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## MCAT

### Course: MCAT > Unit 9

Lesson 22: Thermochemistry- Thermochemistry questions
- Phase diagrams
- Enthalpy
- Heat of formation
- Hess's law and reaction enthalpy change
- Gibbs free energy and spontaneity
- Gibbs free energy example
- More rigorous Gibbs free energy / spontaneity relationship
- A look at a seductive but wrong Gibbs spontaneity proof
- Endothermic vs. exothermic reactions

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# Gibbs free energy example

Determining if a reaction is spontaneous by calculating the change in Gibbs free energy. Also calculates the change in entropy using table of standard entropies. Created by Sal Khan.

## Want to join the conversation?

- This is really amazing! But I still don't understand how energy being released in the reaction can cause it to become spontaneous, because, Sal, you said in the earlier video that when the change in entropy is negative the reaction isn't spontaneous....please help?!?!!!!(23 votes)
- As I understand, if energy is being released, the system doesn't require much energy to do the reaction, so it is easier for the reaction to occur.(17 votes)

- So I don't see how you got -890.3 KJ for the heat change... I looked up the heat of formations of everything. CH4; -74.8 KJ, O2: 0 CO2: -393.5 KJ, and H2O: -241.8.... so you would do heat of products minus heat of reactants.

-241.8*2+-393.5= -877.1

-74.8+0=-74.8

-877.1--74.8= -802.2(11 votes)- That is because you looked up the heat of formation for water as a gas. If you used the heat of formation for water as a liquid ~ -285.8 KJ (see the equation above in the video), you would get -890.3 KJ.(29 votes)

- What if we are not allowed to look up the data on the tables. We get other reactions that we balance to find ΔH. Could you make a video using this method? I'm taking an AP class in high school and the table way won't be on the exam but the balance the equation way will be.(8 votes)
- I took AP chemistry last year and, if I understand your question, you need to add the equations together until you get a balanced equation that matches yours. You must also add the Enthalpies.(5 votes)

- Great video! I do have a question about why water is listed in the liquid state, though. Wouldn't it be in gas form if you were actually burning methane? Is this an artificial construct by stating it's occurring at 25 C? What happens to the heat produced by the exothermic nature of the process?

5 votes, kimromanchuk(4 votes)- It doesn't belong in the comment section, & there is no way for anyone to answer him from the comments. If you'll notice I quoted his name and votes!(2 votes)

- Why does oxygen have an enthalpy value of 205kJ? I thought that Sal said in a previous video that enthalpy of an element in its standard state is 0?(3 votes)
- Those numbers were not enthalpies, they are the standard molar
**entropies**at 298 K and 101.3 kPa. Also, you misread the units.

The standard molar entropy of O₂ at the above conditions is 205.1 J/(mol∙K)(6 votes)

- Hey... Just in relation to the first reaction, I seem to be getting totally different numbers for change of formation for CH4, CO2 and H2O. Where do the numbers 186, 213 and 69 come from....? It's probably a very silly question, apologies, I am just stuck at this... I thought that to calculate the change of enthalpy you needed to add the products' heat of formation C+D and subtract reactants heat of formation -A-B from it?(3 votes)
- Numbers 186, 213 and 69 are entropy values to calculate delta S, not enthapy change of formation. For delta H he already did all the calculations and didn't inlude the original values.(2 votes)

- I thought that if a molecule is in its elemental form that its heat of formation will be 0. Can someone explain to me why the heat of formation of o2 is listed at 249 and not at 0??(2 votes)
- It Isn't.

At1:05, he tells you that the heat of formation of oxygen is zero.

At1:40, he tells you that the values in blue are the**entropies**of the molecules.

You can tell that from the units.

They aren't kJ·mol⁻¹ (energy); theyare J·K⁻¹mol⁻¹ (entropy)(3 votes)

- Hello Thank you for this. I was wondering about O2, for some reason I was under the impression that O2 would be 0 since that is its standard form... At least that's what I thought. Any help in clarifying why it is not would be appreciated. thanks in advance.(2 votes)
- If the temperatures on the surface of the sun are that high the why do we see reactions occurring on the surface of the sun (hydrogen atoms coming together to make helium)?(1 vote)
- First of all the conversion of hydrogen to helium occurs deep inside a star not on the surface. Second, these reactions are nuclear and not chemical. It is the extreme temperatures in the sun that allow the fusion reactions to occur.

Also at the temperature of the surface of the sun chemicals will be ripped apart and the atoms will be stripped of electrons becoming plasma.(3 votes)

- So after the entropy intuition, I have a little trouble with the standard entropy - how can it be defined? Shouldnt it depend also on volume of the system?(2 votes)
- It is standart -molar- entropy, I think there's the rub. It is the certain amount of molecules. And in liquids and solids volume is a constant, in gas it think they use standart molar volume of ideal gas.(1 vote)

## Video transcript

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.