I have this reaction herewhere if I had a mole of methane, and I react that withtwo moles of oxygen, I'll produce a mole of carbondioxide and two moles of water. And what we want to answer inthis video is whether this reaction is spontaneous. And we learned in the lastvideo that to answer that question, we have to turn toGibbs free energy, or the change in Gibbs free energy. And the change in Gibbs freeenergy is equal to the enthalpy change for the reactionminus the temperature at which it is occurring, timesthe change in entropy. And if this is less thanzero, then it's a spontaneous reaction. So I gave us a littlebit of a head start. I just calculated the change inenthalpy for this reaction, and that's right here. And we know how to do that. We've done that severalvideos ago. You could just look up the heatsof formation of each of these products. For water you'll multiplyit by 2, since you have 2 moles of it. And so you have the heats offormation of all the products, and then you subtractout the heats of formation of all the reactants. And of course the heat offormation of O2 is O, so this won't even show up in it, andyou'll get minus 890.3 kilojoules. Well, this tells us that thisis an exothermic reaction. That this side of the equationhas less energy in it-- you could kind of think of itthat way-- is that side. So some energy must havebeen released. We could even put here, youknow, plus e for energy. Let me write, plus some energyis going to be released. So that's why it's exothermic. But our question is, isthis spontaneous? So to figure out if it'sspontaneous, we also have to figure out our delta s. And to help figure out thedelta s I, ahead of time, looked up the standardmolar 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 littlenaught symbol there-- the standard molar entropy-- sowhen we say standard, it's at 298 degrees Kelvin. Actually, I shouldn'tsay degrees Kelvin. It's at 298 Kelvin You don'tuse the word degrees, necessarily, when youtalk about Kelvin. So it's at 298 Kelvin, which is25 degrees Celsius, so it's at room temperature. So that's why it's consideredstandard temperature. So the standard entropy ofmethane at room temperature is equal to this numberright 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, Imultiply that by 2. If I have 3 moles, Imultiply that by 3. So the total change in entropyof this reaction is the total standard entropies of theproducts minus the total standard entropiesof the reactants. Just like what we didwith enthalpy. So that's going to be equal to213.6 plus-- I have 2 moles of water here. So it's plus 2 times-- let'sjust write 70 there. 69.9, almost 70. Plus 2 times 70, and then Iwant to subtract out the entropy of the reactants, orthis side of the reaction. So the entropy of 1 mole of CH4is 186 plus 2 times 205. So just eyeballing it already,this number is close to this number, but this number is muchlarger than this number. Liquid water has a muchlower-- this is liquid water's entropy. It has a much lower entropythan oxygen gas. And that makes sense. Because liquid formed, there'sa lot fewer states. It all falls to the bottom ofthe container, as opposed to kind of taking the shape ofthe room and expanding. So a gas is naturally goingto have much higher entropy than a liquid. So just eyeballing it, we canalready see that our products are going to have a lowerentropy than our reactants. So this is probably goingto 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 tosubtract out-- so 186 plus 2 times 205 is equal to 596. So minus 596, and whatis that equal to? So we put the minus 596, andthen plus the 353.6, and we have minus 242.4. So this is equal to minus 242.4joules per Kelvin is our delta s minus. So we lose that much entropy. And those units might not makesense to you right now, and actually you know these arebut of arbitrary units. But you can just say, hey, thisis getting more ordered. And it makes sense, becausewe have a ton of gas. We have 3 separate molecules,1 here and 2 molecules of oxygen. And then we go to 3 moleculesagain, but the water is now liquid. So it makes sense to methat we lose entropy. There's fewer statesthat the liquid, especially, can take on. But let's figure out whetherthis reaction is spontaneous. So our delta g is equalto our delta h. We're releasing energy,so it's minus 890. I'll just get ridof the decimals. We don't have to bethat precise. Minus our temperature. We're assuming that we're atroom temperature, or 298 degrees Kelvin. That's 28-- I should justsay, 298 Kelvin. I should get in the habitof not saying degrees when I say Kelvin. Which is 25 degrees Celsius,times our change in entropy. Now, this is goingto be a minus. Now you might say, OK, minus242, you might want to put that there. But you have to be very,very, very careful. This right here isin kilojoules. This right here is in joules. So if we want to writeeverything in kilojoules, since we already wrote thatdown, let's write this in kilojoules. So it's 0.242 kilojoulesper Kelvin. And so now our Gibbs free energyright here is going to be minus 890 kilojoules minus290-- so the minus and the minus, you get a plus. And that makes sense, that theentropy term is going to make our Gibbs free energymore positive. Which, as we know, since we wantto get this thing below 0, this is going to fightthe spontaneity. But let's see if it canoverwhelm 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 entropytimes 298, that's our temperature, is minus 72. So this term becomes-- and thenwe put a minus there-- so it's plus 72.2. So this is the entropy termat standard temperature. It turns into that. And this is our enthalpy term. So we can already see that theenthalpy is a much more negative number than ourpositive term from our temperature times ourchange in entropy. So this term is goingto win out. Even though we lose entropy inthis reaction, it releases so much energy that's goingto be spontaneous. This is definitely less than0, so this is going to be a spontaneous reaction. As you can see, these Gibbs freeenergy problems, they're really not too difficult. You just really need tofind these values. And to find these values, it'lleither be given, the delta h, but we know how tosolve for the delta h. You just look up the heatsof formations of all the products, subtract out thereactants, and of course you wait by the coefficients. And then, to figure out thechange in entropy, you do the same thing. You have to look up the standardmolar entropies of the products' weight by thecoefficients, subtract out the reactants, and then justsubstitute in here, and then you essentially have theGibbs free energy. And in this case,it was negative. Now, you could imagine asituation where we're at a much higher temperature. Like the surface of the sun orsomething, 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, thingsbecome interesting. If you could imagine, ifyou had a 40,000 Kelvin temperature here, then all of asudden the entropy term, the loss of entropy, is goingto matter a lot more. And so this term, this positiveterm, is going to outweigh this, and maybe itwouldn't be spontaneous at a very, very, very, veryhigh temperature. Another way to think about it. A reaction that generates heatthat lets out heat-- the heat being released doesn't matter somuch when there's already a lot of heat or kinetic energyin the environment. If the temperature was highenough, this reaction would not be spontaneous, becausemaybe then the entropy term would win out.


You are watching: Calculate the change in gibbs free energy for each of the following sets of δh∘rxn, δs∘rxn, and t.


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But anyway, I just wanted to dothis calculation for you to show you that there's nothingtoo abstract here. You can look up everything onthe web, and then figure out if something is goingto be spontaneous.