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I have this reaction here[br]where if I had a mole of

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methane, and I react that with[br]two moles of oxygen, I'll

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produce a mole of carbon[br]dioxide and

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two moles of water.

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And what we want to answer in[br]this video is whether this

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reaction is spontaneous.

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And we learned in the last[br]video that to answer that

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question, we have to turn to[br]Gibbs free energy, or the

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change in Gibbs free energy.

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And the change in Gibbs free[br]energy is equal to the

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enthalpy change for the reaction[br]minus the temperature

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at which it is occurring, times[br]the change in entropy.

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And if this is less than[br]zero, then it's

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a spontaneous reaction.

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So I gave us a little[br]bit of a head start.

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I just calculated the change in[br]enthalpy for this reaction,

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and that's right here.

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And we know how to do that.

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We've done that several[br]videos ago.

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You could just look up the heats[br]of formation of each of

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these products.

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For water you'll multiply[br]it by 2, since you

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have 2 moles of it.

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And so you have the heats of[br]formation of all the products,

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and then you subtract[br]out the heats of

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formation of all the reactants.

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And of course the heat of[br]formation of O2 is O, so this

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won't even show up in it, and[br]you'll get minus 890.3

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kilojoules.

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Well, this tells us that this[br]is an exothermic reaction.

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That this side of the equation[br]has less energy in it-- you

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could kind of think of it[br]that way-- is that side.

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So some energy must have[br]been released.

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We could even put here, you[br]know, plus e for energy.

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Let me write, plus some energy[br]is going to be released.

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So that's why it's exothermic.

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But our question is, is[br]this spontaneous?

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So to figure out if it's[br]spontaneous, we also have to

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figure out our delta s.

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And to help figure out the[br]delta s I, ahead of time,

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looked up the standard[br]molar entropies for

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each of these molecules.

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So for example, the standard--[br]I'll write it here in a

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different color.

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The standard-- you put a little[br]naught symbol there--

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the standard molar entropy-- so[br]when we say standard, it's

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at 298 degrees Kelvin.

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Actually, I shouldn't[br]say degrees Kelvin.

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It's at 298 Kelvin You don't[br]use the word degrees,

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necessarily, when you[br]talk about Kelvin.

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So it's at 298 Kelvin, which is[br]25 degrees Celsius, so it's

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at room temperature.

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So that's why it's considered[br]standard temperature.

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So the standard entropy of[br]methane at room temperature is

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equal to this number[br]right here.

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186 joules per Kelvin mole.

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So if I have 1 mole of methane,[br]I have 186 joules per

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Kelvin of entropy.

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If I have 2 moles, I[br]multiply that by 2.

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If I have 3 moles, I[br]multiply that by 3.

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So the total change in entropy[br]of this reaction is the total

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standard entropies of the[br]products minus the total

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standard entropies[br]of the reactants.

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Just like what we did[br]with enthalpy.

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So that's going to be equal to[br]213.6 plus-- I have 2 moles of

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water here.

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So it's plus 2 times-- let's[br]just write 70 there.

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69.9, almost 70.

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Plus 2 times 70, and then I[br]want to subtract out the

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entropy of the reactants, or[br]this side of the reaction.

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So the entropy of 1 mole of CH4[br]is 186 plus 2 times 205.

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So just eyeballing it already,[br]this number is close to this

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number, but this number is much[br]larger than this number.

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Liquid water has a much[br]lower-- this is

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liquid water's entropy.

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It has a much lower entropy[br]than oxygen gas.

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And that makes sense.

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Because liquid formed, there's[br]a lot fewer states.

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It all falls to the bottom of[br]the container, as opposed to

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kind of taking the shape of[br]the room and expanding.

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So a gas is naturally going[br]to have much higher

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entropy than a liquid.

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So just eyeballing it, we can[br]already see that our products

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are going to have a lower[br]entropy than our reactants.

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So this is probably going[br]to be a negative number.

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But let's confirm that.

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So I have 200, 213.6 plus--[br]well, plus 140, right?

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2 times 70.

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Plus 140 is equal to 353.6.

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So this is 353.6.

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And then from that, I'm going to[br]subtract out-- so 186 plus

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2 times 205 is equal to 596.

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So minus 596, and what[br]is that equal to?

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So we put the minus 596, and[br]then plus the 353.6, and we

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have minus 242.4.

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So this is equal to minus 242.4[br]joules per Kelvin is our

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delta s minus.

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So we lose that much entropy.

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And those units might not make[br]sense to you right now, and

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actually you know these are[br]but of arbitrary units.

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But you can just say, hey, this[br]is getting more ordered.

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And it makes sense, because[br]we have a ton of gas.

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We have 3 separate molecules,[br]1 here and 2

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molecules of oxygen.

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And then we go to 3 molecules[br]again, but the

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water is now liquid.

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So it makes sense to me[br]that we lose entropy.

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There's fewer states[br]that the liquid,

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especially, can take on.

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But let's figure out whether[br]this reaction is spontaneous.

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So our delta g is equal[br]to our delta h.

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We're releasing energy,[br]so it's minus 890.

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I'll just get rid[br]of the decimals.

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We don't have to be[br]that precise.

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Minus our temperature.

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We're assuming that we're at[br]room temperature, or 298

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degrees Kelvin.

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That's 28-- I should just[br]say, 298 Kelvin.

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I should get in the habit[br]of not saying

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degrees when I say Kelvin.

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Which is 25 degrees Celsius,[br]times our change in entropy.

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Now, this is going[br]to be a minus.

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Now you might say, OK, minus[br]242, you might want to put

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that there.

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But you have to be very,[br]very, very careful.

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This right here is[br]in kilojoules.

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This right here is in joules.

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So if we want to write[br]everything in kilojoules,

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since we already wrote that[br]down, let's write this in

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kilojoules.

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So it's 0.242 kilojoules[br]per Kelvin.

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And so now our Gibbs free energy[br]right here is going to

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be minus 890 kilojoules minus[br]290-- so the minus and the

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minus, you get a plus.

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And that makes sense, that the[br]entropy term is going to make

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our Gibbs free energy[br]more positive.

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Which, as we know, since we want[br]to get this thing below

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0, this is going to fight[br]the spontaneity.

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But let's see if it can[br]overwhelm the actual enthalpy,

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the exothermic nature of it.

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And it seems like it will,[br]because you multiply a

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fraction times this,[br]it's going to be a

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smaller number than that.

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But let's just figure it out.

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So divided by 1, 2, 3.

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That's our change in entropy[br]times 298, that's our

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temperature, is minus 72.

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So this term becomes-- and then[br]we put a minus there-- so

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it's plus 72.2.

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So this is the entropy term[br]at standard temperature.

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It turns into that.

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And this is our enthalpy term.

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So we can already see that the[br]enthalpy is a much more

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negative number than our[br]positive term from our

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temperature times our[br]change in entropy.

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So this term is going[br]to win out.

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Even though we lose entropy in[br]this reaction, it releases so

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much energy that's going[br]to be spontaneous.

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This is definitely less than[br]0, so this is going to be a

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spontaneous reaction.

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As you can see, these Gibbs free[br]energy problems, they're

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really not too difficult.

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You just really need to[br]find these values.

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And to find these values, it'll[br]either be given, the

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delta h, but we know how to[br]solve for the delta h.

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You just look up the heats[br]of formations of all the

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products, subtract out the[br]reactants, and of course you

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wait by the coefficients.

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And then, to figure out the[br]change in entropy, you do the

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same thing.

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You have to look up the standard[br]molar entropies of

0:08:44.020,0:08:46.840
the products' weight by the[br]coefficients, subtract out the

0:08:46.840,0:08:50.110
reactants, and then just[br]substitute in here, and then

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you essentially have the[br]Gibbs free energy.

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And in this case,[br]it was negative.

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Now, you could imagine a[br]situation where we're at a

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much higher temperature.

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Like the surface of the sun or[br]something, where all of a

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sudden, instead of a 298 here,[br]if you had like a 2,000 or a

0:09:06.380,0:09:08.140
4,000 there.

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Then all of a sudden, things[br]become interesting.

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If you could imagine, if[br]you had a 40,000 Kelvin

0:09:14.710,0:09:17.530
temperature here, then all of a[br]sudden the entropy term, the

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loss of entropy, is going[br]to matter a lot more.

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And so this term, this positive[br]term, is going to

0:09:22.480,0:09:25.870
outweigh this, and maybe it[br]wouldn't be spontaneous at a

0:09:25.870,0:09:27.960
very, very, very, very[br]high temperature.

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Another way to think about it.

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A reaction that generates heat[br]that lets out heat-- the heat

0:09:34.560,0:09:37.090
being released doesn't matter so[br]much when there's already a

0:09:37.090,0:09:39.980
lot of heat or kinetic energy[br]in the environment.

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If the temperature was high[br]enough, this reaction would

0:09:42.920,0:09:46.050
not be spontaneous, because[br]maybe then the entropy term

0:09:46.050,0:09:46.980
would win out.

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But anyway, I just wanted to do[br]this calculation for you to

0:09:49.130,0:09:51.360
show you that there's nothing[br]too abstract here.

0:09:51.360,0:09:53.950
You can look up everything on[br]the web, and then figure out

0:09:53.950,0:09:56.320
if something is going[br]to be spontaneous.

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