Chemical Thermodynamics: Entropy - Video Tutorials & Practice Problems

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Entropy represents the quantity of a system's thermal energy that couldn't be converted into mechanical work.

Understanding Entropy

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Entropy

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So when we talk about entropy, we need to realize that entropy is closely related to the second law of thermodynamics. Now this states that molecular systems tend to move spontaneously to a state of maximum randomness or disorder. So all of this is saying is when it comes to our total entropy which is also our entropy of the universe, it is ever increasing. That means that the universe itself is becoming more and more chaotic universes or Galaxies spin out of control, their their stars eventually explode, become um supernovas and eventually become black holes. So the whole nature of the universe is one of degradation and breaking down into basically nothingness and chaos. I know it sounds a bit depressing but that's the general idea behind Second Law, everything moves to a state of disorder and randomness. So here delta S. Of our universe or delta S. Total equals delta S. Of our system, which is our chemical reaction plus delta S. Of the surroundings. And together following the second law of thermodynamics, this will always be greater Than zero because it's always increasing. Now, here we're gonna say again, the disorder, chaotic behavior is classified is entropy, which is our delta S variable. We're gonna say here that we can talk about delta S. In terms of face changes as well. This helps us determine what the sign will be for a particular process. Now here, if we take a look, we're going from a solid to liquid to a gas. If we're going from a solid to a liquid, we say that that phase change is called melting or fusion if we're going from a liquid to a gas, that's vaporization and then we can go straight from a solid to a gas which is sublimation in these processes. We're gonna say the distance between molecules are increasing because remember in a solid the molecules are tightly held together in a liquid, they're around each other but they're able to freely move around and in a gas they're very spread apart. We're gonna say here, we can see that we're going from a more from a more ordered state to a less ordered state as we transition towards gasses. As a result, we're gonna say here that our entropy is increasing. So delta S. Increases. And in these processes because they're all increasing my entropy, we say that delta S. A sign will be positive. Now if we look at it the opposite way, if we're going from gas to liquid to solid, gasses are the most entropic and we're becoming more and more ordered as we move to the right. So we're gonna say here, the distance between molecules are decreasing as they become more tightly packed as they transition to the solid state. In this case we'd say that our entropy is decreasing and if our entropy is decreasing, that means that the sign of our entropy would have to be negative. So again remember entropy deals with chaos or disorder. The second law says that the natural processes or spontaneous process of the universe is to always increase our entropy. So delta S universe will be greater than zero. Remember in phase changes, you can have a positive entropy or a negative entropy depending on what type of phase changes can occur. Now it is possible for us to have an entropy equal to zero, But this is hard to come by. This is our 3rd law of thermodynamics, where the entropy is equal to zero at absolute zero. So the temperature would have to be zero kelvin At zero Kelvin all motion stops within any given object and it's at that instance that the entropy of that object is exactly zero. Now we're going to say here that this zero kelvin is highly theoretical. The average temperature of the universe itself is around two kelvin. This zero kelvin is is just a theoretical number. If we were able to get to that temperature, we assume that all motion would stop. But again, we haven't been able to get to that number. So that's why it's highly theoretical. Now that we understand the basic generalities of entropy will move on to the next video where we talk about comparing the entropy ease of different compounds to one another. So click on the next video and see how we approach comparisons of entropy

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Entropy

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So when we're taking a look at covalin compounds, we'll be able to compare their entropy is to one another. Remember covalin compounds are compounds composed of just non metals bonded to one another. We're gonna say here whenever you need to compare the entropy of different covalin compounds. The guidelines below must be followed. What you do first is you have to compare their phases to one another. We're gonna say here that gasses which have their molecules most spaced out are the most in tropic. After that we have a quiz compounds. These are just compounds that are dissolved in some type of solvent, usually water. Then we're gonna say liquid which would be water or some other type of liquid, maybe hexane solvent or methylene chloride solvent. Then we have finally solids Now solids have their molecules most closely packed to one another so they have the lowest entropy. That's why we can say as we go from gasses to solids, the molecules become more organized and less chaotic and therefore less entropic. Now let's say your compounds that you're comparing are in the same phase. If so, we move on to the second criteria which are micro states. Micro states are just the different ways a molecule can bend and orient itself and they would have similar energies now to make this simple, we're just gonna connect micro states to complexity. Now we're gonna say here the more elements in the compound than the more complex the compound, the higher its micro states and therefore the greater its entropy for example, we're looking at N. 02 gas versus N. 03 gas, they're both covalin compounds. They both are gasses. So we go to the criteria of micro states here, N. 02 has in it one nitrogen and two oxygen for a total of three elements. N 03 has in it, one nitrogen and three oxygen for a total of four elements. Because N. 03 has more elements making it up, it has more complexity. Therefore more micro states, therefore a greater entropy. Now let's say that the carbon compounds are comparing have the same phase, the same number of elements making them up. Then we go to the final factors to look at in order to determine which one is more entropic. We look at their masses, we're gonna say here the greater the mass than the greater the entropy of that compound or element for example, here we have Xenon gas versus helium gas. Both are gasses. Both have the same complexities because of just one element each. So we go to masses. Xenon weighs 131 g approximately from the periodic table. Helium weighs approximately four g from the periodic table. Because Xenon weighs more Zenon would be more entropic. So remember, these are the guidelines we must follow. In this exact order when comparing the entropy of different covalin compounds, click onto the next video and see how we approach comparing the entropy ease of different ionic compounds

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Entropy

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Now when you need to compare the entropy of different ionic compounds, you must utilize lattice energy. Now we're going to say here that remember an ionic compound fundamentally is just the connection between a positive ion which is a cat ion and a negative ion which is an an ion. Remember that cat eye on that positive ion could represent a medal Or the ammonium ion and H four positive. That negative ion will represent a nonmetal. Now we're going to say here that lattice energy represents the energy released when one mole of an ionic crystal is formed from its gaseous ions. So here we have sodium in its gaseous state, it's ionic state and chloride ion in its gaseous state. They combined together the form sodium chloride solid forming bonds means a release of heat or energy is required. That's why our delta H value here is negative. It's an XR thermic process. Now to simplify lattice energy, we can say that lattice energy when simplifying columns law, it's just electrostatic energy. It is equal to the absolute value of cat ion charge times and ion charge divided by cat ion radius plus an ion radius. Now, what do I mean by that? Well, here we have sodium chloride and if we look at this periodic table here, We see sodium here and chlorine here. So this is period 123 that they're in. So they're in row three. Now using this formula, we'd say sodium is plus one chlorine is minus one. So that would be equal to plus one times negative one in absolute terms divided by their radius. on the bottom. We'll just go by the period numbers here. So that would be three plus three. So that comes out to 16. So we can say that this represents the simplified view of its lattice energy. All you have to do is find another compound and compare its lattice energy that sodium chloride and see which one has the greater lattice energy. So if we're comparing for example, let's say we're looking at magnesium oxide. So MG is there and always there magnesium oxide, MG is two plus oxygen is two minus, you'd say magnesium here is in row three. So that's three plus oxygen in row two. So that's two. So that gives us 4/5. And if you punch them into your calculator that give you 0.80 vs 1/6 which is less than 0.80 that's closer to 0.167 or so. So you can say magnesium oxide, it has a higher lattice energy than sodium chloride. A stronger lattice energy means that the connection between the ions is stronger. It's a stronger bond between them. So here mgl solid will be a stronger ionic compound than sodium chloride. Now here we're going to say, the larger the lattice energy than the stronger the ionic bond between the ions. Remember, entropy is chaos or disorder. The stronger the ions connections to one another, the harder it is for us to break them apart. So what does this mean in terms of the entropy of these ionic compounds? Well, we'd say that a stronger lattice energy results in higher boiling point because it takes more energy to be able to break them apart and therefore it also requires a higher melting point. Again, more energy and temperature to break them apart. And the more tightly held they are together, the harder it is for us to break them apart and become more entropic. So this means we're gonna have a lower entropy. So basically when we're comparing covalin compounds together, we use the three steps here, we look at phase than micro states than mass. If we're comparing ionic compounds to one another, then we look at their lattice energies, the greater the lattice energy of the ionic compound, then the lower its entropy will be. So just remember these guidelines when you're faced with grouping compounds together and from there determining which one has the greatest entropy to which one has the lowest entropy

Entropy Calculations

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example

Entropy Calculations 1

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So here, for example, when we're asked to predict the sign of entropy for each of the following processes. So for the first one we have I to it's at 90°C and 5.0 atmospheres of pressure, then it transitions to a new temperature of 50°C and 10 atmospheres of pressure. Now we can see that the change what changes are occurring while our temperature goes from 90°C to 50°C. So that tells us we have a drop in temperature, Then our pressure goes from five atmospheres to 10 atmospheres. So we have an increase in pressure. So we have to think about what the sign of entropy would be here is entropy increasing, therefore making it positive or is entropy decreasing, therefore making it negative. Well, we can see that remember if our temperature is decreasing what's happening to the molecules within any given substance as the temperature is dropping, those molecules become less energetic and therefore huddle closer together, making more bonds, also pressure. If you have these molecules in a closed container and you're increasing the pressure, you're forcing them be closer to one another. This also supports the idea that the entropy is decreasing. So here we say that our entropy would be negative. There'd be a decrease in my entropy. Also, you could say if your pressure is increasing, that means your volume is decreasing. We didn't talk about volume here, but pressure and volume are related to one another and the fact that they're inversely proportional to each other. They're basically opposites whatever happens to one, the opposite happens to the other next here we have ammonium chloride solid, breaking up to give us hydrogen chloride as a gas plus ammonia gas. Now we're gonna say here you have one reactant that is split up into two or more products. So what's happening? We're breaking bonds were causing an increase in our entropy or chaos. So Delta S here would have to be positive. There's an increase in entropy. Finally, we have CH four gas plus 2 +02 gas gives us C +02 gas plus two waters as a liquid. What we need to realize here is that what's going on? We have as a total one mole of gas plus another two. So we have three moles of gas here And then on the product side we only have one mole of gas And we have two moles of liquid. Remember gasses have the most entropy. So by converting those three moles of gas into just one mole of gas and now two moles of liquid. We've decreased our overall chaos or disorder entropy here would be negative. There's a decrease in my entropy. So just remember, are you helping to form bonds or you're helping to break bonds. This is important in understanding what the sign change would be for your delta ask value. Now that you've seen this example, move on to the next. We haven't talked about this one yet. So if you want you can just skip straight to the second video and see how we approach this problem. Or you can attempt to do it on your own. So guys, if you're gonna do it on your own, attempted, If you get stuck, don't worry, just look at the next video.

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example

Entropy Calculations 1

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Alright. So for this example we haven't quite done questions like this yet. But let's give it a look. We're gonna say when an aqueous solution containing aluminum ion at 25 degrees Celsius is mixed with an aqueous solution of hydroxide at 25 degrees Celsius. And immediate precipitate of insoluble aluminum hydroxide is formed. So in our equation we have one mole of aluminum ion reacting with three moles of hydroxide ion to form one mole of aluminum hydroxide here were given the entropy of formation for each one of the compounds. So realize here that all compounds all elements have an entropy greater than zero. The only way a compound on element can have an entropy equal to zero is if the temperature is at absolute zero. So the temperature is zero kelvin. Each of them would have a delta S. A formation equal to zero, as long as the temperature is above that all the delta S. Values for formation will be greater than zero. Alright, so here it asks us uh the standard reaction and delta H of my reaction is negative 61.33 kg per mole. They ask us to calculate the delta S. Total for this reaction. Now remember delta S total is related to delta S. Universe, which is part of our second law of thermodynamics. So delta S. Universe which is the same thing as total, equals delta S. Of my reaction plus delta S. Of my surroundings. So we have to find both the reaction and the surroundings, add them together to get our total. So here we're gonna start off by figuring out what delta S of reaction is. Now, whether it's delta S. Of my reaction, delta F, delta G. Of my reaction or delta H. Of my reaction. All of them equal products minus reactant. So we're gonna use these entry piece of formation values that were given to us. So here our product is aluminum hydroxide, we have exactly one mole of it. So we're gonna say here, one mole of aluminum hydroxide, Which has an entropy of formation of 216 jules over moles times K minus my reactant. So we have one mole of aluminum ion. Each one is negative 3 25 jewels over most times K. Plus three moles of hydroxide. Each one is negative 10.90 jewels or moles times K. So here my moles will cancel out. So my units here will be jewels over kelvin. So here when we do that, we're gonna get 2 16 Michael Jules over kelvin -157.7 jewels over Kelvin. So minus of a minus really means you're adding. So that's gonna come out to 573.7 jules over kelvin from my delta S of reaction. Now that we found that we now figure out what our delta ask of surroundings will be delta S, surroundings is equal to negative entropy of my reaction, divided by my temperature in kelvin. The temperature we're told is 25°C. So we just add to 73.15 to that, Which gives me 2 98.15 Kelvin. So to 98.15 Kelvin here on the bottom and the entropy of our reaction is already given to us. We didn't have to calculate it, but if we did they would give us the entropy of formation for each compound. And again you would just do products minus reactant in the same way you did the of the entropy of our reaction. Alright, so that's negative of a negative 61.33 killer jewels. So any reaction usually is just in jewels or killed jules, you can drop the molds that we have there. Um So here's what we're gonna have here, we're gonna have At the end, we're gonna say here this is gonna give me a positive .205702 killer jewels over Kelvin. But our entropy of reaction is in jewels over kelvin. So we'll change this to jewels. So one killer jewel Is equal to 1000 jewels. So that's 205.702 jewels over Kelvin. And then all we have to do now is add those two together. So delta s total which is my delta s of reaction plus surroundings. That's 5 73.7 Jules over Kelvin plus 205.702 jules over kelvin, That's gonna be approximately 779.402 jewels over Kelvin. Now, this answer because our delta S total or universe is greater than zero, this tells us that this entire process is a spontaneous process. So again, Delta S total or universe is greater than zero. So by the second law of thermodynamics, this is a spontaneous process. So that's how we approach this problem now that you've seen examples one and to attempt to do the practice question left here on the bottom. Once you do that, come back and see how I approach that same exact practice question.

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Problem

Problem

Which of the following has the greatest entropy, S?

A

1.00 mole of liquid water at 30° C

B

1.00 mole of water vapor at 30° C

C

1.00 mole of regular ice at -10° C

D

1.00 mole of 'dry ice' (solid CO_{2}) at -10° C

E

1.00 mole of water under 10 atm of pressure at -10° C