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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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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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temperature here, then all of a
sudden the entropy term, the
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loss of entropy, is going
to matter a lot more.
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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
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being released doesn't matter so
much when there's already a
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lot of heat or kinetic energy
in the environment.
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If the temperature was high
enough, this reaction would
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not be spontaneous, because
maybe then the entropy term
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would win out.
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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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You can look up everything on
the web, and then figure out
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if something is going
to be spontaneous.
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