Download Thermodynamics - Deland High School

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 This means that you can change kinetic energy to potential (and vice-versa), but
you can never make the amount of energy go away.
o 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.
 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.
o 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).
In class practice worksheet about energy, followed by more notes!
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 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.
o 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.
We will talk at great length about how we calculate each of these values during the
next few classes.
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)?
o Use the example of a glass of water and have the students figure out that the
three factors that affect the amount of energy are the size of the glass (m), the
amount you want to heat it (∆T), and the substance you want to heat (Cp).
The enthalpy change that accompanies the heating/cooling of a pure substance is
determined by the equation:
∆H = mCp∆T
where:
o ∆H = the change in enthalpy (positive for heating, negative for cooling)
o m = the mass of the thing being heated (in grams)
o Cp = the specific heat / heat capacity – the amount of energy needed to heat
the thing by 10 C.
o ∆T = the change in temperature (in degrees Celsius).
Sample problems:
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
o 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)
21.30 C
Lab: Calculating the heat capacity of water.
Heating/cooling homework
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?
∆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?
∆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
Do more calculations like this if they wish.
Practice calorimetry worksheet
Calorimetry lab (where they sketch the heating curve, calculate energy that was
required)
Calorimetry homework
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.
Example (book, p. 506-7): Given the following equations:
o S(s) + O2(g)  SO2(g)
o 2 SO3(g)  2 SO2(g) + O2(g)
∆H = -297 kJ
∆H = 198 kJ
What is the heat of reaction for 2 S(s) + 3 O2(g)  2 SO3(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)  SO2(g)
O2(g)_ ∆H = +297 kJ.
∆H = -297 kJ, then SO2(g)  S(s) +
 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)  SO2(g)
∆H = -297 kJ, then 2 S(s) + 2 O2(g)
 2 SO2(g) ∆H = (2)(+297) = 594 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 SO2(g)
∆H = 2(-297 kJ) = -594 kJ
2 SO2(g) + O2(g)  2 SO3(g)
∆H = -198 kJ
2 S(s) + 3 O2(g)  2 SO3(g)
∆H = -792 kJ
Another example:
Find ∆H for the reaction 2 H2O2(l)  2 H2O(l) + O2(g) given the equations:
 2 H2(g) + O2(g)  2 H2O(l)
∆H = -572 kJ
 H2(g) + O2(g)  H2O2(l)
∆H = -188 kJ
Answer: -196 kJ
Another example:
Find ∆H for the reaction 2 CO(g) + 2 NO(g)  2 CO2(g) + N2(g) given:
 2 CO(g) + O2(g)  2 CO2(g)
∆H = -566.0 kJ
 N2(g) + O2(g)  2 NO(g)
∆H = 180.6 kJ
Answer: -746.6 kJ
Practice problems: Hess’s Law
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.
Let’s see how this works with some sample problems…
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
Heats of formation worksheet
Heats of formation race
Heats of formation homework
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.
In lab, we’ll explore why it’s energetically favorable for some processes to be
endothermic and some to be exothermic.
Spontaneity lab
Continue with notes about 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)  NH4OH(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 When ∆G is negative, a process is spontaneous.
o When ∆G is positive, a process is not spontaneous.
o 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)  2 NH3(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:
∆S is
positive
∆S is
negative
∆H is positive
∆G only negative (rxn
only spontaneous) at
high temperatures
∆G always positive (rxn
never spontaneous)
∆H is negative
∆G always negative (rxn
always spontaneous)
∆G only negative (reaction
only spontaneous) at low
temperatures
If requested, give them more sample problems with made up values for ∆H and
∆S to determine the spontaneity of chemical reactions.
Practice worksheet about this
Hwk sheet about this
Quiz review sheet
Quiz