Thermodynamics
Up to this point in the class, we’ve seen lots of evidence of thermodynamics. We’ve heated up liquids to watch them
boil and we’ve even watched them burst into flames. In our daily lives we use freezers, furnaces, air conditioners,
automobiles, and any of a thousand other devices that use thermodynamics. Thermodynamics is everywhere.
Of course, even though it may be everywhere, it’s probably also a term that you’re not familiar with. Here’s a definition:
Thermodynamics: The study of energy (i.e. what it is, how it is transformed from one form to another, how it is
used to get things done).
As is probably obvious, this term is extremely vague because it covers a huge number of processes that take place in the
world. To get an understanding of what this really means, we need to start where we usually do… at the beginning.
Some terms and ideas you need for studying thermodynamics:
As with everything, thermodynamics has specialized terms to describe the things that go on in the real world. Before
moving on to seeing how energy behaves, we need to first understand the terms that are used to describe it.
Energy: The ability to do work or to produce heat. There are two types of energy:


Kinetic energy: The energy something has when it moves. (i.e. moving objects, moving particles, vibrating
molecules, etc)
o Temperature is a measure of the particles in an object. We know this from the KMT, which says that the
amount of energy is proportional to the temperature (in K). The more the particles in an object move
around, the higher the temperature.
Potential energy: Stored energy that’s waiting for its chance to get moving. (i.e. objects that are waiting to fall
off of a shelf, energy stored in chemical bonds, etc).
o Chemical potential energy: The energy that’s stored in chemical bonds.
One of the main laws that describe how energy behaves in the world is called the law of conservation of energy (also
known as the “first law of thermodynamics”):
Energy is never created or destroyed – it can only be converted between potential and kinetic energy.
o
o

This means that you can change kinetic energy to potential (and vice-versa), but you can never make the
amount of energy go away.
Examples:
 You can take the stored gravitational energy of a book on a shelf and make it into the kinetic
energy of the book moving through the air.
 You can use the potential energy of the bonds in gasoline in your car to make your car move.
You can use the energy of moving turbines (kinetic) to charge a battery (potential)How energy moves from one
place to another:
o Heat (q): The movement of energy from one thing to another through the motion of molecules
(thermal energy).
 Heat spontaneously moves from hot things to cold. This is why a hot pan can burn you and you
can’t burn a hot pan – the energy goes only from the pan to you because it’s hotter.
 Heat and temperature are NOT the same thing: Heat is the transfer of energy, temperature is a
measure of the kinetic energy of the object once the energy has finished transferring.

o

Example: If I give you five dollars, the money is what I give you (heat) and your net
value is what goes up after you get the money (temperature). Your wealth is not a five
dollar bill, but the five dollar bill does affect it.
Work (w): The movement of energy from one thing to another through the motion of larger things
(mechanical energy).
 When a weight is lifted up a hill, work has been done on it (this is what you’ll learn in physics).
In chemistry, heat is a lot more useful than work because chemistry generally deals with the movement of
atoms and molecules a lot more often than it deals with the movement of large machines.
Since heat (energy that’s being transferred through thermal motion) is more important than work (for our purposes),
how do we measure it? More definitions of use:

System: Whatever we’re studying.
o This can be practically anything. If we are studying what happens when we heat a pan on the stove, the pan
will be the system we are studying.
o If a system gets energy added to it, the amount of energy it has after the change is positive. Because of this,
an endothermic process is any process in which energy is absorbed by the system we’re talking about.
o If a system has energy taken away from it, the amount of energy it has after the change is negative. Because
of this, an exothermic energy is any process in which energy is given off by the system we’re talking about.

Surroundings: Everything outside the system.
o When studying the pan above, the surroundings will primarily consist of the stove (because it’s putting
energy into the pan), though it technically consists of everything but the pan.

Universe: The system + the surroundings.
o In a general sense, the universe consists of everything that exists anywhere. From a thermodynamic
standpoint, the universe usually consists of whatever the system we’re referring to is as well as whatever is
putting energy into or taking energy away from it.
o Example: If we’re studying a space heater, the heater will be our system, the house will be our surroundings,
and the universe will be the heater and the house together (we ignore the irrelevant rest of the world, since
it doesn’t really play a part in anything).
Quantifying energy:
 The traditional unit of energy is the calorie (cal), which is the amount of energy you need to add to 1 gram of
water to heat it by 10 C.
 Food is measured in units of 1000 calories called kilocalories (kcal), which is more commonly known as the
Calorie (Cal).
 The metric unit of energy is the joule (J). There are 4.184 J/cal.
 Because a joule isn’t very much energy, we usually measure energy in units of 1000 joules called kilojoules (kJ).
Enthalpy: ∆H
Now that we’ve learned what heat is, it’s handy to think about how much heat a system can potentially give off to other
systems. This term is called “enthalpy.”
Enthalpy (H): The amount of heat that a system can potentially give to other systems.
 Unfortunately, it’s impossible to know what the overall enthalpy of a system is – after all, how much heat one
system can give to another depends on a large number of factors, including the nature of the system and the
nature of whatever system it wants to give energy to.
o
o

Instead of talking about how much enthalpy something has, we instead talk about how much the
enthalpy of a system changes when heat is taken away from it or added to it.
This term is given the symbol ∆H, where ∆ represents the change in enthalpy that occurs during a
process.
Types of enthalpy:
o Heat of reaction (∆Hrxn) describes the enthalpy change that occurs as the result of a chemical reaction.
 If ∆Hrxn is positive the reaction is endothermic (it has absorbed energy) and if ∆Hrxn is negative, it
is exothermic (and has given off energy).
 Heat of combustion (∆Hcomb) describes the change in enthalpy when something undergoes a
combustion reaction.
o Enthalpy changes having to do with phase changes:
 Molar heat of vaporization (∆Hvap) is the change in enthalpy when one mole of a compound
boils.
 Molar heat of fusion (∆Hfus) is the change in enthalpy when one mole of a compound melts.
o Standard heat of formation (∆H0f) is the change in enthalpy when a compound is formed from its
elements.
Enthalpy changes caused by heating/cooling:
Not surprisingly, one of the easiest ways to change the amount of energy that a system has (and thus, can give to other
systems) is by heating it or cooling it. It’s this energy we’ll start with first.
How much energy is needed to heat something up (or cool it down)?
The enthalpy change that accompanies the heating/cooling of a pure substance is determined by the equation:
∆H = mCp∆T
where:
o
o
o
o
∆H = the change in enthalpy (positive for heating, negative for cooling)
m = the mass of the thing being heated (in grams)
Cp = the specific heat / heat capacity – the amount of energy needed to heat the thing by 10 C.
∆T = the change in temperature (in degrees Celsius).
Sample problems:
o
o
How much energy will be needed to heat 45 grams of ethanol (Cp = 2.44 J/g0C) from 200 C to 400 C? 2196 J
If burning a single match gives off 250 J, how much hotter can it make a 50 gram block of silver (Cp of silver is
0.235 J/g0C)
Answer: 21.30 C
Lab: Calculating the heat capacity of water. **I will teach more about this after “moles”.
Enthalpy changes during phase changes:
As you might imagine, phase changes also change the enthalpy of a system. As you know, melting a compound causes
the energy of the particles to increase as they break intermolecular forces and boiling it has a similar change. Let’s
investigate what happens during phase changes.
Freezing/melting:
The amount of energy that’s added/removed from a substance during the freezing or melting process is described by the
equation:
∆H = n ∆Hfus
where
o ∆H = the enthalpy change for this process.
o n = the number of moles of the compound melting or freezing
o ∆Hfus = the molar heat of fusion (which is a constant)
Example: How much energy is required to melt 56 grams of frozen water?
Answer:
∆H = (3.11 moles)(6.01 kJ/mol) = 18.7 kJ = 18,700 J
Important: When undergoing the phase transition from liquid to solid (or vice versa), all of the energy goes into
breaking intermolecular forces. As a result, the temperature of the material doesn’t change as it undergoes the
transition!
o This is why ice cubes keep a cold drink at 00 C until they have completely disappeared.
Boiling/condensing:
The amount of energy that’s added/removed from a substance during the boiling or condensing process is described by
the equation:
∆H = n ∆Hvap
where
o ∆H = the enthalpy change for this process.
o n = the number of moles of the compound boiling or condensing
o ∆Hfus = the molar heat of vaporization (which is a constant)
Example: How much energy is required to boil 56 grams of frozen water?
Answer: ∆H = (3.11 moles)(40.7 kJ/mol) = 126.6 kJ = 126,600 J
Important: When undergoing the phase transition from liquid to gas (or vice versa), all of the energy goes into breaking
intermolecular forces. As a result, the temperature of the material doesn’t change as it undergoes the transition!
o
This is why boiling water remains at 1000 C until all of the water has been boiled away rather than increasing in
temperature!
Now that we’ve learned all of this, we have enough information to learn what the energy change is for a compound as
we heat it from a frozen material through both phase changes until it’s a gas.
o
The whole process of figuring out how much energy is required to make a chemical or physical change take
place is called calorimetry. The device we use to do this is called a calorimeter.
To do this, you must first figure out where you
are on this graph:
Once you know where you are and where you
need to end up, you can figure out what
equations you need to figure out the enthalpy
change for the process.
Example: How much energy is needed to heat
55 grams of water ice from a temperature of -150
C to steam at a temperature of 1500 C?
Solution: This problem requires several steps:
o
Step 1: Heat the ice from -150 C to 00 C
o ∆H = mCp∆T = (55 grams)(2.03 J/g0C)(150) = 1,675 J
o
Step 2: Undergo the phase change from a solid to liquid.
o ∆H = n∆Hfus = (3.06 mol)(6.01 kJ/mol) = 18.39 kJ
o
Step 3: Heat the liquid from 00 C to 1000 C
o ∆H = mCp∆T = (55 grams)(4.184 J/g0C)(1000) = 23,012 J
o
Step 4: Undergo the phase change from a liquid to gas.
o ∆H = n∆Hvap = (3.06 mol)(40.7 kJ/mol) = 124.54 kJ
o
Step 5: Heat the steam from 1000 C to 1500 C
o ∆H = mCp∆T = (55 grams)(2.01 J/g0C)(500) = 5,528 J
And when you’re done with all of this, just add up the different values you found above:
∆Htotal = 1.68 kJ + 18.39 kJ + 23.01 kJ + 124.54 kJ + 5.53 kJ
∆Htotal = 173.15 kJ
Hess’s Law:
So far, we’ve talked about how to find the enthalpy change for phase changes and heating/cooling chemical compounds.
Of course, since this is a chemistry class, we’re far more interested in chemical reactions. As a result, most of our time
talking about thermodynamics will be spent finding the elusive value ∆Hrxn.
To find the heats of reaction, we use a few tricks. The first is known as Hess’s Law:
Hess’s Law: You can find the enthalpy of reaction for any process if you can add up two or more known reactions to
describe the process.

Put another way: The overall energy change for a chemical process is equal to the sums of the energy changes
for all of the chemical processes that went into it.
o
An analogy: Let’s say we’re going on a trip in the mountains. If on the first day you go up in altitude by
3,000 feet and on the second day you go down in altitude by 1,800 feet, overall, you have gone a total of
+3,000 – 1,800 = +1,200 feet.
o
Likewise, for chemical reactions, if you have one process with ∆Hrxn = +3,000 kJ and another process with
∆Hrxn = -1,800 kJ, the overall ∆H will be +1,200 kJ.
Given the following equations:
o S(s) + O2(g)
∆H = -297 kJ
2(g)
o 2 SO3(g)
2(g) + O2(g) ∆H = 198 kJ
What is the heat of reaction for 2 S(s) + 3 O2(g)
3(g)?
Solution: Add the given equations together to get the equation we’re looking for. Because simply adding them up
doesn’t always give us the answer we want, we can do the following tricks:

If you want to switch the products and reactants with one another, you need to also change the sign of the
∆Hrxn.
o

For example: If S(s) + O2(g)
2(g)
∆H = -297 kJ, then SO2(g)
(s)
+ O2(g)_ ∆H = +297 kJ.
If you need to multiply the number of times you use an equation, multiply the ∆H value by the same amount.
o For example, If S(s) + O2(g)
∆H = -297 kJ, then 2 S(s) + 2 O2(g)
∆H = (2)(+297) = 594
2(g)
2(g)
kJ
Given these rules, let’s see how we can add the equations together to give us the one we want:
2 S(s) + 2 O2(g)
2(g) ∆H = 2(-297 kJ) = -594 kJ
2 SO2(g) + O2(g)
2 S(s) + 3 O2(g)
3(g)
3(g)
∆H = -198 kJ
∆H = -792 kJ
Another example:
Find ∆H for the reaction 2 H2O2(l)
 2 H2(g) + O2(g)
2O(l)
 H2(g) + O2(g)
2O2(l)
Answer: -196 kJ
2O(l)
+ O2(g) given the equations:
∆H = -572 kJ
∆H = -188 kJ
Another example:
Find ∆H for the reaction 2 CO(g) + 2 NO(g)
2(g) + N2(g) given:
 2 CO(g) + O2(g)
2(g) ∆H = -566.0 kJ
 N2(g) + O2(g)
∆H = 180.6 kJ
(g)
Answer: -746.6 kJ
Finding heats of reaction from heats of formation:
Of course, what we’ve just learned with Hess’s law presents some problems if we don’t have other chemical reaction
data to figure out. Fortunately, even if we don’t have data for each chemical reaction, we can still find the heat of the
reaction we’re looking for from the heats of formation of the products and reactants.
The big equation: ∆H0rxn = Σ∆H0f(products) – Σ∆H0f(reactants)

What the terms in this equation mean:
o ∆H0rxn is the “standard heat of reaction” – The enthalpy change for the process under standard
conditions.

Standard conditions: 1 atm, 250 C
o
∆H0f = “standard heat of formation” - the change in enthalpy that accompanies the formation of one
mole of the compound in standard state.
 Standard state is the form of the compound that exists at 250 C and 1 atm.
 ∆H0f for all elements in their standard state is defined as zero.
o
Σ = “sum of”; just add them all together.
Example: Determine the heat of combustion of methane:
CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g)
Given the following standard heats of formation:
 carbon dioxide: -394 kJ/mol
 methane: -75 kJ/mol
 water: -242 kJ/mol
Solution: Add the heats of formation of the products together and subtract the heats of formation of the reactants:
Heats of formation of the products:
 CO2: -394 kJ
 H2O: 2(-242 kJ/mol) = -484 kJ
Total: -878 kJ
Heats of formation of the reactants:
 methane: -75 kJ
 oxygen: 2(0 kJ/mol) = 0 kJ
Total: -75 kJ
Subtracting, we get ∆H0rxn = (-878 kJ) – (-75 kJ) = - 803 kJ
Another sample problem:
Given the following information:
Compound
nitric acid (aqueous)
water (liquid)
calcium nitrate (aqueous)
calcium hydroxide (solid)
∆H0f (kJ/mol)
-214.4
-285.8
-960.1
-999.1
Find the standard heat of reaction for:
2 HNO3(aq) + Ca(OH)2(s) 2 H2O(l) + Ca(NO3)2(aq)
Answer: -103.8 kJ
Spontaneity
When talking about chemical processes, we often want to know whether something will happen on its own or not. As
we’ll see, whether that happens or not is a little more complicated to predict than you’d think.
Spontaneous processes: Processes that happen by themselves without any intervention.
 These processes may not happen very quickly, but they will eventually happen by themselves.
 Examples:
o The combustion of gasoline: This process is spontaneous but VERY, VERY slow without the addition of
heat.
o The rusting of a nail: It happens slowly under normal conditions but the process can be speeded to be
almost instantaneous under the right conditions.
Why do things happens spontaneously?
 As we’ve seen, there’s a tendency for energy to move from places where there’s a lot of it to places without
much energy. As a result, we can surmise that any process in which the amount of energy for the system
decreases (giving it up to the surroundings) will be favorable.
Entropy and Free energy
The first energetic factor: Entropy (S)


Entropy is a thermodynamic measure of the randomness in the universe.
As you have seen from the lab, it is a general rule of the universe that processes that produce randomness are
favored over processes that decrease the amount of randomness in the universe.
o The second law of thermodynamics: All processes result in an overall increase in randomness for the
universe.



How much the randomness increases or decreases for a system during a process can be measured energetically
and is a thermodynamic quantity.
As with enthalpy, we cannot know the exact entropy for a system. However, we can know how much the
entropy of a system changes during the process. As a result, the most useful term is the change in enthalpy,
called ∆S.
∆S = Sfinal - Sinitial
and this value is positive for any spontaneous process
How to predict whether something will be spontaneous or not:
o Phase changes result in significant changes in enthalpy:
 Phase changes that result in greater molecular freedom have a positive ∆S, and those that result
in less molecular freedom have a negative ∆S.
 Example: For melting, ∆S = +, for freezing ∆S = -.
o Dissolving solids and liquids results in a positive ∆S.
 When you dissolve a solid the particles are no longer locked in place and can now move freely,
leading to greater randomness.
 When liquids are miscible, the mixed liquids are more random than the pure liquids are. As a
result, ∆S is positive.
o Dissolving a gas results in a negative ∆S.
 When you dissolve a gas a lot of its freedom to move around is taken away. As a result, this
makes it have less random.
o By examining the states of products and reactants in a process, you can usually tell if the entropy for a
reaction is positive or negative.
 If a gas is made from nongases, it’ll probably be positive.
 Example: 2 H2O2(l) 2 H2O(l) + O2(g) ∆S = +
 If it’s the other way around, ∆S will be negative
 If solids or liquids are made from gases, it’ll probably be negative.
 Example: HCl(g) + NH3(g)
4OH(s) ∆S =  If it’s the other way around, ∆S will be negative.
 If the number of gas molecules increase, it’ll be positive – if it’s the other way around it will be
negative:
 Example: 2 H2(g) + O2(g) 2 H2O(g) ∆S = -
The second energetic factor: Enthalpy



If a process takes place and the system is at lower energy than it was before, this also fulfills the tendency of the
universe to spread around its energy.
As you saw in the lab, processes that give off energy are more likely to be spontaneous than those that require
energy.
As a result, exothermic reactions are more likely to be spontaneous than endothermic reactions.
Put them together and you get the following equation:
∆G = ∆H – T∆S
where ∆G is called the “Gibbs free energy”: the amount of energy that is available for a system to do work.
o
o
o
When ∆G is negative, a process is spontaneous.
When ∆G is positive, a process is not spontaneous.
When ∆G = 0, the forward rate of reaction and reverse rate of reaction are the same. The system is said
to be at equilibrium.
Because ∆G determines whether a process is spontaneous or not, we can use ∆H and ∆S to determine whether a
process is spontaneous or not.
As with ∆H values, there are several ways of determining ∆G for a process. We will focus on one, which involves
solving the equation:
∆G = ∆H – T∆S
Example: Is the reaction N2(g) + 3 H2(g)
3(g)
spontaneous at 450 K? ∆H0rxn = -91.8 kJ and ∆S0rxn = -197 J/K.
Answer: Let’s solve the equation above using our information:
∆Hrxn = -91.8 kJ = -91,800 J (need to convert to J to get same units!)
T∆Srxn = (450 K)(-197 J/K) = -88,650 J
∆G = ∆H – T∆S = -91,800 – (-88,650) = -3,150 J
The reaction is spontaneous at 450 K!
Postproblem analysis: Does this make sense?

∆H is negative, which means that heat is released by this reaction. This tends to drive reactions toward
spontaneity.
 ∆S is negative, which means that the amount of randomness in the system decreases.
o Problem: Does this violate the second law of thermodynamics (which says that all processes must
undergo an increase of entropy)?
o No! Though the amount of entropy in the system decreases, the energy given off by the system causes
the surroundings to randomize more than enough to make up for it! That’s why the ∆H part of this
equation is important!
 This does make sense!
Another sample problem: At what temperature will the reaction we just investigated be at equilibrium?
Solution: Because all reactions are at equilibrium when ∆G = 0, we set ∆G = 0 and solve for the temperature.
∆G = ∆H – T∆S
0 = -91,800 – (T)(-197 J/K)
T = 466 K
At temperatures hotter than 466 K, this reaction will no longer be spontaneous.
Overall, the relationship between ∆H, ∆S, and ∆G can be expressed by the following chart:
∆H is positive
∆H is negative
∆G only negative (rxn only
∆G always negative (rxn always
∆S is positive
spontaneous) at high
spontaneous)
temperatures
∆G always positive (rxn never
∆G only negative (reaction only
∆S is negative
spontaneous)
spontaneous) at low temperatures