Heat / enthalpy / that H thing

Now that you know the introductory terminology for thermodynamics, it’s time to actually learn something.  Don’t get all frowny – you knew this was coming sooner or later!

sadtutu

Like most people, this girl studies chemistry while wearing a tutu.

We’re going to start with talking about enthalpy, because it’s my website and this is how I want to do it.  If you don’t like it, you should start your own site.


Enthalpy (H)

There’s a big official definition of enthalpy involving pressure and volume and stuff, but as a chemist, I’ve never actually had to use any of it.  As a result, I’ll just give you a nice and easy functional definition of enthalpy:

Enthalpy is the amount of heat that some system can transfer to another system.

This may seem easy to understand, because it’s our personal experience that if you have one thing with a lot of energy (e.g. a hot plate) and something with less energy (e.g. a frozen cat), the hot plate will transfer energy in the form of heat to the cat.

Unfortunately, it’s not simple at all.  For example, let’s say that instead of putting a frozen cat on the hot plate, we put a red hot nickel ball on the hotplate instead.  In this case, heat would probably be transferred into the hot plate, rather than transferred out of it.

In other words, the term H, by itself, doesn’t really tell us much, as it is highly dependent on the surroundings of the system you’re interested in.


Let’s try that again:  Enthalpy

Given that the enthalpy of a system is impossible to determine, we have to settle for finding the change in enthalpy (ΔH) for a process.  This is something we can measure, because we can measure the properties of a system both before and after a process to see what’s happened.

As you’ve undoubtedly seen, there are a bunch of different ΔH terms you need to keep track of. Fortunately, they’re all the same in that all they do is help you to figure out how much heat has either moved into something or moved out of something.  Let’s look at some of our favorite enthalpy terms:

heats table

If ΔH for a process is negative, we refer to the reaction as being exothermic because it has lost energy.  When you perform an exothermic reaction, the surroundings get hot because all of that energy had to leave the system and go somewhere.  That somewhere are the surroundings.

If ΔH for a process is positive, we refer to the reaction as being endothermic because it has gained energy.  Endothermic reactions feel cold, because the stuff that’s reacting is literally pulling heat from your hand (i.e. it’s surroundings) during this reaction.


Finding heats of reaction from heats of formation

Let’s say that we have a chemical process that looks like this:

A + 2 B → 3 C

To figure out the heat of reaction, you need to figure out how much more or less energy the products have than the original reactants.  Fortunately for us, we can do this by simply comparing the sums of the heats of formation for these compounds.

Let’s assume that the heat of formation of A is 50.0 kJ/mol, the heat of formation of B is 35.0 kJ/mol, and the heat of formation of C is -45.0 kJ/mol.  To find the overall heat of reaction, all we have to do is compare the heat of formation of the reagents to those of the products.

  • The reagents:  One mole of A has a heat of formation of 50.0 kJ, and two moles of B have a heat of formation of 70.0 kJ (35.0 x 2).  The total heats of formation of the reagents:  120.0 kJ.
  • The products:  Three moles of C have a heat of formation of -135 kJ (-45.0 x 3).  The total heat of formation is -135 kJ.

If we start off with 120.0 kJ and end up with -135 kJ, you can see that we have gone down in energy by a total of -255 kJ during this reaction (if this isn’t clear, you can find this by subtracting the heats of formation of the reagents from the heats of formation of the products).  As a result, this reaction is exothermic and can be written as:

A + 2 B → 3 C        ΔHrxn = -255 kJ


Hess’s Law:  Many paths to the top

Let’s say that you’re planning on going to a mountain holiday and want to hike from Thermoville to Enthalpytown.  However, before you make the hike, you’d like to figure out the altitude difference between the two towns.  After all, nobody wants to hike uphill more than they have to.

Being a clever person, you decide to ask a villager for the information.  Being an idiot, the villager draws a map but can’t give you the precise information you want:

map

“OK,” you ask the idiot, “what does this mean for me?”

The idiot replies, “You’re here and you want to get to that place with the mean dog.”  I don’t know how high that mean dog’s house is, but I do know that you have to go down 250 meters to see Bobby and that Bobby has to go 680 meters up if he wants the mean dog to bite him.

Though stupid, the villager gave you the information you want.  To get up to Enthalpytown (i.e. the place with the mean dog), you first have to go down 350 meters to the valley where Bobby lives and then go up 680 meters to get to the town.  The total altitude change:  330 meters.

However, one thing still bothers you.  “Why did you include ‘that big rock’ on the map?”  The villager answers, “There’s a big rock on that hill.”  As I said, the villager really was an idiot.

So, what does all this mean?  I’m glad you asked!  It means that if you don’t know what the heat of a reaction is, you can still figure it out if you can find the heats of some other reactions that will take you to the same place.  Sure, they may not take you there in exactly the way you were planning to go, but as long as you know what they are, you should be in good shape.  That’s the basic idea behind Hess’s law, so if you got confused by its phrasing in your textbook, remember that it’s just as simple as an idiot and a mean dog.

Here’s how it works:

Let’s say that we want to find the heat of reaction for the process 2 A + B → C given the following information:

A + B → C          ΔHrxn = 125 kJ

B → C          ΔHrxn = -67 kJ

The trick to solving this problem is to somehow manipulate the equations that we’ve been given to figure out the answer we want.   We can do this in the following ways:

  • Reverse the equations we’re given.  In our example, if A + B → C has a heat of reaction of 125 kJ, the reverse reaction C → A + B will have a heat of reaction of -150 kJ.
  • Multiply the coefficients in the equation we’re given.  If A + B → C has a heat of reaction of 125 kJ, 2 A + 2 B → 2 C will have a heat of reaction twice that, or 250 kJ.

Armed with this information, let’s solve this problem!

  • Looking at our first known equation of A + B → C, you can see that this is the only equation we’ve been given with A in it.  As a result, we can conclude that we need to do whatever we can to get “2A.”  Given that we’re allowed to multiply the coefficients, let’s multiply this reaction by two to get 2 A + 2 B → 2 C with a heat of reaction of 250 kJ.
  • You may notice that the equation we have from the last step has too many moles of both B and C. As a result, we’ll have to change the second equation in such a way that we can cancel out some of each.  This can be done by reversing the equation to change B→ C to C→ B, which will change the heat of reaction from -67 kJ to +67 kJ.

When we add these two equations to each other, we end up with quite a mess:

2 A + 2 B + C→ 2 C + B     ΔHrxn = 317 kJ

However, since we can cancel out terms that are on both sides of the equation:

2 A + 2 B + C → 2 C + B     ΔHrxn = 317 kJ

All turns out just fine, with a final answer of:

2 A + B → C     ΔHrxn = 317 kJ

(Note:  The answer is actually 320 kJ, given the influence of significant figures.  However, given that this problem is meant to illustrate Hess’s law and not significant figures, I decided to leave it the way it is.)


Calorimetry:  Bombing your way to the top

Let’s say that I want to figure out the heat of reaction for something but I don’t want to look up the heats of formation or any heats of reaction.  In fact, I’d rather just do a chemical reaction and measure it for myself.

You’re in luck!  To make this happen, all you need is a bucket of water, a thermometer, a stirrer, and a bomb.

bomb

Chemistry is awesome!

Unfortunately for us, the bomb we use doesn’t look like this, nor will there be any huge fireballs created.  Rather, the bomb is a small container in which you do the reaction.  I’m guessing that it’s called a bomb because you have to “set off” the reaction inside.  But without the destruction.

Once you’ve gotten all of this together, it’s time to do an experiment using bomb calorimetry.  The idea behind bomb calorimetry is that if you do a reaction inside of a container that’s sitting in a big bucket of water, the heat from the reaction will be transferred into the water.  As a result, measuring the temperature increase of the water will allow you to figure out the energy change inside the bomb.

Bomb_Calorimeter_Diagram

In this diagram, the reaction takes place in the bomb and the heat is transferred to the water. The rest of the stuff, while interesting, is there to either measure what’s going on or to make sure things work right.

To find the heat that’s released into the water, you simply use the following equation:

heatchange

As a result, if you want to find the heat of combustion of a compound and place one mole inside the bomb to be burned, all you have to do is measure the temperature change and know the mass of the water next to the bomb to find your answer.  For example, if the mass of the water was 750.0 grams and the temperature change was 23.7 degrees Celsius, the overall heat of combustion would be:

ΔH = (750.0 grams)(4.184 J/gC)(23.7 degrees C)

= 74,400 J

And you didn’t even have to blow anything up!

(Note:  In rare cases, bomb calorimeters actually do blow up.  However, this is very unusual, and generally caused by lack of maintenance.  Link.)


Figure credits:

  • Sad tutu girl:  Image courtesy of num_skyman at FreeDigitalPhotos.net
  • Bomb:  Image courtesy of Idea go at FreeDigitalPhotos.net
  • Actual bomb calorimeter:  By Lisdavid89 (Own work) [CC BY-SA 3.0 (http://creativecommons.org/licenses/by-sa/3.0)%5D, via Wikimedia Commons
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