I have this reaction here where if I had a mole of
methane, and I react that with two moles of oxygen, I'll
produce a mole of carbon dioxide and
two moles of water.
And what we want to answer in this video is whether this
reaction is spontaneous.
And we learned in the last video that to answer that
question, we have to turn to Gibbs free energy, or the
change in Gibbs free energy.
And the change in Gibbs free energy is equal to the
enthalpy change for the reaction minus the temperature
at which it is occurring, times the change in entropy.
And if this is less than zero, then it's
a spontaneous reaction.
So I gave us a little bit of a head start.
I just calculated the change in enthalpy for this reaction,
and that's right here.
And we know how to do that.
We've done that several videos ago.
You could just look up the heats of formation of each of
these products.
For water you'll multiply it by 2, since you
have 2 moles of it.
And so you have the heats of formation of all the products,
and then you subtract out the heats of
formation of all the reactants.
And of course the heat of formation of O2 is O, so this
won't even show up in it, and you'll get minus 890.3
kilojoules.
Well, this tells us that this is an exothermic reaction.
That this side of the equation has less energy in it-- you
could kind of think of it that way-- is that side.
So some energy must have been released.
We could even put here, you know, plus e for energy.
Let me write, plus some energy is going to be released.
So that's why it's exothermic.
But our question is, is this spontaneous?
So to figure out if it's spontaneous, we also have to
figure out our delta s.
And to help figure out the delta s I, ahead of time,
looked up the standard molar 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 little naught symbol there--
the standard molar entropy-- so when we say standard, it's
at 298 degrees Kelvin.
Actually, I shouldn't say degrees Kelvin.
It's at 298 Kelvin You don't use the word degrees,
necessarily, when you talk about Kelvin.
So it's at 298 Kelvin, which is 25 degrees Celsius, so it's
at room temperature.
So that's why it's considered standard temperature.
So the standard entropy of methane at room temperature is
equal to this number right 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, I multiply that by 2.
If I have 3 moles, I multiply that by 3.
So the total change in entropy of this reaction is the total
standard entropies of the products minus the total
standard entropies of the reactants.
Just like what we did with enthalpy.
So that's going to be equal to 213.6 plus-- I have 2 moles of
water here.
So it's plus 2 times-- let's just write 70 there.
69.9, almost 70.
Plus 2 times 70, and then I want to subtract out the
entropy of the reactants, or this side of the reaction.
So the entropy of 1 mole of CH4 is 186 plus 2 times 205.
So just eyeballing it already, this number is close to this
number, but this number is much larger than this number.
Liquid water has a much lower-- this is
liquid water's entropy.
It has a much lower entropy than oxygen gas.
And that makes sense.
Because liquid formed, there's a lot fewer states.
It all falls to the bottom of the container, as opposed to
kind of taking the shape of the room and expanding.
So a gas is naturally going to have much higher
entropy than a liquid.
So just eyeballing it, we can already see that our products
are going to have a lower entropy than our reactants.
So this is probably going to 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 to subtract out-- so 186 plus
2 times 205 is equal to 596.
So minus 596, and what is that equal to?
So we put the minus 596, and then plus the 353.6, and we
have minus 242.4.
So this is equal to minus 242.4 joules per Kelvin is our
delta s minus.
So we lose that much entropy.
And those units might not make sense to you right now, and
actually you know these are but of arbitrary units.
But you can just say, hey, this is getting more ordered.
And it makes sense, because we have a ton of gas.
We have 3 separate molecules, 1 here and 2
molecules of oxygen.
And then we go to 3 molecules again, but the
water is now liquid.
So it makes sense to me that we lose entropy.
There's fewer states that the liquid,
especially, can take on.
But let's figure out whether this reaction is spontaneous.
So our delta g is equal to our delta h.
We're releasing energy, so it's minus 890.
I'll just get rid of the decimals.
We don't have to be that precise.
Minus our temperature.
We're assuming that we're at room temperature, or 298
degrees Kelvin.
That's 28-- I should just say, 298 Kelvin.
I should get in the habit of not saying
degrees when I say Kelvin.
Which is 25 degrees Celsius, times our change in entropy.
Now, this is going to be a minus.
Now you might say, OK, minus 242, you might want to put
that there.
But you have to be very, very, very careful.
This right here is in kilojoules.
This right here is in joules.
So if we want to write everything in kilojoules,
since we already wrote that down, let's write this in
kilojoules.
So it's 0.242 kilojoules per Kelvin.
And so now our Gibbs free energy right here is going to
be minus 890 kilojoules minus 290-- so the minus and the
minus, you get a plus.
And that makes sense, that the entropy term is going to make
our Gibbs free energy more positive.
Which, as we know, since we want to get this thing below
0, this is going to fight the spontaneity.
But let's see if it can overwhelm 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 entropy times 298, that's our
temperature, is minus 72.
So this term becomes-- and then we put a minus there-- so
it's plus 72.2.
So this is the entropy term at standard temperature.
It turns into that.
And this is our enthalpy term.
So we can already see that the enthalpy is a much more
negative number than our positive term from our
temperature times our change in entropy.
So this term is going to win out.
Even though we lose entropy in this reaction, it releases so
much energy that's going to be spontaneous.
This is definitely less than 0, so this is going to be a
spontaneous reaction.
As you can see, these Gibbs free energy problems, they're
really not too difficult.
You just really need to find these values.
And to find these values, it'll either be given, the
delta h, but we know how to solve for the delta h.
You just look up the heats of formations of all the
products, subtract out the reactants, and of course you
wait by the coefficients.
And then, to figure out the change in entropy, you do the
same thing.
You have to look up the standard molar entropies of
the products' weight by the coefficients, subtract out the
reactants, and then just substitute in here, and then
you essentially have the Gibbs free energy.
And in this case, it was negative.
Now, you could imagine a situation where we're at a
much higher temperature.
Like the surface of the sun or something, 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, things become interesting.
If you could imagine, if you had a 40,000 Kelvin
temperature here, then all of a sudden the entropy term, the
loss of entropy, is going to matter a lot more.
And so this term, this positive term, is going to
outweigh this, and maybe it wouldn't be spontaneous at a
very, very, very, very high temperature.
Another way to think about it.
A reaction that generates heat that lets out heat-- the heat
being released doesn't matter so much when there's already a
lot of heat or kinetic energy in the environment.
If the temperature was high enough, this reaction would
not be spontaneous, because maybe then the entropy term
would win out.
But anyway, I just wanted to do this calculation for you to
show you that there's nothing too abstract here.
You can look up everything on the web, and then figure out
if something is going to be spontaneous.