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PRINCIPLES OF ALCHEMY
FIRE

Part Two

Many things influence the rates of chemical reactions.

Equilibrium determines the yield of a chemical reaction.

Enthalpy changes are involved in chemical reactions.

Blame the disorder of the universe on entropy.

Gibbs energy explains all chemical reactions.

Why do some chemical reactions occur fast and others slow?

Ah, the rates of chemical reactions are an important subject to an Alchemist. The rate of the chemical reaction is the speed of it. Some reactions are fast while others are slow. Alchemists study the rates of chemical reactions in order to find the fastest, yet safest, path to the product.

Yeah, but what causes the reaction to go fast or slow?

The rate (the speed) at which the products are formed depends upon the rate at which the reactants collide, interact and are transformed into products.

Sounds complex.

It is, but equations make it easy to understand.
Consider this equation. Oh, these letters represent molecules not atoms!
A + B ---(---> ABCD---)--->C + D

OK. You have reactants A and B on the left. You have products C and D on the right. But, what is this ABCD in the middle?

That is the activated complex. Remember?

Oh, right. That is the middle zone of a reaction. The activated complex.

Yes. The first thing you must do to start a chemical reaction like this is to get the two reactants to collide together.

So to increase the rate of a chemical reaction you should try to increase the rate of collisions among the reactants?

Right. How might you do that?

I would make sure that A and B were mixed together real good. That way they can find each other more easily.

Yes. If the reactants are solids you could grind them up into tiny particles. That would increase the total surface area of the reactants so they would be more likely to collide.

If you increased the amount of reactants would it go faster?

Yes, in most cases it would.

But isn't that adding molecules to the left side of the equation? Won't that unbalance the equation?

No, not at all. You see the chemical equation you balance is a diagram of the FINAL RATIOS of products you get FROM the reactants. If you add ten times more reactant you will just get ten times as much product. Increasing the amount of reactants is a great way to speed up most reactions. Certainly if you don't have enough reactants the reaction will be very slow (if it occurs at all).

OK. What about water? Can you increase the amount of reactants in a reaction that occurs in water?

You sure can. You just increase the concentration of the reactants. If water is a reactant too, then you must see to it that there is enough water. And in correct proportions.

So you can add as much reactants as you want?

Well, eventually the extra reactants will precipitate out of the solution and will probably not be useful to the reaction. So you can over do it. But some of the fastest reactions in water happen when the reactants are near saturation levels (about to precipitate out of solution, but not quite).

Alchemists do a lot of their Alchemy in water, don't they?

Yes, we do. Water is a great way to get the reactants to mix together. The water acts as a carrier, and it can also supply some of the reactants needed to make it go. Water is a good way to mix ions and a great way to get hydrogen ions (H+) into the reaction.

What if the reactants are not solids and don't work in water? What if it is a totally gaseous reaction? You know, like hydrogen and fluorine gas to make hydrogen fluoride.

Good question. Think about how you would get two reactants (A and B or H2 and F2) to react quickly?

I'd mix them well.

Anything else? Anything else you could do to increase the number of collisions of those gas molecules.

Hmmm. Is there a way to push them together? You know, to make the collisions more likely.

Yes. You can increase the pressure! By increasing the pressure you squeeze the reaction, with all those reactants, into a smaller volume.

Yeah! That would push the reactants closer together.

Right. A higher pressure will increase the rate at which the reactant molecules (A's and B's) collide because at higher pressures there are more reactants in a given volume.

Great! Any other tricks to speed up a chemical reaction?

Yes. I think you've missed a very important one. Increase the temperature.

Oh, yeah! But how does that make for more collisions?

Think about it, Arthur. What happens to atoms and molecules when you heat them up?

They wiggle more.

Right. Molecules move around faster when they are heated up. At higher temperatures all molecules move faster and this increases the chance of a collision.

Oh, I see. Do all these tricks to speed up reactions always work? For any reaction?

Oh, goodness no. There can be lots of problems. Maybe you can't make the solution any more concentrated. Maybe at too high a temperature the reactants are destroyed or the products can't be formed.

So how can you know what will work?

Well, you think about what might work and what might go wrong and then you simply give it a try!

Huh? That's it?! You don't really know how to make a chemical reaction go faster? You just "give it a try"?

That's right. However, a good Alchemist knows enough about what tricks to try to make it go faster. He tries a trick and may discover that it doesn't work. Maybe heating the reaction doesn't help. It might even hurt the speed! So then he tries to increase the pressure or maybe the concentrations.

I see what you mean.

Good. It's experimental. Just remember, reactants must collide in spite of the electrostatic repulsion caused by their electron shells. (Even the cations have electron shells.)
The best way to overcome this repulsion is to have lots of reactants (high concentration) in a tiny space (high pressure) moving at high speeds (high temperatures).

Gee, I was kind of hoping for some magic trick to make a reaction go faster.

Well, there kind of is a magic trick to do that. It is called a catalyst.

What's a catalyst? How many cats does it take to make one?

No. No. It has nothing to do with cats! A catalyst (pronounced "cat-ah-list") is a substance that increases the rate of a chemical reaction, but it is not changed or used up in the reaction.

Great! Where do I get one!

Actually, you are full of thousands of different kinds of catalysts that all work together to keep you alive. They each have their own name but together we call them all enzymes (pronounced "N-zimes").

So without these thousands of different enzymes the chemical reactions that make me alive would go much slower?

Yes. As a matter of fact they would go so slow that you won't be able to live at all. Enzymes are required for life (on Earth).

These enzymes are very important!

Yes, they are. Enzymes are made of very complex chemicals called proteins. Some enzymes have metal ions in them too. Regardless, our enzymes very quickly turn reactants (like food) into products (like people)!

Wow! Are enzymes easy to understand?

No. They are an advanced area of a special field of BioAlchemy, but you can learn all about them in some other class.

How does a catalyst work?

That depends on the catalyst. You see, each catalyst usually can catalyze (speed up) only one kind of reaction. It is the speed and specificity of catalysts which make them so "magical".

By "specificity" you mean it only works on special reactions. Right? So you must have the right catalyst to do the right job. Otherwise it won't work.

Right.
Catalysts work by making a special activation complex with a lower transition state.

How?

They do this by grabbing each reactant and holding them together, often in specific positions that help the old bond(s) break and the new bond(s) form.

I bet these catalysts juggle the electrons.

Right. Catalysts are excellent jugglers of electrons. Many of them use transition metal ions with strange outer electron shells to do their work.

Amazing!

Yes, I think so too! Once the product is made it is released, and the catalyst returns unchanged to be used again to make more products. Over and over, again and again.

So you never run out of catalyst!

Eventually they "wear out" but you get a lot of work out of them before they do. The important thing is that catalysts are not turned into products.

I bet catalysts are used a great deal by Alchemists.

You're right. Most Alchemists try to find better ways to make reactions go and catalysts are always welcome. Industries in the 20th century often use simple catalysts (metals, ions, etc.) to make industrial products. Some foods are made with the help of enzymes, and enzymes are even used to clean clothes.

And I'm full of catalysts. Enzymes!

Right!

I see why the rate of a chemical reaction is important. But isn't the amount of products you make also important.

Oh, yes indeed. The yield of a chemical reaction is the amount of product made. The yield depends upon the reaction's equilibrium.

What?

Let's begin at the beginning. When a chemical reaction begins, it turns reactants into products, moving the equation from left to right, because there are only reactants to start with and no products yet formed.
A + B ------> C + D

Yeah. That makes sense. It's obvious.

Yes, it is. We call that reaction, going from reactants to products, the forward reaction. The forward reaction is what most people think about when they write an equation. However, as reactants are used up and products are made, (most) reactions start to go in reverse (!) running from right to left.
A + B <------ C + D
That's the reverse reaction.

I think that's very odd. Are reverse reactions common?

Yes. Chemical reactions are reversible and thus able to go in the opposite direction. At least to some extent.

But, how can that be? You wouldn't get ANY products. Would you?

Oh, you get products. It all depends upon the reaction's equilibrium.

You want to explain that?

Of course. Once the reaction has started, these two reactions (forward and reverse) occur at the same time. Eventually (actually, often very quickly) the forward and reverse rates settle down to the same rate. At this point the overall chemical reaction (both forward and reverse) is said to be in a state of equilibrium, meaning opposite reactions (in opposite directions) are going on at the same rate. At this point (equilibrium) products are made as fast as they are destroyed!

So at equilibrium, the reaction has stopped.

No! That is a common mistake. It would APPEAR as if the reaction has stopped, but in fact the equilibrium reactions are "dynamic". That means there is a continuous interchange of atoms between reactants and products.

So while the reactants are becoming products by the forward reaction, the products are going back to reactants by the reverse reaction.

Right! We often write the reactions with a double arrow like this
A + B <-------> C + D
to remind us that there are two opposite reactions going on. The forward reaction makes products and the reverse reaction turns the products back to reactants!

What a stupid way to do Alchemy!

It isn't a matter of stupidity. It's just the way the laws of the universe work. At least with respect to chemical reactions.

I assume this equilibrium affects the yield.

Aye. The yield of a reaction is the amount of product made and that depends on the "position" of the equilibrium. This equilibrium position is the point at which the forward reaction and the reverse reaction have "canceled each other out", producing a "constant" yield. It is WRONG To assume that ALL the product returns to reactant, otherwise there would be no Alchemy!. Instead, a set amount of product (the yield) is produced at equilibrium.

Can you explain this with some kind of example?

I certainly can. Consider this reaction:
100A + 100B -----> 0C + 0D It starts with all reactant and no product (like most reactions). The numbers in front of the reactants represent the amounts of A and B. They could be 100 molecules of each or 100 tons. It doesn't really matter (to understanding equilibrium).

OK. I'm with you so far.

Good. The forward reaction starts and produces something like this:
20A + 20B --------> 80C + 80D
Do you see what that means?

I think so. That equation represents the idea that 80 As and 80 Bs have gone to make 80Cs and 80Ds. So there are only 20 As and 20 Bs left over.

Right. That is just the forward reaction at work, but as the products begin to accumulate the reverse reaction starts. (I'll switch the directions here to make it more clear.)
80C + 80D -----> 79C + 79D + 1A + 1B (plus the 20A and 20B remaining).
[Note: The algebra in these examples is not meant to be strictly correct. I'm being a little "sloppy" here to make my point. I am not being mathematically accurate.]

So now you are showing me that the yield of products is decreasing.

Right. BUT, the forward reaction is still occurring, so those newly formed A and B can go into the forward reaction and return the products!
21A + 21B ------> 1C + 1D (plus the 79C and 79D from before.)

This is SO confusing! What is going on here?

Both the forward and the reverse reactions are going on here.

But when does it end? What about this "equilibrium"?

It doesn't really end. It just continues. When the RATES of the forward reaction and reverse reactions are equal, equilibrium has been reached.

But both reactions are still occurring?

Aye. However, the yield is now constant. The total amount of product doesn't change once the reaction reaches equilibrium. See?

I think I understand.

Good. Equilibrium often occurs a wee bit past the point at which the reverse reaction begins. Putting the forward and reverse reactions together could give an equilibrium something like this
25A + 25B <------> 75C + 75D

I think I get it. Because the forward and reverse rates are equal (at equilibrium), the amounts of reactant and product will not change.

Right. The reaction will appear to have stopped (or "gone to completion" as Alchemists say) but in fact the reactions continue back-and-forth without changing the overall numbers. At this point the yield will not change. In this example, you get a yield of 75%.

What can you do to increase the yield?

That's a good question. Often you can change the position of the equilibrium, and thus the yield of product, by changing the conditions of the reaction.

The conditions?

Yes. A change in temperature, pressure or concentration can shift the equilibrium to a new position.

Hey, aren't they the same things that can change the rate of a reaction too?

Yes they are, and that's why they also work to shift the equilibrium position.

Hmmm. Is there one perfect temperature, pressure and concentration that works best?

Yes, but only for each kind of reaction. It depends on the particulars of the reaction you are trying to help. There are no "magic" conditions that will increase the yield of ANY reaction.

I suppose you have to try it and see.

Yes, ultimately you have to experiment. Try the reaction under different conditions of temperature, pressure and concentrations to see what gives you the best yield. A good Alchemist can make an educated guess as to what to change and how to change it.

For example?

For example, if the forward reaction is helped by a bit of heat but the reverse reaction is not, then heating the reaction should increase the yield. Do you see why?

Sure. If heating the mixture increases the speed of the forward reaction while the reverse reaction rate is unchanged, then the forward reaction is favored and you'll get more forward reaction products!

Right. But sometimes you can't change the temperature. Some products can be destroyed by higher temperatures. But there are other ways. Here's an example. Suppose you have a reaction in which two molecules come together to form one product molecule. Like this.
A + B <------> C

A combination reaction!

That's right. Very good.
Can you think of a way to increase the forward reaction (the combination reaction) without heating? Think of a condition you might change that would help the reaction go from left to right.

Well, if I could increase the number of collisions between A and B, it would probably increase the forward reaction. Is that right?

Yes, it is.

Then I would increase the pressure.

Good. Why?

Because an increase in pressure would cause more collisions among the A and B molecules. That would increase the forward reaction.

Right. How would an increase in pressure affect the reverse reaction (decomposition), going from C -----> A + B ?

I don't know? I guess it wouldn't affect it at all. I don't see why increasing the collisions among the C molecules would cause them to break up into A and B.

You're right. The higher pressure will not help the reverse reaction but it will help the forward.

I bet if I increased the amount of A and B it would increase the forward reaction, too. Right?

Hmmm... It depends. You can manipulate the concentrations of reactants to increase the yield, but that increase is really a special case.

How do you mean?

Well, if you had this equation
A + B <------> C + D
and you just added more of A and B , you would also get more product to go into the reverse reaction. Your overall yield of products would increase, but that is because you just added more molecules to begin with.

So it does increase the yield!

Yes it does, but that's obvious. If you want to make more bread you just start with more flour. There's nothing particularly special about that. However, there is a nice trick to increase the yield if you can separate the products from the reactants.

What? How can you do that? You can't reach in and "grab" products out of the reaction as it is running. Can you?

With some reactions you can!
What would happen if the reactants (A and B) were gases and the products (C and D) were liquids?

Hmmm. The gaseous atoms would collide together to form the products. Then the products would "rain" out. The liquid products would fall out of the gaseous reactants.

Exactly. If you then collected the "rain" of products you could remove them from the reaction mix and they would not get the chance to go in the reverse direction.

Oh, I see what you mean. I bet that if the products were solids, they would also fall out of the reaction and pile up in the bottom of the container.

Right. If the products produced by the forward reaction are in a different state of matter (solid, liquid or gas) than the reactants, you have an easy way to separate them.
Tell me Arthur, how could you help the products to be in one state of matter instead of the other?

By changing the temperature. If you cooled the container you might be able to make the products change into a different state.

Right.

But won't the reactants also change? I mean, if you change the temperature or pressure of the reaction you change it for both the forward and the reverse reactions.

Aye. They might all change to the same state. It depends on the specific chemicals. It depends on the state of the reactants at the temperature you want to run it at and the state of the products at that temperature.

So it would help if I ran the reaction at a temperature that caused the reactants to be gases but the products to be liquids or solids.

Right. Then you could collect the products or at least make it less likely that they would enter the reverse reaction.

When you use these tricks, changing temperature and pressure, must you always change it so that the products and reactants end up as different states of matter?

Not necessarily. Sometimes a change in the reaction conditions will favor the forward reaction (more than the reverse) without changing states.
A fellow named Le Chatelier did a great deal of experiments about the various conditions necessary to increase the yields. He discovered that a change in temperature, pressure or concentration will upset the equilibrium shifting it to a new equilibrium position. That's called Le Chatelier's principle. He said that by changing those conditions it "stresses the system" to make more (or less) of the products.

Sounds like Le Chatelier was another very important Alchemist.

Yes, he was. I hope you realize that it is useful to understand what influences the rate of chemical reactions and their equilibrium. It helps when planning experiments and producing molecules. Reaction rates and equilibriums occupy the thoughts of many Alchemists.

Yeah, but in the end it all comes down to giving it a try!

True! Any questions?

Yeah, I might be a bit confused here. When I balance an equation I end up with numbers (coefficients) in front of the molecules. Are they the same numbers you've been using here to explain yields?

No. The balanced equations tell you the ratios of the starting reactants and final products by accounting for all the atoms and electrons. A balanced equation is required to understand what you need and what you get, but it doesn't take into account equilibrium. Equilibrium equations tell you the ratios of the final reactants and products by accounting for how all the molecules use the forward and reverse reactions.

Nope, I'm still confused. Try again.

OK. When you balance an equation you are making a "ledger" or an account of all the atoms and electrons. The coefficients in the balanced equation tell you the ratios you need in the reactants and what ratios you will get in the products. Because you cannot make atoms or electrons out of nothing, the balanced equation is required. It helps you understand the reaction better.

But the equilibrium is different. It doesn't tell you what you SHOULD get but what you DO get!

Yes, exactly! What you do get are products from reactants.
You will get the products in the ratios determined by the balanced equation and the reactants will be used up in the same ratio as the ratio from the balanced equation..
On the other hand, the equilibrium tells you how much reactant actually becomes product. You can think of the balanced equation as an ideal situation.

What do you mean?

Well, for example, your balanced equation may tell you that because you have six atoms of a particular element going into a reaction (the reactants) you MUST have six of those same atoms coming out (as products). The balanced equation tells you the ratios you would get if ALL the reactants went to products.

I see. But the equilibrium tells me how many actually DO end up as products or reactants.

Right. I think we need an example to follow this discussion. Recall this equation.
H2 + F2 ------> 2HF

Yeah. It is balanced because there are two hydrogens and two fluorines on both sides. I bet the electrons are balanced too.

Right. This balanced equation tells you the ratios of reactants and products that are possible. Tell me how much HF could be made if you had 100 molecules of H2 and 100 molecules of F2.

Instead of just one molecule of each?

Right. Assume, for the moment, that it is a totally forward reaction.

So all the reactants go to products?

Right. Tell me how many molecules of HF would be made from 100 molecules of both reactants.

OK. Can I just use normal math on this? Can I just make believe that I have 100 of these reactions all occurring at the same time?

Yes, you can.

Then that's easy. I'll just multiply it all by 100. Like this.
100 X H2 + 100 X F2 ------> 100 X 2HF
100H2 + 100F2 ------> 200HF
So I get 200 molecules of HF. Is that right?

Absolutely! Does it still balance?

Yeah, but all the numbers are now much bigger.

Yes, they are. You see, when you first balance an equation you want to get the simplest ratio of coefficients. Once you have that balanced equation, you can (later) multiply it by any number to figure out how much product you get from a certain amount of reactants. The important point is always remember to multiply everything! Otherwise it won't be balanced. (It has to do with algebra.)

OK, I see. So how does equilibrium fit into all this?

Well, so far in this example, we have pictured the reaction as being a 100% forward reaction with no reverse reaction. But that is unusual. Now let's consider the same problem, but assume that at equilibrium, there is a 25% reverse reaction going on. How would that affect the ratio of reactants and products? Start with the 100% forward reaction and try to adjust it to take into account the reverse reaction.

OK, 100H2 + 100F2 ------> 200HF is the 100% forward reaction. That means all the reactants will go to products.
All the H2 and F2 will disappear and only HF will be in the container. Right?

Right. You could write that as 0H2 + 0F2 ------> 200HF, to remind you that there are no H2 and F2 molecules left, only (200) molecules of HF. Notice that is NOT a balanced equation anymore, but that's OK, because we are now talking about equilibrium. It doesn't balance because we have assumed all the reactants have been used up to make the products. By starting with a balanced equation we are only checking to see that all the atoms (and electrons) are accounted for.

I see. But the back reaction means only 75% actually ends up as product. The other 25% stay put as reactants. Right?

Right! Sort of. Remember, reactants don't really "stay put" at equilibrium. They just look like they are not doing anything.

OK, yeah, I forgot.
Anyway, if I understand this equilibrium, the actual ratio will be different from a 100% forward. Instead of
0H2 + 0F2 ------> 200HF, I get some going back like this
H2 + F2 <------- 2HF through a reverse reaction.

Right. But that shows only one molecule of HF going in the reverse reaction. Is that a 25% reverse reaction?

Ah, no it's much less. Can I add them up and see how much?

Yes, give it a try.

OK. 0H2 + 0F2 ------> 200HF is a 100% forward reaction
1H2 + 1F2 <------- 2HF is a balanced reaction but in the reverse direction.. It changes only one product back to reactants, so I'll put them together.
---------------------------
1H2 + 1F2 <------> 198HF
This means I have one molecule of each reactant and 198 molecules of the product at equilibrium.
Is that right?

Well, you are getting there. 1H2 + 1F2 <------> 198HF is the equilibrium reaction when the reverse reaction is only 1% of the forward reaction.

Oh, I see. Well, I'll just multiply the reverse reaction by 25 instead of 1.
0H2 + 0F2 ------> 200HF is a 100% forward reaction
25H2 + 25F2 <------ 50HF is a 25% reverse reaction (and I've kept it balanced to keep track of how much reactant and product are involved).
------------------------------------
25H2 + 25F2 <------> 150HF

Very good. This is the equation at equilibrium.
(Strictly speaking, the algebra is not being shown correctly, but you get the idea.)
Notice that it is NOT a balanced equation, it is an equilibrium equation. Do you know what that means?

Yeah, it means I'm getting more confused!

Now, now Arthur. If you are getting confused, just slow down and think about what you are doing and why.
A balanced equation is a single, simple equation that tells you the amount of reactants needed to make an amount of products. It takes into account all the atoms and electrons. You can think of it as a 100% forward reaction. It tells you the best ratio of reactants that will give you the right ratio of products. It counts the atoms and electrons.

I see. The forward reaction is the ratio of reactants that will give me a ratio of products if it were 100% forward and no reverse. All the reactants would disappear and all the products would form.

Right. But real Alchemy is rarely a 100% forward reaction (even with enzymes). In real life you have to take the reverse reaction into account. Notice that the reverse reaction is the same as the forward, except reversed!

So at equilibrium you have 25H2 + 25F2 <------> 150HF.

Right.

But that's NOT a balanced equation.......

Right and it isn't suppose to be! It's an equilibrium equation.
By taking equilibrium into account, you show that you don't get a yield of 100% but something less. With a reverse reaction of 25%, you get a forward reaction of only 75%. So your yield will be less. Instead of getting 200 molecules of HF for every 100 molecules of H2 and F2, you get only 150 molecules of the product. We say the yield is (only) 75% of the maximum "theoretical" yield.

So the 25H2 and 25F2 just sit there! Oops! NO! They are "dynamic". What I mean to say is that no matter how long I wait around there will always be 25 molecules of both H2 and F2 around. I can't get rid of them.

Right! I think you are getting the hang of it now, but it will take some practice.

It sure will!

This is always a difficult area for Alchemy students so don't feel bad if you don't understand it the first time you hear it. Read it again and again. Try some problems and practice.

OK. Before we leave this problem, I want to know something. IF I had only a 75% yield, because of this equilibrium, could I try to increase the yield by changing temperature, pressure and stuff like that? You know, can I stress the system using Le Chatelier's principle?

You sure can! As a matter of fact an experienced Alchemist would tell you that 75% yield from that reaction is very bad. By altering conditions you could increase the yield, maybe to 99.9%.

Wow! And to find out how to increase the yield, I would alter the conditions and see what works best.

Right. A great deal of Alchemy work involves finding the best conditions to produce the highest yields, quickly and safely. That's a special area of Alchemy called Alchemy Engineering.

Does it require a lot of math?

Yes it does. Alchemy Engineers use very advanced forms of math called calculus and statistics to figure out the best routes to products. Then they go to the lab and do the experiments to see if the idea really works. But you don't need those advanced classes to understand the basis of Alchemy Engineering. The steps you have just gone through are really all you need to know.

So, in the future, have these Alchemy Engineers figured out the best way to make everything?

Well, not quite. There are so many chemicals to make! They have found the "best" conditions to make very high yields for the common chemicals. But then someone discovers a new catalyst or makes a machine that can produce more pressure, and suddenly all these Engineers decide to adopt the new way. Alchemy Engineers are always trying to find a better way to make a molecule.

Tell me, Arthur, what do you think is the important things to know about chemical reactions?

Well, electrons are the real players but you mention energy a lot when you talk about chemical reaction.

That's right. All chemical reactions must follow certain laws about energy. Those are the laws of thermodynamics.

Thermo-what?

Thermodynamics is the way energy moves.

Like the way a hot pot gives up its energy as it cools down?

Yes. Exactly. The way a hot object cools is determined by the laws of thermodynamics. But energy is more than just temperature. Do you recall what temperature really is all about?

Yeah, temperature is the energy that causes molecules to wiggle. The more you heat them up the more molecules wiggle. Right?

Right. But molecules also have a hidden energy in them. That's what causes reactions to give off heat (exothermic) or suck in heat (endothermic). We have a special name for the energy stored in molecules. It's called enthalpy (pronounced "N-thal-P").

What a weird name!

Aye, it's Greek. Just think of it as the internal energy. Enthalpy takes into account the temperature too, because that affects how much the molecules vibrate but enthalpy is mostly a measure of the total bond and shell energies. Enthalpy is molecular energy.

How do you measure enthalpy?

You can't! Not directly. But you can measure the change in enthalpy that occurs during chemical reactions. The difference between the enthalpy of the reactants and the enthalpy of the products is the energy given off or absorbed by the reaction.

How's that? (Without getting too deeply into the math.)

Imagine this.
The enthalpy change (energy given off or absorbed) = enthalpy of the products - enthalpy of the reactants.

Did you write that backward?

No. That's the right way. Why?

Well, it looks backward. You have the products before the reactants. It looks backward.

I see what you mean. In a sense, you could write it the other way around. However, we Alchemists have agreed to use the equation for enthalpy change that way around. Sorry if that confuses you but that is just the way we Alchemists have agreed to calculate enthalpy changes. If you are going to understand Alchemy you should use the same methods we all use.

OK. The change in enthalpy is the enthalpy of the products minus the enthalpy of the reactants.
Products - reactants (even though the equation is written reactants -----> products).
I would have designed Alchemy differently!

I'm sure you would have.
If you can imagine how this equation works, you will see how changes in enthalpy release or absorb heat.

I think I see. Exothermic reactions release heat because the products formed have less enthalpy than the reactants. The leftover energy is given off as heat.

Right! That's an excellent observation. Look at that equation again and tell me, will the change in enthalpy be positive or negative for an exothermic reaction?

Hmmmm,
The enthalpy change (energy given off or absorbed) = enthalpy of the products - enthalpy of the reactants.
You just said that in an exothermic reaction the products have less enthalpy than the reactants. So, the enthalpy change equals the low enthalpy products minus high(er) enthalpy reactants. I can write that as
Enthalpy change (heat given off ) = low enthalpy products - high enthalpy of the reactants.
Hmmm.
If you had less enthalpy in the products than in the reactants you would have a smaller number for the product enthalpy than for the reactant enthalpy.

Right. Now continue that idea and explain it as numbers would behave.

OK, I'll try. I know that when you subtract a big number (the enthalpy of the reactants) from a small(er) number (the enthalpy of the products) you get a negative number.
Does that mean the enthalpy change is negative?

Yes, it does! Well done, lad. Look at that again.
The enthalpy change (heat given off or absorbed) = enthalpy of the products - enthalpy of the reactants.
Let's make up some numbers to show that more clearly. Let's say that the reactants have 20 units of enthalpy. Why don't you make up a number for the products' enthalpy? Remember, we are thinking about an exothermic reaction.

OK. An exothermic reaction gives off heat because the high enthalpy reactants become low(er) enthalpy products. You said the reactants have 20 units of enthalpy, so there must be less than 20 units of enthalpy in the products. Otherwise it wouldn't be exothermic. Right?

Absolutely.

So, any enthalpy less than 20 units would work. I'll choose 15 units of enthalpy for the products.

Good. That guarantees an exothermic reaction. Now do the math!

Ughh!
Actually it's not that hard. The change in enthalpy equals 15 units of product enthalpy minus 20 units of reactant enthalpy.
That's 15 - 20 = -5. So the enthalpy change is -5 units.
When reactants (20 units) are made into products (15 units) there is enthalpy left over (5 units).

Right! Five units of enthalpy are given off so the number is negative. See?

Yeah, but that seems a bit weird. After all, an exothermic reaction gives off heat. I would think you get positive enthalpy out of it. Not negative.

Well, that is a common mistake for young Alchemists. But think of what you just said. "You GET enthalpy OUT OF IT". That expression explains that something is coming out of the reaction.

Yeah, the heat! That's why it is exothermic.

Right. Because the enthalpy LEAVES the molecules during the reaction, the enthalpy change is negative. Enthalpy leaves so it is negative.

I think I understand, but let me try it the other way around.
Endothermic reactions absorb heat from the environment to make the higher enthalpy products from lower enthalpy reactants. Right?

Right.

That means in an endothermic reaction the low energy reactants go to high(er) energy products. The equation looks like this.
The enthalpy change (heat absorbed) = the high(er) enthalpy products - the low(er) enthalpy reactants.
That gives a positive number. So endothermic reactions have a positive change in enthalpy.

Right!

Now I see. The reason the enthalpy change is positive for endothermic reactions is because you have to ADD the enthalpy to make the products. It's making sense now. I see why Alchemists write the equations that way.

Good.

It's funny that you can use math in chemistry. You know, adding and subtracting enthalpy with numbers and signs.

Actually, thermodynamics is very easy to describe with math. You can add and subtract enthalpies from a series of reactions and find out the final enthalpy change. A fellow named Hess proved that the enthalpy change (total energy released or absorbed) as reactions move from reactants to products, is "path independent".

What does that mean?

It means no matter how you "travel" from reactants to product, the total change in enthalpy is always the same. All that matters is the enthalpy of the reactants you started with and the enthalpy of the final products you got out of it. For example, let's say you have a long complex series of chemical reactions in order to make the final products. I'll write that as
A + B ---> C + D ------> E + F -------> G+ H ------> I + J -----> K + L ------> M + N
The total change in enthalpy would be calculated from just the starting reactants (A + B ) and the final products (M + N).

What about all those others in between?

They don't count. Can you tell me the enthalpy change of this series of reactions that begins with A and B and ends with M and N?

Yeah, I suppose it is just the enthalpy of M and N (the products) minus the enthalpy of A and B (the reactants).

Right!

Sounds wrong, though.

But it isn't wrong. It's right! It follows Hess's discovery and it has been proven in the lab many times. It is so "true" that we call it Hess's Law. Have a look at this path.
A + B ---> O + P ------> Q + R -------> S + T ------> U + V -----> W + X ------> M + N
What would be the total enthalpy change?

It's the enthalpy of M and N (the products) minus the enthalpy of A and B (the reactants). You see that is what I mean! I shouldn't be ignoring the O + P , the S + T, and the other stuff in between. Should I?

Yes, you should ignore them! Hess discovered you can ignore the stuff in between. This is exactly what we mean by path independence.

Strange! What about activation energies? Where do they come in to the calculation?

They don't.

What!? But you need the activation energy to start any reaction. That is, if it has an activation energy. Does Hess's Law only work for reactions that don't need activation energy?

No, Hess's Law works for ALL reactions and ALL series of reactions. Recall that the activation energy (the hill between the reactant and product) goes down on the other side (of the hill) as the reaction goes to products. That downhill slope makes up for the enthalpy change as if the hill weren't there at all. It is true you must add energy to over come an activation barrier (if one exists), but you get it back.

Seems like it shouldn't work that way.

I know what you mean. I felt that way too when I first learned Alchemy. But Hess's Law works! Activation energies and any other energies don't come into the final enthalpy change. The change is "independent of path". So the first reaction could be endothermic and require an activation energy. Then the next reaction may be exothermic. And the next reaction could be anything! It doesn't matter because all that up and down cancel out. Only where you start and where you end up really matters.

Because Hess proved that enthalpy changes are "independent of path".

Right! Hess showed that enthalpy changes (like chemical changes) can be added and subtracted to calculate changes in enthalpy. And it doesn't matter what kind of enthalpy you use. You can add and subtract enthalpy from all kinds of energy.

What do you mean? What kinds of enthalpy are there?

There are many different kinds of enthalpy changes (ways to make or absorb energy). You already know some of them. For example, ionization energy is the energy required to remove an electron (endothermic) to form the cation. Or it can be the energy released (exothermic) when an electron is returned to the cation.

Ionization energy is a form of enthalpy!?

Aye, it is. Ionization involves energy changes in a substance so it is part of the enthalpy change. A chemical reaction that involves the ionization of an atom takes in energy to remove the electron. That's endothermic and a positive change in enthalpy. Understand?

Maybe. I'll have to think about that. What about electron affinity? That takes energy. What if an electron is added to an atom. Do you add in its enthalpy?

Yes. Or you subtract it. It depends on the atom. Atoms (elements) with low electron affinity will not readily accept an electron unless energy is provided.

So some atoms suck in energy in order to take on the electron and Hess says you include that energy in the calculation of the final, total enthalpy change?

Absolutely. In this case we are talking about an atom (element) with poor electron affinity. That means you must use energy to make the anion. Therefore, if you make an anion (in the products) from a neutral atom (in the reactants) you must add energy to get the electron to "stick". That single enthalpy change is positive because you must add it into the reaction to make it "go".

OK. If there is an enthalpy change between reactants and products, you must count it in.

Right. When thinking about enthalpy and enthalpy changes just think to the roots of each problem. Ask yourself where the energy (enthalpy) is changing and what it might do?

I see. If the addition of an electron is favorable to an atom, like fluorine, energy is given off.

Right. And that enthalpy change is negative, because some enthalpy is released as fluorine grabs the electron. The electron affinity table is really a table of the enthalpy change to turn an atom into an anion.
Tell me, if you had a fluorine anion (F-) as the reactant but a neutral fluorine atom (F0) as the product, what would be the enthalpy change?

Ah, let's see. That's going in the opposite direction from what we just talked about.
Before, going from neutral fluorine (F0) to an anion (F-), energy was released. That's because fluorine is so good at grabbing electrons. That energy released means there was a negative change in enthalpy.
I guess that to do it the opposite way, to go from anion (F-) back to neutral fluorine (F0) , the enthalpy change will be opposite. Is that right?

Aye! To remove the electron from the fluorine anion (F-) takes energy. It is an endothermic reaction. The enthalpy change is positive and that tells you that you must add energy to pull that electron away from the fluorine anion (F-).

Are there other enthalpy changes?

Yes, and they do not always involve electrons. They can even involve the weak forces.

You mean enthalpy is responsible for things like melting? I thought that has to do with heat. Heating something up makes it melt. Right?

Aye. But the heat is used in TWO ways. Heat causes molecules to wiggle. Actually they vibrate. More heat means more vibration.

Heating a piece of ice causes the water molecules held as crystals to wiggle. I mean vibrate. OK, but what about that second way the heat is used?

That would be the enthalpy of fusion. Sometimes people call it the heat of fusion.

Fusion? I thought we were talking about melting ice, not nuclear reactions.

Indeed! Unfortunately, scientists tend to use the word "fusion" for many different things. In this case (the change of state) fusion means to join together or freeze.

Now I AM confused. Are we talking about ice melting or water freezing?

Well they are the same process. One is the reverse of the other. But they are both described by the enthalpy of fusion.

So melting and freezing are just enthalpy changes?

Aye!

I bet that the enthalpy change of melting has the opposite sign to the enthalpy change of freezing.

Very good! I think you are getting the hang of it. Now tell me, Arthur, which enthalpy change would give you the positive value? Melting or freezing?

Hmmm, reactions with positive changes in enthalpy suck in energy. They are endothermic. So I must figure out the reaction, melting or freezing, that requires energy.
That's easy! Melting requires energy. So melting (the chemical reaction going from ice to liquid water) is a positive change in enthalpy.

Absolutely right, Arthur.
I think water is a great chemical to use in most examples of key Alchemy concepts. Each state of water's matter has its own name: ice, "water" and steam. Even the directions of the reactions, as the water changes states, have names like melting and freezing.

It just seems odd to think of melting as a chemical reaction.

I agree, but it is. And it follows the same kind of enthalpy rules. Why not write down the chemical equations for melting and freezing? Jot down the enthalpy change too.

OK. Melting is "ice" ------> "water" and it's a positive change in enthalpy. That's endothermic, meaning it needs heat added.
Freezing is "water" ------> "ice" and because it is the opposite of melting, it must have a negative enthalpy. That means it's exothermic. Hey, does that mean that freezing gives off heat?

Yes it does! Freezing gives off heat!

You're out of your mind, wizard! How can ice give off heat?

The ice doesn't give off heat but the initial chemical reaction which MAKES the ice does! As water freezes it actually gives off a wee amount of heat but it is so small you don't notice it. Once the ice is made, the reaction is over and no more heat is given off.

That is soooooo weird.

I agree. You see, we tend to think of the surrounding air temperature as the only energy around, but it can be hidden away and behave in ways that simple heat does not.

OK. If the heat of fusion is the enthalpy change that occurs when water and ice change into each other, what about water and steam?

Ah, that would involve the "enthalpy of vaporization".

No, it wouldn't! Vaporization has to do with vapors of water. I'm talking about boiling water. Turning it to steam. Steam is not the same as a vapor is it? You said so!

I see why you are confused. I get confused too!. Unfortunately, the use of the word vapor and vaporization is all mixed up in Alchemy and I truly wish someone would come along and clear it up. It confuses young students like yourself, and it sure confused me when I first learned Alchemy.

So unconfuse me.

OK. You are right. Steam and vapor are two different things. Vapor can be fog or simply the damp that is in the air. Vapor is water molecules floating in air, but they are of low energy. Given a chance, they will stick to other water molecules (of similar low energy) and then they "rain".

Yeah, fogs are cold, not hot! Fogs and clouds are made of low energy water vapor not steam.

Right. Steam, on the other hand, is made of very hot water molecules. So hot that they cannot stick to each other to form water drops.

Exactly my point! The enthalpy change as water turns to steam should be called the heat of boiling, not heat of vaporization. It's boiling, not vaporizing.

I agree! I agree. Unfortunately, there was some unclear thinking in Alchemy when some of these processes got named. And the names have stuck.

So we have to learn to use words we don't mean!?

Afraid so. At least this is a good example of that problem.
Enthalpy (Heat) of vaporization is the heat absorbed (endothermic) to cause a liquid (at the boiling point) to turn into a gas. Heat is given off (exothermic) when the gas (at the boiling point) turns into the liquid.
Can you tell me how the heat of vaporization would work? Like you did for the heat of fusion.

OK. Heat of vaporization is the enthalpy change of boiling and condensing. They just have opposite signs. Right?

Right!

Good. Well it takes energy, heat, to make water boil, so I think boiling water is an endothermic reaction. That is, it requires enthalpy to be put in so the enthalpy change is positive for boiling. That means the enthalpy change is negative for condensation.

Right. Now write that down and explain it as you did for melting and freezing.

OK.
Boiling is just the change of "water" ------> "steam". It is endothermic (positive enthalpy change), so you have to add energy.
Condensation is "steam" ------> "water". It is the opposite of boiling so the enthalpy change is negative, and it is exothermic so energy (enthalpy) is given off.

Right. Notice that condensation gives off a wee bit of heat, but you really don't notice it among the hot steam.

I see. But I have a question. As steam cools to ice, are there only two enthalpy changes to think about? The condensation and the freezing?

No, you also have to take into account the energy in the molecules as well. That's "heat". As steam condenses into water, it gives off (a wee bit of) energy as heat. But then you have water liquid at the high temperature of the steam. It is still at the boiling temperature (boiling point). To freeze, that very hot water must first lose a great deal of "wiggle enthalpy". It does that by simply cooling down.

So before the very hot water can freeze it must lose a lot of heat?

Right! As it gives off heat, the enthalpy of the water is lowered. Water molecules vibrate less as the temperature drops. Eventually they vibrate so slowly that they are cool enough to freeze. Then all they have to do is lose the enthalpy of fusion and they will freeze.

Sounds complex.

Yes, there are a series of enthalpy changes occurring here. Some enthalpy is lost as the temperature cools. That's just simple heat lost as the water cools between the two temperatures. It's nice to think of that as "heat enthalpy". It's just the enthalpy released as the water molecules cool...

which causes them to wiggle less and that makes it easier for the weak bonds to form for the change of state (from liquid to solid).

Right. Notice the two different types of enthalpy. One is based on temperature and the other on states of matter. Hess's Law tells me I can just add them all up to get the final enthalpy change.

Hey, wait a minute! You shouldn't have to count the cooling of the hot water down to the freezing point. That's what Hess says. Right?

Wrong, but I see why you might think that. It depends on how you define the beginning and final enthalpies. The heat of vaporization is the enthalpy change that occurs when water boils or condenses. Either way you have water or steam at the boiling temperature. The heat of vaporization is just the enthalpy change for the change in state. Same is true for the heat of fusion. Neither of them take into account the enthalpy change required to change the temperature. You must include that separately.

Yeah, but the change is path independent. When changing steam to ice, the reactants are steam and the products are ice. That tells me the total enthalpy change of the reaction is just the enthalpy changes due to the changes in state. Right?

Sort of right. The enthalpy changes for those changes in the state of matter are calculated at the temperature at which that change occurs. (By the way, we are ignoring pressure.)

Yeah, so?

So, if you only considered the enthalpy changes caused by the changes in state you would be considering a reaction that turned steam into ice. But the ice would be as hot as the steam was!

That's impossible. You can't have ice as hot as steam. Can you?

It's not possible. Therefore, you must include the changes in temperature too. That affects the whole enthalpy change.

But the temperature change is in between!!! Hess says that enthalpy change is independent of path so we don't include it. It has nothing to do with the final products!

That is where you are wrong. It DOES have to do with the final product. Have you ever found ice at boiling hot temperatures?

No. It's impossible.

Right.

Argghh!!

OK, let's get to the heart of the problem. All enthalpy changes involving changes of state are determined at the temperature at which that change occurs. BUT (!) in the reaction we have been talking about we are thinking about TWO changes of state and they MUST occur at different temperatures. You see, what you are hoping for is a single enthalpy change to do it all. The enthalpy tables (values of enthalpies) are not designed to take into account ALL possible reactions. Instead they are broken down into a series of steps and you pick which steps to add into the calculations.

Why don't these "tables " give a value for the enthalpy changes that occur as steam turns to ice (or ice to steam)?

Because that is too specialized. Few Alchemists are interested in such a chemical reaction. Usually they go one step at a time. That is why the enthalpy values are always tied to a particular temperature. Then all an Alchemist has to do is work out the enthalpy changes as the reactants go to products AND not forget to take into account any changes in temperature that are required.

Hmm, sounds weird. But I think I'm getting the idea. I must carefully consider any temperature changes that occur as well as the more interesting stuff like melting or even ionization.

Right.

So the enthalpy change from a block of ice (at a certain temperature) to steam (at a certain temperature) is always the same. And the reverse is just the same amount but the opposite sign?

Yes! What's really amazing is this - you could choose any path between the reactants (ice at a temperature) and the products (steam at a temperature) to calculate the enthalpy change. It's independent of path.

Hang on, let me get this straight. (I just cannot leave this alone!)
Let's say I take a block of ice at -20oC and calculate the enthalpy change to get it to steam at 200oC. But instead of changing it directly to the 200oC steam, I melt it, then refreeze it, then melt it, then refreeze it and then (finally) go all the way to transform it into the 200oC steam. Will the total change in enthalpy still be the same as if I had just gotten there directly?

Aye, that's what Hess's Law says! I know that sounds strange but it's true. You see, as you go back and forth melting and freezing the water, you are going back and forth with the enthalpy changes too. At one point you add enthalpy, but at another point you subtract enthalpy. Because all you are doing is reversing the sign, each switch is back to the way it was. That causes the water to change back to its old enthalpy.

OK. Let me take a shot at that problem. The total enthalpy change for turning ice at -20oC to steam at 200oC is a series of enthalpy changes involving changes of state AND temperature.

Absolutely! Now tell me what changes you must consider if you did that reaction. You can do the enthalpy math in any order you like, but I like to keep it in order as if I were actually doing the reaction. It helps me to think through it clearly.

I definitely need all the help I can get to think through this clearly! OK, here goes.
First there is the enthalpy change to heat the block of ice up to its melting temperature. That's 0oC, right?

Right. Ice melts at 0oC (ignoring pressure).

Fine. I guess that would be an endothermic reaction because you have to heat the ice to its melting point. So the enthalpy change for the first step is a positive change in enthalpy.

Right. To heat the ice to its melting point you must add heat and that is a positive enthalpy change. We could put numbers to that change (and all enthalpy changes) but let's not get too deep into it for now.

Fine by me.
OK, the second step is to melt the ice (which is now at 0oC). So I must include the enthalpy of fusion, and that will be a positive change in enthalpy. Hey, all these will be positive changes in enthalpy, won't they?

Aye, and that is smart of you to see it. Turning ice at -20oC to steam at 200oC is a reaction that requires heat to be added at each step. That is convenient for us. But, you should realize that is not the case for all chemical reactions. It depends.

But for this one, it is all positive enthalpy changes because I need to be adding energy all the way.

Right. Now you have identified the enthalpy change to warm the ice and melt it. What other changes must occur?

Well, the third step is to warm that cold water at 0oC, to hot water at 100oC. That's the boiling temperature of water. Right?

Right.

Fine, so the third step is the positive enthalpy change to warm the water from 0oC to 100oC.
The fourth step is the boiling of the hot water and that requires the enthalpy of vaporization to be added in.
Then all I have to do is heat the steam from 100oC to 200oC. That final step is just a simple temperature change and will require more positive enthalpy.

Very good, Arthur. You have identified all the enthalpy changes needed to turn ice at -20oC (the reactants) to steam at 200oC (the products).

I see you made a point to remind me about the temperatures of the reactants and products.

Well, that IS important. It affects the final enthalpy.

Yeah, and even the (hidden) temperature change at 0oC and 100oC (the changes of state) must be included because it is impossible to get to the products without them.

Right! Now tell me, what would be the enthalpy change if, instead of going directly from -20oC ice to 200oC steam, you took a detour along the way. Suppose that after you melted the ice you then froze it again to -100oC, before continuing on to the steam at 200oC.

The enthalpy change is still the same as before. That's because the steps you added in the middle only change the enthalpy temporarily. But you reverse them all as you continue towards the final product. So it doesn't count.

Right. If on the path to the 200oC steam you were to freeze it again, it doesn't matter. It's...

... path independent!

Exactly. You could boil and condense it a dozen times. A million times! You could heat it into a plasma and then cool it back to steam! You could strip some electrons away and add them back, but the total enthalpy change will still be the same as if you had gotten there directly. All those other enthalpy changes will cancel themselves out as you work toward the final product. Enthalpy changes are independent of path. Hess says so!

It is very strange how enthalpy works. Sometimes it seems so,... wrong!

I agree. The trick to all enthalpy problems is to think about it at the level of energy going in and out of the system.

System? What system? What's a system?

Good questions. A "system" is a reaction, its container or anything you want to define as being "in the system". It's just a way to think about bits of the universe.

A system can be anything?

Yes.
You "define" a system to help you think about the problem. A system is whatever the Alchemist wants it to be.

The system isn't a real thing is it?

Not really. It's just a definition. A pot of boiling water on the stove can be thought of better (imagined) if you think in terms of systems. You can define the system as the water in the pot. The heat that makes the water boil comes from the pot, outside that system.

Or from the stove that heats the pot.

Yes. We could define the system as both the water and the pot. They are then "in the system". The stove and everything else is outside the system. A system is often defined as containing certain materials and certain energies. Everything outside the system is "the rest of the universe".

Does it matter where you draw this imaginary line?

Not really. You can draw the imaginary line (define the system) anyway you want, but it helps to draw your system boundaries to make it easy to think about the reactions you are working. And you must be sure the system is isolated or at least know where it is not.

Huh?

An "isolated system" cannot exchange heat or anything else with its surroundings.

So I would have to include the stove in the system, because it provides the heat to boil the water. Right?

Right. Systems are just a way to imagine the enthalpy problems. You cannot be right or wrong when defining a system, but you must know how enthalpy enters and leaves the system.

Unless it is an isolated system because nothing goes in or out of an isolated system, not even the heat.

Right. If it is an isolated system, you must account for how the enthalpy is rearranged within the system. For example, let's say you have an exothermic chemical reaction in an isolated system. What happens to the enthalpy?

Hmmm. The enthalpy change will be negative (exothermic) giving off energy (enthalpy). But in an isolated system that energy can't go anywhere. Gee, I don't know. Where does the energy go?

It could "go" into any other reaction that would be described as endothermic.

But that endothermic reaction must occur IN the system. Right?

Very right. What would become of the molecules in that isolated system?

Ah, they would suck in the energy and wiggle more! They would warm up .

Right. The enthalpy given off by the chemical reaction could be used to make the molecules vibrate more. It warms them up.

I see. In the isolated system the enthalpy must go somewhere within the system and temperature is an easy place for it to go.

Yes. Temperature changes are the likely outcome of enthalpy changes, and it is easy to keep track of it.

Could something other than a temperature change balance the enthalpy change?

Yes. You might have another chemical reaction going in a direction that requires the enthalpy change of the first reaction.

And I can use the idea of "systems" to keep track of it.

Yes. Some Alchemy students love to define systems as they work out their enthalpy thoughts. But some others are able to do enthalpy problems with little thought to defining systems.

I'll try to keep track of the systems and see if it helps me to imagine it.

Good idea. Thermodynamics is often described as "the energy of systems".

But all thermodynamics have to do with is changes in enthalpy. Right?

Well, changes in enthalpy are the most important change that occurs. However, thermodynamics also requires that you keep track of entropy too!

Entropy?

Entropy ("N-tro-P") is the amount of disorder in a system.

Does this have anything to do with Alchemy?

It certainly does!
Entropy is a very important part of thermodynamics and therefore important in chemical reactions.

But what is it? What do you mean by disorder in a system?

Good question. Entropy is a kind of universal calculation of order. Or more correctly, disorder.

For example?

Imagine your room as a system.

An isolated system?

Yes, why not? Imagine you are in it.

I imagine it would be a mess!

Good! That's entropy! Your room is disordered. It has a high entropy.

Huh?

Entropy is a measure of disorder. If your room is disordered, it is high in entropy.

It's a mess!

Yes, so your room is disordered and high in entropy. How would you reduce the entropy of your room?

I guess I would have to make it less "disordered". I'd have to clean it up. Straighten it up. Arrange my stuff.

Exactly. You would have to do work to make your room more orderly. It takes energy to do that work doesn't it?

Sure does. That's why I hate doing it!

Right! You expend energy to change your room from a disordered high entropy state to an ordered low entropy state.

Is this "entropy" thing real or is it just like "system"? Something you make up in order to think about the problem clearly?

Oh, entropy is real. Very real! Let's talk about molecules and I'll show you what I mean. Recall that you can add heat to ice in order to make it melt. Right?

Right. That's obvious. That's enthalpy not entropy. Right?

Well, it's both! Up until now we have only discussed the enthalpy changes of melting. Melting requires an input of energy. To melt something you add the enthalpy of fusion.

OK, I'm with you so far. What about entropy?

What about entropy? You and I agree that the enthalpy of liquid water is higher than the enthalpy of the ice right?

Yeah, you have to put energy into the ice to melt it. You add enthalpy. Melting is a positive enthalpy change. But what about the entropy?

Yes, what about the entropy?

Argghhh! I don't know!! What about entropy!? What about the entropy of melting? All this entropy stuff has to do with order and disorder.

Right. Now think about this. What is the order in the ice? What is the order in the (liquid) water? Do they have the same entropy? Do they have the same amount of disorder?

No! Ice is water molecules arranged in repeating patterns. Crystals.
But liquid water molecules don't have any order about them.

Right. Liquid water is more disordered than solid ice. That means liquid water has a higher entropy (a greater disorder) than solid ice.
Do you understand?

I think so. Liquid water is like my room, disordered, so they both have a high entropy.

Right. Notice that entropy is DISORDER not order. Alchemy students sometimes get tripped up on that.

Yeah, it is more natural to think that entropy should be order, not the other way around.

True. So keep track of the meaning of entropy. It means DISorder. Something of high entropy is highly disordered.

Ice is more ordered than water, and if I put my room in order it would be more like ice. I see. So if I were to clean up my room, I would lower its entropy?

Right. By putting your room in order you would lower its entropy. It's the same process as freezing. When water freezes it puts its molecules in order and when you tidy up your room you are putting your stuff in order. Both of those activities lower the entropy (of water or the room).

Do I have to arrange my room like an ice crystal in order to lower the entropy.

No, not really. As a matter of fact, you can arrange it anyway you like and it will lower the entropy. The only thing that would increase the entropy of your room would be to drop your stuff around randomly.

Does ice have an entropy of zero? Because it is so highly ordered?

No. It is difficult to get any system down to zero entropy. It requires more than just a bit of order in the crystals.
Among pure chemicals, crystals have the lowest entropies (greatest order). But they still have a tiny bit of wiggle among the molecules. To get their entropies to absolute zero, you must stop them wiggling by removing all the enthalpy (heat).

This entropy thing sounds strange. And it sounds like enthalpy and entropy are tied together.

They are. As you put heat into a substance, its entropy increases. Many solids, especially crystals, are highly ordered and thus have very low entropy. When the solid melts, its entropy increases because it becomes more disordered. Gas is the state of matter with the highest entropy because it has no order to it at all. The gas molecules just fly around randomly.

Does entropy apply to more than just the states of matter? And my room?

Oh, yes. Entropy is an important part of thermodynamics and thus Alchemy. Entropy is a function of both strong and weak bonds. It's just easier to imagine it with the weak forces responsible for the states of matter.

Does entropy change in "real" chemical reactions? You know when atoms get rearranged into different kinds of molecules.

Aye, entropy changes often occur in most reactions. Melting a crystal increases entropy (increases disorder). Freezing it into a crystal increases entropy (decreases disorder).

Yeah, but how does entropy change when molecules change? Do different molecules have different entropies?

Yes. Large, complex molecules (with lots of bonds between lots of atoms) have highly ordered bonds and therefore have less entropy than small molecules. Reactions that change molecules' sizes change the entropy of the system.

Like how?

Think about it. Decomposition reactions increase entropy because they make smaller (disordered) molecules from larger (ordered) ones.

Does that mean combination reactions decrease entropy because they turn (disorder) small molecules into (more ordered) large molecules?

Aye! Entropy changes occur in opposite pairs, like enthalpy changes. Remember exothermic and endothermic reactions?

Yeah, I get it! Entropy changes can be described in opposite pairs. Combination decreases entropy and decomposition increases it.

Right. Now let's bring in that word "systems" again. Your messy room could be a system. A system of high entropy (high disorder). If you were already in the room you would expend some effort to arrange your room. That effort lowers the room's entropy.

Effort is the right word all right. It takes a lot of work to put my room in order.

Exactly! Imagine you expended energy to order your room, to lower the entropy.

Keep imagining.

What I am describing here is an isolated system, your room, going from a state of disorder to a (slightly) more ordered state. To lower the entropy of the room you had to work at it. Now tell me, Arthur, how did your room become more ordered?

Ah, I did it. I lowered the room's entropy. Hey, does that mean that I increased my own entropy cleaning it?

Yes it does! It must! Your body broke many chemical bonds while you worked. Some of those chemical reactions gave off heat. That was the enthalpy change. Many of those same chemical reactions also changed large molecules like sugar and fat, into smaller molecules like water and carbon dioxide.

So I swapped entropy with my room!

Right. And because you were in the system while you worked, it was an isolated system. If part of an isolated system decreases in entropy, then some other part of the system MUST increase in entropy. Otherwise it is not an isolated system.

Well, I'm never going to clean up my room again!

That assumes you've cleaned it once (ha ha).

Very funny. You know, when I become king I will hire a maid to clean my room. Then I'll keep all my low entropy and high enthalpy to myself.

That's a good idea, but you have to agree that you would no longer have an isolated system. Not anymore. Do you see why?

Yeah. The maid comes into my system, decreases the entropy (by increasing hers) and then she leaves me with my tidy room. I like that idea.

I'm sure you would.

You know this entropy ideal could get me out of a lot of chores. "Sorry, I don't want to decrease my entropy."

In fact, the decrease in your entropy as you order your room is very small. That's because life is VERY low in entropy. That's because life contains billions of highly ordered molecules.
However, I'll let you in on a wee "secret" about entropy. You can't fight it! Eventually entropy wins. It must, that's the law of the universe. It's a law of thermodynamics.

What? You make it sound like there's some kind of magical law of nature. Like there's no hope to make anything that lasts forever.

That's exactly right. A wise Alchemist once said, "The entropy of the universe tends toward the maximum", meaning the whole universe is always moving toward the most disorder. As surely as water rolls downhill, so too does the entropy of the universe constantly head toward randomness.

That sounds so,... spooky!

It is. A little knowledge about entropy puts perspective on the universe. Only tiny parts of the universe (the sun, the earth, me, you, etc.) enjoy low entropy (high order) but that cannot last forever! Most of the universe is at very high entropy already. The path to a higher entropy (more disorder) is a route we all must follow.

Definitely spooky! And this is real science?

Aye!

Well, tell me wizard, how did I come about. I'm a low entropy (highly ordered) form of matter. Right?

Right, you are.

Well, how? If all the universe is heading toward maximum entropy, how can chemicals (like my body) and life (like me) come about and how do I stay alive?

You ask the best questions.
This brings us to a very important part of thermodynamics. Enthalpy and entropy are the two thermodynamic energies that determine whether a chemical reaction will or will not occur at a certain temperature.

Yeah? You already told me that. So?

So (you didn't let me finish), what you didn't know is that chemical reactions use both entropy and enthalpy to make them "go". That means you can exchange entropy and enthalpy. If you have the right chemical reactions, that is.

How do you mean?

Well, you and your mom and all of us eat! Eating is a complex process but the bottom of it all is the fact that we take in high enthalpy molecules when we eat. Then we "burn" them in decomposition reactions. These reactions give off enthalpy (energy). That enthalpy can be put to a lot of different uses. One use is to decrease the entropy of the system.

But you said when you burn food you use decomposition reactions. That increases entropy. It doesn't decrease it!

Right. Those molecules go to higher entropies as they are "burned" into water and carbon dioxide. But other chemical reactions (helped by enzymes) use the entropy and enthalpy changes to do the opposite. That allows you to decrease the entropy of other molecules. Do you see what I mean?

I think so. You make low entropy molecules by stealing entropy (and enthalpy) away from the food you eat.

Right. Over the course of a lifetime, you will eat a mountain of food and burn it. That's what keeps you alive. Your mother ate lots of food in order to make you. Mom's are master wizards at creating very low entropy matter.

Babies!

Right. We are all a bag of complex chemicals. That low entropy comes about by decomposition reactions. It keeps us alive. And mother's use special chemical reactions to make very low entropy matter like you and me!

This is pretty amazing stuff!

And all of it is true! It's real Alchemy.

Life is a huge series of chemical reactions and they can all be described by enthalpy and entropy changes? Just the math of equilibriums and thermodynamics?

Yes! Life is a bunch of chemical reactions, but we Alchemists accept that it is very complex and hard to understand. The important point is that Alchemist don't see anything magical about life. Complex, yes. Magical, no.

Mysterious?

Hmmm, maybe. It would be a great accomplishment to define all the chemical reactions that make us alive. But Alchemy may do it one day. Not by the 20th century. (But someday?)

Amazing. Has anyone figured out how to predict the enthalpy and entropy changes of simple chemical reactions?

Yes. Josiah Willard Gibbs applied the laws of thermodynamics to chemical equilibrium and revolutionized Alchemy. Gibbs produced a mathematical equation that allowed him to predict whether a chemical reaction would go forward or reverse and by how much.

Would it work for life? Can Gibbs tell you how life works? What makes life "go"? In detail?

Ah, in theory, yes. In practice, no. It's just too complex. Besides, Gibbs' equations only work for chemistry that is in equilibrium. Life is usually not at equilibrium. Far from it. Instead, life's chemical reactions are constantly at work heading towards equilibrium but not quite making it (until we die).

Very weird. Tell me more about Gibbs and his energies and equations.

Gibbs energy is the TOTAL amount of chemical energy in a system and includes the entropy, enthalpy and temperature of all the chemicals involved.

Everything?! All at once?

Aye. Chemical reactions always run from highest Gibbs energy to lowest Gibbs energy. That is, any REAL chemical reaction MUST lose Gibbs energy because the reactants (always the substances with higher Gibbs energy) turn into products (always the substances with lower Gibbs energy). Gibbs energy, like all energy, "runs downhill".

So the change in Gibbs energy must be negative for a reaction to occur?

Right! Can you write an equation to show that? I'll give you a clue. It looks like the changes we talked about earlier in our talk about entropy and enthalpy.

Hmmm. The change in Gibbs energy in a reaction should be the Gibbs energy of the products minus the Gibbs energy of the reactants. Is that right?

Aye. Write that as an equation.

OK. Change in Gibbs energy = Products' Gibbs energy - Reactants' Gibbs energy.

Excellent. Notice that a product with higher Gibbs energy than the reactants' will give a positive number for the Gibbs energy change. See?

I think so. If the Gibbs energy of the products is 50 units and the Gibbs energy of the reactants is 40 units, then you have a difference of 10 units of Gibbs energy. Like this.
Products' Gibbs energy - Reactants' Gibbs energy = Change in Gibbs energy
50 units - 40 units = 10 units
Does that mean you need 10 units of Gibbs energy to make it work?

What do you think?

I think you do! I think you need 10 units of Gibbs energy to make it work. Yeah, because you must change the reactants with only 40 units of Gibbs energy into products with 50 units of Gibbs energy. You have to add 10 units of Gibbs energy, otherwise you can't make the higher Gibbs energy products. That's why it's a positive 10 units. You have to add it.

Very good point, Arthur. To make that reaction "go" from reactants (with 40 units) to products (with 50 units) you must add 10 more Gibbs energy units.

Hey, if you have to add energy to a reaction to make it go, is that an endergonic reaction?

Yes, it is! And I'm glad you saw that. The Gibbs energy change accounts for the change in ALL the energy between reactants and products. In this example the products have more total energy (more Gibbs energy) than the reactants. The Gibbs energy change is positive and that means it is an endergonic reaction.

I see. An endergonic reaction won't go because you cannot make products with higher Gibbs energy than the Gibbs energy of the reactants. At least, not unless you add more Gibbs energy to the system!

Yes. Excellent.
How could you do that? How could you add more Gibbs energy units in an isolated system.

Well, if the system is isolated you can't! You would have to get the units from within the system if it is isolated.

Right. We take on food to keep our Gibbs reactions going in the right direction. Our system, life, is never isolated (for very long).

But what about a small, easy to understand chemical reaction. What happens to a chemical reaction in an isolated system if it has a positive change in Gibbs energy? I mean the whole system needs 10 units to work.

You tell me. What happens if the Gibbs energy change is positive?

Nothing! Nothing will happen if the Gibbs energy change is positive!

Well, something will happen, but not what you want. You see, the reaction cannot "go" in that direction. The change in Gibbs energy is positive, so it won't go in that direction. It can't. Thermodynamics says so.

But the opposite reaction can occur! Can't it?

Yes, it can! The only reactions that really "go" (from reactants to products) MUST have a negative Gibbs energy change. Understand?

Sure do! The Gibbs energy change must be negative. It must give off Gibbs energy.
Hey, that means all real reactions, all reactions that "go" must have a negative change in Gibbs energy. And that means all real reactions go in the direction that releases Gibbs energy.
So ALL real REACTIONS are EXERGONIC!

Right! All real chemical reactions are exergonic. An endergonic reaction won't go in the direction it is written (unless energy is supplied from outside the system). The thermodynamics (enthalpy, entropy and temperature) behind Gibbs energy dictate the DIRECTION of a chemical reaction.

Wow! Gibbs energy does it all. It takes into account ALL the energy changes. Whichever direction gives the negative Gibbs energy change is the direction the reaction will go. That would be the exergonic direction.

Right! If you calculate the Gibbs energy change to be negative, exergonic, that reaction will occur as written (as you have defined the products and reactants). Otherwise it runs in the opposite direction.

You would think Alchemy could come up with a more exciting way to describe the most important idea of all! "Gibbs energy change" just doesn't sound important enough!

Well, you can call it "Alchemy's Special Idea" or whatever you like. But we Alchemists call it Gibbs energy equations. Thanks to Gibbs and Hess we now have a way to predict the energy needed to change reactants into products. We just figure out their changes in entropy, enthalpy and temperature. All that math is included in Gibbs energy changes. (The details are covered in an Advanced Alchemy course.)

So, I use balanced equations to count up the atoms and electrons I need. And I use Gibbs energy to calculate which way it will really go.

Right. The equilibrium point is also determined by Gibbs energy, but that gets pretty heavy with math, so we will just leave it here.

Can you give me a hint of what you mean by the Gibbs energy also determining the equilibrium point of the reaction? It sounds important. The equilibrium point determines the yield. Right?

Right. Hmmm.
OK. Assume you have two reactions and they are both exergonic

So they both have negative changes in Gibbs energy and so they will go in the direction we want.

Right. Let's assume one of them releases 40 Gibbs unit and the other releases 100 Gibbs units. Take a guess which one will have the greatest yield. Which one will turn more product to reactant?

Ah, the one releasing 100 Gibbs units?

Right. Why?

I don't know. It just seemed like the right guess. I think that a reaction giving off 100 units of Gibbs energy is more likely to go in the right direction than a reaction that gives up only 40 units. I think the reaction with all the extra Gibbs energy to release will be the one most enthusiastic about producing products.

Well, chemicals aren't "enthusiastic" about their reactions but you are pretty close to a good explanation.
You see, the larger the amount of Gibbs energy released the more the forward reaction is preferred to the reverse reaction. Think about that.

Oh, I get it.
A reaction with a lot of left over Gibbs energy is a reaction that is very exergonic so the forward reaction is its preferred direction. Its reverse direction would be very endergonic and it wouldn't want to go that way.

Exactly. In a reaction starting with reactants with very high Gibbs energy going to products with very low Gibbs energy, there will be lots of Gibbs energy released as it moves from reactants to products. But the reverse reaction has a long climb "uphill" to return the other way.

Hey, how come there's any reverse reaction at all. The reverse reaction is endergonic! It won't go that way.

Right. By itself it won't. But the reverse reaction can steal some of the Gibbs energy from the forward reaction to push itself "uphill". All these molecules, both reactants and products, are in the same system. Naturally, the reverse reaction can't take more Gibbs energy than the forward reaction can supply, so the reverse reaction is limited by the forward reaction. I don't want to get into the math of it, but when the forward reaction has plenty of Gibbs energy released, the reverse reaction doesn't dominate so the forward reaction is favored.

I see. It makes sense to me. The bigger the difference in Gibbs energy from reactants to products the more the reaction will go in a particular direction. The direction is determined by whether the equation (as written) gives off Gibbs energy (exergonic) or takes in Gibbs energy (endergonic). It will go in the exergonic direction. And the more negative the Gibbs energy change the greater will be the yield !

Right. So a reaction that releases 40 units of Gibbs energy will not give as great a yield as a reaction that releases 100 units of Gibbs energy.

Wait a minute. What about Hess and his law.

What about it?

Hess says the difference in energy between reactants and products is independent of path.

Yes, that's right. How does that worry you now?

Well, you just talked about two reactions with different Gibbs energy releases. That doesn't make sense!

Oh, I see what you mean. I failed to explain that those two reactions are different reactions - different equations. You're right. Any equation will have a single Gibbs energy change. It must! The example I gave was for two different reactions. Either their reactants or their products are different. (Maybe both.)

OK, they are different reactions.

Yes, I just wanted to give you an example of two different Gibbs energy changes. In fact, in that example, maybe the same product is formed, but each reaction starts with different reactants.
Hmm, tell me Arthur, if both those reactions, the one releasing 40 units of Gibbs energy and the other releasing 100 units of Gibbs energy, go to make the same product(s), what can you tell me about the Gibbs energy of the different reactants.

I don't know.

Think about Hess's Law.

Hmmm,
We end up with the same product(s) so we end up with materials with the same Gibbs energy in them. Is that right?

Right. I think you're heading in the right direction. So what can you tell me about the starting materials, the reactants, in those two reactions?

Hmm, and this involves Hess's Law?

Actually, Hess's Law is involved here but you don't need to get too specific about it to answer my question.

OK. They both end up making products with the same Gibbs energy, but they give off different amounts of Gibbs energy in the process. Hmmm.
That means they must start with different amounts of Gibbs energy. Right?

Right. Those two reactions have different Gibbs energy in their reactants. Tell me which one has the higher Gibbs energy - the reaction that releases 100 units or the one that releases 40 units.

It would have to be the one that releases 40 units.

Ah, no. Try again.

OK. The one that released 100 Gibbs units has reactants with more Gibbs energy.

OK, right. (Nice guessing.) Can you tell me why?

Hmmm,
The reactants with higher Gibbs energy had to release more energy as they made product. Yeah, that makes sense.

Yes, it does.
Let's put some symbols to this in order to make it clear.
A ------>C gives off 40 Gibbs units
B------->C gives off 100 Gibbs units

Now I see. The product (C) is the same. To make C, B gave off more units than A did.

Right. How much more Gibbs energy did B give off?

B gave off 60 more Gibbs energy units than A did (as C was made).

Right. So how much more Gibbs energy is hidden in B?

60! B has 60 more Gibbs units than A has!

Right. You see, it's just thinking about the math of thermodynamics!

Wow. You know, it has been a long hard push to get this in my head. And I'll study it all a lot more before I'm done. But it seems worth it!

Good. I agree that Alchemy is not easy. It isn't "instinctive". That's' why it's hard (for most students).

Is there more Alchemy to learn? Or am I now an expert Alchemist?

Oh, the learning never ends. There are some areas of Alchemy we haven't touched upon because they can only be understood by using math. And even the topics we have covered could be explained in greater detail by math. To learn more Alchemy (from anyone) you should be prepared to study algebra and feel comfortable using big numbers. But the material we have covered provides a foundation on which to learn more.

OK. I'll see how well I do with the questions you give me.

Good idea. My questions will add some more to your Alchemy knowledge and hopefully they will get you excited about learning more Alchemy.


This work was created by Dr Jamie Love and licensed under a Creative Commons Attribution-ShareAlike 4.0 International License Creative Commons Licence.