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I have this reaction here
where if I had a mole of

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

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00:00:05,970 --> 00:00:08,045
produce a mole of carbon
dioxide and

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two moles of water.

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And what we want to answer in
this video is whether this

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reaction is spontaneous.

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And we learned in the last
video that to answer that

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question, we have to turn to
Gibbs free energy, or the

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change in Gibbs free energy.

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And the change in Gibbs free
energy is equal to the

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enthalpy change for the reaction
minus the temperature

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

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

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a spontaneous reaction.

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So I gave us a little
bit of a head start.

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I just calculated the change in
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
videos ago.

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You could just look up the heats
of formation of each of

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these products.

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For water you'll multiply
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
formation of all the products,

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and then you subtract
out the heats of

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formation of all the reactants.

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

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

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kilojoules.

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Well, this tells us that this
is an exothermic reaction.

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

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could kind of think of it
that way-- is that side.

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So some energy must have
been released.

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

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Let me write, plus some energy
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
this spontaneous?

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

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figure out our delta s.

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

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looked up the standard
molar entropies for

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each of these molecules.

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

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different color.

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The standard-- you put a little
naught symbol there--

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

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at 298 degrees Kelvin.

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Actually, I shouldn't
say degrees Kelvin.

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

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necessarily, when you
talk about Kelvin.

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

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at room temperature.

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So that's why it's considered
standard temperature.

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So the standard entropy of
methane at room temperature is

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equal to this number
right here.

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186 joules per Kelvin mole.

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

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Kelvin of entropy.

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If I have 2 moles, I
multiply that by 2.

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If I have 3 moles, I
multiply that by 3.

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So the total change in entropy
of this reaction is the total

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standard entropies of the
products minus the total

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standard entropies
of the reactants.

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Just like what we did
with enthalpy.

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So that's going to be equal to
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
just write 70 there.

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69.9, almost 70.

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

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entropy of the reactants, or
this side of the reaction.

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

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So just eyeballing it already,
this number is close to this

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number, but this number is much
larger than this number.

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Liquid water has a much
lower-- this is

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liquid water's entropy.

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It has a much lower entropy
than oxygen gas.

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And that makes sense.

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

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It all falls to the bottom of
the container, as opposed to

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kind of taking the shape of
the room and expanding.

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So a gas is naturally going
to have much higher

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entropy than a liquid.

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

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are going to have a lower
entropy than our reactants.

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So this is probably going
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--
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
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
is that equal to?

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So we put the minus 596, and
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
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
sense to you right now, and

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actually you know these are
but of arbitrary units.

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

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

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We have 3 separate molecules,
1 here and 2

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molecules of oxygen.

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And then we go to 3 molecules
again, but the

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water is now liquid.

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So it makes sense to me
that we lose entropy.

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There's fewer states
that the liquid,

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especially, can take on.

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

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So our delta g is equal
to our delta h.

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

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I'll just get rid
of the decimals.

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We don't have to be
that precise.

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Minus our temperature.

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

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degrees Kelvin.

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

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I should get in the habit
of not saying

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degrees when I say Kelvin.

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

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Now, this is going
to be a minus.

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Now you might say, OK, minus
242, you might want to put

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that there.

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But you have to be very,
very, very careful.

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This right here is
in kilojoules.

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This right here is in joules.

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So if we want to write
everything in kilojoules,

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

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kilojoules.

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So it's 0.242 kilojoules
per Kelvin.

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

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be minus 890 kilojoules minus
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
entropy term is going to make

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our Gibbs free energy
more positive.

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

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0, this is going to fight
the spontaneity.

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

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the exothermic nature of it.

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And it seems like it will,
because you multiply a

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fraction times this,
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
times 298, that's our

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temperature, is minus 72.

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So this term becomes-- and then
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
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
enthalpy is a much more

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negative number than our
positive term from our

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temperature times our
change in entropy.

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So this term is going
to win out.

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

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much energy that's going
to be spontaneous.

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This is definitely less than
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
energy problems, they're

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really not too difficult.

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You just really need to
find these values.

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

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

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You just look up the heats
of formations of all the

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products, subtract out the
reactants, and of course you

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wait by the coefficients.

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

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same thing.

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You have to look up the standard
molar entropies of

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the products' weight by the
coefficients, subtract out the

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reactants, and then just
substitute in here, and then

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you essentially have the
Gibbs free energy.

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And in this case,
it was negative.

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

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much higher temperature.

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Like the surface of the sun or
something, where all of a

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

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4,000 there.

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Then all of a sudden, things
become interesting.

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If you could imagine, if
you had a 40,000 Kelvin

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00:09:14,710 --> 00:09:17,530
temperature here, then all of a
sudden the entropy term, the

184
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loss of entropy, is going
to matter a lot more.

185
00:09:20,150 --> 00:09:22,480
And so this term, this positive
term, is going to

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outweigh this, and maybe it
wouldn't be spontaneous at a

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very, very, very, very
high temperature.

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Another way to think about it.

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A reaction that generates heat
that lets out heat-- the heat

190
00:09:34,560 --> 00:09:37,090
being released doesn't matter so
much when there's already a

191
00:09:37,090 --> 00:09:39,980
lot of heat or kinetic energy
in the environment.

192
00:09:39,980 --> 00:09:42,920
If the temperature was high
enough, this reaction would

193
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not be spontaneous, because
maybe then the entropy term

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would win out.

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00:09:46,980 --> 00:09:49,130
But anyway, I just wanted to do
this calculation for you to

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show you that there's nothing
too abstract here.

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00:09:51,360 --> 00:09:53,950
You can look up everything on
the web, and then figure out

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if something is going
to be spontaneous.