so this lesson is going to be on molar enthalpy and calorimetry so we're not going to be just talking but enthalpy changes we added this word molar so what is the enthalpy change per mole of chemical that reacts now we're going to have the same starting point to get to molar enthalpy as we do enthalpy calorimetry is the heart of of both okay so first i start up a statement that's really important to ultimately getting to the formula we're going to use okay the change in enthalpy so that is delta h increases as more limiting reagent is added okay we're going to measure that limiting reagent in moles and that hopefully is an obvious statement okay that's a chemical version if i put more wood in a fire you're going to get more heat that's all this statement is saying i put in more limiting reagent i get a i get a increased enthalpy change so both of those are variables they change every question they change even within one question you might have i might tell you how many moles and what the enthalpy change is and then i could ask you what if you tripled the moles what would the enthalpy change be and you would just triple it however okay the ratio of enthalpy change now i didn't write the word change here but i mean the ratio of change in enthalpy two moles is a constant okay and we're going to learn how to predict these constants and and use them okay now enthalpy 2 moles that were 2 implies division so now we're building one of the most important formulas we're going to use this unit the ratio of change in enthalpy 2 moles i'm highlighting both of these are variables equals molar enthalpy now i use a subscript and delta h m change in enthalpy will there's going to be an energy unit we're going to use kilojoules most of the time n is moles and our units of molar enthalpy is just the division of those kilojoules per mole every chemical for a given reaction type is going to have a constant molar enthalpy if you look at methane that is heating your house or heating your hot water tank methane releases so many kilojoules for every mole that reacts okay that is never going to change all you can do is put in more moles of methane or less moles of methane and you'll get more or less energy and that can relate to more or less temperature change in your water or the the air that's being heated so as we work our way through this is going to be a super important link to get through multi-step questions that's not going to change so we're going to calculate molar enthalpy on the next slide so what steps am i going to take to calculate the molar enthalpy and we have to do this for the limiting reagent for the chemical that runs out okay now you're not going to be asked to do limiting excess reagent calculations like in chem 20 the question will will tell you or imply what runs out so the first thing we're going to do is the same thing we did before determine the amount of heat or thermal energy that is transferred into uh into the surroundings the water okay heat which is q is m c delta t we'll assume we have an isolated environment and whatever that amount of heat is came from our chemicals next same as before we're going to think of what sine we should have we'll put a positive sign if the reaction is endothermic the surroundings cool down we'll put a negative if it's exothermic the surroundings warmed up okay that these two pieces are the same as before they give us delta h and now we have this new step we're going to divide by divide that enthalpy change that we just got by the moles of the limiting reagent before i go on to an example i'll just remind you of the two common ways you're going to have to calculate moles if you have a pure substance this might look like 10 grams of calcium metal reacts okay well that 10 grams is is pure okay you need to take the mass of that substance and divide it by the molar mass as opposed to a solution where you have a solute and a solvent you get moles is equal to concentration times volume okay you're not going to see gas calculations like you did in chemistry 20 to solve for moles so let's solve for molar enthalpy in the following reaction so again like the previous question when you get on a test or your workbook question the last line tense of the question so we're asked to determine not enthalpy change like we saw before but molar enthalpy so that is delta h m so find the question make a bit of a plan of how you tackle that question and then go back and and read through the data so to get molar enthalpy change we need to find the delta h and then we've got to divide it by the moles of the limiting react reagent and we're told to do this for ethanol so that must totally run out to solve for the enthalpy change in a calorimetry experiment we need m c delta t and we need to figure out what the surroundings is it's often water but every now and then it's not so jumping to the middle the most the surroundings uh ethanol was combusted but the heat went into the thermal energy went into 1.63 liters of water okay so the surroundings is water so i need the mass of water the c of water the delta t of water so i don't care about this ethanol mass for the mc delta t that's not where the energy went i need to just focus there that is the only mention of the surrounding water so i have to do a little bit of conversion we have 3.63 liters but i want that in milliliters because i know every milliliter of water is a gram now your data booklet has the meaning of all the prefixes milli is one one thousandth of a liter there's a thousand milliliters in a liter so this is three thousand six hundred and thirty milliliters or because it's water that's 3630 grams because a milliliter is exactly a gram just for water so that is our mass 3630 grams times this is water 4.19 joules per gram per degree celsius times i need the change in temperature so what is the difference in temperature 26.18 minus 19.88 that is a difference of 6.30 degrees celsius my joules and sorry my grams and grams cancel degrees celsius degrees celsius cancel i'm left with just joules and we tend to get really big numbers of joules and i'll probably convert it to kilojoules i have three measured digits in all these calculations but you'd want to keep extra digits till you get to the very end so my calculator spit out 95 821 joules which i'm not done so i'm not going to round that down yet or 95.8 to 1 kilojoules counting by thousands of joules and we always have to get joules first okay so that's my first step of three get the thermal energy the q then think of the sine and please always write your sign in even if it's positive i'd like you to write it in to convey to me you know what sign it is the final temperature is warmer so energy was released to make it warmer so this is a negative change in enthalpy okay then i just need to get the moles and the chemical that's all running out is the ethanol so i need moles of ethanol so f means two carbons so you should be able to figure out what the formula is of ethanol and get the molar mass moles of this pure substance is mass over molar mass this isn't a solution the ethanol is a pure three and a half grams divided by the molar mass of ethanol which is 46.08 grams of ethanol per mole grams and grams cancel we're left with moles on the top and you get zero 0.0760 moles so much less than a mole less than a tenth of a mole released that 95 kilojoules again you would want to keep all the digits in your calculator for both of both the red number and the green number and then at the very end i'm going to round to three sig figs because that was the the limiting uh significant figures in both pieces of data so i'm after that uh division um write that minus down first i see students forgetting to put the minus in their answer sometimes 1200 and i'm going to round it to 60 kilojoules per mole my 1 two three sig figs so that zero is just a placeholder okay and that's our molar enthalpy okay this is a value we could look up in a data table and compare because ethanol the combustion of ethanol should always have the exact same molar enthalpy this is a constant the two values we just used to solve these are variables but the ratio is always the same okay so that's how you calculate molar enthalpy from calorimetry data