Bond enthalpy and enthalpy of reaction | Chemistry | Khan Academy

We're going to be talking about bond enthalpy and how you can use it to calculate the enthalpy of reaction. Bond enthalpy is the energy that it takes to break one mole of a bond. So, one mole of a bond. Different types of bonds will have different bond enthalpies.

As an example, we can talk about a carbon-hydrogen bond, or a carbon-hydrogen single bond. So, this carbon is probably attached to some other stuff because carbons usually have more than one single bond. But we're going to ignore everything else attached to the carbon. We're just going to represent it as a big blob, like popcorn. Maybe it's a protein; it could be a sugar molecule; it could be a lot of things. But we're ignoring that blob.

One other thing I forgot to say earlier is that this is the energy it takes to break one mole of a bond in the gas phase. So it's a pretty specific definition. In the case of our carbon-hydrogen bond, the bond enthalpy of this bond, so if we break this bond, let's do a sort of dotted line. If we break this bond, we have to add energy, and what we'll get as our products is we'll get our popcorn.

What happens is when we break this bond, the two electrons that originally made up the bond—one of the electrons will go to the carbon, and the other electron will go to the hydrogen. We usually represent single electrons like that using a single dot, sort of like when you write Lewis structures. You can write lone pairs with two dots. So, here's our carbon with one dot, or one electron, and our hydrogen with one electron. These are both still in the gas phase.

So the ΔH of this reaction is the bond enthalpy, which I will abbreviate as BE. Some important things to remember about bond enthalpy are that bond enthalpy is always positive. So, it's always going to take energy; you're always going to have to add energy to break a bond. If we take the reverse of the bond, if we take the reverse of the bond enthalpy, another way to think about this is to flip this reaction. If we take the reverse of this reaction, that means we're making a bond.

Since we know that breaking a bond always takes energy, that means making a bond always releases energy. It will always be negative to make a bond, and that's another way of saying it will always release energy. The third thing that we're going to discuss about bond enthalpy is that you can use it to estimate ΔH of reaction.

ΔH of reaction is—or the enthalpy of reaction—is something that chemists are often interested in. We want to know if it's exothermic or endothermic. You might know that there are lots of other ways of calculating ΔH of reaction, such as using Hess's Law, or another way is using ΔH of formation. Then there are other ways too. So this is just another way that we can use to calculate ΔH of reaction using bond enthalpies.

We're going to go through an example of that next. The example reaction is taking propine, which is C₃H₄ gas, and reacting it with hydrogen—so hydrogen gas—to get propane, C₃H₈ gas. I don't know about you; I'm pretty bad at looking at a chemical formula like this and knowing exactly what the molecule looks like. So, I'm going to draw out the Lewis structures.

The Lewis structure for propine has three carbons and one carbon-carbon triple bond, and then it has four hydrogens. So that's propine. We also have hydrogen gas, and our product is propane. So propane has all single bonds, so three carbons with single bonds and eight hydrogens bound to the carbons. That's the reaction we are interested in. What we want to know here is what is ΔH of reaction and how can we calculate it using bond enthalpies?

We said earlier that bond enthalpy is the energy it takes to break a bond. So what we're going to do next is look at our reaction in terms of what bonds are broken and what bonds are formed. This is a lot easier to do using the Lewis structures. First, let's talk about which bonds are broken. If we compare our reactants and our products, we're breaking this carbon-carbon triple bond, and we're also going to break this hydrogen-hydrogen bond.

One thing we forgot to do earlier, which is super important, is we actually need to make sure our reaction is balanced. We have four plus two equals six hydrogens on our reactant side, and we have eight hydrogens on our product side. That's not balanced, so we actually need two hydrogen molecules on the reactant side.

So, let's draw one more in. Yes, we said we are breaking a hydrogen-hydrogen bond. We're actually breaking two hydrogen-hydrogen bonds. It's important to keep track of how many of each type of bond we're breaking because the bond enthalpy is per mole. So, if you have twice as many moles, it'll take twice as much energy to break all of those bonds.

Then we can look at the bonds that are formed. Since we broke this carbon-carbon triple bond, that means we needed to make a new bond, and the new bond we made in our product molecule is this carbon-carbon single bond. Not only did we form a new single bond between these two carbons, but now these carbons are attached to a bunch of hydrogens.

So, we made four new carbon-hydrogen bonds. Let's write that out so that we can keep track of them when we do our final calculation of ΔH of reaction. If we just look at the bonds broken, we have a carbon-carbon triple bond and a couple of hydrogen-hydrogen bonds.

Let's also just write down how many of each we have because we'll need that for our calculation. So we have one carbon-carbon triple bond and we have two hydrogen-hydrogen bonds that are broken. Then, we can also look up their bond enthalpies, which are in kilojoules per mole. Bond enthalpies you can typically look up in your textbook or online, and they usually come in a table of bond enthalpies. The units can be kilojoules per mole.

Sometimes you'll also see calories or kilocalories per mole. I already looked up these bond enthalpy values. So, carbon-carbon triple bonds have a bond enthalpy of 835 kilojoules per mole, and hydrogen-hydrogen bonds have a bond enthalpy of—sorry—436 kilojoules per mole.

Next, if we look at the bonds that are broken, we have a carbon-carbon single bond, and we have one of those bonds forming. The bond enthalpy for that, which is also in terms of kilojoules per mole, is 346. Last but not least, we have the carbon-hydrogen bonds that we're forming, and we have four of those. Each of those, the bond enthalpy is 413 kilojoules per mole.

Now we can take all of this information and put it together to calculate ΔH of reaction. So, ΔH of reaction, if we're thinking about it in terms of bonds made and broken, it's the total energy change during a reaction. It's just the energy it takes to break all of our bonds in the reactants—so to break this carbon-carbon triple bond and the two hydrogen-hydrogen bonds—plus the energy it takes to make new product bonds.

We said earlier that you always have to add energy to break bonds, so bond enthalpy is always positive. We know this part of our calculation should always be a positive number. What that means is that it always releases energy to make new bonds. When energy is released, ΔH becomes more negative.

So, this number here, when we're talking about adding up the energy it takes to make new bonds, these should be negative numbers. Now let's plug in the values we have for bond enthalpy for all of these bonds that are made and broken in our reaction. Let's start with the bonds that are broken.

We have our carbon-carbon single bond that will require 835 kilojoules per mole, and we have only one of them. We also have to break two hydrogen-hydrogen bonds, so 2 times 436 kilojoules per mole, which is the bond enthalpy of that bond. So, that's all of the bonds we break. Now we have to add up the energy that's released when we make the new bonds.

We have this carbon-carbon single bond, so that is 346 kilojoules per mole, and that's negative because that energy is released. Then the last bond is the carbon-hydrogen bond—also negative because the energy is released, and we have four of them. Each of them will release 413 kilojoules per mole.

If we stick all of this into our calculator to get our final answer, what I got was that the ΔH of reaction for this hydrogenation reaction between propine and hydrogen gas is negative 291 kilojoules per mole. We can see that this overall reaction releases energy because ΔH is negative, so it's exothermic.

That's how you can use bond enthalpies to calculate ΔH of reaction.