Gibbs Free Energy

Hey, guys, In this new video, we're gonna take a look at Gibbs Free Energy. Now, we've been talking about the system and the surroundings, and we've been talking about entropy and to a lesser extent, delta H Entropy. Now we're gonna take a look at the last major variable R Delta G value. We're going to say, we know that according to the second law of thermodynamics ah, spontaneous reaction will be when Delta as total or Delta s universe is greater than zero. But now we're gonna bring into the mix Delta G. Besides looking at Delta as total, it's also great to look at Delta G because knowing the sign for Delta G will be a great way to determine if your reaction is spontaneous or not. Overall, Now, to do this, we use two different equations here. We'll take a look at the one on the left side. It says Delta G zero equals Delta H zero minus T delta s zero. Remember these zero means that were under standard conditions and when it comes to thermal dynamics, standard conditions means one atmosphere and 25 degrees Celsius. And then on the other side we have Delta G equals Delta G zero plus r T l N Q. Here, this would be Delta G under nonstandard conditions. So we're not at one atmosphere or our temperature is not 25 degrees Celsius. Here it equals Delta G zero again, our here's our gas constant, which is 8.314 Jules over moles times K t is our temperature in Calvin que is known as our reaction quotient. Okay, we're gonna see our reaction quotient is just products overreact INTs, we're gonna say when it comes to queue, it ignores two states of matter. It ignores solids and liquids. So if it's a solid or liquid, it ignores it. If it's acquis or gashes or gas, it will look at them. Okay, so ignore solids and liquids. But look at a quiz and gasses. Now we're gonna say here if Delta G under nonstandard conditions is less than zero, then our reaction is spontaneous. If it's greater than zero than its non spontaneous and if it's equal to zero, then our reaction is at equilibrium. Now let's take a look at example one. And when it comes to this particular type of example, this is a standard type of question for any exams that you take that deal with Delta, G, Delta, H and Delta s. So remember our techniques that we use here in order to solve this particular type of question. Since we're gonna need room to see everything, I'm going to remove myself from the image guys. So we have more room to work with. Okay, The question says which of the following statements is true for the following reaction. So we have n 204 gas breaks down to give us to go to gas. We're going to stay here. The Delta H value is negative. The Delta s value is positive, and then we'll see that our options are the reaction could be spontaneous or non spontaneous, that petting on the types of temperatures we're talking about now in your book, there's gonna be a long large chart where you have to memorize different things. Now I say, Look, you can look at the chart, but there's a much easier way to handle the question like this That has to do with the little square thing that we have to the right side. So here, we're gonna use this image to help us figure out how to answer this question easily without having to learn a huge chart. So here we're gonna say we have our delta as values on top there positive and negative on the sides. We have our Delta H values positive and negative. Now, here's the thing we're gonna say when both Delta H and Delta s are positive, then our reaction is spontaneous at high temperatures at high temperatures. If Delta H is positive and Delta S is negative, then it's non spontaneous at all temperatures. So non at all temperatures. If Delta H is negative and Delta asked, is positive, then we're gonna say they're spontaneous at all temperatures. So spawn at all temperatures. Then finally, if Delta H and Delta s are both negative, then it's on Lee spontaneous at low temperatures. So it's gonna be high non spawn low, easy way to remember now if we go back to the question we have a Delta H, that's negative. A Delta s. That's positive. So when Delta H is negative and Delta s is positive, they said the reaction is spontaneous at all temperatures. Ah, question like this is very basic on your exam. So it's important you guys remember this quick, easy technique Instead of having toe sit down and memorize an entire graph an entire chart in your book. Just simply use this diagram. You can easily answer the question. The hard part really is just remembering who goes on the top who goes on the side. Just remember, Delta s values are on top. Delta H is are on the side. Then it's high, non spawn low and that's all. It iss an easy way to break down. Ah, very complicated issue. So just remember the steps. Remember how to draw this diagram and you'll be fine when it comes to a question like this.

Hey, guys, In this new video, we're gonna take a look at the calculations behind Gibbs Free Energy. So if we take a look, at example, when it says the reduction of iron three oxide with hydrogen produces iron metal and keep the richness follows. So we have one mole of iron, three oxide, solid reacting with three moles of hydrogen gas to give us two moles of iron solid plus three moles of water vapor. I give you the Delta H or entropy value and the Delta s entropy value. The question says, Is this reaction spontaneous under standard state conditions at degrees Celsius? If not at what temperature will it become Spontaneous. So this is a two part question. So they're asking us if it's spontaneous or not. Remember, the best variable to calculate spontaneity is Delta G. So I gave you Delta H I gave you. Delta asked, and we have temperature so we can use Delta G zero equals Delta H zero minus t Delta s zero. And we know we should use this version because I'm saying standard state conditions. So we're supposed to use Delta G with a zero. So what we need to do next is we have to make sure that our Delta H and R Delta s values agreeing Units here, Delta H uses killer jewels, but Delta s uses jewels now. Killer jewels are the default type of units for these types of questions. So let's just convert the Delta s tequila jewels as well. So remember, we're gonna say, for every 1000 jewels there is one killer Joel. So this is equivalent to 0.141 five killer jewels over K. So we're gonna plug in the 98.8 killer. Gul's minus temperature has to be in Kelvin as well. So I had to 73.15 to 25. That gives us 2 98 15 kelvin times a Delta s value we just calculated. Okay, so the Kelvin's will cancel out. So to be killer jewels minus killer jewels. When we do that, we get a answer off 56.633 Killah jewels. Now, remember, if Delta G is negative, then it's going to be spontaneous. The fact that we got a positive answer means that this reaction is not spontaneous. So now it's up to us. to figure out what temperature will it become? Spontaneous. Now, anytime they ask you, at what temperature does it become Spontaneous. That means you assume that Delta G zero is equal to zero. Okay, so we're gonna use this this same exact equation here, except now, Delta G is zero, and that's still equals the same value for Delta H. The same value for Delta s here. We're not gonna know what temperature is. That's what we're trying to solve. So zero equals 98.8 killed. Jewels minus T times 0.1415 Killer jewels over K. What we're gonna do First, we're going to subtract out the 98.8 killer Joel's. And that equals negative tee times. Point won't 0.1415 killer jewels over K. We need to isolate temperature by itself. So we're gonna divide out negative 0.1415 killer jewels over K on both sides. So this cancels out with this, and the negative sign also cancels out. So we have a negative getting divided by another negative. Kelly jewels. Cancel out. Our answer will be in Kelvin. So the temperature here equals 6. 98.233 temperature. Now, technically, if we take that number and plug it back in, then Delta G will equal zero. Remember, when Delta G equals zero were at equilibrium or technically, still not spontaneous. So what you would say is temperatures above this answer would give you a spontaneous reaction. Okay, so we're gonna say temperatures above this answer would be spontaneous temperatures below it would be non spontaneous. So make make careful. Sure. Like when you're looking at your exam, what exactly are they're asking you in terms of temperature? Are they looking for an exact answer or they asking, um, at what temperature will it be? Spontaneous. Just remember, when you work it out, the temperature you get, that's technically the temperature at equilibrium. So all you have to say is any temperature above that, whether it be like 0.1 uh, degrees higher, it's still gonna be spontaneous. It'll just be a little bit spontaneous. Okay, so remember the answer. You get temperatures above that answer will make your reaction spontaneous. Now that we've done example one together, I want you guys to attempt to do practice one here were saying, If Delta G is small and positive, which of the following statements is true? And I'll give you guys a huge hint. Remember, we have to think of Delta G in three situations when it's less than zero equal to zero or greater than zero. What do each of those mean? What do I mean by the word small? Also, remember, if you're spontaneous one way, then you'll be non spontaneous the other way. Just because Delta G is positive doesn't mean it can't be spontaneous, period. It would be non spontaneous in the Ford. What about the reverse? Ask yourself these questions and you'll be able to answer this question really quickly. Good luck, guys.

Hey guys in this new video, we're going to continue with further calculations dealing with Gibbs free energy. So in this example, it says for mercury, the entropy of vaporization equals 58.5 kilojoules per mole. And then the entropy of vaporization equals 92.9 Jews over Kelvin's times moles at 25 degrees Celsius. Now I'm asking you does mercury boil at 3300 and 50 degrees Celsius and one atmosphere pressure. Now this question is similar to something we've seen earlier. What I'm really asking you for is calculating the normal boiling point of mercury. If we can find out the normal boiling point of mercury, we can see if this temperature is high enough for mercury to begin to boil. So remember if we're looking for the normal boiling point of mercury, we're going to assume that delta G zero is equal to zero. And doing this, we're going to say delta G zero equals delta H zero minus T delta S zero. Since this is equal to zero, we can now plug in all, we know we can plug in the delta H value as well as the delta S value So it's similar to something we've seen earlier. So this will be 58.5 kilojoules over moles minus t we don't know temperature. We're looking for it times, remember, enro um entropy of vaporization has to be in kilojoules just like delta H is in kilojoules. So divide that by 1000 gives us 10000.929 kilojoules over K times moles. So we're gonna subtract 58.5 kilojoules over moles. So negative 58.5 kilojoules over moles equals negative temperature times 0.929 kilojoules over K times moles, we need to isolate our temperature. So we're going to divide out negative zero 929 kilojoules over K times moles. OK? So when we do that, this cancels out with this, the negative sign cancels out. We're gonna say that kilojoules and moles will cancel out with these kilojoules and moles leaving us with what units left Kelvin? So temperature here equals 6 29.7 Kelvin. OK. So we're gonna say here that is the normal boiling point of water. So that means the temperature has to be at minimum, this temperature for mercury to begin to boil. So we go back to the 350°C, we have to change it to Kelvin. So add 2, 73.15 we get 6 23.15 Kelvin. And we can see that this temperature here is not high enough. It has to be a minimum of 6:29.7. The temperature we calculated earlier. So we're gonna say no, it does not boil, not hot enough has to be a little bit hotter than that.

for this one. It says the chemical reaction to a no to be our gas breaks down to give us to a no to gas plus br to gas. It has an entropy value of 1 35 jewels over moles times K and an entropy value of 9. 26 killed. Jewels over moles, I asked. Calculate the temperature when k q equals 4.50 times 10 to the five. Now, what we need to realize here is where we have to use two equations. I give you Delta s and I give you a Delta H. So what equation connects them together? Delta G zero equals Delta 80 minus t Delta s zero. At the same time, I'm introducing our equilibrium. Constant K. What equation connects, connects delta G decay or that will be Delta G zero equals negative rt Alan Kay. Now we're gonna say that both of these equations are equal to Delta G zero, which would mean that this portion is equal to this portion. So let's make them equal to one another. We're gonna say here Delta H zero minus t Delta s zero equals negative rt lnk. Now, since we're gonna need some room to work this out. I'm gonna remove myself guys from the image so we have more room to work with. So let's plug in what we know. We know what Delta H is. It's 9 26 killer jewels over moles minus temperature, which we don't know here. You need to convert your delta s value in to kill a Jules. So divide this by 1000. So that gives us point 135 killer jewels over most times. K are is your gas constant, which is 8. But actually, we need to convert it toe from that number because it's 8.314 jewels over moles times K. That's what our equals. It's that constant number. But we needed to be in kill jewels because this is in kill jewels as well as this. So we divide that number by 1000 to get this new number temperature. We don't know what temperature is here either. So we just put t again and then Alan Kay, which is 4.50 times 10 to the five. Okay, Now let's combine like terms, so we're gonna multiply this times, Ellen of this answer. So that's gonna give us negative 0.1082 to 3 t Because remember our temperatures still there as a variable and that equals the other side that equals 9 26 minus 260.135 t. We're gonna add 0.135 t to both sides. So 9 26 equals 0.2677 70. We wanna isolate temperature, so divide out the 0. And that will give us a temperature off 3.46 times 10 to the four. Kelvin, Hopefully, you guys were able to follow along when doing this question. And just remember, Delta G zero equals more than just one type of equation here. It equals these two particular types in order to connect them together to find our temperature, you need to know what those equations are in the first place. And after that it would just be simple algebraic methods in order to find our final answer

Hey, guys, In this new video, we're gonna continue with our discussion on calculations dealing with Gibbs Free Energy. But now look at it under nonstandard conditions. So an example. One, it says. Calculate Delta G of reaction. You'll notice that there isn't a small circle on top, meaning that this is under nonstandard conditions. So calculate Delta G reaction at 25 degrees Celsius under the conditions shown below for the following reaction here I tell you, Delta G, under standard conditions is 31.6 Killer Joel's. Now we know we're dealing with non standard conditions here, so it's gonna be Delta G equals Delta G zero plus r T l N Q. Now what you need to remember is that Q is our reaction quotient and it equals products. Overreact INTs. It ignores solids and liquids. Luckily, neither one of those these compounds is a solid or liquid. So we're gonna look at both of them. So we're gonna take you here equals C l three. The coefficient is a to that. To me, this is going to be the power now divided by l two. It's coefficient is three. So its power will be three is. Well, okay, The partial pressures of seal to unseal three, given as these numbers, respectively. So those are the values were gonna plug in, So C l three is 4.9 squared C l to his 0.83 cubed when we plugged this sin. It gives us 41.9912 So that's what our Q equals. So we're gonna say now Delta G is 31.6 killer jewels plus 0. killer jewels over Calvin times Moles time here will have to be in Kelvin, so add to 73. 15 toe are 25 degrees Celsius. That gives us 2 98 Kelvin and then Allen off 41.9912 when we plug all that in. So we're gonna do Eleanor 41.9912 times to 98.15 times 0. plus 31.6. When you do all that, that gives you 40 0.86 killer jewels. So what we can say about this reaction? It is highly non spontaneous Because remember, if your Delta G is greater than zero, it's a non spontaneous reaction

So now that we've seen example one, let's take a look. At example, too. So, for example, to were saying for the reaction and two plus 02 giving us to 102 gas, we have our delta G R first Delta G equal to 75,550 jewels over moles at that temperature of 1 75 Calvin. And then we're gonna have a second Delta G equal to 41,875 jewels over moles at 225 Kelvin's. So it's asking us calculate Delta S and Delta H for the reaction. So this is what we gonna do for this particular one? Since we're dealing with two types of Delta G here, we're gonna say Delta G equals Delta H zero minus t Delta s zero. That will be for our first Delta G values. But remember, I gave you a second one. So we're gonna say minus delta G zero equals Delta H zero minus t Delta s zero. Now that negative sign is gonna get distributed toe everything within the brackets. So it becomes now. And since we're gonna need some room to do these calculations, guys, I'm going to remove myself from the image so we have more room to work with. So we're gonna now plugging what we know we have 75,550 for one Delta G in that equals Delta H minus 1 75 Calvin times, Delta s minus 41,000, 875. Remember, this negative gets distributed. Okay, so this delta H now becomes negative. Delta H here. We had a negative sign. A negative times a negative. Gives me a positive the Delta H values. Now cancel out. Then we're gonna get 33,000, 675 equals 50 times Delta s We want his delta s. So we divide by 50 Calvin on both sides. So now Delta s equals 6 73. Jules over moles, times K. So that will be our entropy value. Now, if we want to find Delta H, we simply now do. Delta G zero equals Delta H zero minus t Delta s zero. We can choose either one of these Delta G values Doesn't really matter which one we choose. So we're gonna say 75,550 equals Delta 80 which we're looking for minus 1 75 Kelvin times the six. 73 5 jewels over moles times K we calculated from earlier. Okay, now, when you multiply these two together, it's gonna give you 117863 The Calvin's canceled out so the units will be left jewels of Rommel's. So we're gonna add 117863 to both sides. So that's gonna give us at the end. Delta H zero equals 1. times 10 to the five Jules over moles. So this is what the technique you should approach when they give you two Delta jeez to temperatures and they ask you basically to find Delta S and Delta H. This is the exact method that you should use.