- Gibbs free energy and spontaneity
- Gibbs free energy and spontaneity
- Gibbs free energy example
- More rigorous Gibbs free energy / spontaneity relationship
- A look at a seductive but wrong Gibbs spontaneity proof
- Changes in free energy and the reaction quotient
- Standard change in free energy and the equilibrium constant
- 2015 AP Chemistry free response 2c
Gibbs free energy example
Determining if a reaction is spontaneous by calculating the change in Gibbs free energy. Also calculates the change in entropy using table of standard entropies. Created by Sal Khan.
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- This is really amazing! But I still don't understand how energy being released in the reaction can cause it to become spontaneous, because, Sal, you said in the earlier video that when the change in entropy is negative the reaction isn't spontaneous....please help?!?!!!!(23 votes)
- As I understand, if energy is being released, the system doesn't require much energy to do the reaction, so it is easier for the reaction to occur.(16 votes)
- So I don't see how you got -890.3 KJ for the heat change... I looked up the heat of formations of everything. CH4; -74.8 KJ, O2: 0 CO2: -393.5 KJ, and H2O: -241.8.... so you would do heat of products minus heat of reactants.
-877.1--74.8= -802.2(11 votes)
- That is because you looked up the heat of formation for water as a gas. If you used the heat of formation for water as a liquid ~ -285.8 KJ (see the equation above in the video), you would get -890.3 KJ.(28 votes)
- What if we are not allowed to look up the data on the tables. We get other reactions that we balance to find ΔH. Could you make a video using this method? I'm taking an AP class in high school and the table way won't be on the exam but the balance the equation way will be.(7 votes)
- I took AP chemistry last year and, if I understand your question, you need to add the equations together until you get a balanced equation that matches yours. You must also add the Enthalpies.(4 votes)
- Great video! I do have a question about why water is listed in the liquid state, though. Wouldn't it be in gas form if you were actually burning methane? Is this an artificial construct by stating it's occurring at 25 C? What happens to the heat produced by the exothermic nature of the process?
5 votes, kimromanchuk(4 votes)
- It doesn't belong in the comment section, & there is no way for anyone to answer him from the comments. If you'll notice I quoted his name and votes!(2 votes)
- Why does oxygen have an enthalpy value of 205kJ? I thought that Sal said in a previous video that enthalpy of an element in its standard state is 0?(3 votes)
- Those numbers were not enthalpies, they are the standard molar entropies at 298 K and 101.3 kPa. Also, you misread the units.
The standard molar entropy of O₂ at the above conditions is 205.1 J/(mol∙K)(6 votes)
- Hey... Just in relation to the first reaction, I seem to be getting totally different numbers for change of formation for CH4, CO2 and H2O. Where do the numbers 186, 213 and 69 come from....? It's probably a very silly question, apologies, I am just stuck at this... I thought that to calculate the change of enthalpy you needed to add the products' heat of formation C+D and subtract reactants heat of formation -A-B from it?(3 votes)
- Numbers 186, 213 and 69 are entropy values to calculate delta S, not enthapy change of formation. For delta H he already did all the calculations and didn't inlude the original values.(2 votes)
- If the temperatures on the surface of the sun are that high the why do we see reactions occurring on the surface of the sun (hydrogen atoms coming together to make helium)?(1 vote)
- First of all the conversion of hydrogen to helium occurs deep inside a star not on the surface. Second, these reactions are nuclear and not chemical. It is the extreme temperatures in the sun that allow the fusion reactions to occur.
Also at the temperature of the surface of the sun chemicals will be ripped apart and the atoms will be stripped of electrons becoming plasma.(3 votes)
- So after the entropy intuition, I have a little trouble with the standard entropy - how can it be defined? Shouldnt it depend also on volume of the system?(2 votes)
- It is standart -molar- entropy, I think there's the rub. It is the certain amount of molecules. And in liquids and solids volume is a constant, in gas it think they use standart molar volume of ideal gas.(1 vote)
- At1:09you gave your response for change in H. How did you get -890.3 KJ? I'm a missing a simple detail or should i go watch another one of your videos?(2 votes)
- Watch the videos. Otherwise it the Enthalpy formation energy or products (C and D) minus the enthalpy formation energy of reactants (A and B): delta H(C)+delta H(D)-deltaH(A)-deltaH(B). The values must be looked up from tables.(1 vote)
- hey when sal writes that little c is it actually an = sign?? thanks!(1 vote)
- Yes, it is an = sign. He probably didn't release the pressure on his stylus between the two strokes.(3 votes)
I have this reaction here where if I had a mole of methane, and I react that with two moles of oxygen, I'll produce a mole of carbon dioxide and two moles of water. And what we want to answer in this video is whether this reaction is spontaneous. And we learned in the last video that to answer that question, we have to turn to Gibbs free energy, or the change in Gibbs free energy. And the change in Gibbs free energy is equal to the enthalpy change for the reaction minus the temperature at which it is occurring, times the change in entropy. And if this is less than zero, then it's a spontaneous reaction. So I gave us a little bit of a head start. I just calculated the change in enthalpy for this reaction, and that's right here. And we know how to do that. We've done that several videos ago. You could just look up the heats of formation of each of these products. For water you'll multiply it by 2, since you have 2 moles of it. And so you have the heats of formation of all the products, and then you subtract out the heats of formation of all the reactants. And of course the heat of formation of O2 is O, so this won't even show up in it, and you'll get minus 890.3 kilojoules. Well, this tells us that this is an exothermic reaction. That this side of the equation has less energy in it-- you could kind of think of it that way-- is that side. So some energy must have been released. We could even put here, you know, plus e for energy. Let me write, plus some energy is going to be released. So that's why it's exothermic. But our question is, is this spontaneous? So to figure out if it's spontaneous, we also have to figure out our delta s. And to help figure out the delta s I, ahead of time, looked up the standard molar entropies for each of these molecules. So for example, the standard-- I'll write it here in a different color. The standard-- you put a little naught symbol there-- the standard molar entropy-- so when we say standard, it's at 298 degrees Kelvin. Actually, I shouldn't say degrees Kelvin. It's at 298 Kelvin You don't use the word degrees, necessarily, when you talk about Kelvin. So it's at 298 Kelvin, which is 25 degrees Celsius, so it's at room temperature. So that's why it's considered standard temperature. So the standard entropy of methane at room temperature is equal to this number right here. 186 joules per Kelvin mole. So if I have 1 mole of methane, I have 186 joules per Kelvin of entropy. If I have 2 moles, I multiply that by 2. If I have 3 moles, I multiply that by 3. So the total change in entropy of this reaction is the total standard entropies of the products minus the total standard entropies of the reactants. Just like what we did with enthalpy. So that's going to be equal to 213.6 plus-- I have 2 moles of water here. So it's plus 2 times-- let's just write 70 there. 69.9, almost 70. Plus 2 times 70, and then I want to subtract out the entropy of the reactants, or this side of the reaction. So the entropy of 1 mole of CH4 is 186 plus 2 times 205. So just eyeballing it already, this number is close to this number, but this number is much larger than this number. Liquid water has a much lower-- this is liquid water's entropy. It has a much lower entropy than oxygen gas. And that makes sense. Because liquid formed, there's a lot fewer states. It all falls to the bottom of the container, as opposed to kind of taking the shape of the room and expanding. So a gas is naturally going to have much higher entropy than a liquid. So just eyeballing it, we can already see that our products are going to have a lower entropy than our reactants. So this is probably going to be a negative number. But let's confirm that. So I have 200, 213.6 plus-- well, plus 140, right? 2 times 70. Plus 140 is equal to 353.6. So this is 353.6. And then from that, I'm going to subtract out-- so 186 plus 2 times 205 is equal to 596. So minus 596, and what is that equal to? So we put the minus 596, and then plus the 353.6, and we have minus 242.4. So this is equal to minus 242.4 joules per Kelvin is our delta s minus. So we lose that much entropy. And those units might not make sense to you right now, and actually you know these are but of arbitrary units. But you can just say, hey, this is getting more ordered. And it makes sense, because we have a ton of gas. We have 3 separate molecules, 1 here and 2 molecules of oxygen. And then we go to 3 molecules again, but the water is now liquid. So it makes sense to me that we lose entropy. There's fewer states that the liquid, especially, can take on. But let's figure out whether this reaction is spontaneous. So our delta g is equal to our delta h. We're releasing energy, so it's minus 890. I'll just get rid of the decimals. We don't have to be that precise. Minus our temperature. We're assuming that we're at room temperature, or 298 degrees Kelvin. That's 28-- I should just say, 298 Kelvin. I should get in the habit of not saying degrees when I say Kelvin. Which is 25 degrees Celsius, times our change in entropy. Now, this is going to be a minus. Now you might say, OK, minus 242, you might want to put that there. But you have to be very, very, very careful. This right here is in kilojoules. This right here is in joules. So if we want to write everything in kilojoules, since we already wrote that down, let's write this in kilojoules. So it's 0.242 kilojoules per Kelvin. And so now our Gibbs free energy right here is going to be minus 890 kilojoules minus 290-- so the minus and the minus, you get a plus. And that makes sense, that the entropy term is going to make our Gibbs free energy more positive. Which, as we know, since we want to get this thing below 0, this is going to fight the spontaneity. But let's see if it can overwhelm the actual enthalpy, the exothermic nature of it. And it seems like it will, because you multiply a fraction times this, it's going to be a smaller number than that. But let's just figure it out. So divided by 1, 2, 3. That's our change in entropy times 298, that's our temperature, is minus 72. So this term becomes-- and then we put a minus there-- so it's plus 72.2. So this is the entropy term at standard temperature. It turns into that. And this is our enthalpy term. So we can already see that the enthalpy is a much more negative number than our positive term from our temperature times our change in entropy. So this term is going to win out. Even though we lose entropy in this reaction, it releases so much energy that's going to be spontaneous. This is definitely less than 0, so this is going to be a spontaneous reaction. As you can see, these Gibbs free energy problems, they're really not too difficult. You just really need to find these values. And to find these values, it'll either be given, the delta h, but we know how to solve for the delta h. You just look up the heats of formations of all the products, subtract out the reactants, and of course you wait by the coefficients. And then, to figure out the change in entropy, you do the same thing. You have to look up the standard molar entropies of the products' weight by the coefficients, subtract out the reactants, and then just substitute in here, and then you essentially have the Gibbs free energy. And in this case, it was negative. Now, you could imagine a situation where we're at a much higher temperature. Like the surface of the sun or something, where all of a sudden, instead of a 298 here, if you had like a 2,000 or a 4,000 there. Then all of a sudden, things become interesting. If you could imagine, if you had a 40,000 Kelvin temperature here, then all of a sudden the entropy term, the loss of entropy, is going to matter a lot more. And so this term, this positive term, is going to outweigh this, and maybe it wouldn't be spontaneous at a very, very, very, very high temperature. Another way to think about it. A reaction that generates heat that lets out heat-- the heat being released doesn't matter so much when there's already a lot of heat or kinetic energy in the environment. If the temperature was high enough, this reaction would not be spontaneous, because maybe then the entropy term would win out. But anyway, I just wanted to do this calculation for you to show you that there's nothing too abstract here. You can look up everything on the web, and then figure out if something is going to be spontaneous.