# Thermochemical Equations Practice Problems

I'm going to show how to calculate how much heat is released or absorbed with chemical

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so here I have a chemical equation that shows what happens when methane also known as natural gas burns in the air so methane ch4 combines with oxygen to make carbon dioxide and water but as you probably know if you burn methane it releases a ton of heat and that's what this Delta H thing is over here this number is telling us how much heat we're dealing with it's eight hundred ninety point four kilojoules which is just a unit that we use to measure Heat now just in case you weren't sure whether burning natural gas releases heat or absorbs it we have this negative sign here the negative is telling us that this is an exothermic reaction which means that heat is released so we want to calculate how much heat is released by burning 27 point five grams of methane or natural gas when I'm doing calculations that use chemical### equations

like this use a lot of math numbers measurements that are in grams like twenty seven point five grams are not particularly useful I really want moles so the first thing that I'm going to do is I'm going to take twenty seven point five grams of ch4 and I'm going to convert that to moles here's how I'm going to do it I'm going to use a conversion factor that includes the molecular weight or the molar mass of ch4 I'm not going to talk about I'm not going to talk much about moles here but there are other...videos on that topic so I have gram ch4 up here I'd have gram ch4 down here their top and bottom which means they're going to cancel out and when I do the math I'm going to get one point seven two moles of ch4 so now I know how many moles of ch4 I have at this point I can go up and take a look at this equation look at the methane here there's no number in front of it which means that we have one ch4 you have one if there's not a number right we have one ch4 or one mole of ch4 one mole of ch4 and negative a ninety point four kilojoules what this means is that one mole of ch4 makes this amount of heat okay so if I had exactly one mole of ch4 I know exactly much how much heat gets released but I don't have one mole I have one point seven two moles so what I got to do is I've got to write this as a conversion factor and use this conversion factor to turn one point seven two moles of ch4 into an amount of heat all right so here's how I can do this there are two conversion factors that I can write using this information one of them looks like this it's one mole ch4 over negative eight ninety point four kilojoules now I can also flip this and I can write negative 890 point four kilojoules over one mole CH for both of these conversion factors are equally correct but I'm going to be starting with one point seven two moles ch4 and so I want to use a conversion factor that's going to cancel out moles of ch4 and is going to leave me with units...

of kilojoules I'm going to want to use this one instead of this one so I bring this over you can see that moles up here cancels out moles down here and now I'm going to do this math which is this times this divided by one and I'm going to get 1530 kilojoules this is how much total heat is released when we burn 27 point five grams let's take a minute to look at the steps that we went through because we're going to have to use them for future

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okay the first thing that we did was we took grams of ch4 and we converted it to moles of ch4 we used a conversion factor and that included the molecular weight in order to do this step okay that's once we had moles we use that to get Oh joules of heat and we made use of a conversion factor that included this information here that one mole of ch4 made this much heat multiplied them and that's how we got the total amount of heat let's do another example here we're going to calculate how much heat is created by seventy nine point two grams of oxygen let's think about the steps we're going to take real quick okay first we're going to go from grams of o2 to moles of o2 we use a molar mass for that and then we'll take moles of o2 and use that to get kilojoules of heat okay so the first step is we'll do this really fast seventy nine point two grams o to multiply that by a conversion factor that includes the molecular weight grams of o2 cancels out because grams of o2 is down here...we do the math and we end up with a two point four eight moles of o2 now we can go up and look at this equation now this is really important pay it pay attention here this is important in the previous example we said that one mole of ch4 made 890 point four kilojoules of heat okay and we were able to use this information to write this conversion factor but it's different with oxygen check this out there's a two in front of the o2 okay and so that means that two moles of this makes this much heat so here is the statement 402 get it one mole of ch4 but two moles for o2 two moles make this much heat so when I'm talking about oxygen I need to write a conversion factor that looks like this two moles of o2 making this much heat okay don't confuse the two of these it all depends on the number that you have in front of the chemical one here two here two and two okay so we will use this conversion factor of two moles o2 or I can also flip it to get negative 890 point four kilojoules over two moles o2 all right now which of these am I going to want to use well I'm going to be starting here within moles of o2 two point four eight moles o2 I will want to multiply it by the one that's going to cancel these units out so I'm going to choose this one now I have moles o2 up here cancels out moles o2 down here and now the math I'm going to do is this times this divided by two is going to give me 1102 odd eul's and this is my final answer now let's do two...

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## problems

where we go the other way we start with a certain amount of heat and we figure out how much chemical we need order to get that amount of heat here's an equation for nitrogen gas combining with hydrogen gas to make ammonia this also releases a bunch of heat as you can tell because it Delta H number here is negative we want to find out how many grams of n2 we're going to need in order to make this much heat so here are the steps that we're going to go through to figure this out the first thing that we'll do is we'll take the amount of heat that we want and use that to figure out how many moles of n2 we're going to need we use a chemical equation and this number to figure that out once we've found out how many moles of n2 we're going to need we can then use the molecular weight to figure out how many grams of n2 so let's start up here with the equation there is nothing in front of n2 which means that we're talking about one or one mole and this is how much heat we're dealing with so the equation is telling us that one mole of n2 makes negative ninety two point six kilojoules of heat okay so there are two conversion factors that we can write with this information one with n two on the bottom one with N two on the top and one with N two on the bottom I'm going to start with negative five five zero point zero kilojoules and choose the conversion factor that's going to get rid of kilojoules and leave me with moles...it's going to be this one kill the jewels on the top kilojoules on the bottom they cancel out and when I do the math negative five fifty times one divided by this I get five point nine four moles of n2 so now I have my moles of n2 and to take the final step from moles of n2 to grams of n2 I'm going to want to use a conversion factor that makes use of the molecular weight here twenty eight point zero grams moles and two cancels their moles and two cancels there and I end up with 166 grams of nitrogen in order to make this much heat here we're going to figure out how many moles of h2 we need to get negative 155 kilojoules we're going to use the same plan of attack that we did earlier where I can go from kilojoules to moles using the information that's in this equation and then once we have moles we'll use a molecular weight to go from moles to grams okay so the first thing we do to go from kilojoules to moles is we look up here at the equation okay where are concerned with h2 with hydrogen here and notice there's a three in front of hydrogen in this equation which means that three moles of hydrogen make this much heat so we'll write it like this three moles h2 make negative ninety two point six kilojoules of heat so I'll be starting with negative 155 kilojoules and I'll multiply it by a conversion factor that says this information okay here it is three moles on top ninety two point six kilojoules on the bottom kilojoules kilojoules they...

cancel out I do the math this times this divided by this is going to give me five point zero two moles of hydrogen now I'm halfway there I'll take my five point O two moles of hydrogen and I will multiply that by a conversion factor with a molar mass I've actually done all of the work right here moles of hydrogen times conversion factor with molar mass in order to give me ten point zero grams of hydrogen which is the total that I need so I start with kilojoules go to moles and then go to grams so two things that are important to keep in mind you first when you start with grams that's not particularly useful you're going to want to use moles okay so the first thing I want to do is you'll want to convert grams to moles the second thing it's probably even more important is to keep in mind that you want to look carefully at the number the coefficient that's in front of the chemicals here in the equation if it says 1 that means one mole of nitrogen makes this if it says 3 here it means you need 3 moles of this to make this much heat and so the conversion factor that you use is going to depend on the number in front of the reactants or products in the equation