hello bobcats in this video we will be discussing constant pressure calorimetry now um this is slightly different than the last video where we talked about the calorimetry used to measure the transfer of energy from a hot substance to a cool substance such as the metal transferring its energy the hot metal transferring its energy to cold water and the change in temperature that is two substances of different temperatures this calorimetry constant pressure calorimetry is usually used for measuring the energy involved in reactions or we can say enthalpy so let's define calorimetry one more time so calorimetry is an experimental technique used to measure the amount of heat transferred to and from a system measure the amount of heat and that's q [Music] measured transferred sorry amount of heat transferred to and from a system okay now most classrooms use what's called a coffee cup calorimeter styrofoam calorimeter so most classrooms use a styrofoam cup calorimeter i'm to measure this heat transferred by reactions heat transferred by reactions in aqueous solutions so remember aqueous solutions contain the compounds dissolved in water so really we're looking at heat transferred from reactions to the water which is the surroundings now so what it was a coffee cup counter under kind of consist of we have a our styrofoam cup calorimeter i'm gonna do the best i can to draw it so you take styrofoam cups uh usually two of them and put them into each other but i'm just gonna draw one so i have a styrofoam cup and reason we're using styrofoam cups is because it's the simplest way to create an insulated system or as much as we can possibly get of an insulated system in which the energy and the matter uh cannot go from back and forth from surroundings and um system or leave a certain amount of system in other words i want to keep all the energy inside this cup and all the energy um being either absorbed or released by the reactants into our surroundings and we want the surroundings to be the solution so we don't want energy getting outside now coffee cup are not the best thing because there is some error here but it is as good as we can get without spending a whole lot of money and so and then you have a thermometer because you have to measure the change in temperature of the the water in the solution and i'm just going to say solution from now on so um we got a thermometer in here there's my thermometer and i'm going to go ahead and label it thermometer thermo meter we also have this insulation to keep as much energy in as we possibly can so it's all being absorbed by the water and none of is being is um being absorbed by the rest of the surroundings like the air the the cup itself and so forth and we use styrofoam for that and then inside you have the aqueous solution now remember when i say aqueous solution that is containing that's reactants dissolved in water so i'll write that down active reactants dissolved in water so really it's the water we want the water to be the surroundings so from now on i'm just going to call it a solution but really where the reactants are the the system and water is the surrounding now the important thing here is that this cup calorimetry is carried out at atmospheric pressure atmospheric pressure because it's not um completely isolated um the pressure or the pressure outside equals the pressure inside so inside the cup is the same pressure as the outside which is atmospheric pressure so because it's done at atmospheric pressure that means that it is constant so atmospheric pressure is constant constant so it's constant pressure that's why we call it constant pressure calorimetry and because it's at constant pressure therefore we know we can say that the delta h of the reaction is equal to q or the heat transfer um remember at constant pressure okay so that was an important concept that we learned last time about um energy is that the enthalpy of reaction is really just the heat transferred at constant pressure now for this to help out really understand what's going on because it's the water that's changing temperature in the solution we're going to say that the heat exchange or energy q of the solution is equal to the mass of the water of that solution times the specific heat of water of that solution times the change in temperature of the water after the reaction okay so we're really measuring the change in temperature of the water the water is the surrounding now so if the water is the surrounding we can also say that the q of the solution is equal to a negative q of the reaction so as the solution gains is because the reaction is losing or vice versa but it's opposite and equal to what's happening with the reaction the reaction is the system we're looking at the solution as the surroundings now remember up here we because the delta h reaction is also equal to q we can say that um that what we'll say here that the delta h of the reaction is equal to the q of the reaction because it's at constant pressure and so again if this is um q q the solution is m c delta t we can say that here we can say that the delta h of the reaction is equal to because remember we said it's a q here the delta h reaction is a negative m c delta t okay and really it's the m c delta t of the solution so when we measure this heat exchange of the solution we're getting the the enthalpy change of the reaction or the heat exchange of the reaction so to make make sure we understand this we're really measuring the energy that's being absorbed by the solution or given off by the solution but we also know that the energy absorbed by the solution is also equal to the energy released by the reaction which is the system and if the reaction the energy of the reaction is also known as the delta h or enthalpy of the reaction because it's at constant pressure we can say that the delta h of this reaction is a negative m c delta t of the solution and this is what we really need to know so when we're looking at the delta the calorimetry of constant pressure we're really trying to find the delta h of a reaction now so let's look at some oh well let's make a couple other statements before we look at a problem so another statement we want to look at is that to make sure we understand it the system is the reactant products of the reaction and products of the reaction and the surroundings that we were really measuring or looking at include the calorimeter the thermometer and the water in the solution include the calorimeter which we are hoping does not absorb much of the energy because it's styrofoam the thermometer which absorbs just a tiny bit just enough to get a reading but most of the energy absorbed from the reaction is from the surroundings of the water in the solution okay but it's the water that we're really hoping it's all uh absorbs the energy or releases energy so we can say that when the temperature of the solution increases which really we're looking at the water but i'm just going to write solution now when this temperature of the solution increases it is because energy is being released by the reaction which is the system into the surroundings or by released by the reaction into the surroundings or into the solution which i'm going to say here is water just to make sure okay now because the water is absorbing the energy it is going to increase in temperature and it is the reaction that releases the energy released by the reaction so we call this an exothermic reaction and you should get a negative delta h and when the temperature of the solution decreases um it is due to the reaction absorbing energy from the surroundings reaction which is the system absorbing energy oops not a q energy sorry from the surroundings which is really the solution or the water of the solution and because the water is actually losing its energy to the reaction the water decreases in temperature and we call this an exothermic reaction no not exothermic sorry endothermic i think i'm getting tired endothermic reaction and you have a positive delta h okay so let me say again because i kind of messed up when the temperature of the solution decreases it is due to the fact that the reaction which is the system is absorbing energy from the surroundings which is the solution really mostly water and the water then goes down in temperature because it's losing energy and we say that the reaction itself is endothermic because it is absorbing energy so in this case we're looking at the reaction absorbing the energy okay now let's do a practice problem and i'm going to write the equation one more time so that you guys really have it down what we really look at in this case is that the delta h of a reaction or the enthalpy change or energy of the reaction is equal to a negative m c delta t of the solution okay and that's the equation we really want to look at now let's do a practice problem let me pull it out where did i put it all right i thought i had it right here available here we go so this is the problem from the powerpoint and so let's go through it we've got um 25 mils of a strong acid reacting with 25 is added to 25 ml of strong base into the coffee cup when they're added to each other they do they have a reaction it's called a neutralization reaction they produce water and salt now the solution temperature rises it goes from 25 20.5 to 25 degrees so because the water is rising in temperature of the solution i should say we know that this has to be an exothermic reaction another reaction is releasing energy to the solution and the solution gains temperature because the solution is mainly water we're going to say that it has the same specific heat as water and because it's mainly water we're going to say that has the same density as water okay so now what we want to do is we want to calculate the enthalpy change of the reaction so here we're going to calculate the delta h of the reaction and in here in b assuming that the strong acid has a concentration of 1.5 molar calculate the enthalpy change per mole so we really want to know how many kilojoules per mole okay so that's what we're looking for in this problem okay enthalpy change per mole of the acid we're looking at the acid okay so when i do this part a let's look work through part a a couple of things we know that the equation is delta h of the reaction is equal to a negative m c delta t of the solution so i had to find the mass now in this case mass of the reaction is equal to [Music] well we have 50 mil 25 mils plus 25 ml so that's total of 50 mils but because uh it's mostly water we're gonna say it's one gram per mil so for if we have 50 mils we have 50 grams so i'll write it here as 25 grams plus 25 grams of the water so we actually have 50 grams of solution right here okay now it's mass my c or specific heat is 4.184 joules per gram degrees celsius and then my change in temperature of the solution which is mainly water is going to be final minus initials that's 25.0 minus 20.5 so we get a 4 positive 4.5 degrees celsius change and so that's the other one we're looking for so i'm just going to substitute up here and we're going to say a negative 50 grams times the specificity of 4.184 joules per gram degree celsius times that change in temperature which is 4.5 degrees celsius that's my enthalpy of the reaction or the heat involved in the reaction and that will equal let's look make sure all our units cancel grams cancels here celsius cancels there so we're left with joules so my enthalpy change of the reaction or heat exchange of the reaction is equal to a negative but 940 calculator says 41.4 but we only want there's three three three three sig figs so 940 joules of energy and of course we know that it's exothermic because it's negative exothermic because it's negative and it makes sense that it's exothermic because it's the water is increasing in temperature because the water is gaining the energy from the reaction so that's the energy involved in what specifically we have up here the 25 mils and 25 mils but for b we want to find out how much energy per mole okay so i'm gonna i'm looking for kilojoules per mole and we know that we have a negative 941 joules so i'm just going to convert that to kilojoules real quick so that's 1000 joules in one kilojoule and then we have to have the moles so how do i get my moles now well let's go ahead and write this out and say this is going to be a negative 0.941 kilojoules of energy and then i need to figure out my moles now remember we had 25 ml so that's 0.025 liters and the the the concentration was 1.5 molar so that's 1.5 moles per liter so i give you moles so that comes out to be 0.0375 moles so again if i just want kilojoules per mole i'm just going to take this and divide by that so let me start here again one more time kilojoules per mole is equal to 0.941 kilojoules over 0.0375 moles and that comes out to be a negative 25 kilojoules per mole so i just want to make sure i got kilojoules per mole and that's how we work that problem um i will uh our next video is going to be on stoichiometry of thermal chemical reactions and i'll see you then