Rank each set of substances in order of increasing standard molar entropy (S°). Explain your reasoning. a. NH₃(g); Ne(g); SO₂(g); CH₃CH₂OH(g); He(g)

Video transcript

Hey, everyone. And welcome back to another video rank each set of substances in order of increasing standard molar entropy, explain your reasoning. The substances are ammonia in his gas phase organ in his gas phase, selenium dioxide in his gas phase propanol in its gas phase. And krypton gas we're given for answer choices for. So we're going to solve this problem and then we will analyze them. So first of all, let's remember that the standard molar entropy depends on primarily we factors. Number one, that's a phase and all of them are gasses. So we cannot really compare their phases, meaning this is not the standpoint that we want to use to solve this problem. So they're all gasses, we're going to skip this step. Number two, that's the complexity of the molecule. How many atoms do we have in each molecule? So this is our second criterion to consider. And what we notice is that first of all, we have ammonia, that's a total of four atoms, right? We have three hydrogens and one nitrogen organ with one atom selenium dioxide with three atoms. Now, for propanol, we can see that we have a total of 12 atoms and krypton has one. So now let's remember that complexity is really important for us. We want to understand that the fewer the number of atoms, the less complex the molecule is. So now we have a choice between krypton and organ followed by the next molecule which has a greater number of atoms. So that would be selenium dioxide. So we're going to right down selenium dioxide followed by ammonia because it has four atoms and finally propanol. So we're going to state C three H seven oh. Now, we almost have our ranking. But now we need to determine which one between the two krypton or argon would have a lower standard molar entropy. And for that purpose, let's remember that molecular weight comes into play, the lower the molecular weight, the lower the entropy. So if we use the periodic table, we can say that argo is lighter than krypton. So Argan goes first, meaning our ranking would be organ followed by krypton because krypton is heavier, it would have a greater standard molar entropy. And then we just continue with the complexity of the molecules, selenium dioxide ammonia. And eventually after ammonia, we can say that we would have C three H 70 which is propanol. And looking at the answer choices, we can conclude that the ranking of interest would be option. A organ followed by krypton, followed by selenium dioxide, ammonia and propanol. Now, what would be the reason? Well, essentially as the answer suggests, a more complex substance will have greater entropy. When two substances have similar complexity, the substance with greater mass will have greater entropy. Now, for the other choices, definition would be a substance with greater mass will have greater entropy regardless of the complexity of the substance. That's incorrect, right? Because we said that first of all, we are considering states followed by complexity and only then we consider molecular weight C states that a less complex substance will have greater entropy, right, which is incorrect. It also states that when two substances have similar complexity, the substance with greater mass will have a lower entropy. So in this case, we can also say that it's incorrect just based on the first part of the statement because a more complex substance will have a greater entropy because it has a greater disorder. And our d is also incorrect as states a substance with lower mass will have greater entropy regardless of the complexity of the substance. Let's remember that the complexity of the substance is considered before considering the molecular weight. And therefore our final answer in the correct choice would be a thank you for watching.