Absolute entropy and entropy change | Applications of thermodynamics | AP Chemistry | Khan Academy

Entropy can be measured on an absolute scale, which means there is a point of zero entropy. 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 the 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. At 0 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 0 at 0 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 could 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'll use these units for entropy. Now that we understand the concept of zero entropy, let's look at the entropy of a substance. 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. 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 0 Kelvin. So really, it's a change in entropy, and therefore it would be 5.7 minus 0, with 0 being the entropy of a hypothetical crystal at 0 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 knot 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 1 atmosphere. For gases, the standard state is referring to the pure gas at a pressure of one atmosphere. 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 1 mole of a pure solid under a pressure of 1 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, gases have higher standard molar entropies than liquids, as we can see comparing the values. 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 gases. 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, have 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 dot 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. 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. 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.

Therefore, we're going to 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.

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. 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 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, and in our balanced equation, we have one mole of calcium oxide, so we're going to 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. In our balanced equation, once again there's a one in front of carbon dioxide, so we're going to multiply the standard molar entropy by one mole and moles will cancel. So we're going to add those two together, and from that, we're going to subtract the standard molar entropy of our reactant, which is calcium carbonate.

The standard molar entropy of calcium carbonate is 92.9 joules per Kelvin mole, and in our balanced equation, there's a 1 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 the coefficients happen to be one. However, if one of our reactants or products in a different reaction happens to have a coefficient of 2, we would multiply the standard molar entropy by 2 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. 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 the mole of reaction is just talking about how the balanced equation is written. The one mole of carbon dioxide cancels out, and that leaves us with joules per Kelvin mole of reaction. Using these units is simply saying for the decomposition of one mole of calcium carbonate, there will 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.