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Main content
Current time:0:00Total duration:5:44
AP.Chem:
ENE‑3 (EU)
,
ENE‑3.C (LO)
,
ENE‑3.C.1 (EK)
,
ENE‑3.D (LO)

Video transcript

- [Instructor] Hess's law states that the overall change in enthalpy for a chemical reaction is equal to the sum of the enthalpy changes for each step. And this is independent of the path taken. So it doesn't matter what set of reactions you use. If you add up those reactions and they equal the reaction that you're trying to find, you can also sum the enthalpies to find the enthalpy change for the reaction. As an example, let's say we're trying to find the change in enthalpy for the reaction of carbon with hydrogen gas to form C2H2, which is acetylene. We can calculate the change in enthalpy for the formation of acetylene using these three reactions below. Our approach will involve looking at these three reactions and comparing them to the original reaction to see if we need to change anything. For example, if we look at reaction one, there's one mole of acetylene on the left side of the equation. And if we compare that to the original reaction, there's one mole of acetylene on the right side of the equation. So we need to reverse equation one to make it look more like our original reaction. To save time, I've gone ahead and reversed equation one. So you can see, I did that down here. Looking at the original equation for equation one, here where the products and now we've made those products the reactants. And what were the reactants over here for equation one have now become the products. The change in enthalpy for equation one is -1,299.6 kilojoules per mole of reaction. Kilojoules per mole reaction just means how the reaction is written in the balanced equation. And since we reversed equation one, we also need to reverse the sign for Delta H. So instead of this being a negative, instead of this being a negative, we're gonna go ahead and change this into a positive. And also let's go ahead and cross out the first equation. So we don't get confused. Next, we look at equation two and we compare it to our original. For equation two there's one mole of solid carbon on the left side and looking at our original reaction, there's two moles of carbon on the left side. So to get equation two, to look like our original equation we need to multiply everything through, by a factor of two. So we're gonna multiply everything in equation two by a factor of two. To save some time, I have gone ahead and written out what we would get. We would get two carbons plus two O2s goes to 2CO2. The change in the enthalpy for the formation of one mole of CO2 was -393.5 kilojoules per mole of reaction. But now we're forming two moles of CO2. And since we multiplied the equation through by a factor of two, we also need to multiply the change in enthalpy by a factor of two as well. And also let's go ahead and cross out this first version here because now we're forming two moles of CO2. Next, we look at equation three and we can see there's one mole of hydrogen gas on the left side of the equation which matches the original reaction which also has one mole of hydrogen gas on the left side. So we don't need to do anything to equation three. And since we're not doing anything to the equation, we're also not gonna do anything to the change in the enthalpy. So it's gonna stay -285.8 kilojoules per mole of reaction. Next we add up all of our reactants and products. So we have two CO2 plus H2O plus 2C plus 2O2 plus H2 plus one half O2. So those are all written down here for our reactants. And then for the products, let me just change colors here. We have C2H2 plus 5O2 plus 2CO2 plus H2O. And so those are written over here for the products. Next we see what we can cancel out. There's 2CO2 on the left side and there's 2CO2 on the right side. So those cancel out. There's one water on the left and one water on the right. And there's 2O2s plus one half O2 which is 2.5O2s or five halves O2s. So the oxygen's cancel out on both sides as well. And we can see we're left with two carbons plus one hydrogen goes to form one C2H2 which is the same as our original equation. Since we were able to add up our equations and get the overall equation, according to Hess's law, we should also be able to add the changes in enthalpies for these steps to get the change in the enthalpy for the overall reaction. If we look at the changes in enthalpy for the individual steps, we had +1299.6 for the first equation. And so that's up here. For the second equation we had negative 393.5 times two, which is -787. And for our third equation, we had -285.8. So that's -285.8. When we add everything together we get +226.8 kilojoules per mole of reaction. So for the formation of one mole of acetylene from two moles of carbon and one mole of hydrogen the change in enthalpy for this reaction is equal to +226.8 kilojoules per mole of reaction.
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