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Worked example: Using Hess's law to calculate enthalpy of reaction

AP.Chem:
ENE‑3 (EU)
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ENE‑3.C (LO)
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ENE‑3.C.1 (EK)
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ENE‑3.D (LO)

Video transcript

this problem is from chapter 5 of the Kotz treichel and Townsend chemistry and chemical reactivity textbook so they tell us suppose you want to know the enthalpy change so the change in total energy for the formation of methane ch4 from solid carbon as graphite that's right there and hydrogen gas so we want to figure out the enthalpy change of this reaction how do we get methane how much energy is absorbed or released when methane is formed from the reaction of solid carbon is graphite and hydrogen gas so they tell us the enthalpy change for this reaction cannot be measured in the laboratory because the reaction is very slow so normally if you could measure it you would have this reaction happening and you would kind of see how much heat or what's the temperature change of the surrounding solution maybe this is happening so slow that it's very hard to measure that temperature change or you can't do it in any meaningful way we can however measure enthalpy changes for the combustion of carbon hydrogen and methane so they're giving us the enthalpy changes for these combustion combustion reactions as combustion of carbon combustion of hydrogen combustion of methane and they say use this information to calculate the change in enthalpy for the formation of methane from its elements so anytime you see this kind of situation where they're giving you the enthalpies for a bunch of reactions and they say hey we don't know the enthalpy for some other reaction and that other reaction seems to be made up of similar things your brain should immediately say hey maybe this is a Hess's law problem Hess's Hess's law and all Hess's law says is that if a reaction is the sum of two or more other reactions then the change in enthalpy of this reaction is going to be the sum of the change in enthalpy of those reactions now when we look at this this and this tends to be the confusing part is how can you construct this reaction out of these reactions over here and what I like to do is just start with the end product so I like to start with the end product which is methane in a gaseous form and when we look at all of these equations over here we have the combustion of methane so this actually involves methane so let's start with this but this one involves methane as a reactant not a product but what we can do is just flip this arrow and write it as methane as a product so if we just write this reaction we flip it so now we have carbon dioxide gas carbon dioxide gas let me write it down here carbon dioxide gas plus plus I'll do this in another color plus two plus two waters plus two if we're thinking of these as moles or two molecules of water you could even say two molecules of water in its liquid state that can I guess you can say that you know this would not happen spontaneously because would require an energy but if we just put this in the reverse direction if you go in this direction you're going to get two waters I'll do that in a different order two oxygens I should say I'll do that in this pink color so two two oxygens plus and that's in its gaseous state plus a gaseous methane plus a gaseous methane so plus a gaseous methane ch4 ch4 in a gaseous state and all I did is I wrote this this third equation but I wrote it in reverse order I'm going from the reactants to the products so if I'm going from the rear when you go from the products to the reactants it will release 890 point three kilojoules per moles of the reaction going on but if you go the other way it will need 890 kilojoules so the Delta H here I'll do this in a neutral color so the Delta H the Delta H of this reaction right here is going to be the reverse of this so it's 800 it's positive 890 point three kilojoules kilojoules per mole of the reaction all I did is I reversed the order of this reaction right there the good thing about this is I now have something that least ends up with what we eventually want to end up with this is this is where we want to get this is where we want to get eventually now how if we want to get there eventually we need at some point have some carbon dioxide we have to have at some point some water to deal with so how can we get carbon dioxide and how can we get water well these two reactions right here this combustion reaction gives us carbon dioxide this combustion reaction gives us water so we could just rewrite those let me just rewrite them over here and I'll I will I will let me use some colors so if I start with graphite carbon in graphite form so carbon in its graphite form plus I already have a color for oxygen plus oxygen plus oxygen and it's gaseous state it will produce carbon dioxide in its gaseous form it will produce carbon that's a different shade of green it will produce carbon dioxide in its gaseous form and this reaction so when you take the the enthalpy of the carbon dioxide and from that you subtract the enthalpy of these reactants you get a negative number which means this had a lower enthalpy which means energy was released because there's now less energy in this system right here so this this is essentially how much is released but our change in enthalpy here our change in enthalpy of this reaction right here that's reaction 1 I'll just rewrite it - 393 point 5 kilojoules kilojoules kilojoules per mole for or per mole of the reaction occurring so the reaction occurs on mole times it would this would be the amount of energy that's essentially released this is our change in enthalpy so if this happened we'll get our carbon dioxide now we also have and so we would release this much energy and we'd have this product to deal with but we also now need our water so then this reaction right here gives us our water the combustion of hydrogen so we have and I had I haven't done hydrogen yet so let me do hydrogen in a new color that's not a new color let me do blue so right here you have hydrogen gas I'm just rewriting that reaction hydrogen glass plus 1/2 o to pink is my color for oxygen 1/2 1/2 o - gas will yield will give us some water will give us h2o will give us some liquid water we'll give us some liquid water now before before I just write this number down let's think about whether we have everything we need to make this reaction occur because this gets us to our final product this gets us to the gaseous methane we need a mole or we can even say a molecule of carbon dioxide and this reaction gives us exactly one molecule of carbon dioxide so that's a check and we need two molecules of water now this reaction only gives us one molecule of water so let's multiply both sides of the equation to get two molecules of water so this is a two we multiply this this by two so this essentially just disappears you multiply one half by two you just get a one there and then you put a two over here so I just multiplied this second equation by two so I just multiplied this this becomes a 1 this becomes a 2 and if you're doing it twice as much of it because we multiplied it by two the Delta H now the change in enthalpy of the reaction is now going to be twice this so let's get the calculator out it's going to be it's now going to be negative two eighty five point eight times two because we just multiplied the whole reaction times two so negative five hundred seventy one point six so it's negative five hundred and seventy one point six kilojoules per mole of the reaction now let's see if the combination if the sum of these reactions actually is this reaction up here actually is this reaction up here and to do that actually let me just copy and paste this top one here because that's kind of the order that we're going to go in but you don't have to but it just makes it hopefully a little bit easier to understand so let me just copy and paste this actually I could cut and paste it cut and then let me paste it down here that first one and let's see now what's going to happen let's see what's going to happen and to see whether the sum of these reactions really does end up being this top reaction right here let's see if we can cancel out reactants and products or let's see what would happen so this produces carbon dioxide but then this mole or this molecule of carbon dioxide is then used up in this last reaction so this produces it this uses it so those cancel out me do it in the same color so it's in the screen this reaction produces it this reaction uses it now this reaction right here produces the two molecules of water and now this reaction down here I want to do that same color there's two molecules of water now this reaction down here uses those two molecules of water now these this reaction right here it requires one molecule of molecular oxygen this one requires another molecule of molecular oxygen so these two combined these two combined are two molecules of molecular oxygen so those are the reactants and in the end those end up as the products of this last reaction so those actually they go into the system and then they leave out the system or out of this the sum of reactions unchanged so they cancel out with each other so we could say that and that we cancel out and so what are we left with what are we left with in the reaction well we have some solid carbon as graphite plus two moles or two molecules of hydrogen molecular hydrogen yielding all we have left on the product side is some methane so it is true that the sum of these reactions is exactly what we want all we have left on the product side is the graphite the solid graphite plus the molecular hydrogen plus the gaseous hydrogen do it in that color plus two hydrogen gas and all we have left on the product side all we have left on the product side is the methane all we have left is the methane in the gaseous form so it is true that the sum of these reactions remember we have to flip this reaction around and change its sign and we have to multiply this reaction by two so that the sum of these becomes this reaction that we really care about so since this is the sum of these reactions it's change in enthalpy its change in enthalpy its change in enthalpy of this reaction is going to be the sum of these right here that is Hess's law so this is the fun part so we just add up these values right here so we have negative 393 point now that's not what I wanted to do let me just clear it so I have negative three 93.5 so that step is exothermic and then we have minus five seventy one point six that is also exothermic those were both combustion reactions which are as we know very exothermic and then we have the endothermic step the reverse of that last combustion reaction so plus 890 point three plus 890 point three gives us negative seventy four point eight it gives us negative seventy four point eight kilojoules for every mole of the reaction occurring or the reaction occurs a mole time so there you go we figured out the change in enthalpy and it is reasonably it is reasonably exothermic nowhere near as exothermic as these combustion reactions right here but it is going to release it is going to release energy and we're done
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