Worked example: Using bond enthalpies to calculate enthalpy of reaction | Khan Academy

• [Educator] Bond enthalpies can be used to estimate the standard change in enthalpy for a chemical reaction. Let's use bond enthalpies to estimate the enthalpy of combustion of ethanol.

Looking at our balanced equation, we have one mole of ethanol reacting with three moles of oxygen gas to produce two moles of carbon dioxide and three moles of water in the gaseous state. To find the standard change in enthalpy for this chemical reaction, we need to sum the bond enthalpies of the bonds that are broken. From that, we subtract the sum of the bond enthalpies of the bonds that are formed in this chemical reaction.

To figure out which bonds are broken and which bonds are formed, it's helpful to look at the dot structures for our molecules. So let's start with the ethanol molecule. We're gonna approach this problem first like we're breaking all of the bonds in these molecules. And we're also not gonna worry about units until the end, just to save some space on the screen.

So looking at the ethanol molecule, we would need to break a carbon-carbon bond. So let's go ahead and write this down here. Right now, we're summing the bond enthalpies of the bonds that are broken. So we have one carbon-carbon bond. So let's write in here the bond enthalpy for a carbon-carbon bond.

Next, we have five carbon-hydrogen bonds that we need to break. So to this, we're going to write in here a five, and then the bond enthalpy of a carbon-hydrogen bond. Next, we have to break a carbon-oxygen single bond. So we write a one, and then the bond enthalpy for a carbon-oxygen single bond.

And then for this ethanol molecule, we also have an oxygen-hydrogen single bond. So we'll write in here a one, and the bond enthalpy for an oxygen-hydrogen single bond. We saw in the balanced equation that one mole of ethanol reacts with three moles of oxygen gas.

So to represent the three moles of oxygen gas, I've drawn in here three molecules of O2. And we can see in each molecule of O2, there's an oxygen-oxygen double bond. So we would need to break three oxygen-oxygen double bonds. So to this, we're going to add a three times the bond enthalpy of an oxygen-oxygen double bond.

And this now gives us the sum of the bond enthalpies for all the bonds that need to be broken. It takes energy to break a bond. So the summation of the bond enthalpies of the bonds that are broken is going to be a positive value.

And since it takes energy to break bonds, energy is given off when bonds form. So next, we're gonna sum the bond enthalpies of the bonds that are formed. And notice we have this negative sign in here because this energy is given off. So we're gonna write a minus sign in here, and then we're gonna put some brackets because next we're going to sum the bond enthalpies of the bonds that are formed.

In our balanced equation, we formed two moles of carbon dioxide. So to represent those two moles, I've drawn in here two molecules of CO2. And we can see that in each molecule of CO2, we're going to form two carbon-oxygen double bonds. So that's a total of four carbon-oxygen double bonds.

So down here, we're going to write a four times the bond enthalpy of a carbon-oxygen double bond. We also formed three moles of H2O. And in each molecule of water that's drawn here, we form two oxygen-hydrogen single bonds. And since we have three moles, we have a total of six oxygen-hydrogen single bonds.

So to this, we're going to add six times the bond enthalpy of an oxygen-hydrogen single bond. The next step is to look up the bond enthalpies of all of these different bonds. For example, the bond enthalpy for a carbon-carbon single bond is about 348 kilojoules per mole.

You might see a different value if you look in a different textbook. However, we're gonna go with 348 kilojoules per mole for our calculation. And we're gonna multiply this by one mole of carbon-carbon single bonds. Next, we look up the bond enthalpy for our carbon-hydrogen single bond.

And that's about 413 kilojoules per mole of carbon-hydrogen bonds. And we're multiplying this by five. And we continue with everything else for the summation of the bond enthalpies of the bonds broken. When we do this, we get positive 4,719 kilojoules.

Next, we do the same thing for the bond enthalpies of the bonds that are formed. So the bond enthalpy for our carbon-oxygen double bond is 799 kilojoules per mole, and we multiply that by four. The bond enthalpy for an oxygen-hydrogen single bond is 463 kilojoules per mole, and we multiply that by six.

When we add these together, we get 5,974. So for the final standard change in enthalpy for our chemical reaction, it's positive 4,719 minus 5,974, which gives us negative 1,255 kilojoules. Notice that we got a negative value for the change in enthalpy.

And that means the combustion of ethanol is an exothermic reaction. And 1,255 kilojoules of energy are given off for the combustion of one mole of ethanol. Also notice that the sum of the bond enthalpies of the bonds formed, which is 5,974, is greater than the sum of the bond enthalpies of the bonds broken, which is 4,719.

And since we're subtracting a larger number from a smaller number, we get that negative sign for the change in enthalpy. If the sum of the bond enthalpies of the bonds that are broken, if this number is larger than the sum of the bond enthalpies of the bonds that have formed, we would've gotten a positive value for the change in enthalpy.

And that would be true for an endothermic reaction. We did this problem, assuming that all of the bonds that we drew in our dot structures were broken and all of the bonds that we drew in the dot structures were formed. However, if we look closely to the dot structures or just look closely to what we wrote here, we show breaking one oxygen-hydrogen single bond over here, and we show the formation of six oxygen-hydrogen single bonds over here.

So we could have just canceled out one of those oxygen-hydrogen single bonds. So we could have canceled this out. And instead of showing a six here, we could have written a five times the bond enthalpy of an oxygen-hydrogen single bond. We still would have ended up with the same answer of negative 1,255 kilojoules.

So if you look at your dot structures, if you see a bond that's the same on the reactant side and the same on the product side, you don't have to show the breaking and forming of that bond. You can make the problem a little bit shorter, if you want to.

Finally, let's show how we get our units. So to get kilojoules as your final answer, if we go back up to here, we wrote a one times 348. The one is referring to breaking one mole of carbon-carbon single bonds. And the 348, of course, is the bond enthalpy for a carbon-carbon single bond.

So this was 348 kilojoules per one mole of carbon-carbon single bonds. When you multiply these two together, the moles of carbon-carbon single bonds cancel and this gives you 348 kilojoules. And so, that's how to end up with kilojoules as your final answer.

You also might see kilojoules per mole of reaction as the units for this. And kilojoules per mole of reaction means how the reaction is written. So for the combustion of one mole of ethanol, 1,255 kilojoules of energy are released.

To get kilojoules per mole of reaction as our units, the balanced equation had a one as the coefficient in front of ethanol. Therefore, you're breaking one mole of carbon-carbon single bonds per one mole of reaction. So we can use this conversion factor.

Now, when we multiply through the moles of carbon-carbon single bonds, cancel and this gives us 348 kilojoules per mole of reaction.