Thermochemical Equations Practice Problems

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

equations

, so here I have a chemical equation that shows what happens when methane, also known as natural gas, is burned in air, so CH4 methane is combines with oxygen to form carbon. dioxide and water, but as you probably know, if you burn methane a ton of heat is released and that's what this Delta H is here, this number tells us how much heat we're dealing with, it's 89.4 K, which is just one unit We used to measure heat now in case you weren't sure whether burning natural gas releases heat or absorbs it.

We have this negative sign here. The negative tells us that this is an exothermic reaction, meaning that heat is released, so we want to calculate how much heat is released by burning 27.5 G of methane or natural gas when I do calculations using chemical

equations

like this I use a lot math numbers measurements that are in GRS like 27.5 G are not particularly useful I really want moles So the first thing I'm going to do is take 27.5 g of CH4 and convert it to moles. This is how I will do it. I will use a conversion factor. that includes the molecular weight or molar mass of CH4.

I'm not going to talk too much about moles here, but there are other videos on that topic, so I have grams of CH4 up here and I have grams of CH4. down here they are up and down, which means they will cancel out and when I do the math I will get 1.72 moles of CH4, so now I know how many moles of CH4 I have at this point. up and take a look at this equation, look at the methane here, there is no number in front of it, which means we have a CH4, you have one, if there is no number, we have a CH4 or a mole of CH4, a mole of CH4 and 89.4.

K, what this means is that one mole of CH4 produces this much heat, so if I had exactly one mole of CH4, I know exactly how much heat is released, but I don't have one mole, I have 1.72 moles, so what? What I have to do is write this as a conversion factor and use this conversion factor to convert 1.72 moles of CH4 into an amount of heat. So this is how I can do this. There are two conversion factors I can write using this information. one of them looks like this, it's a Mo ch 4 over 89.4 K. Now I can also reverse this and I can write 89.4 K over 1 Mo CH4.

Both conversion factors are equally correct, but I will start with 1.72 moles of CH4 and so I want to use a conversion factor that will cancel out the moles of CH4 and leave me with units of kles. I'm going to want to use this one instead of this one, so I bring this in, you can see. The moles up here cancel the moles down here and now I'm going to do these calculations which are multiplied by this division by 1 and I'm going to get 1,530 K. This is the amount of total heat that is released when we burn 27.5 G. let's take a minute to see the steps we followed because we are going to have to use them for future

problems

.

Well, the first thing we did was take grams of CH4 and we converted them to moles of CH4. We used a conversion factor that included the molecular weight to do this step, well, once we had moles, we used that to get kilos of heat and we made use of a conversion factor that included this information here that one mole of CH4 produced This quantity. the heat multiplied them and that's how we got the total amount of heat. Let's do another example. Here we are going to calculate how much heat is created with 79.2 grams of oxygen. Let's think about the steps we're going to take very quickly.

Well, first. We're going to go from grams of O2 to moles of O2, we're going to use a mass m for that and then we're going to take moles of O2 and use them to get klas of heat, so the first step is to do this real quick 79.2 G O2 multiply that by the conversion factor that includes the molecular weight, the grams of o2 cancel out because the grams of o2 is down here, we do the math and we end up with uh 2.48 moles of o2, now we can go up and look at this equation. is really important, pay attention here, this is important in the previous example we said that one mole of CH4 produced 89.4 K of heat and we can use this information to write this conversion factor, but it is different with oxygen, look at this, there's a two in front of O2, okay, that means two moles of this generates that much heat, so here's the statement for O2, get one Mo of CH4, but two moles of O2, two moles generates that much heat, so that when I talk about oxygen I need to write a conversion factor that looks like these two moles of o2 that generate so much heat.

Okay, don't confuse the two, it all depends on the number you have in front of the chemical one here two here two and two, okay, so we'll use this conversion factor of 2 moles of O2 or I can also invert it to get 89.4 K over 2 moles of O2. Which of these will I want to use? Well, I'm going to start here with moles of O2 2.48 moles of O2 I'm going to want to multiply it by that will cancel out these units, so I'm going to choose this one now that I have moles of O2 up here cancel out the moles of O2 down here and now the math What I'm going to do is that this multiplied by two will give me 1,100 kles and this is my final answer.

Now let's do two

practice

problems

where we go the other way, we start with a certain amount of heat and we calculate how much chemical we need to get that amount of heat here is an equation for nitrogen gas combining with hydrogen gas to produce ammonia, this also releases a lot of heat as you can see because the Delta H number here is negative, we want to know how many grams of N2 we are going to need to generate that much heat, so these are the steps we are going to take to solve this. The first thing we will do is take the amount of heat. that we want and we use it to calculate how many moles of N2 we are going to need, we use a chemical equation and this number to calculate it once we find out how many moles of N2 we are going to need, then we can use the molecular weight to calculate how many grams of N2, like this So let's start here with the equation: there's nothing in front of N2, which means we're talking about one or one mole and this is the amount of heat we're dealing with.

The equation tells us that one mole of N2 produces 92.6 K of heat, so there are two conversion factors that we can write with this information, one with N2 at the bottom, one with N2 at the top, and one with N2 at the bottom. I'm going to start with 55.0 K and choose the conversion factor that will remove the kles and leave me with moles, it will be this kles at the top K at the bottom they cancel out and when I do the calculations -550 * 1 / this I get 5 .94 moles of 2, so now I have my moles of N2 and to make the final step from moles of N2 to grams of N2, I'm going to want to use a conversion factor that uses the molecular weight here. 28.0 G moles n two cancellations there moles n two cancellations there and I end up with 166 G of nitrogen to generate that much heat here we are going to calculate how many moles of H2 we need to get -55 K We are going to use the same plan of attack that we did before, let's to go from KJ to moles using the information that is in this equation and then once we have the moles, we will use a molecular weight to go from moles to grams.

The first thing we do to go from kles to moles is to look at the equation here. Well, here we are concerned about H2 with hydrogen and we notice that there is a three in front of the hydrogen in this equation, which means that three moles of hydrogen make up this quantity. heat, so we'll write it like this: 3 moles of H2 produces 92.6 K of heat, so I'll start with 15 K and multiply it by a conversion factor that tells this information. Well, here are three moles on top, 92.6 KJ. the lower kles kles cancel out. I do the math, this multiplied by this divided by this will give me 5.02 moles of hydrogen.

Now halfway through I'll take my 5.02 moles of hydrogen and multiply it by a conversion factor with an M. Mass In fact, I've done all the work here: moles of hydrogen multiplied by the conversion factor with molar mass to get 10.0 G of hydrogen, which is the total I need, so I start with kles, go to moles and then to grams. So first it's important to keep two things in mind: When you start with grams that aren't particularly useful, you'll want to use moles. So the first thing you'll want to do is convert grams to moles. The second thing that's probably even more important is to keep in mind that you have to look carefully at the number, the coefficient that's in front of the chemicals here in the equation, if it says one, that means one mole of nitrogen does this, if it says three here.

It means that you need three moles of this to generate that much heat, so the conversion factor you use will depend on the number in front of the reactants or products in the equation.