Understanding Hess's Law and Cycles

In this video, I'm going to explain what Hess's law is, what a Hess cycle is, and I'll show you the easiest method for using Hess cycles to calculate an unknown enthalpy change, with three different examples. In order to understand Hess's law, you'll need to first understand what an enthalpy change is. The simplest way of explaining an enthalpy change is that it's the amount of heat energy released or taken in per mole of substance during some kind of physical or chemical change. If heat energy is given out overall during a change, the temperature of the surroundings increases and the process is described as exothermic. If heat energy is taken in overall during a change, the temperature of the surroundings decreases and the process is described as endothermic. Examples of enthalpy changes include the enthalpy change of formation, which is the change in energy when one mole of a compound is formed from its elements in their standard states, the enthalpy change of combustion, which is the change in energy when one mole of a substance is burnt completely in oxygen, and the bond enthalpy, which is the change in energy when one mole of gaseous covalent bonds is broken. Now that we know what an enthalpy change is, we can take a look at Hess's law. Hess's law states that the enthalpy change during a chemical change is independent of the steps taken. In other words, as long as you start at the same reactants and end at the same products, it makes no difference how you get there and what intermediates you go by, the overall enthalpy change will be the same. A Hess cycle is just a visual representation of Hess's law, and we can use Hess cycles to work out unknown enthalpy changes. If we know these two enthalpy changes, we can calculate this one, because remember it doesn't matter how we get there as long as we start in the same place and finish in the same place overall. Hess cycles most commonly use either formation enthalpies, combustion enthalpies, or bond enthalpies. Let me show you how they work. The first type of Hess cycle I'm going to show you is one that uses formation enthalpies. Remember the definition of the formation enthalpy is the energy change when one mole of a compound is formed from its elements in their standard states. Let's take a look at this reaction showing the complete combustion of methane in oxygen. You might be asked to calculate the enthalpy of combustion for methane. We can do this by setting up a Hess cycle with given enthalpies of formation. Try not to get confused here. We're being asked to calculate the enthalpy of combustion for methane, but we've been given enthalpies of formation, so we'll set this Hess cycle up using formation enthalpies. To do this, I'm going to put a box underneath the reaction which contains the elements used to form all of my reactants and all of my products. I could balance the elements in here but it's really not important. I'm just going to use this box to visually represent the elements which form the compounds in the reaction. Now, since the formation enthalpy describes the formation of a compound from its elements, I'm going to draw arrows going from my box of elements up to my reactants and up to my products. The arrow on the left represents the total formation enthalpies of all of my reactants, and the arrow on the right represents the total formation enthalpies of all of my products. So on this arrow I'm going to put the formation enthalpies for the reactants. I'll need the enthalpy of formation for methane, which is minus 75 kilojoules per mole. Notice that there's no formation enthalpy for oxygen. This is because oxygen is just an element. so it has no formation enthalpy. On the other arrow, I'm going to put the formation enthalpies for the products. The enthalpy of formation for carbon dioxide is minus 394 kilojoules per mole, so I'll add this, and the enthalpy of formation for water is minus 286 kilojoules per mole. But remember, the enthalpy of formation is the energy change for the formation of one mole of a compound. But in this equation, There are two moles of water, so I'll need to double the formation enthalpy for water when I add it to this arrow. Here's where it can get confusing. At school, you were probably taught that to find an alternative route, You should start at the reactants, go down to the box at the bottom, and then up the other side. You then need to find the total enthalpy change for both roots and equate them in order to find the missing enthalpy change. The problem with this method is that since you're going against the direction of this arrow, you'll need to flip the signs of the enthalpies attached to this arrow, and this is where a lot of mistakes can be made, especially when there are lots of enthalpies involved here. I've always found that a much easier method for solving Hess cycles is to find two roots which both follow the direction of your arrows. They don't have to start at the reactants and end at the products. For this Hess cycle, I can start both roots at my box underneath the reaction. One root will go from the box up to the reactants and across to the products. The other root will go from the box directly to the products. This still obeys Hess's law because we're starting and ending at the same place, so the overall enthalpy change will be the same for both roots. And if you follow both of these roots, neither one goes against the direction of any arrow, so I can just use the enthalpies as they're written, no need to change any of the signs at all. So for this root, it'll be the enthalpy of formation for methane. minus 75 kilojoules per mole plus the enthalpy of combustion for methane which we're trying to calculate. According to Hess's law that will all be equal to the other root which is the enthalpy of formation for carbon dioxide minus 394 kilojoules per mole plus two times the enthalpy of formation for water or two times minus 286 kilojoules per mole since there are two moles of water. Now, with a bit of rearranging, the enthalpy change of combustion for methane will be equal to this, which comes out to minus 891 kilojoules per mole. So when you look for two roots which both follow the arrows, Hess cycles are much easier to solve. Now, I'll show you an example of a Hess cycle that uses combustion enthalpies. Remember, the definition of the enthalpy of combustion is the energy change when one mole of a substance is burned completely in oxygen. Let's take a look at this reaction, showing the formation of benzene. We could be asked to calculate the enthalpy change of formation for benzene with given enthalpies of combustion. Again, don't get confused here. We're being asked to calculate the enthalpy of formation for benzene, but we've been given enthalpies of combustion. So we'll set this Hess cycle up using combustion enthalpies. To do this, I'll put a box underneath the reaction and in it I'll put the combustion products for everything in this reaction. Again, I could balance what's in the box, but it's really not necessary. I'm just going to use this box to visually represent the products which form from combusting the substances in our reaction. Since the combustion enthalpy describes the formation of these combustion products. I'm going to draw arrows from my reactants down to the box and from my products down to the box. This sets up my Hess cycle. On this arrow, I'm going to put the enthalpies of combustion for my reactants. Remember, the enthalpy of combustion describes the combustion of one mole of a substance. Since there are six moles of carbon in this reaction, I'll need to multiply the enthalpy of combustion for carbon by six. Similarly, since there are 3 moles of hydrogen, I'll need to multiply the enthalpy of combustion for hydrogen by 3. On this arrow, I just need the enthalpy of combustion for benzene. Now, just like before, I just need to find two routes which follow the direction of the arrows. One route will go from the reactants directly down to the box, and the other route will go from the reactants to the products. and then down to the box. So, according to Hess's law, these two roots are equal, because we're starting and ending in the same place. Again, since I'm following the direction of the arrows for both roots here, I can just use the enthalpies as they're written, no need to change the signs. With a bit of rearranging, we can calculate the enthalpy of formation for benzene, which comes out to plus 45 kilojoules per mole. The final example I'm going to show you is a Hess cycle that uses bond enthalpies. Remember, the definition of the bond enthalpy is the energy change when one mole of gaseous covalent bonds is broken. Let's take a look at this reaction, showing the complete combustion of ethanol. We could be asked to calculate the enthalpy of combustion for ethanol using bond enthalpies. Remember, we set the Hess cycle up using whatever types of enthalpies we've been given, so we'll set this one up using bond enthalpies. To do this, underneath I'll put a box of all of the gaseous atoms that would be formed if we broke all of the bonds in our reaction. Since the bond enthalpy essentially describes the formation of these atoms, I'm going to draw arrows going from my reactants down to the box, and from my products down to the box. Now, just like before, I can start attaching the relevant enthalpies to the relevant arrows. You'll notice that the enthalpy of vaporization for ethanol has also been given here. The definition of the enthalpy of vaporization is the energy change when one mole of a liquid is boiled to form a gas. The reason this has been given is that the bond enthalpy is always quoted for gaseous molecules, but in our reaction, ethanol is in the liquid state. This means that we first have to vaporize our ethanol before we can break the bond. bonds. So both of these enthalpies will be required for ethanol in our Hess cycle. So on this arrow I'll put the enthalpy of vaporization for ethanol, as well as the relevant bond enthalpies for the reactants. If we look at the bonds in our reactant molecules, there is a single C-C bond, a single C-O bond, five C-H bonds, and a single O-H bond in ethanol. Oxygen contains an oxygen-oxygen double bond, but there are three moles of oxygen in this reaction, so we'll need to multiply this by three. Now on this arrow I'll put the relevant bond enthalpies for the products. If we look at the bonds in our product molecules, there are two carbon-oxygen double bonds in a molecule of carbon dioxide, but there are two moles of carbon dioxide. so that's a total of 4 moles of carbon-oxygen double bonds. Each water molecule contains two OH bonds, and there are 3 moles of water, so that's a total of 6 moles of OH bonds. Now it's just a case of finding my two roots by following the arrows. In this case, one root goes from my reactants directly down to the box, and the other route goes from my reactants across to the products and then down to the box. Just like before, I can equate these two routes thanks to Hess's law, and I don't need to worry about changing any of the signs. Now with a bit of rearranging, we can calculate the enthalpy of combustion for ethanol, which comes out to minus 1015 kilojoules per mole. So in summary, when setting up your Hess cycles, always try and find two roots which follow the direction of your arrows. That way you won't have to worry about changing any of the signs of the enthalpy changes and it will make everything a lot simpler when it comes to your calculations. I hope you enjoyed this video, please subscribe for more and let me know in the comments if you have any questions.

If heat energy is given out overall during a change, the temperature of the surroundings increases and the process is described as exothermic. If heat energy is taken in overall during a change, the temperature of the surroundings decreases and the process is described as endothermic. Examples of enthalpy changes include the enthalpy change of formation, which is the change in energy when one mole of a compound is formed from its elements in their standard states, the enthalpy change of combustion, which is the change in energy when one mole of a substance is burnt completely in oxygen, and the bond enthalpy, which is the change in energy when one mole of gaseous covalent bonds is broken. Now that we know what an enthalpy change is, we can take a look at Hess's law.

Hess's law states that the enthalpy change during a chemical change is independent of the steps taken. In other words, as long as you start at the same reactants and end at the same products, it makes no difference how you get there and what intermediates you go by, the overall enthalpy change will be the same. A Hess cycle is just a visual representation of Hess's law, and we can use Hess cycles to work out unknown enthalpy changes.

If we know these two enthalpy changes, we can calculate this one, because remember it doesn't matter how we get there as long as we start in the same place and finish in the same place overall. Hess cycles most commonly use either formation enthalpies, combustion enthalpies, or bond enthalpies. Let me show you how they work. The first type of Hess cycle I'm going to show you is one that uses formation enthalpies.

Remember the definition of the formation enthalpy is the energy change when one mole of a compound is formed from its elements in their standard states. Let's take a look at this reaction showing the complete combustion of methane in oxygen. You might be asked to calculate the enthalpy of combustion for methane.

We can do this by setting up a Hess cycle with given enthalpies of formation. Try not to get confused here. We're being asked to calculate the enthalpy of combustion for methane, but we've been given enthalpies of formation, so we'll set this Hess cycle up using formation enthalpies.

To do this, I'm going to put a box underneath the reaction which contains the elements used to form all of my reactants and all of my products. I could balance the elements in here but it's really not important. I'm just going to use this box to visually represent the elements which form the compounds in the reaction. Now, since the formation enthalpy describes the formation of a compound from its elements, I'm going to draw arrows going from my box of elements up to my reactants and up to my products. The arrow on the left represents the total formation enthalpies of all of my reactants, and the arrow on the right represents the total formation enthalpies of all of my products.

So on this arrow I'm going to put the formation enthalpies for the reactants. I'll need the enthalpy of formation for methane, which is minus 75 kilojoules per mole. Notice that there's no formation enthalpy for oxygen. This is because oxygen is just an element.

so it has no formation enthalpy. On the other arrow, I'm going to put the formation enthalpies for the products. The enthalpy of formation for carbon dioxide is minus 394 kilojoules per mole, so I'll add this, and the enthalpy of formation for water is minus 286 kilojoules per mole. But remember, the enthalpy of formation is the energy change for the formation of one mole of a compound. But in this equation, There are two moles of water, so I'll need to double the formation enthalpy for water when I add it to this arrow.

Here's where it can get confusing. At school, you were probably taught that to find an alternative route, You should start at the reactants, go down to the box at the bottom, and then up the other side. You then need to find the total enthalpy change for both roots and equate them in order to find the missing enthalpy change. The problem with this method is that since you're going against the direction of this arrow, you'll need to flip the signs of the enthalpies attached to this arrow, and this is where a lot of mistakes can be made, especially when there are lots of enthalpies involved here.

I've always found that a much easier method for solving Hess cycles is to find two roots which both follow the direction of your arrows. They don't have to start at the reactants and end at the products. For this Hess cycle, I can start both roots at my box underneath the reaction.

One root will go from the box up to the reactants and across to the products. The other root will go from the box directly to the products. This still obeys Hess's law because we're starting and ending at the same place, so the overall enthalpy change will be the same for both roots. And if you follow both of these roots, neither one goes against the direction of any arrow, so I can just use the enthalpies as they're written, no need to change any of the signs at all. So for this root, it'll be the enthalpy of formation for methane.

minus 75 kilojoules per mole plus the enthalpy of combustion for methane which we're trying to calculate. According to Hess's law that will all be equal to the other root which is the enthalpy of formation for carbon dioxide minus 394 kilojoules per mole plus two times the enthalpy of formation for water or two times minus 286 kilojoules per mole since there are two moles of water. Now, with a bit of rearranging, the enthalpy change of combustion for methane will be equal to this, which comes out to minus 891 kilojoules per mole.

So when you look for two roots which both follow the arrows, Hess cycles are much easier to solve. Now, I'll show you an example of a Hess cycle that uses combustion enthalpies. Remember, the definition of the enthalpy of combustion is the energy change when one mole of a substance is burned completely in oxygen.

Let's take a look at this reaction, showing the formation of benzene. We could be asked to calculate the enthalpy change of formation for benzene with given enthalpies of combustion. Again, don't get confused here. We're being asked to calculate the enthalpy of formation for benzene, but we've been given enthalpies of combustion.

So we'll set this Hess cycle up using combustion enthalpies. To do this, I'll put a box underneath the reaction and in it I'll put the combustion products for everything in this reaction. Again, I could balance what's in the box, but it's really not necessary.

I'm just going to use this box to visually represent the products which form from combusting the substances in our reaction. Since the combustion enthalpy describes the formation of these combustion products. I'm going to draw arrows from my reactants down to the box and from my products down to the box. This sets up my Hess cycle.

On this arrow, I'm going to put the enthalpies of combustion for my reactants. Remember, the enthalpy of combustion describes the combustion of one mole of a substance. Since there are six moles of carbon in this reaction, I'll need to multiply the enthalpy of combustion for carbon by six. Similarly, since there are 3 moles of hydrogen, I'll need to multiply the enthalpy of combustion for hydrogen by 3. On this arrow, I just need the enthalpy of combustion for benzene. Now, just like before, I just need to find two routes which follow the direction of the arrows.

One route will go from the reactants directly down to the box, and the other route will go from the reactants to the products. and then down to the box. So, according to Hess's law, these two roots are equal, because we're starting and ending in the same place. Again, since I'm following the direction of the arrows for both roots here, I can just use the enthalpies as they're written, no need to change the signs. With a bit of rearranging, we can calculate the enthalpy of formation for benzene, which comes out to plus 45 kilojoules per mole.

The final example I'm going to show you is a Hess cycle that uses bond enthalpies. Remember, the definition of the bond enthalpy is the energy change when one mole of gaseous covalent bonds is broken. Let's take a look at this reaction, showing the complete combustion of ethanol. We could be asked to calculate the enthalpy of combustion for ethanol using bond enthalpies.

Remember, we set the Hess cycle up using whatever types of enthalpies we've been given, so we'll set this one up using bond enthalpies. To do this, underneath I'll put a box of all of the gaseous atoms that would be formed if we broke all of the bonds in our reaction. Since the bond enthalpy essentially describes the formation of these atoms, I'm going to draw arrows going from my reactants down to the box, and from my products down to the box.

Now, just like before, I can start attaching the relevant enthalpies to the relevant arrows. You'll notice that the enthalpy of vaporization for ethanol has also been given here. The definition of the enthalpy of vaporization is the energy change when one mole of a liquid is boiled to form a gas.

The reason this has been given is that the bond enthalpy is always quoted for gaseous molecules, but in our reaction, ethanol is in the liquid state. This means that we first have to vaporize our ethanol before we can break the bond. bonds.

So both of these enthalpies will be required for ethanol in our Hess cycle. So on this arrow I'll put the enthalpy of vaporization for ethanol, as well as the relevant bond enthalpies for the reactants. If we look at the bonds in our reactant molecules, there is a single C-C bond, a single C-O bond, five C-H bonds, and a single O-H bond in ethanol.

Oxygen contains an oxygen-oxygen double bond, but there are three moles of oxygen in this reaction, so we'll need to multiply this by three. Now on this arrow I'll put the relevant bond enthalpies for the products. If we look at the bonds in our product molecules, there are two carbon-oxygen double bonds in a molecule of carbon dioxide, but there are two moles of carbon dioxide.

so that's a total of 4 moles of carbon-oxygen double bonds. Each water molecule contains two OH bonds, and there are 3 moles of water, so that's a total of 6 moles of OH bonds. Now it's just a case of finding my two roots by following the arrows. In this case, one root goes from my reactants directly down to the box, and the other route goes from my reactants across to the products and then down to the box.

Just like before, I can equate these two routes thanks to Hess's law, and I don't need to worry about changing any of the signs. Now with a bit of rearranging, we can calculate the enthalpy of combustion for ethanol, which comes out to minus 1015 kilojoules per mole. So in summary, when setting up your Hess cycles, always try and find two roots which follow the direction of your arrows. That way you won't have to worry about changing any of the signs of the enthalpy changes and it will make everything a lot simpler when it comes to your calculations.

I hope you enjoyed this video, please subscribe for more and let me know in the comments if you have any questions.

Transcript for:Understanding Hess's Law and Cycles

Transcript for:
Understanding Hess's Law and Cycles