hello bobcats in this video we will be discussing thermochemical equations and stoichiometry so let's discuss what a thermochemical equation looks like so thermo chemical equations there's two ways to write thermal chemical equations you can include the energy in the equation itself or you can write the equation and then write the delta h or the enthalpy change of the equation to the side let me show you what i mean so thermochemical equation in general what you can say is that we can say that if we have reactants going to products plus heat so if we put heat on the product side that is an exothermic reaction and we know that delta h will be a negative value and we can say that's exothermic now if we have reactants plus heat produces product as a thermochemical reaction the heat is now on the written on the reactant side and so the delta h will be a positive value and it'll be endothermic so if heat is written the heat value is written on the product side it's an exothermic reaction and delta h is a negative if the heat value is written on the reactant side delta h is positive and it's endothermic so for a couple examples that blank black pin is running out let me grab another black pen so for another for example if we take two moles of aluminum solid plus three half moles and i'll explain why we have a fraction here in just a minute i know even though we haven't had fractions before as coefficients so two moles three half moles produces one mole of al2o3 plus 1776 kilojoules per mole okay so that's energy per mole and if you'll notice i wrote this on the product side so this is an exothermic reaction and if we are including the value as part of the equation we don't write the negative sign here or we could write it as two moles of aluminum plus three half moles of oxygen gas will produce one mole of aluminum oxide that's supposed to be solid aluminum oxide solid and then to the side we can say delta h is equal to a negative 1776 kilojoules per mole of reaction okay so you'll notice that i either can include the value as part of the equation but we don't put the sign it depends on what side you put it if it's on the product side it's an it's exothermic if it's on reactant side it's endothermic this case is on the product side i can also write it as just the reaction and into the side right the delta h is equal to the negative 1776 kilojoules per mole now and this is understood as per mole of reaction so this um up here what we have to interpret this is is that one mole of reaction based on how it's written here is said to occur when two moles of aluminum reacts with 1.5 moles of oxygen to produce one mole of aluminum oxide and 1776 kilojoules of energy of heat now so this is written so that it works with these coefficients now let me explain why we have three halves here so if you wrote the reaction let's look at the reaction really so if i have aluminum solid plus oxygen gas to produce aluminum oxide solid we would balance it so that we would have um [Music] let's see we have two aluminums here one aluminum we have two oxygens three so i need to put a two here 2 times 6 2 times 3 is 6 so i need a 3 here and then a 4 here now that's the original balance but based on how this is written we're looking at this um 1776 kilojoules per mole of reaction based on one mole of product being produced so i have to therefore go through and divide each of these by two so i get one mole here so that it really ends up written as two moles of aluminum plus three half moles of oxygen produces one mole of aluminum oxide and then that would be um with a delta h value of a negative 176 kilojoules per mole based on how this is written okay and in this case per one mole of product produced now if i were to write the delta h up here we would for for the balanced reaction up here without dividing then i would have to double this and so double 1776. this would be a negative three five five two kilojoules per mole based on this reaction before you you divide by two okay but most reactions try to base it on either a mole of product produced or one mole of one of the reactants so in this case that's why we have the three halves because uh to get to one mole of something here we divide by two to get one mole of product now but really we just need to really pay attention and say that this value is based on these coefficients this value would be based on the four three two coefficients okay and much the time and i should say mole per reaction down here rxn multiple reaction here rxn and multiple reaction so most of the time this uh writing it as mole per reaction is not included in the enthalpy value because it is understood to be mole of reaction as is written with the coefficients as it is written okay so a lot of times you'll just say negative 1776 kilojoules instead of kilojoules per mole but it is understood that it's per mole of reaction based on the coefficients as it is written okay so let's do some practice problems with stoichiometry of thermochemical equations so thermal chemical stoichiometry with thermal chemical equations thermo boy can't write it correct can i thermo chemical so let's look at the stoichiometry with thermochemical equations sorry about the mess there uh so let's look at we're going to do uh four different uh practice problems here and the equation that we're going to use the thermochemical equation that we're going to use is one mole of methane gas plus two moles of oxygen gas will produce one mole of carbon dioxide gas plus two moles of h2o gas and then we're going to say that the delta h is equal to a negative 8 91 kilojoules per mole of this reaction okay um so that means 891 kilojoules if you burn one mole of methane 891 kilojoules if you use two moles of oxygen so forth okay so the first problem let's do is let's determine the amount of heat produced by the combustion of 3.30 mole of methane moles of methane okay so again i'm if you'll notice here as i go through this i want to know determine the amount of heat and so i've got i don't know how many kilojoules of energy if not in this this is kilojoules of energy per one mole of methane but i'm actually wanting to know 3.30 moles of methane so i have more methane than just the one mole but we do know that for every one mole of methane based on the coefficient from the reaction one mole of methane will produce this much energy so that's a negative 8 91 kilojoules and so that's all there is to that one because now i have kilojoules so that's a three point three times eight ninety one and we want one two three sig figs so my answer is going to be a negative two thousand 2940 kilojoules of energy now in this case design energy per mole it's per 3.3 moles so we don't write per mole here because we're not looking at the whole reaction we're just looking at these values and everything all answers should be in kilojoules from here on out when we're looking at energy not just joules but kilojoules the next one we're going to look at is this question number two so determine the amount of heat produced so we're still looking for heat produced but in this time instead of looking at the methane i want to know how much is produced when 54.00 grams of oxygen gas is reacted so this is how much oxygen i have to react so i want to know how much heat i can produce okay so let's look at this one and again i'm looking for amount of heat so okay i want question mark kj kilojoules we know how many kilojoules of energy and now i start with the 54.0 grams of oxygen now hopefully by now you realize in in this sense this is in moles i know i have to get to a mole but if i don't have the mole of something if you have the mass go to the mole by using the molar mass conversion so one mole of o2 is 32.0 grams o2 now remember i put the grams on bottom because they're on top over here now i have moles of o2 and i know that for every two moles of o2 i produce that much energy so for every two moles of o2 i get a negative 891 kilojoules of energy this is fairly straightforward stoichiometry problems let's see if it looks it looks kind of weird i'm gonna try to strain this out a little bit okay hopefully that works a little bit better anyway so when i calculate this multiply and then divide i get a negative 750 to as three sig figs kilojoules of energy produced so by burning this much 54 grams of oxygen is how much energy gets used okay so let's look at another uh sample number three determine the mass so instead of looking at energy let's say i wanted to determine the mass of water h2o gas produced during an enthalpy change of negative 423 kilojoules so what this is really saying is if i have this much energy available neither 432 or i have this much change in energy how much oxygen can i produce when this much energy is produced so again we start in i'm looking for grams so in this case i'm determining the mass of oxygen when i have the amount of energy so what is how many grams of oxygen keeps an oxygen i meant water sorry how many grams of water are produced when this much energy is used or produced sorry when this much energy gets produced so again we know that this reaction is negative 891 kilojoules for every two moles of water because we're looking at water so i'm going to write a negative 8 91 kilojoules for every two moles of h2o now i don't want moles of h2o i want i want the mass of h2o so that's going to be one mole of h2 i put it on bottom because it's on top here has a mass of 30 i mean not 32 but um 16 plus 2.02 is 18.02 grams of h2o and when i multiply and divide i should be able to produce 17 17.1 grams of water from that much energy change okay so hopefully you understand these are really straightforward the last one i want to look at is what if we double and i already introduced this but let's just say um let me use the last one here black number four determine enthalpy for the reaction below determine the enthalpy for the reaction instead of just one we're going to say two moles of ch2 plus 4 moles of o2 this is a gas this is a gas produces 2 moles of co2 gas plus four moles of h2o gas and in this case all we did is we doubled the coefficients remember this is one mole here this is now two moles here so we just doubled the coefficients so my delta h for this reaction is equal to 2 times the original value and we get a negative 17 82 kilojoules per mole okay oops let me move that up so you can see sorry about that and that's the basic uh basically what you need to do for stoichiometry and we will stop this video here