[Script Info]
Title: 
[Events]
Format: Layer, Start, End, Style, Name, MarginL, MarginR, MarginV, Effect, Text
Dialogue: 0,0:00:00.41,0:00:02.73,Default,,0000,0000,0000,,I have this reaction here\Nwhere if I had a mole of
Dialogue: 0,0:00:02.73,0:00:05.97,Default,,0000,0000,0000,,methane, and I react that with\Ntwo moles of oxygen, I'll
Dialogue: 0,0:00:05.97,0:00:08.04,Default,,0000,0000,0000,,produce a mole of carbon\Ndioxide and
Dialogue: 0,0:00:08.04,0:00:09.68,Default,,0000,0000,0000,,two moles of water.
Dialogue: 0,0:00:09.68,0:00:12.38,Default,,0000,0000,0000,,And what we want to answer in\Nthis video is whether this
Dialogue: 0,0:00:12.38,0:00:14.76,Default,,0000,0000,0000,,reaction is spontaneous.
Dialogue: 0,0:00:14.76,0:00:17.29,Default,,0000,0000,0000,,And we learned in the last\Nvideo that to answer that
Dialogue: 0,0:00:17.29,0:00:19.81,Default,,0000,0000,0000,,question, we have to turn to\NGibbs free energy, or the
Dialogue: 0,0:00:19.81,0:00:21.62,Default,,0000,0000,0000,,change in Gibbs free energy.
Dialogue: 0,0:00:21.62,0:00:25.32,Default,,0000,0000,0000,,And the change in Gibbs free\Nenergy is equal to the
Dialogue: 0,0:00:25.32,0:00:29.30,Default,,0000,0000,0000,,enthalpy change for the reaction\Nminus the temperature
Dialogue: 0,0:00:29.30,0:00:32.82,Default,,0000,0000,0000,,at which it is occurring, times\Nthe change in entropy.
Dialogue: 0,0:00:32.82,0:00:35.84,Default,,0000,0000,0000,,And if this is less than\Nzero, then it's
Dialogue: 0,0:00:35.84,0:00:39.24,Default,,0000,0000,0000,,a spontaneous reaction.
Dialogue: 0,0:00:39.24,0:00:41.86,Default,,0000,0000,0000,,So I gave us a little\Nbit of a head start.
Dialogue: 0,0:00:41.86,0:00:46.07,Default,,0000,0000,0000,,I just calculated the change in\Nenthalpy for this reaction,
Dialogue: 0,0:00:46.07,0:00:47.53,Default,,0000,0000,0000,,and that's right here.
Dialogue: 0,0:00:47.53,0:00:48.46,Default,,0000,0000,0000,,And we know how to do that.
Dialogue: 0,0:00:48.46,0:00:50.23,Default,,0000,0000,0000,,We've done that several\Nvideos ago.
Dialogue: 0,0:00:50.23,0:00:52.90,Default,,0000,0000,0000,,You could just look up the heats\Nof formation of each of
Dialogue: 0,0:00:52.90,0:00:54.88,Default,,0000,0000,0000,,these products.
Dialogue: 0,0:00:54.88,0:00:56.95,Default,,0000,0000,0000,,For water you'll multiply\Nit by 2, since you
Dialogue: 0,0:00:56.95,0:00:58.18,Default,,0000,0000,0000,,have 2 moles of it.
Dialogue: 0,0:00:58.18,0:01:00.51,Default,,0000,0000,0000,,And so you have the heats of\Nformation of all the products,
Dialogue: 0,0:01:00.51,0:01:02.50,Default,,0000,0000,0000,,and then you subtract\Nout the heats of
Dialogue: 0,0:01:02.50,0:01:03.78,Default,,0000,0000,0000,,formation of all the reactants.
Dialogue: 0,0:01:03.78,0:01:07.27,Default,,0000,0000,0000,,And of course the heat of\Nformation of O2 is O, so this
Dialogue: 0,0:01:07.27,0:01:11.04,Default,,0000,0000,0000,,won't even show up in it, and\Nyou'll get minus 890.3
Dialogue: 0,0:01:11.04,0:01:11.59,Default,,0000,0000,0000,,kilojoules.
Dialogue: 0,0:01:11.59,0:01:15.09,Default,,0000,0000,0000,,Well, this tells us that this\Nis an exothermic reaction.
Dialogue: 0,0:01:15.09,0:01:18.49,Default,,0000,0000,0000,,That this side of the equation\Nhas less energy in it-- you
Dialogue: 0,0:01:18.49,0:01:20.45,Default,,0000,0000,0000,,could kind of think of it\Nthat way-- is that side.
Dialogue: 0,0:01:20.45,0:01:22.45,Default,,0000,0000,0000,,So some energy must have\Nbeen released.
Dialogue: 0,0:01:22.45,0:01:25.40,Default,,0000,0000,0000,,We could even put here, you\Nknow, plus e for energy.
Dialogue: 0,0:01:25.40,0:01:27.98,Default,,0000,0000,0000,,Let me write, plus some energy\Nis going to be released.
Dialogue: 0,0:01:27.98,0:01:29.75,Default,,0000,0000,0000,,So that's why it's exothermic.
Dialogue: 0,0:01:29.75,0:01:32.01,Default,,0000,0000,0000,,But our question is, is\Nthis spontaneous?
Dialogue: 0,0:01:32.01,0:01:34.10,Default,,0000,0000,0000,,So to figure out if it's\Nspontaneous, we also have to
Dialogue: 0,0:01:34.10,0:01:36.14,Default,,0000,0000,0000,,figure out our delta s.
Dialogue: 0,0:01:39.45,0:01:42.69,Default,,0000,0000,0000,,And to help figure out the\Ndelta s I, ahead of time,
Dialogue: 0,0:01:42.69,0:01:46.09,Default,,0000,0000,0000,,looked up the standard\Nmolar entropies for
Dialogue: 0,0:01:46.09,0:01:48.13,Default,,0000,0000,0000,,each of these molecules.
Dialogue: 0,0:01:48.13,0:01:50.46,Default,,0000,0000,0000,,So for example, the standard--\NI'll write it here in a
Dialogue: 0,0:01:50.46,0:01:51.71,Default,,0000,0000,0000,,different color.
Dialogue: 0,0:02:00.51,0:02:04.98,Default,,0000,0000,0000,,The standard-- you put a little\Nnaught symbol there--
Dialogue: 0,0:02:04.98,0:02:08.39,Default,,0000,0000,0000,,the standard molar entropy-- so\Nwhen we say standard, it's
Dialogue: 0,0:02:08.39,0:02:11.04,Default,,0000,0000,0000,,at 298 degrees Kelvin.
Dialogue: 0,0:02:11.04,0:02:12.73,Default,,0000,0000,0000,,Actually, I shouldn't\Nsay degrees Kelvin.
Dialogue: 0,0:02:12.73,0:02:15.68,Default,,0000,0000,0000,,It's at 298 Kelvin You don't\Nuse the word degrees,
Dialogue: 0,0:02:15.68,0:02:17.44,Default,,0000,0000,0000,,necessarily, when you\Ntalk about Kelvin.
Dialogue: 0,0:02:17.44,0:02:21.02,Default,,0000,0000,0000,,So it's at 298 Kelvin, which is\N25 degrees Celsius, so it's
Dialogue: 0,0:02:21.02,0:02:21.88,Default,,0000,0000,0000,,at room temperature.
Dialogue: 0,0:02:21.88,0:02:24.69,Default,,0000,0000,0000,,So that's why it's considered\Nstandard temperature.
Dialogue: 0,0:02:24.69,0:02:29.70,Default,,0000,0000,0000,,So the standard entropy of\Nmethane at room temperature is
Dialogue: 0,0:02:29.70,0:02:30.98,Default,,0000,0000,0000,,equal to this number\Nright here.
Dialogue: 0,0:02:30.98,0:02:37.88,Default,,0000,0000,0000,,186 joules per Kelvin mole.
Dialogue: 0,0:02:37.88,0:02:42.32,Default,,0000,0000,0000,,So if I have 1 mole of methane,\NI have 186 joules per
Dialogue: 0,0:02:42.32,0:02:43.93,Default,,0000,0000,0000,,Kelvin of entropy.
Dialogue: 0,0:02:43.93,0:02:46.05,Default,,0000,0000,0000,,If I have 2 moles, I\Nmultiply that by 2.
Dialogue: 0,0:02:46.05,0:02:48.78,Default,,0000,0000,0000,,If I have 3 moles, I\Nmultiply that by 3.
Dialogue: 0,0:02:48.78,0:02:54.87,Default,,0000,0000,0000,,So the total change in entropy\Nof this reaction is the total
Dialogue: 0,0:02:54.87,0:02:58.95,Default,,0000,0000,0000,,standard entropies of the\Nproducts minus the total
Dialogue: 0,0:02:58.95,0:03:00.61,Default,,0000,0000,0000,,standard entropies\Nof the reactants.
Dialogue: 0,0:03:00.61,0:03:02.45,Default,,0000,0000,0000,,Just like what we did\Nwith enthalpy.
Dialogue: 0,0:03:02.45,0:03:11.65,Default,,0000,0000,0000,,So that's going to be equal to\N213.6 plus-- I have 2 moles of
Dialogue: 0,0:03:11.65,0:03:12.19,Default,,0000,0000,0000,,water here.
Dialogue: 0,0:03:12.19,0:03:17.81,Default,,0000,0000,0000,,So it's plus 2 times-- let's\Njust write 70 there.
Dialogue: 0,0:03:17.81,0:03:20.09,Default,,0000,0000,0000,,69.9, almost 70.
Dialogue: 0,0:03:20.09,0:03:24.72,Default,,0000,0000,0000,,Plus 2 times 70, and then I\Nwant to subtract out the
Dialogue: 0,0:03:24.72,0:03:28.92,Default,,0000,0000,0000,,entropy of the reactants, or\Nthis side of the reaction.
Dialogue: 0,0:03:28.92,0:03:42.90,Default,,0000,0000,0000,,So the entropy of 1 mole of CH4\Nis 186 plus 2 times 205.
Dialogue: 0,0:03:42.90,0:03:45.43,Default,,0000,0000,0000,,So just eyeballing it already,\Nthis number is close to this
Dialogue: 0,0:03:45.43,0:03:48.22,Default,,0000,0000,0000,,number, but this number is much\Nlarger than this number.
Dialogue: 0,0:03:48.22,0:03:50.62,Default,,0000,0000,0000,,Liquid water has a much\Nlower-- this is
Dialogue: 0,0:03:50.62,0:03:51.65,Default,,0000,0000,0000,,liquid water's entropy.
Dialogue: 0,0:03:51.65,0:03:54.97,Default,,0000,0000,0000,,It has a much lower entropy\Nthan oxygen gas.
Dialogue: 0,0:03:54.97,0:03:55.73,Default,,0000,0000,0000,,And that makes sense.
Dialogue: 0,0:03:55.73,0:04:00.15,Default,,0000,0000,0000,,Because liquid formed, there's\Na lot fewer states.
Dialogue: 0,0:04:00.15,0:04:02.63,Default,,0000,0000,0000,,It all falls to the bottom of\Nthe container, as opposed to
Dialogue: 0,0:04:02.63,0:04:04.68,Default,,0000,0000,0000,,kind of taking the shape of\Nthe room and expanding.
Dialogue: 0,0:04:04.68,0:04:06.60,Default,,0000,0000,0000,,So a gas is naturally going\Nto have much higher
Dialogue: 0,0:04:06.60,0:04:08.23,Default,,0000,0000,0000,,entropy than a liquid.
Dialogue: 0,0:04:08.23,0:04:11.83,Default,,0000,0000,0000,,So just eyeballing it, we can\Nalready see that our products
Dialogue: 0,0:04:11.83,0:04:13.93,Default,,0000,0000,0000,,are going to have a lower\Nentropy than our reactants.
Dialogue: 0,0:04:13.93,0:04:15.40,Default,,0000,0000,0000,,So this is probably going\Nto be a negative number.
Dialogue: 0,0:04:15.40,0:04:19.43,Default,,0000,0000,0000,,But let's confirm that.
Dialogue: 0,0:04:19.43,0:04:30.61,Default,,0000,0000,0000,,So I have 200, 213.6 plus--\Nwell, plus 140, right?
Dialogue: 0,0:04:30.61,0:04:31.27,Default,,0000,0000,0000,,2 times 70.
Dialogue: 0,0:04:31.27,0:04:35.54,Default,,0000,0000,0000,,Plus 140 is equal to 353.6.
Dialogue: 0,0:04:35.54,0:04:39.93,Default,,0000,0000,0000,,So this is 353.6.
Dialogue: 0,0:04:39.93,0:04:47.58,Default,,0000,0000,0000,,And then from that, I'm going to\Nsubtract out-- so 186 plus
Dialogue: 0,0:04:47.58,0:04:54.79,Default,,0000,0000,0000,,2 times 205 is equal to 596.
Dialogue: 0,0:04:54.79,0:04:58.90,Default,,0000,0000,0000,,So minus 596, and what\Nis that equal to?
Dialogue: 0,0:04:58.90,0:05:06.78,Default,,0000,0000,0000,,So we put the minus 596, and\Nthen plus the 353.6, and we
Dialogue: 0,0:05:06.78,0:05:10.91,Default,,0000,0000,0000,,have minus 242.4.
Dialogue: 0,0:05:10.91,0:05:18.22,Default,,0000,0000,0000,,So this is equal to minus 242.4\Njoules per Kelvin is our
Dialogue: 0,0:05:18.22,0:05:21.82,Default,,0000,0000,0000,,delta s minus.
Dialogue: 0,0:05:21.82,0:05:23.80,Default,,0000,0000,0000,,So we lose that much entropy.
Dialogue: 0,0:05:23.80,0:05:26.02,Default,,0000,0000,0000,,And those units might not make\Nsense to you right now, and
Dialogue: 0,0:05:26.02,0:05:28.78,Default,,0000,0000,0000,,actually you know these are\Nbut of arbitrary units.
Dialogue: 0,0:05:28.78,0:05:30.75,Default,,0000,0000,0000,,But you can just say, hey, this\Nis getting more ordered.
Dialogue: 0,0:05:30.75,0:05:33.71,Default,,0000,0000,0000,,And it makes sense, because\Nwe have a ton of gas.
Dialogue: 0,0:05:33.71,0:05:36.08,Default,,0000,0000,0000,,We have 3 separate molecules,\N1 here and 2
Dialogue: 0,0:05:36.08,0:05:38.23,Default,,0000,0000,0000,,molecules of oxygen.
Dialogue: 0,0:05:38.23,0:05:40.84,Default,,0000,0000,0000,,And then we go to 3 molecules\Nagain, but the
Dialogue: 0,0:05:40.84,0:05:42.31,Default,,0000,0000,0000,,water is now liquid.
Dialogue: 0,0:05:42.31,0:05:45.77,Default,,0000,0000,0000,,So it makes sense to me\Nthat we lose entropy.
Dialogue: 0,0:05:45.77,0:05:47.33,Default,,0000,0000,0000,,There's fewer states\Nthat the liquid,
Dialogue: 0,0:05:47.33,0:05:48.79,Default,,0000,0000,0000,,especially, can take on.
Dialogue: 0,0:05:48.79,0:05:51.65,Default,,0000,0000,0000,,But let's figure out whether\Nthis reaction is spontaneous.
Dialogue: 0,0:05:51.65,0:05:57.44,Default,,0000,0000,0000,,So our delta g is equal\Nto our delta h.
Dialogue: 0,0:05:57.44,0:06:01.14,Default,,0000,0000,0000,,We're releasing energy,\Nso it's minus 890.
Dialogue: 0,0:06:01.14,0:06:02.51,Default,,0000,0000,0000,,I'll just get rid\Nof the decimals.
Dialogue: 0,0:06:02.51,0:06:04.49,Default,,0000,0000,0000,,We don't have to be\Nthat precise.
Dialogue: 0,0:06:04.49,0:06:06.35,Default,,0000,0000,0000,,Minus our temperature.
Dialogue: 0,0:06:06.35,0:06:09.22,Default,,0000,0000,0000,,We're assuming that we're at\Nroom temperature, or 298
Dialogue: 0,0:06:09.22,0:06:10.34,Default,,0000,0000,0000,,degrees Kelvin.
Dialogue: 0,0:06:10.34,0:06:13.40,Default,,0000,0000,0000,,That's 28-- I should just\Nsay, 298 Kelvin.
Dialogue: 0,0:06:13.40,0:06:14.59,Default,,0000,0000,0000,,I should get in the habit\Nof not saying
Dialogue: 0,0:06:14.59,0:06:15.96,Default,,0000,0000,0000,,degrees when I say Kelvin.
Dialogue: 0,0:06:15.96,0:06:22.75,Default,,0000,0000,0000,,Which is 25 degrees Celsius,\Ntimes our change in entropy.
Dialogue: 0,0:06:22.75,0:06:25.05,Default,,0000,0000,0000,,Now, this is going\Nto be a minus.
Dialogue: 0,0:06:25.05,0:06:27.76,Default,,0000,0000,0000,,Now you might say, OK, minus\N242, you might want to put
Dialogue: 0,0:06:27.76,0:06:28.53,Default,,0000,0000,0000,,that there.
Dialogue: 0,0:06:28.53,0:06:30.28,Default,,0000,0000,0000,,But you have to be very,\Nvery, very careful.
Dialogue: 0,0:06:30.28,0:06:33.16,Default,,0000,0000,0000,,This right here is\Nin kilojoules.
Dialogue: 0,0:06:33.16,0:06:35.19,Default,,0000,0000,0000,,This right here is in joules.
Dialogue: 0,0:06:35.19,0:06:37.54,Default,,0000,0000,0000,,So if we want to write\Neverything in kilojoules,
Dialogue: 0,0:06:37.54,0:06:39.66,Default,,0000,0000,0000,,since we already wrote that\Ndown, let's write this in
Dialogue: 0,0:06:39.66,0:06:40.37,Default,,0000,0000,0000,,kilojoules.
Dialogue: 0,0:06:40.37,0:06:46.62,Default,,0000,0000,0000,,So it's 0.242 kilojoules\Nper Kelvin.
Dialogue: 0,0:06:52.37,0:06:55.79,Default,,0000,0000,0000,,And so now our Gibbs free energy\Nright here is going to
Dialogue: 0,0:06:55.79,0:07:00.88,Default,,0000,0000,0000,,be minus 890 kilojoules minus\N290-- so the minus and the
Dialogue: 0,0:07:00.88,0:07:02.76,Default,,0000,0000,0000,,minus, you get a plus.
Dialogue: 0,0:07:02.76,0:07:05.77,Default,,0000,0000,0000,,And that makes sense, that the\Nentropy term is going to make
Dialogue: 0,0:07:05.77,0:07:08.48,Default,,0000,0000,0000,,our Gibbs free energy\Nmore positive.
Dialogue: 0,0:07:08.48,0:07:11.48,Default,,0000,0000,0000,,Which, as we know, since we want\Nto get this thing below
Dialogue: 0,0:07:11.48,0:07:13.98,Default,,0000,0000,0000,,0, this is going to fight\Nthe spontaneity.
Dialogue: 0,0:07:13.98,0:07:18.84,Default,,0000,0000,0000,,But let's see if it can\Noverwhelm the actual enthalpy,
Dialogue: 0,0:07:18.84,0:07:20.47,Default,,0000,0000,0000,,the exothermic nature of it.
Dialogue: 0,0:07:20.47,0:07:22.48,Default,,0000,0000,0000,,And it seems like it will,\Nbecause you multiply a
Dialogue: 0,0:07:22.48,0:07:24.01,Default,,0000,0000,0000,,fraction times this,\Nit's going to be a
Dialogue: 0,0:07:24.01,0:07:25.09,Default,,0000,0000,0000,,smaller number than that.
Dialogue: 0,0:07:25.09,0:07:27.56,Default,,0000,0000,0000,,But let's just figure it out.
Dialogue: 0,0:07:27.56,0:07:31.83,Default,,0000,0000,0000,,So divided by 1, 2, 3.
Dialogue: 0,0:07:31.83,0:07:37.17,Default,,0000,0000,0000,,That's our change in entropy\Ntimes 298, that's our
Dialogue: 0,0:07:37.17,0:07:40.41,Default,,0000,0000,0000,,temperature, is minus 72.
Dialogue: 0,0:07:40.41,0:07:43.71,Default,,0000,0000,0000,,So this term becomes-- and then\Nwe put a minus there-- so
Dialogue: 0,0:07:43.71,0:07:47.36,Default,,0000,0000,0000,,it's plus 72.2.
Dialogue: 0,0:07:47.36,0:07:50.30,Default,,0000,0000,0000,,So this is the entropy term\Nat standard temperature.
Dialogue: 0,0:07:50.30,0:07:51.37,Default,,0000,0000,0000,,It turns into that.
Dialogue: 0,0:07:51.37,0:07:52.97,Default,,0000,0000,0000,,And this is our enthalpy term.
Dialogue: 0,0:07:52.97,0:07:55.90,Default,,0000,0000,0000,,So we can already see that the\Nenthalpy is a much more
Dialogue: 0,0:07:55.90,0:07:58.32,Default,,0000,0000,0000,,negative number than our\Npositive term from our
Dialogue: 0,0:07:58.32,0:08:00.48,Default,,0000,0000,0000,,temperature times our\Nchange in entropy.
Dialogue: 0,0:08:00.48,0:08:04.53,Default,,0000,0000,0000,,So this term is going\Nto win out.
Dialogue: 0,0:08:04.53,0:08:08.33,Default,,0000,0000,0000,,Even though we lose entropy in\Nthis reaction, it releases so
Dialogue: 0,0:08:08.33,0:08:10.84,Default,,0000,0000,0000,,much energy that's going\Nto be spontaneous.
Dialogue: 0,0:08:10.84,0:08:14.34,Default,,0000,0000,0000,,This is definitely less than\N0, so this is going to be a
Dialogue: 0,0:08:14.34,0:08:17.36,Default,,0000,0000,0000,,spontaneous reaction.
Dialogue: 0,0:08:17.36,0:08:19.52,Default,,0000,0000,0000,,As you can see, these Gibbs free\Nenergy problems, they're
Dialogue: 0,0:08:19.52,0:08:20.63,Default,,0000,0000,0000,,really not too difficult.
Dialogue: 0,0:08:20.63,0:08:23.90,Default,,0000,0000,0000,,You just really need to\Nfind these values.
Dialogue: 0,0:08:23.90,0:08:27.23,Default,,0000,0000,0000,,And to find these values, it'll\Neither be given, the
Dialogue: 0,0:08:27.23,0:08:29.67,Default,,0000,0000,0000,,delta h, but we know how to\Nsolve for the delta h.
Dialogue: 0,0:08:29.67,0:08:31.72,Default,,0000,0000,0000,,You just look up the heats\Nof formations of all the
Dialogue: 0,0:08:31.72,0:08:35.21,Default,,0000,0000,0000,,products, subtract out the\Nreactants, and of course you
Dialogue: 0,0:08:35.21,0:08:38.11,Default,,0000,0000,0000,,wait by the coefficients.
Dialogue: 0,0:08:38.11,0:08:40.49,Default,,0000,0000,0000,,And then, to figure out the\Nchange in entropy, you do the
Dialogue: 0,0:08:40.49,0:08:40.91,Default,,0000,0000,0000,,same thing.
Dialogue: 0,0:08:40.91,0:08:44.02,Default,,0000,0000,0000,,You have to look up the standard\Nmolar entropies of
Dialogue: 0,0:08:44.02,0:08:46.84,Default,,0000,0000,0000,,the products' weight by the\Ncoefficients, subtract out the
Dialogue: 0,0:08:46.84,0:08:50.11,Default,,0000,0000,0000,,reactants, and then just\Nsubstitute in here, and then
Dialogue: 0,0:08:50.11,0:08:51.85,Default,,0000,0000,0000,,you essentially have the\NGibbs free energy.
Dialogue: 0,0:08:51.85,0:08:54.64,Default,,0000,0000,0000,,And in this case,\Nit was negative.
Dialogue: 0,0:08:54.64,0:08:56.57,Default,,0000,0000,0000,,Now, you could imagine a\Nsituation where we're at a
Dialogue: 0,0:08:56.57,0:08:57.72,Default,,0000,0000,0000,,much higher temperature.
Dialogue: 0,0:08:57.72,0:09:00.89,Default,,0000,0000,0000,,Like the surface of the sun or\Nsomething, where all of a
Dialogue: 0,0:09:00.89,0:09:06.38,Default,,0000,0000,0000,,sudden, instead of a 298 here,\Nif you had like a 2,000 or a
Dialogue: 0,0:09:06.38,0:09:08.14,Default,,0000,0000,0000,,4,000 there.
Dialogue: 0,0:09:08.14,0:09:10.70,Default,,0000,0000,0000,,Then all of a sudden, things\Nbecome interesting.
Dialogue: 0,0:09:10.70,0:09:14.71,Default,,0000,0000,0000,,If you could imagine, if\Nyou had a 40,000 Kelvin
Dialogue: 0,0:09:14.71,0:09:17.53,Default,,0000,0000,0000,,temperature here, then all of a\Nsudden the entropy term, the
Dialogue: 0,0:09:17.53,0:09:20.15,Default,,0000,0000,0000,,loss of entropy, is going\Nto matter a lot more.
Dialogue: 0,0:09:20.15,0:09:22.48,Default,,0000,0000,0000,,And so this term, this positive\Nterm, is going to
Dialogue: 0,0:09:22.48,0:09:25.87,Default,,0000,0000,0000,,outweigh this, and maybe it\Nwouldn't be spontaneous at a
Dialogue: 0,0:09:25.87,0:09:27.96,Default,,0000,0000,0000,,very, very, very, very\Nhigh temperature.
Dialogue: 0,0:09:27.96,0:09:29.16,Default,,0000,0000,0000,,Another way to think about it.
Dialogue: 0,0:09:29.16,0:09:34.56,Default,,0000,0000,0000,,A reaction that generates heat\Nthat lets out heat-- the heat
Dialogue: 0,0:09:34.56,0:09:37.09,Default,,0000,0000,0000,,being released doesn't matter so\Nmuch when there's already a
Dialogue: 0,0:09:37.09,0:09:39.98,Default,,0000,0000,0000,,lot of heat or kinetic energy\Nin the environment.
Dialogue: 0,0:09:39.98,0:09:42.92,Default,,0000,0000,0000,,If the temperature was high\Nenough, this reaction would
Dialogue: 0,0:09:42.92,0:09:46.05,Default,,0000,0000,0000,,not be spontaneous, because\Nmaybe then the entropy term
Dialogue: 0,0:09:46.05,0:09:46.98,Default,,0000,0000,0000,,would win out.
Dialogue: 0,0:09:46.98,0:09:49.13,Default,,0000,0000,0000,,But anyway, I just wanted to do\Nthis calculation for you to
Dialogue: 0,0:09:49.13,0:09:51.36,Default,,0000,0000,0000,,show you that there's nothing\Ntoo abstract here.
Dialogue: 0,0:09:51.36,0:09:53.95,Default,,0000,0000,0000,,You can look up everything on\Nthe web, and then figure out
Dialogue: 0,0:09:53.95,0:09:56.32,Default,,0000,0000,0000,,if something is going\Nto be spontaneous.