in this video i'm going to explain what hesse's law is what a hes cycle is and i'll show you the easiest method for using hair 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 hesse cycle is just a visual representation of hess's law and we can use his 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 hair 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 -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 of the products for this hair 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 routes 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 hair cycles are much easier to solve now i'll show you an example of a hair 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 hair 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 hair 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 four carbon by six similarly since there are three moles of hydrogen i'll need to multiply the enthalpy of combustion for hydrogen by three on this arrow i just need the enthalpy of combustion for benzene now just like before i just need to find two roots which follow the direction of the arrows one roots will go from the reactants directly down to the box and the other roots 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 bonds so both of these enthalpies will be required for ethanol in our hair 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 cc bond a single co bond five ch bonds and a single oh 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 four moles of carbon oxygen double bonds each water molecule contains two oh bonds and there are three moles of water so that's a total of six moles of o-h 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 root goes from my reactants across to the products and then down to the box just like before i can equate these two roots 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 hair 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