Time for Gibbs Free Energy, the most important equation for understanding reaction favorability! It’s going to be important that we understand the significance of all the terms of this equation.
Concept: Breaking down the different terms of the Gibbs Free Energy equation.6m
We know that the thermal dynamics or spontaneity of a reaction is directly related to the Gibbs free energy. So it's going to be really important that we understand how to break down the Gibbs free energy equation and know how to use it. In this video what I'm going to do is I'm basically just going to describe every term and help you guys see how it relates to reactions.
The first thing, just going all the way back, Gibbs free energy or delta G is going to predict the favorability of the reaction. Remember that another word for favorability is spontaneity. And it's comprised of three terms. It's comprised of the delta H, the delta S and the T. So what I want to go through is just each one, one at a time and talk about what they mean.
The delta H is the enthalpy. Enthalpy can get confusing because delta H and delta G, a lot of times they happen to be the same. A lot of times if you have a negative enthalpy you'll also have a negative spontaneity. And a lot of students get confused thinking that they're actually the same thing. They're not. The delta H is just one component of the spontaneity. But we also have to take into account the temperature and the delta S or the T and the delta S.
Let me just start right there. What is the enthalpy? It's simply the sum of the bond association energies for the reaction. What that means is I'm going to be making bonds and I'm going to be breaking bonds. All of those reactions require that I'm putting in energy or I'm receiving some energy. When I add all those together, whatever my end number is, that's my enthalpy. That's it. That's all it has to do with. And later on, we're going to learn how to actually add and subtract that stuff. But for right now I'm just trying to tell you guys the big picture.
If something has a negative enthalpy, a negative delta H, that's what happens when you make bonds. When you make bonds, you are getting some energy back. Why? Because I already taught you guys. Remember that I showed you guys in the free energy diagram how you could gain energy by putting two atoms together. That's what we call exothermic. Exothermic doesn't always mean that the reaction is favored. Remember favorability has to do with exergonic or edergonic. But it is one component of it.
Well, what if your enthalpy is positive? If it's positive that means I'm breaking bonds because it requires energy to break bonds. Once you break bonds that's going to be endothermic because of the fact that it requires that I'm putting energy into the system to make those atoms separate. That's all that enthalpy is. It's just the sum of the endothermic parts and the exothermic parts together and then at the end you see if the overall number is negative or positive and that determines your delta H.
So now let's talk about one that's actually a little bit more difficult to understand is the entropy, the delta S. The entropy is a measure of disorder in the system. And this can seem very confusing because it's like how does the system know how disordered it is. I'm going to explain this in more depth later when I actually talk about entropy all on its own. But for right now I just want you guys to know what the signs mean.
And if it's negative entropy that means that I'm going to a more ordered state. That means that I'm basically – entropy means how disordered something is, so what's the opposite of disorder? That's order.
If it's positive, that means it's more disordered. A state always wants to be in its most disordered arrangement possible. So that's going to be a good thing. So when delta S is positive, that means my reaction is going to be a little bit more favored. Hopefully, you guys understand that negative means ordered, positive means disordered. Positive is the one that is actually favored.
Then finally we have out last variable which is temperature. And notice where temperature is in the equation. Temperature is going to be a coefficient of delta S. What that means is that as temperature goes up, it's going to amplify the effect of entropy on my overall favorability. What that means is that as I jack up the temperature of my reaction, the entropy of the reaction is going to matter a whole lot more in determining the overall fate of my reaction, whether it's going to be favored or not.
On the other hand, as I reduce my temperature down to zero kelvin or absolute zero, it's going to mean that my entropy becomes less and less important. Eventually, as you approach absolute zero, entropy doesn't matter at all anymore because there's actually no more movement. And all that matters is the heat of dissociation. That's kind of theoretical but I hope that makes sense to you guys. We're going to be learning and talking about that more later when we discuss entropy all on its own.
1. Enthalpy (ΔH°) is the sum of bond dissociation energies for the reaction.
2. Entropy (ΔS) is the tendency of a system to take its most probable form.
3. Temperature (T) amplifies the effect of entropy on the overall favorability.
Concept: Intermediates vs. Transition States7m
So it turns out that delta G doesn't just describe a one-step reaction it can also describe multi-step reactions, alright? If you have more than one step to go to completion the delta G would just be the sum of all the steps, alright? Now what I want to do is I want to show you guys a really common example of a two-step reaction and show you what kind of species that generates, OK? It turns out that here I have my two-step reaction shown, you guys don't know this reaction yet so you don't need to actually understand it but here is basically the reaction the free energy diagram and the reaction coordinate, what I start off with here is an alkyl fluoride, OK? Then what I get is this transition state, OK? See how it says TS transition state where the fluorine is being partially broken right now, OK? So that means it's this like middle place where the fluorine is like not fully dissociated but it's not also not fully associated so it's like in the middle, OK? Then what I get is the species where the fluorine is completely gone and now I'm missing the octet of this carbon this is called the Carbocation, this happens to be what we call an intermediate, OK? Then what we get is another transition state where now we're going to try to put a bromine on here, right? Because the end product is that you have an alkyl bromide so now my bromine comes here and it's partially making a bond, OK? That's another transition state and then finally we have the last product which is my alkyl bromide, OK? What I want to show you guys is the way that this actually relates to what transition states are and what intermediates are, OK? Transition states are high or very energetic points of the reaction that cannot be isolated, OK? What does that mean in terms of can't be isolated? It means that if have a test tube and I try to get a bunch of this thing inside that test tube it just wouldn't happen and the reason is because it's a very very high energy state it's almost like if I was jumping into the air and falling back down the transition state is me being up in the air, does that actually mean that I could ever just exist in the air for any given period of time? Not really that's just a period of time that I have to go through in order to get to my and end destination but it's not actually something that you could isolate and just say OK I want to see just Johnny up in the air for like 5 minutes that's not going to happen, OK? Then an intermediate is actually something that is of a higher energy but it can be isolated, OK? It's a molecule that is simply at a higher energy state than normal but it actually can exist or it can be isolated so for example that would be like me jumping onto a stool, OK? Now I have greater potential energy I'm a little bit I don't have I'm not quite as stable because now I have more potential energy but I can stay there as long as I want, OK? And that would be like this Carbocation over right here, this Carbocation it's not as stable as the beginning as you can see the energy for the Carbocation is greater than the energy from my starting reagent which is the Alkyl Fluoride but it's still stable enough that it could be isolated and it could just exist there for any given period of time until I want to do something to it, OK? So I just want to show you guys the difference between a transition state and an intermediate, a transition state is always going to be indicated by this little dagger sign at the top with brackets whenever you see that that means that this is a transition state, OK? An intermediate is usually not going to have any kind of notation like that at all it would just look like that, it would just the dotted line around it has nothing to do with it I'm just saying it might be like a positive charge or whatever, OK? Now how are you going to notice these on a free energy diagram? Because on a free energy diagram these words that I gave you or these letters are not always going to be given to you, just know the transition states are the highest energy states possible so those are going to be your highest points on the graph, OK? So your transition states are always going to relate to the very highest energy points, the ones that you have to just pass through very quickly in order to get to the end, OK? An Intermediate is going to refer to any dip on the graph, OK? Any time you have a dip that it's between two higher points that's going to be an intermediate, OK? Now notice that remember that we used to measure or we talked about how the rate of reaction had to do with the activation energy but in this case, I actually have two different activation energies I have the activation energy 1 that represents the amount of energy it takes to get first past my first transition state, OK? But then I also have a second activation energy here which represents the amount of energy it takes to get over my second transition state, OK? So which of these is going to tell me what the rate of the reaction is? Are they both do I just add them up and then that's the rate? Trick question absolutely not, OK? What we're going to do is we're just going to go with the highest activation energy so whichever one that is whether it's activation energy 1, 2, 3 or 10 let's say it's a 10-step or whatever you just go with the very highest activation energy and that is going to be your rate determining step, OK? Or what we call our slow step so this would be my slow step right here and what that means is that my rate determining step is going to be forming this intermediate right there, why is that? Because you might notice that this activation energy is that high this activation energy is tiny in comparison so which one's going to have a greater effect on the overall rate of the reaction? This one right here, OK? And it turns out that in order to get to the end the most determining thing is just going to be how quickly can I go through the slow step if I can make the slow step a little bit faster that's good for my reaction in terms of rate but it doesn't matter what I do to this activation energy it's so small that it doesn't matter by comparison, I'm always going to be waiting for the first one to form before I do the second one anyway, OK? So I hope that that makes sense to you guys I just want to show us the difference between a transition state and intermediate once we get into actual reactions this is going to matter a whole lot because we have to talk about that all the time in terms of what is a transition state and what is an intermediate, alright? So hopefully it makes sense let's go on to the next topic.
Predict the sign of ΔG for an endothermic reaction with a decrease in entropy.
a) cannot predict
d) no change
Calculate the enthalpy of reaction for following reaction. If possible, estimate ΔG for the reaction.
In an acid base reaction where the products predominates, ΔG° < 0 ( True or False )
For the following reaction, indicate by circling the correct answers for ΔG and K, whether this reaction lies to the right or left.