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19. Chemical Thermodynamics

**Entropy** is the disorder or chaos associated with a system’s inability to convert thermal energy into mechanical work.

Entropy
and Spontaneity

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Hey, guys, In this new video, we're gonna take a look at a new variable, the variable of entropy, and we're gonna see how does the entropy relate to the spontaneity of a chemical reaction? Now, what we're gonna say here is we're going to say that the second law of thermodynamics states that molecular systems tend to move spontaneously toe a state of maximum disorder or randomness. Now, what does this mean? This means that the universe does not want to get ordered. Doesn't wanna come together in a nice little package. The universe naturally wants to go into a chaotic mode. Everything is spinning out of control. Everything is breaking down, becoming less ordered. This is the natural process of our universe. So simply put, the second law says that the universe is going to shit. Okay, so just think of it like that. The universe is getting crappy. Er, by the day, this is the natural state of the universe. This is all the second law of thermodynamics is saying now. On the opposite end of this, the third Law of thermodynamics says that at absolute zero, absolute zero is zero kelvin at absolute zero. The entropy of all objects is equal to zero. So basically, everything is always gonna be in disorder. Always trying to get more and more disordered. Unless it's at zero. Kelvin at zero. Calvin, entropy or chaos will stop. And remember, this is the third law of thermodynamics. So it goes against the second law of thermodynamics. Now we're gonna say that again. Entropy is chaos, disorder, randomness, shiftiness, whatever you want to say, it's called, and the variable we use for entropy is s.

**The 2 ^{nd} Law of Thermodynamics** states systems (chemical reactions) spontaneously move to state of disorder.

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Now let's think about this in a molecular image wise. So we're gonna say in general, as we move from a solid to a liquid toe, a gas, what's happening to the molecules of any compound? So remember, a solid all the molecules are tightly packed together on top of each other. Then, as we become a liquid, the molecules air still around each other. But because it's a liquid there, naturally sliding over each other, moving around one another. And then as they become a gas, they become very spaced out, so they're bouncing all over the place. So we're gonna say in this natural progression, when we go from solid liquid to gas, we're gonna say entropy or chaos will increase because the molecules get farther and farther apart and of entropy is increasing in the sign of entropy will be positive. Now think about it in the opposite way. In the opposite way. We have gasses which are very spaced out. They condense into a liquid. Now the molecules are even closer together. They're not frozen in place yet. We haven't gotten to a solid yet, but they're sliding over one another. Then finally, when we get to a solid. They're frozen in place. So in this direction, we're gonna say that entropy decreases because order is increasing. And if entropy is decreasing, the sign of entropy will be negative.

During a phase change as our molecules grow farther apart then entropy will increase.

During a phase change as our molecules grow closer together then entropy will decrease.

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now taking that into consideration. Let's look at example one, it says, which should have the highest Moeller entropy at 25 degrees Celsius. So here I give you gallium and it's three different phases. We have liquid solid and gas. Now remember, we're just saying who has the highest entropy? We just said Who has the most chaos, whose molecules are most spread out. This is a very easy question. We're gonna say simply the gas phase would have the most entropy because it's the most spread out.

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we're gonna say which substance, for example, to which substance has greater mola entropy. So this will be we're gonna compare these two, these two and these two. So within those three choices, we're gonna choose the best out of each one. So the first one we have ch four gas or CCL four liquid. Now, the way you look at these types of questions is you first look at their faces, Okay, so we first look at phases and are you going to say here is gasses have higher entropy, then liquids and liquids have higher entropy than solids. Now, if they're tied in the same phase, then we move on to complexity. And what the heck do I mean by complexity? All I'm saying is the Mawr elements that make up the compound the more entropy. So the mawr elements that make up the compound than the mawr Entropy. So, for example, if I gave you co two gas versus S eight gas, you would say that CO two is made up off three different atoms. One carbon to oxygen's. An essay is made up of eight atoms of sulfur. So it has mawr elements that make it up. So it has more entropy, so essay would have more entropy. Now, if they have the same number of elements that make them up, then we go to overall mass. And here for three, you're gonna say more mass equals more entropy. Because think of it like this. We said that entropy is chaos, destruction, disorder, randomness. So if we want to think about it, think of it as two cars on the highway. You have a little mini Cooper versus Hummer. Okay, You're gonna say the Hummer is much larger, and if it gets in an accident, it's gonna cause more destruction, more disorder. So we're gonna say the Hummer has a greater mass compared to the smaller car compared to the Mini Cooper. So just think of it. In those terms, more mass equals bigger sides equals mawr chaos that we can have So here, their different phases. So we can easily say it's the gas now for the next one. They're both gasses, so the first one doesn't help us. Let's look at the second criteria. We're gonna say complexity. Here. Both have the same number of atoms. One, there's only one neon Wonsan on so that doesn't help us. So then we move on to the third one mass. We're gonna say that Zen on on our periodic table has a higher atomic mass than neon. And because of that, Zen in has a higher entropy. Then finally, let's look at the last one. We have ch 30 h liquid versus C six h 50 h liquid. So they both have the same face again. Sold. The first rule doesn't help us. What we do next is we look at complexity. We're gonna say C six h 50 h definitely has mawr elements that make it up. So it's more complex, more moving parts mawr, chances of chaos and entropy. So we're gonna say C six h five has mawr entropy Now you could have also simply said that C six h five has a greater mass and because of that, it should have the greater entropy. So remember, Delta s is just chaos or disorder. The farther molecules are from each other in their phases than the greater the entropy. So gashes compounds have mawr entropy than liquids or solids. And we're going to say that the natural tendency of the universe is to be in disorder. So the second law of thermodynamics is saying that the universe is going to shit. Or you could just say that the second law of thermodynamics is saying that the entropy of the universe must always be greater than zero. Because if entropy is always increasing in our universe, there's no chance for it to be less than zero. Always has to be greater than zero. Just remember that for the second law of thermodynamics

Entropy
and Phases

When comparing the entropy of different compounds then we must follow a set of guidelines in the following order.

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Hey guys, in this new video, we're going to continue with our discussion of entropy and the way it relates to spontaneous reactions. So let's take a look. At example, one an example. One, it says arranged the following substances in order of increasing entropy at 25 degrees Celsius. So I want in order of increasing entropy. So that means least entropy to most entropy. Now, remember, we talked about the criterias we have to look at to determine who has the most entropy. First you have to look at the faces and remember, gasses have the most, then liquids, then solids. Now, your professor might be a little crafty in terms of this. Just remember, they might use the word acquis instead of liquid. But we're gonna say right now that those words are interchangeable. Liquid is just the solvent that's dissolving the compound. Acquis means that it's dissolving within our liquid. So we're gonna basically say they're similar to one another. Next, if they have the same phase, then we look at complexity. Remember, the more elements that you have within the compound that make it up, the more complex it is, and therefore, the more entropy it has. So let's say, for example, we're looking at two compounds that had the same face. Let's say we're looking at C 286 at the gas and C three h eight propane gas. They're both gasses. So then we have to look at complexity. We're gonna say propane, which is C three h eight is made up off 11 atoms. Three carbons, eight hydrogen ethane, on the other hand, is only see two age six. It only has eight elements making it up. The more elements within your compound, the more complex it ISS and therefore the higher it's entropy. So C three h eight will be higher. Next. We're gonna say if they have the same number of atoms, then we look at mass. Are you going to say the greater the mass, the greater the entropy? Remember the whole analogy? We have two cars out of control on a highway. Who's gonna cause more damage? The Mini Cooper or the Hummer? The Hummers, much larger, has greater mass, therefore can cause greater damage. So let's take a look here. We have to put them in order. We have a gas here and we have a gas here. So those two, we expect to have the highest entropy. So because they're both gasses, we go on to the next criteria. Complexity H. I is made up of two different elements. H two is also made up of two elements, so they are tied for complexity. So the tie breaker has to be mass h. I has greater mass or h I should have greater entropy. Now that we're done with gas is we can move on to acquiesce and liquids. Here we have liquid and liquid, so we're gonna say H g is liquid and B R two is liquid. Br two has greater complexity because there's too bro means together and then murky would be next. Next we're gonna look at is Zenon Tetra Florida X E F four vs b o. They're both solids, right? So they're tied and phase. So we have to look at complexity. The one with more complexity will have greater entropy. So Zen in Tetra Florida's XY F four. That's five elements that make it up be Ailes Only two elements that make it up. So it be Xena tetra fluoride solid and then finally, barium oxide would have the least entropy because it's ah solid, which has the lowest and also it's only made up of two elements. Therefore, its complexity is much lower than XY F four. Hopefully, guys were able to follow along with these three criterias, so just remember, look at face first, then look at complexity. If that all fails, look at mass. The greater the mass, the greater the entropy. Now that we've seen this example, let's move on to example, too.

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So example to says, we have container A which is this one and container B. They have two different gasses and we allow both of these gasses to funnel into a new container container. See, just let you know it's not drawn to scale. All these gasses in here are in here. Even though I don't show all of them, they're considered all of them to be in there. All the gasses here are also in here now, based on the image of containers, see, what is the sign off entropy? Delta s. And here this little zero here just means standard conditions. When we say standard conditions, we mean one atmosphere and 25 degrees Celsius. That's our standard state. When we're talking about thermodynamics, don't get confused. We also dealt with standard conditions. When we deal with gasses, we say STP now STP for gas is is different under those standard conditions that pressures the one atmosphere. But the temperature is zero degrees Celsius. When we're talking about spontaneity and thermal dynamics, the new temperature now becomes 25 degrees Celsius. Now just think about it like this Delta s chaos, the more disorder, the better. So we have one container filled with gas is we have a second container filled with gas is we? Take those two containers which are very chaotic, and dump them all together into even bigger container where they can just mix around even Mawr bouncing all over the place. So ask yourself, if I take both gasses and I put them together, will the entropy increase or decrease? I have a party of 100 people have another party of 100 people. I take them and now I have a party of 200 people. Will they get even crazier together than apart? That's how you have to think of it. We're gonna say here, the more gas is you put in the crazier things they're gonna get because they're gonna be bouncing around the wall because they're even more of them. So we're gonna stay here. Entropy is positive. So we should get a positive entropy when we mix gasses together because we're causing greater disruption. Greater chaos. Now that we've done examples one and examples to I want you guys to at least attempt to do this following practice question and I'll give you guys a few hints. So here. We're saying an ideal gas is allowed to expand it. Constant temperature. What are the signs for Delta H. Delta S and Delta G? First of all, for Delta s remember as his chaos and entropy here the gasses and A are becoming like the gasses and be they transition. And now they've become like the picture And be ask yourself, since the gasses have more room to bounce around, will their entropy increase or decrease? Next? Delta G. I'll tell you this. Dull Taji. When it's negative, it's spontaneous. When it's positive, it's non spontaneous. So ask yourself this. An ideal gas is a gas that wants enough room for itself. Based on that information, is Delta G positive or negative? Finally, Delta H is entropy. Entropy is just the measurement off a reaction if it absorbs energy or if it releases energy. And when we talk about energy, we're talking about heat energy in this case, So if it's absorbing energy, it's gonna be positive. If it's releasing energy, it's going to be negative, I say. Constant temperature. So what's the sign of Delta H When we realized that were a constant temperature visa, little help, hints will help you figure out what the answer is. And don't worry. If you can't figure it out, just come back and click on the explanation button. You'll see a video of me explaining how to approach this problem. Good luck.

7

Problem

An ideal gas is allowed to expand at constant temperature. What are the signs of ∆H, ∆S & ∆G.

A

∆H = +∆S = +∆G = +

B

∆H = -∆S = +∆G = -

C

∆H = 0∆S = +∆G = -

D

∆H = 0∆S = +∆G = +

Whenever a phase change occurs, first determine if bonds are broken or formed to figure out the signs of enthalpy and entropy. Afterwards, determine if the reaction is spontaneous to determine the sign of Gibbs Free energy.

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Hey, guys, In this new video, we're gonna continue with our discussion off entropy and spontaneous reactions. But now introduce two new variables. Entropy, Delta H or Gibbs Free Energy, Delta G. Now, let's take a look at this example. It says consider the spontaneous fusion off ice at room temperature for this process. What are the signs for Delta H. Delta S and Delta G? Before we even take a look at this question, I'm gonna give you guys some notes to help you realize what the sign of Delta H and Delta s can be. So we're gonna say here if we're going from gas toe liquid to solid, what's happening to our molecules are molecules of coming closer and closer together until they reached a solid state in which they're frozen in place. We're gonna say technically here were forming bonds or forming inter molecular bonds. Molecule A is coming closer molecule B. So they're forming an attraction to each other. It's not a real bond. It's more of an inter molecular force bond. When we're gonna say here, if you're forming bonds, then when we must be releasing energy in order to form those bonds, so Delta H would be negative on the opposite end. We're gonna say Delta s what's happening to the entropy. We're becoming mawr ordered. Bonds are coming closer and closer together. Now you are solid, so you're entropy should also be negative. Now if we go the opposite way, we go from solid toe liquid to gas. What's happening or breaking these inter molecular bonds? And remember, to break bonds, you have to first absorb energy or heat and use that energy you just absorbed in order to break those bonds. If you're absorbing energy, your Delta H will be positive. At the same time. Our Delta s what's happening to the molecules. They're becoming more and more like gas is gas is have tons of chaos. So chaos is increasing, so Delta s should be positive. So this is the way you have to understand it. Whether you're forming bonds or breaking bonds will help you figure out. Are you having a Delta H and Delta S? That's positive, Or are they negative now? If we go back to the question it says fusion. Now the name could be a little bit deceiving. Scientifically fusion means melting when it comes to face changes. Fusion just means melting. And if you're melting, what faces are you going between? So remember what melts we're gonna see. If you go from a solid to a liquid, you're melting. And if you're going from a solid to a liquid, what are you doing to bonds? You're breaking them. Therefore, Delta H and Delta s should both be positive. And because of that, that takes out B, C and E. So our answer is either a or D in this question. Now, the next important piece of information I give to you is the word spontaneous because I say it's spontaneous. Remember, we've said this earlier. If it's a spontaneous reaction, Delta G has to be negative. If it's spontaneous, Delta G is negative. So that one word that I used makes it answer choice. D. Okay, so just remember, are you forming bonds or breaking bonds? This will help you figure out Delta H or Delta s for Delta G. You have to think in your head. Does this naturally occur? Doesn't make sense for this to occur. If it does, then it's a natural process. And if it's a natural process, it's spontaneous. If it's spontaneous, Delta G will be negative. Now. Knowing this, I want you guys to attempt to do the next question on your own. So here we say, Consider the freezing of liquid water at 30 degrees Celsius. So just remember, does freezing mean you're making or breaking Bonds? This will help you figure out what Delta H and Delta s are in terms of signs. Then ask yourself this question. Freezing of water at 30 degrees Celsius. Is that a spontaneous reaction, which is natural, or is it non spontaneous? If it's spontaneous, Delta G will be negative. If it's non spontaneous, Delta G will be positive. Knowing all this, try to answer this question as best as you can. Once you're done, click on the Explanation button and take a look at a video of me explaining how to approach this problem. Good luck, guys.

9

Problem

Consider the freezing of liquid water at 30^{°}C. For this process what are the signs for ∆H, ∆S, and ∆G?

A

∆H = +∆S = -∆G = +

B

∆H = -∆S = +∆G = 0

C

∆H = -∆S = +∆G = -

D

∆H = -∆S = -∆G = +

E

∆H = -∆S = -∆G = -

If bonds are broken then the entropy of a reaction increases, but if bonds are formed then the entropy of a reaction decreases.

10

Problem

Predict the sign of ∆S in the system for each of the following processes:

a) Ag^{+} (aq) + Br^{ -} (aq) → AgBr (s)

b) CI_{2} (g) → 2 CI^{ -} (g)

c) CaCO_{3} (s) → CaO (s) + CO_{2} (g)

d) Pb (s) at 50^{°}C → Pb (s) at 70^{°}C

A

a) ∆S = + b) ∆S = + c) ∆S = + d) ∆S = +

B

a) ∆S = - b) ∆S = + c) ∆S = - d) ∆S = +

C

a) ∆S = - b) ∆S = + c) ∆S = - d) ∆S = -

D

a) ∆S = - b) ∆S = + c) ∆S = + d) ∆S = +

11

Problem

For each of the following reactions state the signs of ∆H (enthalpy) and ∆S (entropy):

a) **Fusion** of ice.

b) **Sublimation** of CO_{2}

c) **Vaporization** of aqueous water.

d) **Deposition** of chlorine gas.

e) **Condensation** of water vapor.

A

a) — b) — c) + d) — e) +

B

a) — b) — c) — d) — e) —

C

a) + b) + c) + d) — e) —

D

a) — b) — c) — d) + e) +

Entropy
and Calculations

The 2^{nd} Law of Thermodynamics states that the entropy of the universe is always increasing and so it must always be greater than zero.

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Hey, guys, In this new video, we're gonna take a look at calculating entropy for a system. Now, first of all, before we begin, just remember, we have the term system. When we say the words system, we mean that this is the portion of the universe that we're concerned with when we say system, we're really talking about the balanced chemical equation. So a system is just are balanced equation that we're looking at. Now we're saying that the second law of thermodynamics talks in terms of the system, and it says that the entropy of a system increases spontaneously. Remember, the entropy of the universe is gradually increasing. The world is getting mawr and more chaotic. Now we're going to say that we can't Onley focus on our system. Besides our system, we have our surroundings. We're gonna say our systems, plus our surroundings gives us our entropy total or our universe. Okay, so system plus surroundings gives us our total universe. Now we're gonna say thus, to calculate the total entropy change Delta s total. We use the following equation. So we're just gonna say Delta s total is the same thing as Delta s universe and that just equals Delta s of my system. Plus delta s of my surroundings. Okay, remember when we say system, we really mean reaction. So Delta s system could also mean Delta s off our reaction.

The entropy of the universe takes the look at the entropy of our system (the chemical reaction) and of the universe.

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Now, we're going to say here in terms of spontaneity, we're going to say if our delta S total is greater than zero, then the reaction is spontaneous. If it's less than zero, then it's non spontaneous. And if it's equal to zero, then our reaction is simply at equilibrium. OK. So just remember that we have to take into consideration both our system and our surroundings as a whole to determine if a reaction is spontaneous overall or not. Because sometimes your delta S system may be highly spontaneous, meaning that the, it's delta S value is very positive. But if we add in the surroundings and at the end, we get a total that's negative, then overall it would be non spontaneous. Have to take into consideration both aspects our system and our surroundings.

If the entropy is greater than zero then we classify the process as spontaneous.

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So if we take a look here, it says, calculate the standard entropy in kilojoules of a reaction at 25 degrees Celsius for the following reaction. So we have one mole of nitrogen gas, combining with three moles of hydrogen gas to give us two moles of ammonia gas as product. And here I give you the standard molar entropy of N two H two and N H three as these numbers respectively. Now any time your professor asks you to find the delta s of our system, they're really saying again, the delta s of our reaction and that is just products, the summation of products minus the summation of reactants. So products minus reactants. And remember this also goes for delta G and delta H. Sometimes you'll be asked to calculate the standard entropy delta H or the standard gives free energy. Delta G. In those cases, delta G system del delta H system are products minus reactants. So here we're just gonna say it equals we have, remember the coefficients 23 and one. So we're gonna say according to our balance equation, we have two moles of N H three, Each NH 3 is this total right here. That's our products minus our reactants. We have one mole and two, each one is 1 91.5 Jews over K times moles plus the other reactant, which is three moles of H two times 1 30.6 Jews over K times moles. Now, we can see that the moles will cancel out and our units here will be in jewels over Kelvin. So when we multiply the two times the 1 92.3, that's gonna give us 3 84.6 Jews over Kelvin minus everything inside of our reactance boxes that comes out to 5 83.3 Jews over Kelvin. Now just subtract those two that gives me -1 90 98.7 jewels over Kelvin. But remember I asked for the answer in killer Jews not in jewels. So we have to do one more conversion. We want kilojoules, one kilojoule, one kg is 10 to the three jewels and remember 10 to the three is the same thing as 1000. So jewels cancel out. So our units at the end will be kilojoules over Kelvin. So it's gonna be negative 0.1987 kilojoules over Kelvin. So that'll be our answers. Just remember if they ask you to find the standard entropy of our reaction. It's just products minus reactants. Now, one more thing we need to realize here just before we move on to the next set of videos. Just remember here, we're gonna say that N two and H two are both in their natural states. Remember your natural state means you're by yourself. You're an element by yourself. Example, carbon graphite or sodium solid or you're connected to copies of yourself such as N two or H two or S eight or P four. So if you're by yourself, you're connected to copies of yourself, that's your natural state. Now we're gonna say here, although N two and H two are both in their natural states, their delta S values are not equal to zero. Remember delta S um uh value is only equal to zero if the temperature is at absolute 00, Kelvin. Now we're gonna say that for an element in its natural state. Delta H and delta G both equal zero. So if they give us the values for N two and H two here, they both would be zero when we're looking at the delta H and their delta G values. Why? Because they're in their natural state. So I just want you guys to remember this for something in its natural state. It means it's by itself or it's connected to copies of itself. In either case, delta G and delta H equals zero. Delta S can only equal zero for those compounds if we're at zero. Kelvin, remember that's the third law of thermodynamics. Just remember that difference. And you guys can answer those types of theory questions when you see them on exams.

The entropy, enthalpy and Gibbs Free energy of a reaction is equal to products minus reactants.

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Hey, guys. So here we're gonna take a look at the entropy of the system here. We're gonna have to calculate the entropy of not only our system, but also of our surroundings at it. And at the end, with all this information will have to determine is our overall reaction of spontaneous one or non spontaneous one. So if we take a look at party, it says, calculate the entropy of my system Now here we could have also said, What's the entropy of my reaction? Because your reaction in your system are the same thing here. The standard Moeller entropy is of each of the compounds are given in these respective values now realize everything around us has entropy. No matter if it's in its standard state or if it's in its nonstandard state. The only way something can have an entropy equal to zero is if it's at absolute zero, which is zero Kelvin. And since our temperature here is not a zero Kelvin, we're going to say that they all have real values, really entropy values. Now, whether it be entropy of system or entropy of reaction, it's just products minus react. It's and here remember coefficients. We'll play a part in our answer. We're gonna need room to do all these calculations, guys. So let me take myself out of the image. So we're going to say when we plug everything in. So our product, we have two moles off iron three oxide, And that's because the coefficient of to each one has a value off 87 0. Jules over Kelvin's times molds minus your reacted. So we have four moles. Iron solid. Each one is 27.3 jewels over Kelvin's Times mold. Plus, we have three moles of +02 Each one is 0.0 jewels over Kelvin's times moles. So moles cancel out so we'll have our answer in jewels over Kelvin. So that comes out to negative 5 49.4 jewels over Kelvin. Here it's customary to have this intelligible sold to one more, converting. So for every 1 kg, it's 10 to the three. So that's negative. 0.5494 killer jewels over Kelvin as our final answer for Part B. We have to figure out the entropy off our surroundings so he remember entropy of your surroundings is equal to negative delta H of reaction or system divided by temperature and Kelvin. So to get Calvin, we had to 73.15 toward degrees Celsius, so that's to 98.15 Kelvin. Now remember whether it's entropy of system or surround of system or reaction, Whether it's envelop off system or reaction, it's still products minus reactions. Okay, so notice the similarities now realize here in entropy, I gave us all the our centerpiece of formation, but when it comes to envelop, I only gave us the Mueller envelop of one compound. Why didn't I give us the entire piece off iron and 02? That's because when something is in its standard state, when an element is in its standard state, it's Delta H, and it's Delta G gives free energy. They're both equal to zero. So here I would be zero and oxygen will be zero for both Delta H and Delta G entropy, though everything has entropy unless where zero kelvin. So it be products minus reactions. So we only care about products and there's two moles off iron three oxide in our equation, which is negative. 8 24.2 kill jewels promote. Both are reactions are in their standard states, so we don't worry about them. So that gives me negative 1648.4 killing jewels, so I'll plug it up here. I remember there was already a negative sign, So this is a negative of a negative. So that comes out to 5.53 killer jewels over Kelvin. And now that we found those two parts you can find in two parts See, we need to figure out Delta s of our universe or total. So that's just adults s system, plus Delta s surroundings. So we just plug in the values that we found in part A and part B. So that gives me 4.98 killer jewels over Kelvin. And we're gonna say here because our adults as total is greater than zero. That would mean that this reaction is spontaneous. So that's how we find each of the components necessary to figure out our overall entropy. And again, as long as entropy overall is greater than zero, it's spontaneous. If it's less than zero, it's non spontaneous. And if it's equal to zero, then we're at equilibrium.

16

Problem

Diethyl ether (C_{4}H_{10}O_{2}, MW = 90.1 g/mol) has a boiling point of 35.6^{o}C and heat of vaporization of 26.7 kJ/mol. What is the change in entropy (in kJ/K) when 3.2 g of diethyl ether at 35.6^{o}C vaporizes at its boiling point?

A

∆S = 3.07 x 10^{-3}

B

∆S = 3.07 x 10^{3}

C

∆S = 9.48 x 10^{-1}

D

∆S = 9.48 x 10^{-3}

17

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Hey, guys, we're not gonna continue with our discussion on solving questions that deal with spontaneity and entropy. So let's take a look at this example. It says the normal boiling point of liquid propane is 231. Kelvin, what is the entropy of vaporization of liquid propane Now, first of all, remember, they don't have to tell you explicitly entropy of vaporization. Instead of saying entropy, they could have said heat of vaporization. In either case, it just means Delta H. So all they're telling me here is the normal boiling point of liquid propane. And from that I'm supposed to figure out what my delta H is. Now. This question is a little bit sneaky, but just realize it follows a major piece of theory. When it comes to this chapter, this is called Trout owns rule. So Trotman's rule says many substances when they go from one more of liquid 21 mole of gas, they're gonna have an entropy value, an entropy off vaporization that is equal to 88 Jules over moles Times K. So just remember, we're no, we're dealing with Charlton's rule because here we're talking about boiling point. Remember remember, if you're boiling your vaporizing your substance, so if you're boiling, you're going from liquid to gas. So again, that deals with the trial tins rule. So all we're gonna stay here now, using Charlton's rule, the formula is Delta s equals Delta H over T temperature and Kelvin. And if you look hard enough, you'll see that this is similar to the entropy off surroundings because entropy of surroundings equals negative delta h of reaction divided by our temperature. So Trotman's rule takes a little spin on that same exact equation to give us this one. So now we know that the Delta S is going to be 88 because we're talking about normal boiling point. So normal means we're dealing with one mole of a liquid. And we're talking about his transition from the liquid state to the gaseous state. And again, Troughton rule says as we make the transition from one mole of liquid toe, one mole of gas, the entropy of vaporization of that substance is about 88 jewels over moles times K. So we know this is 88 Jules over moles times k. We're looking for Delta H. Then we're gonna say the temperature is the temperature I gave to you to 31 Kelvin. So we have to do now is just isolate Delta H. So multiply both sides by 2 31 Kelvin Kelvin's cancel out, and at the end, your unit for Delta H will be in jewels over moles. So that will be 20,328 jewels over moles. And if you're professor wants it in killing jewels, which is the standard units, just remember one killer jewels equal to 1000 jewels. So remember, boiling Point has to do with vaporization. Normal means we're dealing with one mole, one mole evaporation or vaporization of liquid to gas means we're following Trotman's rule, which means your Delta as value is 88 jewels over most times. K. You should know that value because you just plug it in in order to solve For Delta H. Now that we've seen this first example, let's move on to the second example. So this one says, What is the entropy change? So what is Delta s associated with the expansion of one mole of an ideal gas from 2.5 leaders to 6. leaders at a constant pressure of 1.25 atmospheres. So what we need to realize here is we have a change in volume here, and they're talking about pressure on what you should realize is any time we're dealing with pressure and change in volume, we're finding work because work equals negative pressure. Times the change in volume, we're gonna say pressure has to be in atmospheres, and then volume is going to be in leaders. And so what equation connects work with Delta s? We're gonna say it's the Bolts men equation and Bolton equation is Delta s equals K times Ln of work. Now we're gonna say that K is a constant and K comes about after we divide our our constant buy avocados number and our our constant here is 38.314 Jules over moles Times K, remember avocados number of 6.22 times 10 to the 23. So when we divide those two numbers, it gives me 1.38 times 10 to the negative Jules over K. So that is what our K Constant's value is. And since we're gonna need some room guys to do this question, I'm going to remove myself from the image so we have more room to work with. So we know what K is. We're gonna plug that in. That's a constant. But now we just have to find out what work is. So come back over here. Work equals negative pressure. Times change in volume. So that equals negative. 1.25 atmospheres. Times final volume minus initial volume. So 6.3 leaders minus 2.5 leaders. When we work that out, that gives me negative 4.75 leaders, times atmospheres. But remember, here we have jewels. So we're gonna need to convert these units as well. This is also gonna require us to know, uh, conversion factor. We're gonna say for everyone leaders, times, atmospheres, we have one a 1.3 jewels that's gonna give me negative 4. 81 point 175 jewels. Now, here's the thing. We cannot use the negative value. You have to make it positive because if you take the Ellen of a negative number, you'll get an error in your calculator. So when you take this w that you just found it. Move it over here, make sure it's positive. Then all you have to do is do the Ellen of that value we just found for work. Multiply that times our k constant. And you'll get 8.52 times 10 to the negative. 23 Jules over. Kelvin. So what? We've seen in example one an example to our two important rules that tie into spontaneity and deal directly with entropy. So don't forget Trouten rule. And don't forget the bolts, man Constant and the bolts hman equation. These two are vital sometimes in finding Delta s by different means.

Additional resources for Entropy

PRACTICE PROBLEMS AND ACTIVITIES (81)

- Two different gases occupy the two bulbs shown here. Consider the process that occurs when the stopcock is ope...
- Which of the following reactions has ∆Ssys> 0? (a) N2(g) + 3H2(g) -> 2NH3(g) (b) Ag+(aq) + Cl-(aq) ->...
- If energy can flow in and out of the system to maintain a constant temperature during the process, what can yo...
- What is the change in entropy (∆S) when 1.32 g of propane (C3H8) at 0.100 atm pressure is compressed by a fact...
- Predict the signs of ΔH and ΔS for this reaction. Explain your choice.
- Isomers are molecules that have the same chemical formula but different arrangements of atoms, as shown here f...
- Calculate ∆Stotal, and determine whether the reaction is spon-taneous or nonspontaneous under standard-state c...
- Indicate whether each statement is true or false. (a) ΔS is a state function. (b) If a system undergoes a reve...
- The normal boiling point of Br21l2 is 58.8 °C, and its molar enthalpy of vaporization is ΔHvap = 29.6 kJ>mo...
- The element gallium (Ga) freezes at 29.8 °C, and its molar enthalpy of fusion is ΔHfus = 5.59 kJ>mol. (b) C...
- The element gallium (Ga) freezes at 29.8 °C, and its molar enthalpy of fusion is ΔHfus = 5.59 kJ>mol. (a) W...
- Indicate whether each statement is true or false. (c) In a certain spontaneous process the system undergoes an...
- Consider the gas-phase reaction of AB3 and A2 molecules: (a) Write a balanced equation for the reaction. (b) ...
- (a) Does the entropy of the surroundings increase for spontaneous processes?
- (c) During a certain reversible process, the surroundings undergo an entropy change, ΔSsurr = -78 J>K. What...
- (b) In a particular spontaneous process the entropy of the system decreases. What can you conclude about the s...
- Ideal gases A (red spheres) and B (blue spheres) occupy two separate bulbs. The contents of both bulbs constit...
- Rank the situations represented by the following drawings according to increasing entropy.
- (a) What sign for ΔS do you expect when the pressure on 0.600 mol of an ideal gas at 350 K is increased isothe...
- Two systems, each composed of two particles represented by circles, have 20 J of total energy. Which system, A...
- Two systems, each composed of three particles represented by circles, have 30 J of total energy. How many ener...
- (a) In a chemical reaction, two gases combine to form a solid. What do you expect for the sign of ΔS?
- Without doing any calculations, determine the sign of ΔSsys for each chemical reaction. b. CH2 “ CH2( g) + H2(...
- Without doing any calculations, determine the sign of ΔSsys for each chemical reaction. a. 2 KClO3(s) ¡ 2 KCl(...
- For each of the following pairs, predict which substance possesses the larger entropy per mole: (a) 1 mol of O...
- Predict the sign of the entropy change of the system for each of the following reactions: (d) Al2O31s2 + 3 H21...
- Predict the sign of the entropy change of the system for each of the following reactions: (c) 3 C2H21g2¡C6H61g...
- Predict the sign of the entropy change of the system for each of the following reactions: (b) CaCO31s2¡CaO1s2 ...
- Predict the sign of the entropy change of the system for each of the following reactions: (a) N21g2 + 3 H21g2¡...
- Given the values of ΔH° rxn, ΔSrxn ° , and T, determine ΔSuniv and predict whether or not each reaction is spo...
- Given the values of ΔH° rxn, ΔSrxn ° , and T, determine ΔSuniv and predict whether or not each reaction is spo...
- Define entropy, and give an example of a process in which the entropy of a system increases.
- Predict the sign of ΔSsys for each of the following processes: (d) Calcium phosphate precipitates upon mixing ...
- Predict the sign of ΔSsys for each of the following processes: (c) Gaseous CO reacts with gaseous H2 to form l...
- Predict the sign of ΔSsys for each of the following processes: (a) Molten gold solidifies.
- Predict the sign of ΔSsys for each of the following processes: (b) Gaseous Cl2 dissociates in the stratosphere...
- Given the values of ΔH rxn, ΔSrxn, and T, determine ΔSuniv and predict whether or not each reaction is spontan...
- Given the values of ΔH rxn, ΔSrxn, and T, determine ΔSuniv and predict whether or not each reaction is spontan...
- Predict the sign of the entropy change in the system for each of the following processes. (a) A solid sublimes...
- Cyclopropane and propylene are isomers that both have the formula C3H6. Based on the molecular structures sh...
- The standard entropies at 298 K for certain group 4A elements are: C(s, diamond) = 2.43 J>mol@K, Si1s2 = 18...
- How does the molar entropy of a substance change with increasing temperature?
- Three of the forms of elemental carbon are graphite, diamond, and buckminsterfullerene. The entropies at 298 K...
- Using S° values from Appendix C, calculate ΔS° values for the following reactions. In each case, account for t...
- For each pair of substances, choose the one that you expect to have the higher standard molar entropy (S°) at ...
- Rank each set of substances in order of increasing standard molar entropy (S°). Explain your reasoning. b. H2O...
- Rank each set of substances in order of increasing standard molar entropy (S°). Explain your reasoning. a. NH3...
- Which state has higher entropy? Explain in terms of probability. (a) A perfectly ordered crystal of solid nit...
- Rank each set of substances in order of increasing standard molar entropy (S°). Explain your reasoning. c. C(s...
- Use data from Appendix IIB to calculate ΔSrxn ° for each of the reactions. In each case, try to rationalize th...
- Use data from Appendix IIB to calculate ΔSrxn ° for each of the reactions. In each case, try to rationalize th...
- For a certain chemical reaction, ΔH° = -35.4 kJ and ΔS° = -85.5 J>K. (b) Does the reaction lead to an incre...
- Use data from Appendix IIB to calculate ΔSrxn ° for each of the reactions. In each case, try to rationalize th...
- Which state in each of the following pairs has the higher entropy per mole of substance? (d) CO2 at STP or CO...
- Which state in each of the following pairs has the higher entropy per mole of substance? (a) H2 at 25 °C in a...
- Which state in each of the following pairs has the higher entropy per mole of substance? (d) Water vapor at 1...
- Which state in each of the following pairs has the higher entropy per mole of substance? (b) N2 at STP or N2 ...
- Which state in each of the following pairs has the higher entropy per mole of substance? (a) Ice at -40 °C or...
- Find ΔS° for the formation of CH2Cl2( g) from its gaseous elements in their standard states. Rationalize the s...
- What is the entropy change when the volume of 1.6 g of O2 increases from 2.5 L to 3.5 L at a constant temperat...
- An isolated system is one that exchanges neither matter nor energy with the surroundings. What is the entropy ...
- Give an equation that relates the entropy change in the surroundings to the enthalpy change in the system. Wha...
- Phosphorus pentachloride forms from phosphorus trichloride and chlorine: (a) Use data in Appendix B to calcula...
- For the vaporization of benzene, ∆Hvap = 30.7 kJ/mol and ∆Svap = 87.0 J/(K*mol). Calculate ∆Ssurr and ∆Stotal ...
- Determine the sign of ΔSsys for each process. b. water freezing
- Determine the sign of ΔSsys for each process. a. water boiling
- Consider a twofold expansion of 1 mol of an ideal gas at 25 °C in the isolated system shown in Figure 18.1. ...
- For each of the following processes, indicate whether the signs of ΔS and ΔH are expected to be positive, nega...
- The reaction 2 Mg1s2 + O21g2¡2 MgO1s2 is highly spontaneous. A classmate calculates the entropy change for t...
- (a) For each of the following reactions, predict the sign of ΔH° and ΔS ° without doing any calculations. (i) ...
- Indicate and explain the sign of ΔSuniv for each process. a. 2 H2( g) + O2( g) ¡ 2 H2O(l ) at 298 K
- In chemical kinetics, the entropy of activation is the entropy change for the process in which the reactants ...
- The following data compare the standard enthalpies and free energies of formation of some crystalline ionic su...
- When most elastomeric polymers (e.g., a rubber band) are stretched, the molecules become more ordered, as illu...
- Hydrogen gas has the potential for use as a clean fuel in reaction with oxygen. The relevant reaction is 2 H21...
- What does entropy measure?
- For a process to be spontaneous, the total entropy of the system and its surroundings must increase; that is Δ...

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