# Video: Predicting the Change in Entropy in a Balanced Chemical Reaction

In which of the following processes is the change in the entropy of the system most negative? [A] 2Na(s) + 2HCl(g) ⟶ 2NaCl(s) + H₂(g) [B] Zn(s) + 2HNO₃(aq) ⟶ Zn(NO₃)₂(aq) + H₂(g) [C] 2C₆H₁₄(l) + 19O₂(g) ⟶ 14H₂O(g) + 12CO₂(g) [D] 2H₂O(l) ⟶ 2H₂(g) + O₂(g) [E] PbS(s) ⟶ Pb²⁺(aq) + S²⁻(aq)

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### Video Transcript

In which of the following processes is the change in the entropy of the system most negative? A) 2Na solid plus 2HCl gas reacting to form 2NaCl solid plus H₂ gas. B) Zn solid plus 2HNO₃ aqueous reacting to form Zn(NO₃)₂ aqueous plus H₂ gas. C) 2C₆H₁₄ liquid plus 19O₂ gas reacting to form 14H₂O gas plus 12CO₂ gas. D) 2H₂O liquid reacting to form 2H₂ gas plus O₂ gas. Or E) PbS solid reacting to form Pb²⁺ aqueous plus S²⁻ aqueous.

The entropy is related to the number of microscopic configurations that are available to the system given its current conditions, such as its energy, its temperature, its pressure, the volume, etcetera. We call these microscopic configurations microstates. The more microstates the system has, the greater its entropy will be.

As an example, let’s compare these two systems. In one, we have a substance as a solid. And in the other, we have the substance as a gas. In a solid, the particles often have to be arranged in a specific way, like they are in a crystal. But in a gas, the particles have much more freedom to move around the container. This means that a gas will have many more microstates than a solid will. I’ve drawn a few of the different microstates for our system that contains seven gas particles here, but there would be many, many more.

If our system contained just one mole of gas, we would have more than 10 to the 10 to the 23 microstates available to our system, which is a mind-bogglingly huge number of microstate. Since more microstates means more entropy, a gas will have a much greater entropy than a solid. In general, the entropy of a gas will be greater than the entropy of an aqueous solution which will be greater than the entropy of a liquid which will be greater than the entropy of a solid. We can use this rule of thumb to answer the question.

As an example, let’s consider a process where we have a species A as a solid forming species B as a gas. The change in entropy is equal to the entropy of the products minus the entropy of the reactants. Since the product is a gaseous species, it will have more entropy than the reactant, which is a solid. So, the change in entropy for this process will be positive.

Now, let’s consider a process where we have 2C as a gas forming D as a gas. Both the products and the reactions this time our gaseous. But there’s more moles of gas on the reactant side. This means that the entropy change for this process would be negative. With all this in mind, let’s take a look at the answer choices to determine which one will have the most negative change in entropy.

In the first process, we go from two moles of solid and two moles of gas in the reactants to two moles of solid and one mole of gas in the products. The amount of solid stayed the same, but the amount of moles of gas decreased from products to reactants. So, we can predict that this process will have a negative change in entropy because the moles of gas decreased.

In the next process, we have one mole of a solid and two moles of an aqueous species in the reactants, and one mole of an aqueous species and one mole of a gas in the products. Since we gained a mole of gas in the products, we can guess that this process has a positive change of entropy.

In answer choice C, we have two moles of liquid and 19 moles of gas in the reactants forming a total of 26 moles of gas in the products. Since the amount of gas increased so much in the products, we would also expect this process to have a positive change in entropy.

In answer choice D, we have two moles of liquid in the reactants and a total of three moles of gas and the products. Since we have three moles of gas in the products and no gas in the reactants, we would also expect this one to have a positive entropy change.

In the final answer choice, we have one mole of a solid in the reactants and two moles of aqueous species in the products. Again, according to our rule of thumb, we would also expect this answer choice to have a positive change in entropy because aqueous species have more entropy than solids. This means that the process described in answer choice A, where we have 2Na solid plus 2HCl gas forming 2NaCl solid plus H₂ gas is the process that will have the most negative change in entropy.

If we wanted to calculate the entropy changes of these reactions numerically, we could do that by using standard molar entropies. Standard molar entropies are tabulated entropy values for most chemical species that are determined at a specified temperature and pressure. We could calculate the standard molar entropy change for each process by taking the sum of the standard molar entropies for the products times their stoichiometric coefficients minus the sum of the standard molar entropies for the reactants times their stoichiometric coefficients.

Doing this calculation for each of the processes, we can see that answer choice A indeed has the most negative change in entropy. However, our rule of thumb led us astray for a few of the answer choices, which that’s why it’s just a rule of thumb.