constant
pressure more resistant to a temperature change, than a system at constant volume?
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In monoatomic ideal gases, the only way to store energy is through translational
motion.
In molecular gases, rotational and vibrational modes of motion are also available to
store energy. This leads to a greater heat capacity (more heat is needed to raise the
temperature by a given amount.)
This shows that temperature of a gas is really a function of translational kinetic
energy.
Heat capacity of water vapor as a function of temperature and pressure:
https://www.epjconferences.org/articles/epjconf/pdf/2016/09/epjconf_efm2016_02133.pdf
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The Wikipedia entry on molar heat capacity gives a good explanation of modes of
motion and degrees of freedom: https://en.wikipedia.org/wiki/Molar_heat_capacity
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https://www.ems.psu.edu/~bannon/moledyn.html
For linear molecules the moment of inertia about the molecular axis is very small. So
rotation about that axis is not a useful way of storing energy.
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A degree of freedom is a way of storing energy. The more ways that a substance has
of storing energy, the more resistance it has to temperature change and therefore
therefore the greater the heat capacity.
In this discussion, we’re really only considering the component of the internal energy
that is due to the random motions of the molecules and atoms including
translational, rotational, and vibrational motion. We learned earlier that this is called
thermal motion and the associated energy is called thermal energy. There other
contributions to the internal energy, such as bond energy and nuclear energy, but, as
long as the temperature is not too high, these remain constant as energy is
transferred into the system as heat, i.e., they don’t contribute any degrees of
freedom. So they don’t represent ways of storing energy as it is transferred into the
system. In other words, they don’t contribute to the heat capacity.
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C includes all the other components of the internal energy such as bond energy and
nuclear energy. These are all roughly independent of temperature.
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Clicker 1C.3
https://chemistry.stackexchange.com/questions/49602/how-does-carbon-dioxidehave-a-degree-of-freedom-of-6
There are always 3N total independent degrees of freedom for a molecule, where N is
the number of atoms. These come about because when each atom moves, it has
three independent degrees of freedom: its position in each of the x, y, and z
directions.
Now, having independent degrees of freedom for each atom isn’t all that useful.
Instead, we can make combinations of different degrees of freedom. The important
thing when doing so is that the number of independent degrees of freedom are
preserved: it’s just that what a particular degree of freedom does to the atoms
changes.
The standard breakdown of degrees of freedom subtracts out global movement in
each of the three directions. So you have 3N total degrees of freedom, but you can
set aside 3 of them as translation of the whole molecule in each of the x, y and z
directions, leaving (3N-3) degrees of freedom.
Likewise, it’s standard to subtract out the whole molecule rotation. For most larger
molecules, there’s three different degrees of rotational freedom: rotation around
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each of the x, y, and z directions. But for linear molecules like CO2, one of those
rotations (around the axis of the molecule) doesn’t actually change the position of
the atoms. Therefore it’s not a “degree of freedom” which counts against the 3N
total. So while for non-linear molecules there are (3N-3-3) = (3N-6) degrees of
freedom which are independent from the global rotational and translational ones, for
linear molecules there are (3N-3-2) = (3N-5) degrees of freedom which are
independent from the global rotational and translational ones. — A quick clarification.
The reason why we ignore this rotation is not because the center of mass doesn’t
move. The center of mass doesn’t move for any of the global rotations: in the typical
assignment of degrees of freedom the axis of rotation goes through the center of
mass. Instead, the reason the rotation is ignored is that none of the atoms move due
to the “rotation”.
So since CO2 has three atoms and is linear, it has 3*3 – 5 = 4 degrees of freedom
which are independent of the global rotation and translation. We call these
the vibrational modes. There’s different ways you can decompose them, but the most
useful one is 1) the symmetric stretch (both oxygens going out and in at the same
rate while the carbon sits still), 2) the asymmetric stretch (the carbon going back and
forth while the oxygens sit more-or-less still), and 3&4) two different out-of plane
bending modes. (If the molecule is aligned along the x-axis, one each where the
carbon moves in the y and z directions.)
That’s it for CO2: any movement you make – any position you put the atoms in – can
be thought of as a combination of those nine movements (three global translations,
two global rotations, and four internal vibrations/bending).
Data on the heat capacity of hydrogen:
https://pubs.acs.org/doi/pdf/10.1021/ja01390a004
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Clicker 1C.4
For lighter gases, vibrational activation doesn’t occur until much higher temperatures.
For example, for H2, it’s much greater than 1000K.
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Clicker 1C.5
In monoatomic ideal gases, the only way to store energy is through translational
motion.
In molecular gases, rotational and vibrational modes of motion are also available to
store energy. This leads to a greater heat capacity (more heat is needed to raise the
temperature by given amount.
This shows that temperature of a gas is really a function of translational kinetic
energy.
Slight departure of Xenon from equiparition value could be due to experimental error.
But it could also reflect non-ideal behavior.
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How can we tell by looking at the definition of H that it must be a state function?
U is the energy required to create the system, while PV is the energy required to
make room for the system!
The word enthalpy comes from the Greek word for warmth or heat.
Note that for processes occurring at constant pressure in condensed systems (liquids
or solids), ∆H is almost the same as ∆U. Why should this be?
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During the phase transitions, all the heat is being used to weaken or break noncovalent bonds. None is being used to increase the kinetic energy and so no
temperature increase occurs.
Why are the slopes for each phase different?
How can we tell from looking at this curve that it is the melting curve of water?
What accounts for the large heat of vaporization of water?
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The temperature at the standard state is not standardized. In the case of this table
each value was determined at a different temperature (the temperature of the phase
change). Frequently, standard values are determined at 25 ˚C.
He has 2 electrons, while Ar has 18.
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1
The above equation is true because enthalpy is a state function. We only need to now
the initial and final enthalpies. The path between the initial and final states is
irrelevant.
2
A thermochemical equation is just the chemical equation together with an indication
of the reaction enthalpy.
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Ø
If a reaction leads to a net increase in the amount of gas, the system
expands and pushes against the surroundings. So some of of the
system’s internal energy is used up to do work on the surroundings,
and therefore ∆U < ∆H. Ø If a reaction leads to a net decrease in the amount of gas, the system contracts as the surroundings do work on the system. This increases the internal energy of the system, and therefore ∆U > ∆H.
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Remember that a state function is any property that only depends on the current
state of a system and not the path taken to reach that state.
Strictly speaking, Hess’s law refers to the enthalpy change of a chemical reaction, but
the principle applies to the change in any state function associated with any physical
or chemical transformation.
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Of course it’s often not possible to conduct the reaction with all the reactants and
products in the standard state. For example, for a gas phase reaction between two
gases, you would generally have to mix the two gases together before they would
react. But then you are not at standard state because the reactants are not pure. So
standard reaction enthalpy often describes an idealized situation that cannot actually
be achieved. But it still provides useful insights into the nature of the reaction.
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Based on this definition, it must be true that the ∆Hf˚ of an element in its most stable
form is 0.
To build up a table of the standard enthalpies of formation of ions, we define the
standard enthalpy of formation of the aqueous H+ ion as 0. The following youtube
video has a good explanation of how this works:

Starting on page 5-8 of this pdf:
http://www.chemistry.uoguelph.ca/educmat/chm105_w2012_baker/lecture%204.pd
f, there is another good explanation of how we determine enthalpies of formation of
ions.
Look at appendix 2A for examples.
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1D. 1
As long as a species is neutral, it contains just one element, and it has a standard
enthalpy of formation of 0, you can be sure that it represents the reference state of
the element it contains.
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Hess’s law: The overall reaction enthalpy is the sum of the reaction enthapies of
the steps into which the reaction can be divided.
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1D.2
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Hess’s law: The overall reaction enthalpy is the sum of the reaction enthalpies of
the steps into which the reaction can be divided.
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In the above scheme, notice that reactions 2 and 3 are also the formation reactions in
addition to being combustion reactions.
This approach works for any hydrocarbon if we know its enthalpy of combustion.
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https://docs.google.com/document/d/1DKAzdWGDicuo3wU7YYp6wIf9WohUfNz9oqOclygmXQ/edit?usp=sharing
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1
https://web.mit.edu/12A.unified/www/FALL/thermodynamics/notes/node49.html
Our ability to predict the direction of the above two transitions really just depends
on our experience of the universe
• We know from experience that when we bring two bodies into contact with one
another, heat will always move from from the warmer body to the cooler body.
• We know from experience that gases will always expand to fill the available space.
As we’ll see, what we do in classical thermodynamics is to define a new state
function called entropy, and then we rely on our experience about what processes
will occur spontaneously to prove that entropy always increases in the course of a
spontaneous process. We assume that this proof is general and then we can then
predict spontaneity for processes where it’s not so obvious by calculating the entropy
changes and relying on the assumption that spontaneous processes always increase
the entropy. Calculating the entropy change is made easier by the fact that entropy is
a state function. So all we need to know to calculate the entropy change are the
initial and final states – the path is not important.
What spontaneous processes all have in common is that they increase disorder, and
2
so, on this basis, we conclude that entropy is a “measure of disorder”. This is
admittedly vague and not very satisfying. Later, when we talk about the statistical
definition of entropy, the way in which entropy and disorder are related should
become clear.
*If we have two snapshots of an isolated system, we can always be sure that the one
with the greater entropy is the later snapshot. So entropy measurements are what
allow us to distinguish the past from the future.

Why is entropy "the energy unavailable to do work"? What does that mean??
by u/chlorine_kelsey in askscience

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y_unavailable_to_do_work/, As entropy increases, the amount of energy available to
do work decreases. So sometimes we say that entropy is a measure of how much
energy is unavailable to do work. Here are some examples of what this means:
In the example, on the left, the block of metal is much hotter than the surroundings.
In theory, we could transfer this energy to the surroundings by first using it to heat
the gas inside a piston, which could then expand to do work. The excess heat, can
then be released to the surroundings. This can only continue until the metal has
cooled to the temperature of the surroundings. Since energy is conserved, the
amount of energy in the system plus the surroundings has not changed, but because
the system and surroundings are now at thermal equilibrium, the thermal energy is
no longer available to do work.
In the example on the right, when we initially open the stop cock, the random
thermal motions of the gas molecules allow the gas to expand and therefore do
expansion work, but once mechanical equilibrium is reached (i.e., once the pressure
in both compartments is equal), no more expansion work can be done. Once again,
energy must be conserved (by the first law), but the ability of the energy to do work
has decreased.
As we’ll see, entropy is defined is such a way that as the ability of the energy in a
system to do work decreases, the entropy increases.
2
The term “classical thermodynamics” is used to distinguish what we are talking about
in this PowerPoint from “statistical thermodynamics”, which is the subject of the next
PowerPoint.
3
Clicker 2A.1
The second law of thermodynamics only applies to isolated systems. If a system is
able to exchange energy or matter with the surroundings, then the entropy of the
system can decrease during a spontaneous process. People sometimes ignore this
fact, leading to misunderstanding.
4
https://docs.google.com/document/d/17bQIOhr8Bya1Xza5hrcb4eD_TToxL5J/edit?usp=sharing&ouid=110386635534945116488&rtpof=
true&sd=true
Another verbal statement of the second law coined by Clausius: There is no device
that can transfer heat from a colder to a warmer reservoir without net expenditure of
work.
5
“Proof” of the Clausius Inequality
Combining premises 2 and 3 gives −𝑤!”!,$%& > −𝑤!”!,’$$%& . Multiplying both sides of
this inequality by –1 reverses the inequality and thus 𝑤!”!,$%& < 𝑤!"!,'$$%& . Since internal energy is a state function, we know that its change for an irreversible path will be equal to its change for a reversible path. Thus, premise 1 leads to: 𝑞!"!,$%& + 𝑤!"!,$%& = 𝑞!"!,'$$%& + 𝑤!"!,'$$%& Rearranging gives: 𝑞!"!,$%& − 𝑞!"!,'$$%& = 𝑤!"!,'$$%& − 𝑤!"!,$%& Since 𝑤!"!,$%& < 𝑤!"!,'$$%& , the right side of the above equation is positive. The left side of the equation will only be positive if 𝑞!"!,$%& > 𝑞!”!,’$$%& . Dividing both sides of
this last inequality by T yields:
𝑞!”!,$%& 𝑞!”!,’$$%&
>
𝑇
𝑇
To get the inequality as shown in the slide, we can just drop the “sys” subscripts since
thermodynamic functions always refer to the system unless otherwise specified and
invoke premise 4 (the definition of entropy).
6
Objection: if q is always 0 for an isolated system, how can the entropy in an isolated
system (such as the universe) change?
Answer: You are assuming that the entropy change equals q/T. But, that is only true if
the process is reversible. If the process is irreversible (as is the case for all
spontaneous processes), then the entropy change is greater than q/T, so that means
for any irreversible process occurring in an isolated system the entropy change of the
isolated system is positive.
For a process occurring in a system that is not isolated, the entropy change of the
system can be negative, but, if so, it will always be less negative than q/T. It can also
be positive, but, if so, it will always be more positive than q/T. The entropy change of
the system for a spontaneous process can also be 0 as long q/T for the system is
negative.
7
8
9
The proof that entropy is a state function can be generalized to all systems by:
Accepting the idea that heat always travels from a warmer body to cooler body and
never the other way around, and then applying this assumption to a cyclic
thermodynamic process called the Carnot cycle. The details of this are beyond the
scope of this class.
But we don’t really need to prove that entropy is a state function. We can just accept
it as a part of the 2nd law of thermodynamics recognizing that this law never fails.
10
Above equation can only be applied directly if the temperature remains constant
throughout the process. If the temperature is changing, then we need to account for
the temperature change by breaking the process down into a series of steps in which
each step involves only an infinitessimal temperature change. We’ll see how later.
Why does it make sense that the change in the disorder of the system should be
directly proportional to the heat transfer, but inversely proportional to the
temperature?
It’s because as we transfer heat into the system, the atoms and molecules gain kinetic
energy. This results in an increase in the random thermal motions of the molecules,
i.e., an increase in disorder.
If the system has a higher T, then it is already more disordered and so the same
amount of heat transfer will result in a smaller increase in disorder. If you stand up
and start dancing around in the middle of a crowded football stadium, this will have
very little impact on the disorder, but if you do the same in the middle of the library,
this will result in a big increase in the disorder.
11
So, assuming the above system is isolated, the second law of thermodynamics
predicts gases will expand spontaneously to fill the available space.
When we learn about statistical entropy, we’ll learn that the entropy is really a
measure of the number of microstates that a system can assume. A microstate, is a
state in which all the atoms and molecules in the system have a particular position,
angular momentum, vibrational state, and so on. So if we increase the volume of the
system, there are more possible microstates because each atom or molecule can
occupy more positions and therefore the entropy increases.
12
Notice that an increase in temperature results in an increase in entropy. Does this
make sense? Why?
13
Relationships in red are only valid for ideal gases. The others are generally
applicable.
*This one hasn’t been proved previously. The proof is as follows: ∆H = ∆U + ∆(PV) =
nCv,m∆T + nR∆T = n(CV,m + R)∆T = nCP,m∆T
**Why is ∆H 0 for the isothermal expansion. It’s because PV = nRT. If the temperature
and composition of the system are constant, PV does not change. Therefore, ∆H = ∆U
+ ∆(PV) = 0 + 0 = 0
14
For the calculation of entropy change in step 1, we can use any pressure units
because it’s a ratio.
Rankine scale
298 K = 536 R
118.6 K = 213.5 R
15
Clicker 2A.3
16
17
18
1
2
k = 1.381 x 10-23 J K-1
So now we have a definition of entropy with an intuitive and well defined link to the
idea of disorder. The more states that are available to a system the more disordered it
is.
Another way of thinking about this is by reference to knowledge. W can be thought of
as a measure of the lack of knowledge. The following is from the Wikipedia entry on
statistical entropy:
“We can view Ω as a measure of our lack of knowledge about a system. As an
illustration of this idea, consider a set of 100 coins, each of which is either heads up
or tails up. The macrostates are specified by the total number of heads and tails,
whereas the microstates are specified by the facings of each individual coin. For the
macrostates of 100 heads or 100 tails, there is exactly one possible configuration, so
our knowledge of the system is complete. At the opposite extreme, the macrostate
which gives us the least knowledge about the system consists of 50 heads and 50 tails
in any order, for which there are 100,891,344,545,564,193,334,812,497,256 (100
choose 50) ≈ 1029 possible microstates.”
3
So macrostate B is FAR more probable than macrostate A. More generally, if we just
took 100 coins and threw them up into the air, we would discover that macrostates
with roughly equal numbers of heads and tails were FAR more likely than macrostates
with a large excess of heads or a large excess of tails. That’s because there are many
more microstates associated with the former than the latter type of macrostate.
4
5
In Vienna
6
This equation tells us that entropy is proportional to the log of the multiplicity, which
is the number of microstates that are compatible with a systems macrostate.
Knowing this equation, we can understand why spontaneous processes always
maximize the total entropy. It simply reflects the fact that high entropy states are
more probable.
7
As you’ll soon see, the above formula gives the exact same results that are
obtained using the classical definition of the change in entropy:
8
9
10
*When I say equal, I mean equal to within our ability to accurately measure. On a
time averaged basis, they are exactly equal, but after equilibrium is reached, if we
shut the stopcock, there could be a slight excess of molecules on one side or the
other. However, this difference will be far too small to measure.
For 10 molecules, there are 210 (1024) ways of distributing the molecules.
For 100 molecules, there are 2100 (~1030 or ~one nonillion) ways of distributing the
molecules.
For 1000 molecules, there are 21000 (~10301 or ~10 x googol3) ways of distributing the
molecules.
I used the following online binomial distribution calculator to generate these
distributions: https://www.di-mgt.com.au/binomial-calculator.html
11
12
The number 500,000 comes from noticing that there are 10 covalent bonds with 3
possible conformations each and three covalent bonds with 2 possible conformations
each. 310×23 = ~500,000
13
The number 500,000 comes from noticing that there are 10 covalent bonds with 3
possible conformations each and three covalent bonds with 2 possible conformations
each. 310×23 = ~500,000.
Experimentally determined value for average entropy of bp formation = ~-88 J/Kmol
14
This accords with the statistical definition of entropy. In a perfect crystal at
absolute 0, there is only one possible arrangement of the molecules. So: S = k
lnW = k ln(1) = 0
15
Note that the heat capacity is a function of temperature and phase and so to
determine standard molar entropies, it’s necessary to measure Cp,m over a wide
range of temperatures between 0 and 298.15 K and evaluate the integral for each
phase.
16
The Entropy of Water and the Third Law of Thermodynamics. The Heat Capacity of
Ice from 15 to 273°K.
W. F. Giauque and J. W. Stout
Journal of the American Chemical Society 1936 58 (7), 1144-1150
DOI: 10.1021/ja01298a023
Experimentally measured value is 69.91 J/(K∙mol). The residual entropy of ice
accounts for the difference with the value determined in the slide.
Residual entropy is the entropy a substance has at absolute 0. Note that the third
law only applies to perfect crystals in which all molecules have the same
orientation, so that there is no disorder (meaning a multiplicity of 1). As
explained below, ice does not form perfect crystals.
Residual entropy of ice is explained by Pauling in
https://pubs.acs.org/doi/pdf/10.1021/ja01315a102
Also see this video: https://www.youtube.com/watch?v=D-Y4vVKzzmc
Each water molecule contains two hydrogens atoms, so there are 2N hydrogen
17
atoms where N is the number of water molecules. Each of these hydrogens is
bridging two oxygens – it’s covalently bonded to one oxygen and hydrogen
bonded to the other. So as a result, each hydrogen can occupy two different
positions along the axis between its two oxygens, depending on which oxygen it is
covalently bonded to. This would result in 22N possible configurations of the ice
crystal.
However, not all configurations are available. Each oxygen in ice is surrounded by
four hydrogens with each hydrogen making a covalent bond to one oxygen and a
hydrogen bond to the other oxygen. But since oxygen wants to make exactly two
covalent bonds (thus filling its valence shell without gaining charge), only 6 of the
possible 16 arrangements of 4 hydrogen atoms around each oxygen atom are
available. The other two possible arrangements require the oxygen atom to make
0, 1, 3, or 4 covalent bonds to hydrogen, which is not allowed (hydroxide and
hydronium ions do form, but they are much less stable than water and so we can
neglect them in this calculation). Thus, to get the actual multiplicity, we multiply
22N (the total number of possible configurations) by (6/16)N the fraction of the
configurations that give water exactly two covalent bonds to hydrogens. This
yields:
6 ‘
&
𝑊= 2
= 1.5’
16
Plugging this value into Boltzmann’s equation and setting N equal to Avogrado’s
constant gives a residual entropy of 3.4 J/(K mol). Adding this to the value
calculated in the slide yields a result that is essentially the same as the
experimentally measured standard molar entropy of water.
Here is a great video about residual entropy:

17
18
Why is this spontaneous?
19
20
1
2
3
4
The final inequality gives us a way of judging spontaneity without explicitly
considering the entropy change of the surroundings.
5
Entropically driven: at high enough temperature, the entropy decrease of the
surroundings is smaller than the entropy increase of the system
Enthalpically driven: at low enough temperature, the entropy increase of the
surroundings is greater than the entropy decrease of the system
6
7
1
2
Gibbs free energy allows us to predict if a process will occur spontaneously without
explicitly thinking about the entropy change in surroundings. This assumes that the
pressure and temperature are constant.
3
4
Clicker 2D.1
5
When we determine standard free energy change using these formulas, we’ll get an
answer in units of joules if we give n units of moles, and we’ll get an answer in units
of joules/mole if we give n as a unitless quantity. While the former approach may
seem more logical because it recognizes the nature of free energy as an extensive
variable, the latter approach is the preferred approach. This is because other
approaches for determining standard free energy change that we’ll be learning later
in the course automatically give an answer in joules/mole.
6
Clicker 2D.2
7
∆G˚ (at 25 ˚C) = [–394.36 + (2)(–237.13) – (–50.72)] kJ/mol]= –817.90 kJ/mol
∆S˚ = 213.74 + 2 (69.91) – 186.26 – 2(205.14) = –243
Why is this reaction entropically unfavorable?
8
9
10
Question: why do the negative slopes decrease as we go from vapor to liquid to
solid?
11
Density of liquid water is 1.0 g/cm3
Density of ice is 0.92 g/cm3
For most substances, the line between the solid and the liquid phase has a positive
slope. This means that as the pressure increases, the melting temperature increases.
This is because most substances expand as they melt and so, as the pressure
increases, the substance is expanding against a greater pressure and so more energy
(i.e., a higher temperature since temperature is proportional to kinetic energy) is
required to melt the substance since more of the energy needs to be used to push
against the atmosphere.
However, if you look at the water phase diagram, you see that the line between the
solid and the liquid has a negative slope meaning the melting temperature actually
decreases as the pressure increases. This is because the volume of water actually
decreases as water melts. This is a peculiarity of water. I don’t know of other
substances that decrease in volume as they melt. Therefore, the greater the pressure,
the less the energy required to melt ice and therefore the lower the melting point.
Think of it this way: at a higher pressure the atmosphere is pushing on the ice with
12
greater force and so the ice has a greater tendency to melt and thereby occupy a
smaller volume. So it melts at a lower temperature.
This is really another demonstration of the equivalence of work and heat. To melt the
ice, we need to increase its internal energy. This internal energy increase is used to
weaken the intermolecular forces that hold the water molecules together in the rigid
arrangement that we observe in ice crystals. We can supply this energy by increasing
the temperature, resulting in the flow of energy from the surroundings into the water
as heat. But if we increase the pressure of the atmosphere pushing down on the ice
then we end up supplying more of the energy as work (since work equals −Pex∆V) and
therefore the ice melts at a lower temperature.
12
13
14
Proof that ∆G is the maximum non-expansion work that a system can do on the
surroundings.
𝑑𝐺 = 𝑑𝐻 − 𝑇𝑑𝑆; At constant P: 𝑑𝐻 = 𝑑𝑈 + 𝑃𝑑𝑉; 1st law: 𝑑𝑈 = 𝑑𝑞 + 𝑑𝑤
Combining the above three relationships gives: 𝑑𝐺 = 𝑑𝑞 + 𝑑𝑤 + 𝑃𝑑𝑉 − 𝑇𝑑𝑆
For a reversible process and given that 𝑑𝑞#$% = 𝑇𝑑𝑆
𝑑𝐺 = 𝑑𝑞#$% + 𝑑𝑤#$% + 𝑃𝑑𝑉 − 𝑇𝑑𝑆 = 𝑑𝑤#$% + 𝑃𝑑𝑉
= 𝑑𝑤#$%,$'()*+,-*+ + 𝑑𝑤#$%,*-*$'()*+,-* + 𝑃𝑑𝑉
= −𝑃𝑑𝑉 + 𝑑𝑤#$%,*-*$'()*+,-* + 𝑃𝑑𝑉 = 𝑑𝑤#$%,*-*$'()*+,-*
The reversible work is the maximum non-expansion work that a system can do
on the surroundings: 𝑑𝐺 = 𝑑𝑤*-*.$'()*+,-*,/)’
For a finite change in free energy, this becomes: ∆𝑮 = 𝒘𝒏𝒐𝒏.𝒆𝒙𝒑𝒂𝒏𝒔𝒊𝒐𝒏,𝒎𝒂𝒙
15
16
The data on the concentration of ammonium in yeast comes from:
https://www.ncbi.nlm.nih.gov/pubmed/27120628.
The pKa of ammonium is 2D.4. So at pH 5, assuming we are at equilibrium, virtually
all the ammonia will be protonated.
17
18
Constants and Conversions
23
−1
𝑁𝐴 = 6. 022 𝑥 10 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 𝑢𝑛𝑖𝑡𝑠 𝑚𝑜𝑙
−23
𝑘𝐵 = 1. 381 𝑥 10
−1
−1
𝐽𝐾
−1
𝑅 = 8. 314 𝐽 𝑚𝑜𝑙
𝐾
−1 −1
= 0. 08206 𝐿 𝑎𝑡𝑚 𝑚𝑜𝑙 𝐾
4
−1
= 1. 987 𝑐𝑎𝑙 𝑚𝑜𝑙
−1
𝐾
−1
𝐹 = 9. 6485 × 10 𝐶 𝑚𝑜𝑙
1 𝑐𝑎𝑙 = 4. 184 𝐽
1 𝐿 · 𝑎𝑡𝑚 = 101. 3 𝐽
1 𝑎𝑡𝑚 = 1. 01325 𝑏𝑎𝑟 = 760 𝑇𝑜𝑟𝑟 = 760 𝑚𝑚𝐻𝑔
5
1 𝑏𝑎𝑟 = 10 𝑃𝑎
3
1 𝐿 = 0. 001 𝑚
𝑇 (𝐾) = 273. 15 + 𝑇 (˚𝐶)
Equations (Note: It is your responsibility to know when you can apply each equation.)
∆𝑈 = 𝑞 + 𝑤
∆𝑆 =
∆𝑈 = 𝑛𝐶𝑉,𝑚∆𝑇
𝑞𝑟𝑒𝑣
𝑇
𝑉
1
𝐶𝑉,𝑚 = (# 𝑜𝑓 𝑑𝑒𝑔. 𝑜𝑓 𝑓𝑟𝑒𝑒𝑑𝑜𝑚) × 2 𝑅
𝑞 = 𝑚𝐶𝑠∆𝑇
𝐶𝑃,𝑚 = 𝐶𝑉,𝑚 + 𝑅
𝑞 = 𝑛𝐶𝑚∆𝑇
𝑞𝑐𝑎𝑙 = 𝐶𝑐𝑎𝑙∆𝑇
𝑇
∆𝑆 = 𝑛𝑅𝑙𝑛[ 𝑉𝑓 ]
∆𝑆 = 𝑛𝐶𝑚𝑙𝑛[ 𝑇𝑓 ]
𝑖
𝑖
𝑊𝑓
𝑆 = 𝑘𝐵𝑙𝑛𝑊
∆𝑆 = 𝑘𝐵𝑙𝑛[ 𝑊 ]
𝑖
𝑤 =− 𝑃𝑒𝑥∆𝑉
𝑃𝑉 = 𝑛𝑅𝑇
𝑉𝑓
𝑃𝑎 = χ𝑎𝑃
𝑤 =− 𝑛𝑅𝑇𝑙𝑛[ 𝑉 ]
𝑖
𝐻 = 𝑈 + 𝑃𝑉
𝐺 = 𝐻 − 𝑇𝑆
∆𝐻 = ∆𝑈 + (∆𝑛𝑔𝑎𝑠)𝑅𝑇
∆𝑆𝑟𝑥˚ = Σ𝑛𝑆𝑚˚(𝑝𝑟𝑜𝑑) − Σ𝑛𝑆𝑚˚(𝑟𝑒𝑎𝑐𝑡)
−
𝐾𝑊 = 𝐾𝑎𝐾𝑏 = [𝐻3𝑂 ][𝑂𝐻 ]
−14
at 25˚C
𝑝𝐻 ≅ 𝑝𝐾𝑎 + 𝑙𝑜𝑔(
[𝐻𝐴]𝑖𝑛𝑖𝑡
𝑇
𝑜
∆𝐺𝑒𝑞 =− 𝑅𝑇𝑙𝑛𝐾
𝑜
)
𝑖𝑑𝑒𝑎𝑙
𝐸
− 𝑅𝑇𝑎
𝑘 = 𝐴𝑒
𝐸
𝑙𝑛𝑘 = 𝑙𝑛𝐴 − 𝑅𝑇𝑎
𝑙𝑛
𝑜
∆𝑆
+
𝑅
𝑘𝑓
𝑘𝑖
=
𝐸𝑎
𝑅
(
1
1
−
)
𝑇𝑖
𝑇𝑓
∆𝐺 =− 𝑛𝐹𝐸𝑐𝑒𝑙𝑙
[𝐴]𝑡 =− 𝑘𝑡 + [𝐴]0
𝑙𝑛𝐾 = 𝑛𝐹𝐸𝑐𝑒𝑙𝑙˚/𝑅𝑇
𝑙𝑛[𝐴]𝑡 =− 𝑘𝑡 + 𝑙𝑛[𝐴]0
𝑅𝑇
𝑙𝑛𝑄
𝑛𝐹
0.05916
𝑜
𝐸𝑐𝑒𝑙𝑙 = 𝐸𝑐𝑒𝑙𝑙 −
𝑙𝑜𝑔𝑄 at 25 ˚C
𝑛
𝑜
−
[𝐴 ]𝑖𝑛𝑖𝑡
∆𝐺𝑠𝑦𝑠
∆𝐺𝑟𝑥𝑛 = ∆𝐺𝑟𝑥𝑛 + 𝑅𝑇𝑙𝑛𝑄
−∆𝐻
𝑙𝑛𝐾 =
𝑅𝑇
𝑉
𝑍 = 𝑉 𝑟𝑒𝑎𝑙
𝑛
𝑜
∆𝐺𝑟𝑥˚ = Σ𝑛∆𝐺𝑓˚(𝑝𝑟𝑜𝑑) − Σ𝑛∆𝐺𝑓˚(𝑟𝑒𝑎𝑐𝑡)
+
𝑛𝑎
χ𝑎 =
∆𝑆𝑢𝑛𝑖𝑣𝑒𝑟𝑠𝑒 = −
∆𝐻𝑟𝑥˚ = Σ𝑛∆𝐻𝑓˚(𝑝𝑟𝑜𝑑) − Σ𝑛∆𝐻𝑓˚(𝑟𝑒𝑎𝑐𝑡)
𝐾𝑊 = 1. 0 × 10
𝑃𝑎𝑉 = 𝑛𝑎𝑅𝑇
𝐸𝑐𝑒𝑙𝑙 = 𝐸𝑐𝑒𝑙𝑙 −
𝑡1/2 =
𝑙𝑛2
𝑘
1
1
= 𝑘𝑡 +
[𝐴]𝑡
[𝐴]0