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Problem 19

Without using Appendix $\mathrm{B}$ , predict the…


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Problem 18

Without using Appendix $\mathrm{B}$ , predict the sign of $\Delta S^{\circ}$ for
(a) $\mathrm{CaCO}_{3}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{CaCl}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}_{2}(g)$
(b) $2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)$
(c) $2 \mathrm{KClO}_{3}(s) \longrightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g)$

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Video Transcript

Chapter 20 Problem 18 is asking us to use a reaction to predict the sign of entropy. So first thing to remember is that in tribute is the freedom of particle motion. And this, looking at the dispersion of energy when you have an increase in entropy, that can happen as you go from a solid to the liquid to gas phase. So gas phase particles have ah greater dispersion of energy and the particle motion. Um, there particles have a greater freedom of motion and therefore it has a large entropy. Then you have a liquid which doesn't have as much free particle motion as gas, but still more than a solid and finally have a saw it where there's not much particle motion at all. So when you're looking at an increase, entropy you can look at is what phases it changing too. The change in entropy is too noted by having the final state minus the initial state. So you want to subtract the front inter Pete from the initial entropy to see what that front of entropy is. If the final entropy is larger than that initial inter Pete in the initial to be a small we never end up with the positive value. Put L. A here for large s m here for small. If your final entropy is smaller than your initial interest initial entropy and therefore you're subtracting a large number from a smaller number. You'll end up with a negative change of entropy. So look at several reactions and see how they change and determine what the predicted entropy of that system would be. Let's say we start off with a substance A which is a solid and react this solid with substance. B. There's two most of substance B and that substance B is a quiz. So we know that this solid has a low entropy in a quiz or liquid phase has a higher entropy than that of a solid This forms substance see, which is a curious as well. It also forms substance d, which is a liquid, and it forms substance E which is a gas. So if we're comparing here, we start off with solid and a liquid, and we formed a liquid and I never liquid and a solid I'm sorry and a gas. So at the end, the substances there each have a larger dispersion of energy than what we started off with, which is what which is a solid. So in this case, our final entropy is greater than our initial entropy. So we're doing a large number minus a small member, and we're expecting to end up with a positive sign here. And this is because the entropy at the end of the end was large and the entropy at the beginning that show this is all the entropy here at the at the end compared to the initial entropy, which was small, large, minus small will give us a positive sign. Let's compare that to another substance for another reaction where we're starting off with two moles of a gas, two moles of A and A is a gas. We're combining that with one mole of B. B is also gas. In our final product is two more of C, and in this case he is also a guest. So, as I described previously, when you're going from a solid to a liquid to gas, you're increasing the entropy. However, in this case, we start off with the gas and we finished with the gas. So the fact that we're just changing phases. Now it is important we have to think also about how much do we have? In the beginning, we started off with three moles of a substance to of a and one of beat two plus one uses three. At the end, we only end up with two moles of see So what we've done is we've actually combined parts of a B to give us that final product of C. So we've decreased the freedom of particle motion and this case there were three molecules that were able to move around, and at the end, we only have two molecules that are able to move around. So therefore we have decreased that freedom of part of commotion, and the entropy at the end is going to be small, then the entropy at the beginning. And again. That is because here there were three particles that had the freedom to move around. In this case, there's only two molecules or particles have the freedom to move around, so that end is smaller than the initial smaller minus the initial. We're going to end up with a negative entropy here. Fine. Let's look at another example where we start off with two moles of a se two moles of a B and this is a solid and we break this down We break it down into two moles of a which is a solid as well as three moles of B, which is a gas. And to make sure that this looks balanced, I'll put it three here. So we start off with two moles of a solid. We know that has solids have low entropy and we finished with two most of a solid, but also three most of the gas. So gases have large entropy, which means the entropy here at the end is larger because he had three months of a gas as well as to multiple solid compared to the beginning, where we on Lee had to most of the solace. This is a small entropy, therefore a large our final minus. Our initial, which is small, is going to give us a positive delta s so you can use the phases of a reaction as well as a number of most to predict the entropy. And you're just comparing it to whether the entropy at the end was larger than the initial entropy that'll give you a positive value. But of the interview at the end, we'll small the initial injury. You end up with a negative. Don't s.

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