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Course: AP®︎/College Chemistry > Unit 9
Lesson 1: EntropyAbsolute entropy and entropy change
The standard molar entropy of a substance is the absolute entropy of 1 mole of the substance in the standard state. For any chemical reaction, the standard entropy change is the sum of the standard molar entropies of the products minus the sum of the standard molar entropies of the reactants. Created by Jay.
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I wondered if some of these factors influencing entropy (moles, volume, state) are more dominant than others, e.g:
If I had a reaction, which results in a change of state from solid to gasseous, but a decrease in moles:
X(s) + Y(s) --> XY(g) |↓n ↑states
Would we predict an increase or decrease in entropy?(3 votes)
If I'm not mistaken that depends on the quantity of moles and other factors. If the decrease in moles is greater than the increase in other factors like temperature, then there are less microstates and thus; entropy decreases. But if the increase in the other factors is greater than the decrease in moles overwhelming it then that decrease will slightly affect the entropy; increase will be in the majority thus there will be more microstates and an increase in entropy.(1 vote)
How do you measure W in S = K*ln(W) when calculating the standard molar entropy of a substance?(1 vote)
Since W is the number of energetically equivalent ways to arrange the particles of a system we can view this as a combinatorics problem. We can use the combination formula: n!/(r!*(n-r)!), where n is the number of particles and r is the number of ways to place those particles in the container.
It should be noted that even for relatively small molecules, W becomes very large for molar quantities. For example the standard molar entropy of H2 is 130.7 J/(mol*K). Using Boltzmann’s equation where k Boltzmann’s constant set at, 1.380649 x 10^(-23) J/K, we can calculate the number of microstates (technically microstates per mole). In which case W is calculated as e^(9.466 x 10^(24)), an extremely large number which we would expect dealing with a mole amount of particles.
Hope that helps.(2 votes)
what exactly is absolute entropy? Is it the entropy at 0 kelvin?(1 vote)
Video transcript
- [Instructor] Entropy can be
measured on an absolute scale, which means there is a
point of zero entropy. And that point is reached for
a pure crystalline substance when the temperature
is equal to zero kelvin or absolute zero. At zero kelvin, the entropy of the pure
crystalline substance, S, is equal to zero. We can think about why the
entropy is equal to zero by looking at the equation
developed by Boltzmann, that relates entropy, S, to
number of microstates, W. A microstate refers to a
microscopic arrangement of all of the positions and energies of all of the particles. For our pure crystalline
substance at absolute zero, all of the particles are perfectly ordered in their lattice states. And at zero kelvin
there's no thermal motion. So all of the particles are perfectly ordered
in their lattice states with no thermal motion. That means there's only
one possible arrangement for these particles. And that means there's
only one microstate. So when we think about our equation, if we plug in the number of
microstates is equal to one, the natural log of one is equal to zero, which means that the entropy
is equal to zero at zero kelvin for this pure crystalline substance. This is called the third
law of thermodynamics. Next, let's think about what happens to our hypothetically perfect crystal if we increase the temperature. The increased temperature
means the particles gain energy and have motion around
their lattice states. Therefore, there's an increase in the
number of possible microstates. And if there's an increase
in the number of microstates, according to the equation
developed by Boltzmann, that also means an increase in entropy. Since we started with zero
entropy at zero kelvin, and the entropy increases, at all temperatures that are
greater than zero kelvin, the entropy must be greater than zero, or you can say the entropy is positive. We can get the units for entropy from the Boltzmann constant, K. K is equal to 1.38 times 10 to the negative
23rd joules per kelvin. So we will use these units for entropy. Now that we understand the
concept of zero entropy, let's look at the entropy of a substance. And let's say that we
have one mole of carbon in the form of graphite. Standard entropy refers to the absolute
entropy of a substance at a pressure of one atmosphere and a specified temperature. Often that temperature
is 25 degrees Celsius. The standard entropy of
graphite at 25 degrees Celsius is equal to 5.7 joules per kelvin. And since we're dealing
with one mole of graphite, we could write the units
as joules per kelvin mole. And often this is called
standard molar entropy. The standard molar entropy
of graphite is positive because it's being compared to a hypothetically
perfect crystal of graphite at zero kelvin. So really it's a change in entropy and therefore it would be 5.7 minus zero with zero being the entropy
of a hypothetical crystal at zero kelvin. However, when we write standard molar entropies, we don't include the delta sign and we reserve the
delta sign for processes such as phase changes or
also chemical reactions. The superscript, not, refers to the standard
state of the substance. By convention, the standard state of a solid or liquid is referring to the pure
solid or pure liquid under a pressure of one atmosphere. For gases, standard state
is referring to the pure gas at a pressure of one atmosphere. And for solutions, we're
talking about a solution with a concentration of one molar. In our case, we're talking
about one mole of carbon in the form of graphite
in the solid state. Therefore, the standard molar entropy of graphite is referring to the entropy value for one mole of a pure solid under a pressure of one atmosphere. Next let's look at a table showing standard molar entropies
of different substances at 25 degrees Celsius. We just saw that the standard
molar entropy of carbon in the form of graphite is equal to 5.7 joules per kelvin mole. Let's compare that solid to two other standard molar entropies. For example, liquid water has a standard
molar entropy of 69.9, and methane gas has a standard
molar entropy of 186.3. Looking at these numbers, in general, gasses have higher standard
molar entropies than liquids, as we can see comparing the values, and liquids, in general, have higher standard molar
entropies than solids. The reason why this is generally true has to do with the number
of microstates available to solids, liquids, and gasses. Solids are held together
by either chemical bonds or by intermolecular forces. Liquids are held together
by intermolecular forces. And gases, if we assume ideal gases, there are no intermolecular
forces between the particles. So in general, as we go from
a solid to a liquid to a gas, there's an increase in the number of possible
arrangements of particles. There's an increased freedom of movement. And therefore going from
solid to liquid to gas means an increase in the number
of possible microstates. And an increase in the number
of possible microstates means an increase in entropy. Next, let's compare the standard
molar entropies of two gases. Methane is 186.3, whereas ethane is 229.6. Looking at their dots structures, methane is on the left and
ethane is on the right. They're both composed
of carbon and hydrogen. However, ethane has more
carbons and more hydrogens. Single bonds allow free rotation. And since ethane has
more bonds than methane, there's more ways for the
ethane molecule to rotate. Bonds can also stretch and
compress like a spring. So we could imagine
this carbon-carbon bond having some vibrational motion. And since ethane has
more bonds than methane, ethane has more ways to
vibrate than methane does. More ways to rotate and
more ways to vibrate mean that there are more possible ways to distribute the energy
in ethane than in methane. And an increased number of ways
of distributing energy means an increase in the number
of available microstates, which means an increase in the entropy. So in general, as there's an increase
in the number of atoms, or an increase in the molar mass, from say methane to ethane, there's also an increase in entropy. Next, let's calculate the
standard change in entropy for a chemical reaction. For our reaction, let's look at the decomposition of one mole of calcium carbonate turning into one mole of calcium oxide and one mole of carbon dioxide gas. To calculate the standard
change in entropy for this chemical reaction, we need to sum up the
standard molar entropies of the products. And from that, we subtract the sum of the
standard molar entropies of the reactants. So first we think about our products, which are calcium oxide
and carbon dioxide. So we need to sum the
standard molar entropies of these two substances. Next, we think about our reactants. And for this reaction, we
have only calcium carbonate. So therefore, we're gonna plug in the
standard molar entropy of calcium carbonate. Before we plug in our values
for standard molar entropies, let's predict the sign for
the standard entropy change for this reaction. In our reaction, we go from one mole of a solid, to one mole of another
solid and one mole of a gas. We already know that in general gases have higher values
for entropy than solids. And since we have zero moles
of gas on the reactant side, and one mole of gas on the product side, we would predict that the sum of the standard
molar entropies of the products would be greater than the sum
of the standard molar entropy of the reactants. In this case, only one reactant. And since we'd be
subtracting a smaller number from a larger number, we would predict that the
standard change in entropy for this reaction would be positive. Now that we've made our prediction, let's plug in our
standard molar entropies, and these are for a 25 degrees Celsius. And see if our prediction was correct. The standard molar
entropy of calcium oxide is equal to 39.8 joules per kelvin mole. In our balanced equation, we have one mole of calcium oxide. So we're gonna multiply
one mole of calcium oxide by the standard molar entropy
and moles will cancel. Next, we look up the standard molar
entropy of carbon dioxide, which is equal to 213.6
joules per kelvin mole and our balanced equation, once again, there's a one in front of carbon dioxide. So we're gonna multiply
the standard molar entropy by one mole and moles will cancel. So we're gonna add those two together. And from that, we're going to subtract
the standard molar entropy of our reactants, which
is calcium carbonate. So the standard molar
entropy of calcium carbonate is 92.9 joules per kelvin mole. And in our balanced equation, there's a one as a coefficient in front of calcium carbonate. So multiply that by one mole. And once again, the moles will cancel out. For this particular reaction,
in the balanced equation, all of the coefficients happen to be one. However, if one of our reactants or
products in a different reaction happened to have a coefficient of two, we would multiply the standard
molar entropy by two moles. After we do the math we find that the standard change in
entropy for this reaction is equal to positive
160.5 joules per kelvin. Therefore one mole of calcium carbonate decomposing to form one
mole of calcium oxide and one mole of carbon dioxide involves an increase in entropy just as we had predicted earlier on. Sometimes you'll see the
units as joules per kelvin, and sometimes, you might see joules per
kelvin mole of reaction. To understand how to get those units, let's just consider the standard molar
entropy of carbon dioxide. And that value is 213.6 joules per kelvin per one mole of carbon dioxide. And looking at the balanced equation, there's a one as a coefficient
in front of carbon dioxide. Therefore, there's one
mole of carbon dioxide per one mole of reaction. So molar reaction is just talking about, how the balanced equation is written. So the one mole of carbon
dioxide, it cancels out, and that leaves us with joules
per kelvin mole of reaction. So using these units is simply saying, for the decomposition of one
mole of calcium carbonate, there'll be a change in entropy of positive 160.5 joules per kelvin. Therefore, if we were talking about the decomposition of two moles of calcium carbonate, the change in entropy
would be twice this value that we already calculated.