# Understanding Spontaneous Reactions: Exploring Gibbs Free Energy with Methane

Heads up!

This summary and transcript were automatically generated using AI with the Free YouTube Transcript Summary Tool by LunaNotes.

Generate a summary for freeIf you found this summary useful, consider buying us a coffee. It would help us a lot!

## Introduction

In the realm of thermodynamics, understanding whether a chemical reaction is spontaneous is crucial. This article delves into the reaction of methane with oxygen, using Gibbs Free Energy as a fundamental concept to analyze spontaneity. We will explore the necessary calculations of enthalpy and entropy changes to determine the spontaneity of the reaction in detail.

## The Methane Reaction

The reaction we're examining involves the combustion of methane (CH4) with oxygen (O2). In the course of this reaction, one mole of methane reacts with two moles of oxygen, producing one mole of carbon dioxide (CO2) and two moles of water (H2O). The balanced chemical equation can be represented as:

[ \text{CH}_4 + 2 \text{O}_2 \rightarrow \text{CO}_2 + 2 \text{H}_2\text{O} ]

## What is Gibbs Free Energy?

Gibbs Free Energy (G) is a thermodynamic potential that indicates the spontaneity of a reaction at constant temperature and pressure. The change in Gibbs Free Energy (( \Delta G )) is determined using the equation:

[ \Delta G = \Delta H - T \Delta S ]

where:

- ( \Delta H ) is the change in enthalpy
- ( T ) is the absolute temperature in Kelvin
- ( \Delta S ) is the change in entropy

A reaction is spontaneous if ( \Delta G < 0 ).

## Calculating Enthalpy Change (( \Delta H ))

The first step in determining spontaneity is calculating the change in enthalpy for the reaction. We look up the heats of formation of the products and reactants.

**Heats of formation**:- ( \Delta H_{f} ) of CO2 = -393.5 kJ/mol
- ( \Delta H_{f} ) of H2O (liquid) = -285.83 kJ/mol
- ( \Delta H_{f} ) of CH4 = -74.87 kJ/mol
- ( \Delta H_{f} ) of O2 = 0 kJ/mol (as it is a diatomic elemental gas)

For our reaction:

- Products: 1 mol CO2 + 2 mol H2O = -393.5 + (2 × -285.83) = -965.16 kJ
- Reactants: 1 mol CH4 + 2 mol O2 = -74.87 + 0 = -74.87 kJ

Thus, the change in enthalpy is:

[ \Delta H = (-965.16) - (-74.87) = -890.29 \text{ kJ} ]

This indicates the reaction is exothermic, as it releases heat.

## Calculating Entropy Change (( \Delta S ))

Next, we need to calculate the change in entropy for the reaction. Standard molar entropies at 298 K are:

- ( S^{\circ} ) of CH4 = 186 J/(mol·K)
- ( S^{\circ} ) of O2 = 205 J/(mol·K)
- ( S^{\circ} ) of CO2 = 213.7 J/(mol·K)
- ( S^{\circ} ) of H2O (liquid) = 69.91 J/(mol·K)

So, we calculate the total entropy change as follows:

- For products:
- ( S_{products} = 1 \times 213.7 + 2 \times 69.91 = 353.52 ext{ J/K} )

- For reactants:
- ( S_{reactants} = 1 \times 186 + 2 \times 205 = 596 ext{ J/K} )

Thus, the change in entropy is:

[ \Delta S = 353.52 - 596 = -242.48 ext{ J/K} ]

## Applying the Gibbs Free Energy Equation

Now, we can insert our obtained ( \Delta H ) and ( \Delta S ) values into the Gibbs Free Energy equation:

- Convert ( \Delta S ) from J to kJ:
- ( -242.48 \text{ J/K} * (1 \text{ kJ}/1000 \text{ J}) = -0.2425 \text{ kJ/K} )

Assuming standard temperature (298 K), we compute:

[ \Delta G = -890.29 - (298 \times (-0.2425)) ] [ \Delta G = -890.29 + 72.25 ] [ \Delta G = -818.04 \text{ kJ} ]

### Interpretation of Results

Since ( \Delta G = -818.04 < 0 ), the reaction is spontaneous under standard conditions.

## Conclusion

In conclusion, we explored the combustion of methane with oxygen to determine its spontaneity by calculating Gibbs Free Energy. The reaction is exothermic (indicated by a negative ( \Delta H )) and demonstrates a decrease in entropy (a negative ( \Delta S )). Despite losing entropy, the substantial release of energy ensures that the reaction is spontaneous. Understanding these calculations is fundamental for analyzing thermodynamic spontaneity in various reactions.

### Final Thoughts

This example illustrates the essential principles of thermodynamics and Gibbs Free Energy for evaluating chemical reactions. Future considerations could involve assessing how variations in temperature might impact the spontaneity of the reaction, particularly in environments with significantly higher temperatures.

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?

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

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.