okay continue from our example the other day that we had 50 milliliter of one molar naoh reacting with 50 milliliter of one molar hcl okay and we found out that this reaction actually released negative 2.87 kilojoule of energy heat to the environment which was the solution and the container okay yeah and once you've figured out the delta edge and we mentioned it a little bit at the end of the lecture um that is you can't just simply put the calculated delta h value together with the solute with the chemical equation like this so this is the balanced chemical equation for the neutralization b between sodium hydroxide and hydrochloric acid okay and you can't just say okay because i calculated it is negative 2.87 kilojoules so i'm gonna put it together and i'm gonna call this a thermochemical equation okay this is not gonna work why because this amount of heat okay as you can see it's produced by this much of acid and base okay look at your equation how do i know that you used and if you calculated it's point zero five zero mole of an aoh and same amount of hcl well if you look at this equation how do i know that you use this much of hcl and think about it what if you use one mole of naoh shouldn't you produce more energy than that and when i read this equation how do i read this chemical equation we talked about this when we first introduced chemical equations this really should be read as one mole of naoh reacting with one mole of hcl producing one mole of sodium chloride and one mole of water right and if i have one mole of each of these things it's definitely not going to only produce me this much heat point zero five zero mole can already give me this one mo is 20 times of that okay so how should i get this and that's why in our thermochemical equation let me write this down in thermal chemical equation okay we use delta h naught or delta h standard following the balanced chemical equation and this knot or this standard thing here this little circle here means that stoichio metric amount of reactant all reactance and products are involved in reaction so whatever number of moles of reactants and products are showing in that balanced chemical equation then you use that amount then the energy involved will be your standard delta x value okay so in this situation how should i find the delta h value okay so i point zero five zero mole of naoh i also use point zero five zero mole of hcl you have the volume you have a concentration we talked about solution chemistry already so you should be able to figure out the most of naoh and hccl used okay and you got to sink it like this for each negative 2.87 kilojour of energy it's coming from point zero five zero mole of n a wedge okay and for this reaction okay you need to use one mole of naoh okay and the result is negative 57.4 kilojoules and once you do this calculation you can you get the dot edge standard you can now put it together with the balanced chemical and now you have the thermochemical equation okay because this means if you react one mole of naoh with one mole of hcl and you produced one mole of sodium chloride and one more water you will have 57.4 kilojoules of heat or 57.4 kilojoules of energy being released from the reaction to the environment and for a long time okay this particular method using the calorimetry having the chemical reaction happening in a container and then measure containers temp solutions temperature and final initial initial and final temperature and doing calculation of delta h plus q equals zero this was the method for scientists okay to figure out the thermochemical equations of reactions that we know okay and what's the benefit of getting these thermochemical equations well let's take a look at our next example let's say i have a thermochemical equation 2 n2 plus o2 guess producing 2 m2o also guess now delta edge standard equal to positive one 163.14 kilojoules so that's my thermostat thermochemical equation and here my here is my question if 200 grams of n2o is produced from the reaction how much heat is absorbed or released okay so i have a thermochemical equation and i have a gram of my product okay i'm looking for how much heat how much energy is involved during the process well how should i handle this well obviously you have a numerical condition of 200 grams of n2o okay no and chances are we're going to probably have to start from there because that's kind of the numerical condition that's provided but how do i use this thermochemical equation okay now again no the thermochemical equation means and the energy following it means when stoichiometric amount of the reactants and products meaning this case two moles of n2 one mole of o2 producing two moles of n2o then this much energy is involved and in this case it's positive it's endothermic so it's absorbing energy from the environment okay so how should i use this thermochemical equation with the energy data if it's just the equation we already know how to use use it okay we've done plenty of stoichiome stoichiometry calculations before to know how do we use a balanced chemical equation okay but how do i use it with the thermochemical data okay what's the edge standard data well the answer is exact same way okay what do you mean well you're having grams of n2o how do we do stoichiometry before okay i need to turn this into most okay i don't need grams of n2o and i want this to be one mole of n2o so i can use the balanced chemical equation each one mole of n2o is calculate the molar mass it's 44.02 grams okay and then now you have mohs well the next thing you've got to have two modes of n2o on the bottom right okay if this is a stoichiometry you would put like two moles of n2 or one mole of oxygen on top and you get the moles of oxygen or n2 and then you can further calculate it into grams but this time remember these things are all together in one equation okay this means each two moles of n2o made 163.14 kilojoules of heat is being absorbed so why not put your 163.14 kilojoules right there just like you're using a molar ratio okay because well they showed up in the same equation okay now view the delta edge now if the delta h is positive then you can view the amount of energy as part of the reactant because your reaction has to absorb that energy to happen if the edge standard has a negative value you can view it as a product because that's how much heat is going to be produced when the reaction happened and if it's a reactant or product then you can always put them into a mole ratio type of setup okay to do stoichiometry calculations so cancel your mo of n2o now you got energy unit and that's exactly what you wanted and that calculation gave you positive 37.1 kilojoules okay awesome no not 36 this is way more than one mode so 371 sorry 371 kilojoules of heat and let's make a choice here this is positive that means absorbed during this reaction all right so this is an example of how we use a thermal chemical equation to calculate to do to make calculations involving energy during the process of a reaction okay now and you found that we use thermo chemical equation exactly the same way that we use balanced chemical equations before okay which is using the coefficients well now we are we're also using the delta h value as part of a ratio okay before it's molar ratio now it could be energy mole ratio okay and use that if i tell you you have a certain amount of reactant used a certain amount of product produced then you can figure out how much energy is involved in in the reaction and vice versa if i tell you okay this here's a thermochemical equation and i know this reaction happened and it absorbed released this much of energy can you tell me how much reactant is used or how much product is produced and you should be able to do that just like what we did here okay and this is just like what we did previously in stoichiometry all right now all right the next example well so we use we learned to use calorimetry to figure out the delta edge standard okay and we know why do we want the delta edge standard to do stoichiometry type of calculations now this next example okay is a little different you so all right so if two moles of sulfur reacting with three moles of oxygen producing two moles of sulfur trioxide the delta edge is negative 79 791.4 kilojoules and also one mole of sulfur reacting with one mole of oxygen producing one mole of sulfur dioxide okay the delta etch standard is negative 296.8 kilojoule now knowing those two things what is the delta h of 2so2 plus o2 producing two moles of so2 reacting with one mole of o2 gas producing two moles of so3 sulfur trioxide okay so we learned one way of getting dot edge of a chemical reaction and that's through stoichiometry okay and that one we need experimental data we need temperature measurement and heat measurement okay thermal properties of you know possible calorimeter excuse me no this question we're not getting any of that we want a delta edge okay but instead of giving me experimental data i was provided with two other thermal chemical equations and i mean they seems very much related okay see this whole thing is about sulfur oxygen sulfur trioxide sulfur dioxide and both the two provided thermo chemical equation and the thermal chemical equation that i wanted okay involve these similar things but they're definitely not the same reaction okay so how should i do this and before we solve this problem first of all let's make it very clear okay a question like this is very much solvable okay and it is solvable because the fact that i think we talked about this before delta edge is a state function a state function is that where that we can we only care about the differences between different states okay we don't have absolute values of them okay and being state function okay to figure out to figure out that edge we actually have now a rule that we can use to our advantage so before we do this calculation let me introduce this thermal chemical rule that we call the hess law okay heslogo says it's actually a pretty simple way to represent it and the overall dot edge off a reaction is equal to the sum of all and the sum of sorry dot h of all individual steps of the reaction okay what do you mean by this like you you kind of make this sound complicated okay well let's put it this way imagine that you're trying to um figure out okay the delta edge of a chemical reaction and let's say this chemical reaction just kind of don't happen okay like let's say this chemical reaction just don't happen very easily in a lab condition okay let's say why because well let's say this chemical reaction release a lot of heat okay and when you release a lot of heat in a very short time period say in a calorimeter you're turning the calorimeter into a bomb i can't do measurement i can't measure temperature my thermometer is going to be blown away okay and let's say i know this is my reactants potential energy-wise this is my reactants okay and this is my product okay i know there's a huge energy difference between the reactants and products and because this is reactants going down in potential energy to product so a lot of energy will be released right so but i don't know how much how can i figure this out well let's just put it this way if i can't directly do this experiment okay to do the measurements then i can maybe break this reaction down into little steps okay what do you mean i can find another reaction where i have a smaller energy difference let's say this is our product one okay from reactant to product one see this is a smaller energy drop okay and this can happen in let's say this can happen in our calorimeter i can measure this i can get the delta edge okay and then if i can also find a different way another step of the reaction let's say from product one to your original product okay now this is also a small smaller energy drop so we'll call this dot h2 okay so let's say we just found a different way to do this this is a one-step reaction which release a lot of energy very quickly cause the explosion so i can't carry out the experiment but these are two very smooth and slow chemical reactions which i could carry out easily by and marrying to measuring temperature and whatever in a calorimeter okay and now what well then i don't need to make this reaction happen in a calorimeter to get measurement to calculate delta edge okay instead i can measure this in a calorimeter i can then measure this in a calorimeter and because my original reactant and original product are still the same thing then i can very easily using use the hash law and say the delta h i want is actually the total of the two data edge of the individual steps because they gave me the same reactance initial point and end point so whether you go there from one step of reaction or two step of reaction don't matter eventually the energy difference need to be the same because the other edge is a state function okay so hopefully everybody is following me when i make this explanation okay and once we explained it like this you can see okay yeah this reaction kind of do look like steps what do you mean well think about it from sulfur and oxygen to sulfur dioxide okay that seems like one step and then from sulfur dioxide with oxygen to sulfur trioxide that looks like the next step but if you put them together what essentially did you do in this it's like you're putting sulfur and a whole bunch of oxygen together and you eventually got so3 right so they do have that relationship okay so this type of problem where you have two or more what do we call the known equations and you're looking for dotted edge of a unknown the unknown dead edge of another equation we call this type of examples the hassle problems okay so now how do i solve as well problems and let's see we actually have a standard way of solving hassle problems okay i don't want you to go the way of okay let me figure out which reactions are the steps and which reactions are which reaction is the overall reaction are we looking for the overall reaction dot h or are we looking for individual step dot h no doing that is too much trouble okay doing that is too confusing and if you paid attention okay to our class at this point you know how how much i like fixed method something that i can fall back on if i can stick to the rules okay during an exam i always have a better chance of success okay so that's going to be how we deal with hassle problems okay now and here we go let's shrink the page a little bit come on trying to create a little bit more space here so we can do our calculation so here's our way of handling has law problems we're always going to for focus on this equation the one that we want and we're going to call it the target equation okay and the known thermochemical equations okay we're going to call them known and we're going to label them and we're going to call them known equation 1. and known equation 2. so a has law problem always gave you some known equations and make you work for our target equation and the known equations could be one two or even more and the target equation of course is only the one that you wanted and the method we have is that we're always going to focus on the known equation and on the target equation and we're going to identify every single component okay of the target equation okay from the known equation okay find each target equation component from a known equation not where not only we're gonna find it okay we're also going to manipulate the known equation so that component that particular component match the target equation okay i don't need you to memorize this okay and actually if you if you're reading this right now you're probably a little confused on like exactly what are you trying to do okay here's what i'm trying to do okay i'm going to look at this first component of the target equation so2 okay and i'm gonna try to find it in one of the known equation and right there that's your so2 okay so so2 is in known equation number two but right now your so2 doesn't match the so2 in your target equation why what do you mean match well in the target equation your so2 is on the reactant side in the known equation this guy is on the product side so that part doesn't match also in the target equation okay i have two moles of so2 and in the known equation number two you only have one mole of so2 so number of moles don't match either so i'm going to do two things to my known equation number two so that the so2 match my target equation so2 number one step number so the first thing i'm going to do is i'm going to reverse target equation number two what do you mean flip it product to reactant and reactant to product okay you can do that well why not okay a chemical reaction could have a reverse reaction okay if your reactant molecules can hit each other and make product who's to say your product molecules won't just hit each other and break it break back down into your reactant so reverse this and now as you manipulate the known equation okay make sure that you also manipulate your delta h at the same time how do you how do i manipulate the delta h well the delta edge standard of the early known equation number two is sulfur reacting with oxygen producing sulfur dioxide and 296.8 kilojoules of energy will be released during that process well let me ask you this if now i reverse the reaction what happened to my delta edge okay and i think it's pretty logical for you to think for you to say well before the forward reaction okay making so2 release energy to the environment and now i didn't do anything else except for i'm breaking down so2 if you release this much of energy before when you reverse reaction shouldn't you be absorbing the same amount of energy now and if you're thinking like that i mean that's the only logical thought i can think of okay and if you do think like that you're absolutely right if that reaction has a negative dot h this one has a positive dot edge because you reversed it you didn't change most of stuff at all and the amount that doesn't change but since the direction of the reaction change the sign for your delta edge changes okay so i reversed two and i got so2 to the left side which kind of matched the so2 in the target equation but not exactly because i have two modes of so2 there now how do i get that well how do we turn this from one more to two more this is one mode dissociating in one mole of this dissociating into one mole of sulfur and one more o2 or what if i have two moles of so2 dissociating the next step i'm going to double this up 2s and 202 and you may go well this is not a good balanced chemical equation we're not talking about a good balanced chemical equation right now i know that i'm not having the smallest whole number ratio between them okay the point is i have to match the target equation the target equation need to so hell i'm getting 2. well if i multiply the equation by 2 what happened to my delta h well if one mole of this dissociating absorb this much heat how much heat is going to be absorbed when you when you're dissociating two moles that answer should be pretty straight right this need to be timed by two so positive 593.6 kilojoules so by doing these two steps i matched the first component of my target equation okay i'm done there moving on the next component is o2 now when we get to o2 we have a little bit of dilemma what do you mean well because i have o2 in known equation number one i also have o2 in knowing equation number two dang what do i do do i manipulate one or do i manipulate two my suggestion to you at this point is well i'm not gonna do either okay if a component of your target equation shows up in multiple known equations we'll leave it alone we'll try to deal with the next one okay and the next one is so3 did so3 show up in two different equations nope so3 is right here in known equation number one and oh we looked out here look i need two so3 on the right side guess what i have two excuse me i have two so3 on the right side so i don't need to do anything target equation number one nothing needed 2s 302 2so3 dot edge negative 791.4 kilojoule all right so what do i do now oh you tell me now i got both my equation number one and number two manipulated okay and i also got both my so2 and so3 in the right position i did everything i can do like remember o2 showed up in two different known equations we're not dealing with it okay but everything else is taken care of now i want you to add these two equations up left side to left side right side to right side just like what we did in balancing redox equations all right so what what do you see i have 2 so2 okay i have 2s but wait if i'm adding left to left right to right guess where else it has 2s here and here right both solids same exact same thing cancelled so 302 yes and then hold on that's 202 that's 302 oh i can cancel two and have one left okay that makes things easier one or two left and two through does this thing look familiar let's go back up what do we want what was our target equation 2 so2 plus o2 equal to so3 right didn't we just make the target equation you just did okay so what about that delta h because i don't really care about the target equation i just wanted to write delta h well how did you get the target equation you get the target equation by add these two equations up right what should your dot edge of the target equation equal to do add them up okay so add these two things up you get negative 197 thermal chemical equation for the target equation the delta edge standard value achieved so this is how we use has law okay to manipulate different unknown different known equations trying to match to the components of target equation and then eventually trying to combine the known equation to manipulate the known equation and to get the target equation doesn't matter how many known equations you have doesn't matter how many components your target equation have we're basically repeating the same things again and again okay along the way let's write down this and you know put this on your notes whatever done to the known equation due to its dot edge value what do you mean well remember that we reversed one of the known equation you reverse the sign okay remember we multiply one equation by two well multiply your delta h by two if in one situation during your chemical reaction you need to divide okay equation by 3 so be it divide your delta h by 3 as well okay so whatever done to the known equation due to its delta delta value okay and try to avoid components that showed up in multiple known equations try just to deal with target equation components that only showed up in known equation once and try to manipulate the knowing equation to match those components and eventually you're going to be able to add them up so this is how we use has law to solve problems and you will probably want to look for further practice on this well i'll try to put another example in on our next lecture but you definitely want to practice more on this particular topic we'll continue from here on our next lecture and that's it for today