Video: Identifying the Reaction for Which the Entropy of the System Will Decrease in a Set of Balanced Chemical Reaction Equations

For which of the following reactions will the entropy of the system decrease? [A] CACO₃(s) ⟶ CaO(s) + CO₂(g) [B] CO₂(s) ⟶ CO₂(g) [C] 4PCl₃(g) ⟶ P₄(g) + 6Cl₂(g) [D] N₂(g) + 3H₂(g) ⟶ 2NH₃(g) [E] None of the other answers are correct.

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Video Transcript

For which of the following reactions will the entropy of the system decrease? A) CACO₃ solid reacting to form CaO solid plus CO₂ gas. B) CO₂ solid reacting to form CO₂ gas. C) 4PCl₃ gas reacting to form P₄ gas plus 6Cl₂ gas. D) N₂ gas plus 3H₂ gas reacting to form 2NH₃ gas. Or E) None of the other answers are correct.

The entropy of a system has to do with the number of microscopic configurations that are available to the system given its properties, such as its current, its pressure, temperature, or energy. These microscopic configurations or microscopic arrangements of our system are called microstates. If a system has more microstates, that means it will have higher entropy. This definition of entropy is a little abstract. So let’s take a look at some examples.

Let’s compare two different systems. In one, we have a substance that is a solid. And in the other, we have a substance that is a gas. Both of these systems have the same number of particles. We want to determine which of these systems will have a higher entropy. Which we know is related to the number of microscopic configurations or arrangements of these particles. Whichever one has a greater number of arrangements will have a greater entropy. For the substance that is a gas, we can arrange the particles in the container pretty much however we want and the system will still be a gas. Each of these different arrangements or configurations that I’ve drawn are called microstates. I have drawn but a few for our system of seven gas particles. But we could certainly imagine many more.

For a real system of gas particles, which would have about one mole of particles, that system would have about 10 to the 10 to the 23 microstates. Which is a mind bogglingly huge number of microstates. We can arrange the particles of the gas almost however we went to in a container. And that substance will still be a gas. This means that a gas will have many microstates. A solid, in comparison, will have far fewer microstates because there’s fewer ways that we can arrange the particles where that substance will still be a solid. After all, solids have to be stuck together to be a solid, which isn’t true for a gas. Since, in general, gases have more microstates available to them than solids do and since more microstates mean more entropy, gases will have greater entropy than solids do.

We can use this information to create a general rule of thumb to help us decide which substances will have more entropy than others. The entropy of a gas will be the greatest, followed by the entropy of an aqueous solution. Which will be greater than the entropy of a liquid, leaving the entropy of the solid as the smallest. Now that we know, in general, what substances will have greater entropy than others, let’s think of a process where we have a change in entropy. Such as a change where we go from a solid to a gas, like in answer choice B, the sublimation of CO₂.

The change in entropy for this process would be the entropy of the gas minus the entropy of the solid. Since the entropy of a gas is generally greater than the entropy of a solid, we would expect this process to have a positive change in entropy. I do want to note one more time that we’re using a rule of thumb here to guess the sign for the change in entropy for our system. We can’t guarantee that every time a system goes from a solid to a gas, it will have a positive change in entropy. This rule of thumb, however, is really great for solving these types of problems and will get you in the right ballpark most of the time.

Now that we understand a little bit about entropy and how to determine if a reaction will have a positive or negative change in entropy, let’s return to our question. This question is asking us to determine which reaction will have a decrease in the entropy of the system. So we’re looking for a reaction where the change in entropy will be negative. So let’s take a look at our first answer choice. Here we have one mole of a solid forming a mole of a solid and a mole of a gas. Since we’re forming a gas in the products and we have no gaseous reactants, we would expect that this process would have a positive change in entropy. Since gases, in general, have much more entropy than solids do.

Since we’re looking for a reaction where the entropy change for the system will be negative, answer choice A is not the right answer. We used answer choice B as our example. As we’ve already discussed, we would expect this process to have a positive change in entropy because a gas has more entropy than a solid does. In our third answer choice, all of our products and reactants are gaseous. In our reactants, we have four moles of gas. And in our products, we have seven moles of gas. Since we have more moles of gas in the products and gases have so much entropy, we would expect that the entropy change for this process would be positive. Again, we’re looking for a decrease in entropy. So answer choice C is not correct.

In answer choice D, both our products and our reactants, again, are all gaseous. But this time, we start off with four moles of gas, and we form two moles of gas. Since the number of moles of gas is decreasing, we would expect the entropy of this reaction to be negative. That means that answer choice D is the answer we’re looking for. This reaction would have a decrease in the entropy of the system.

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