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.

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.