Hess's law | Thermodynamics | AP Chemistry | Khan Academy

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. 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 were the products, and now we've made those products the reactants. What were the reactants over here for equation one have now become the products. The change in enthalpy for equation one is negative 1299.6 kilojoules per mole of reaction. Kilojoules per mole of reaction just means how the reaction is written in the balanced equation. Since we've reversed equation one, we also need to reverse the sign for delta H. So instead of this being a negative, we're going to go ahead and change this into a positive.

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.

To save some time, I've gone ahead and written out what we would get: 2 carbons plus 2 O2s goes to 2 CO2. The change in the enthalpy for the formation of one mole of CO2 was negative 393.5 kilojoules per mole of reaction, but now we're forming two moles of CO2. 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. 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 going to do anything to the change in the enthalpy, so it's going to stay negative 285.8 kilojoules per mole of reaction.

Next, we add up all of our reactants and products. We have 2 CO2 plus H2O plus 2 C plus 2 O2 plus H2 plus 1/2 O2, so those are all written down here for our reactants. Then for the products, let me just change colors here: we have C2H2 plus 5 O2 plus 2 CO2 plus H2O, and so those are written over here for the products.

Next, we see what we can cancel out. There's 2 CO2s on the left side and 2 CO2s on the right side, so those cancel out. There's one water on the left and one water on the right, and there's 2 O2s plus 1/2 O2, which is 2.5 O2s or five halves O2s, so the oxygens cancel out on both sides as well. We can see we're left with 2 carbons plus 1 hydrogen going to form 1 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 positive 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 negative 787.

For our third equation, we had negative 285.8, so that's negative 285.8. When we add everything together, we get positive 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 positive 226.8 kilojoules per mole of reaction.