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Combustion Reactions

In chemistry, combustion is a high-temperature exothermic redox chemical reaction between a fuel and an oxidant, usually atmospheric oxygen. Combustion in a fire produces a flame, and the heat produced can make combustion self-sustaining. Combustion is often a complicated sequence of elementary radical reactions. Solid fuels, such as wood, first undergo endothermic pyrolysis to produce gaseous fuels whose combustion then supplies the heat required to produce more of them. Combustion is often hot enough that light in the form of either glowing or a flame is produced. A simple example can be seen in the combustion of hydrogen and oxygen into water vapor, a reaction commonly used to fuel rocket engines. The bond energies in the play only a minor role, since they are similar to those in the combustion products.

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Video Transcript

So in our last video, we cover the differences between combination and decomposition reactions as well as I saw memorization. Eso for this video will be talking about a different class of reactions called combustion reactions, So compassion reactions are interesting because they require specific conditions and reactions for it to be a combustion reaction. So promotion reactions typically require specific conditions to occur. So usually these happen with heat, like fire or elevated temperatures, and you also need oxygen to be one of your reactions. So a common reaction that you'll find is one between a few and oxygen to create some kind of offside product, and usually for the products that you make, you create carbon dioxide and water eso. To illustrate this. I have a couple of reactions that happened in the real world to help illustrate these types of reactions and whatever you write reactions, you want to make sure that you write out the chemical formulas, and you also write out the phases. Eso this case methane is in the gas phase office and gas. It's gas. Carbon dioxide is gas and you produce liquid water. So this is the first type of combustion reaction that will be looking at. And right now it looks OK, but usually when you write a reaction, you want to make sure that your reaction is balanced. So as of now, the way that it is written it is not balanced. So we can go ahead and right in coefficients to make sure that the number of atoms on the left side equal the number of atoms on the right side for each element in the reaction eso it start. We can look at the number of carbon atoms. So on the left side we have one carbon atom and on the right side we have one. Um S O right now we're okay with the carbon count for both sides. Now, if you look at hydrogen, we actually have two more hydrogen on left side than the right side. So there are four versus too. So that means we need to add a coefficient off to to make sure that the number of hydrogen atoms are the same on both sets. Eso now looking at our oxygen count, we can see that there are two oxygen's on the left side, but there are four on the right side now. So we need to add a coefficient of to on the left side to make sure that there is the same number of Adams a lot. Right? So now if we look at our equation, we can see that there are the same number of atoms of carbon, hydrogen and oxygen on both sides. And we can see that again for a combustion reaction. You have your fuel. Um, which in this example is methane. You have oxygen as one of your reenactments, and you can create co two and water as your products. So, for my next example, will be looking at something similar to methane on DSO. If you have something like ethanol, which has a chemical formula of ch 40 reacting with auction gas, you again create some amount of carbon dioxide as well as some amount of water. So again, whenever you're looking at chemical reactions, you want to make sure that the equation is bounced. So right now it doesn't look balanced. So using the same procedure as before, we'll figure out how to balance this equation. Eso here, let's start off with oxygen so we can see that on the left side. we have three oxygen's and on the right side, we also have three. If we look at the number of hydrogen atoms, we can see that there are four on the left side and only two on the right side. And so we need to add a coefficient of to. Right now, the carbon count is the same s so we can go back to the make sure we have the right number of auction Adams. And right now we see that there are again three oxygen atoms on the left side. But now there are four on the right side. And so to deal with this, we have thio oxygen on the left side. So the thing about this is that we could add one half as a coefficient to make sure that the number of auction atoms are the same. But we simply can't do that because we can't have half a molecule. And so it's deal with this. We can actually multiply the whole equation by two so that we have a whole number issue between all the molecules. And so what I mean is that now we can do you write this so that we have two of the use methanol items and write this so that we have one option. And, um and we have to carbon dioxides and for water molecules. Let's rewrite this. Okay, so now if we double check our current count, we have two carbons on the left side and the right side. We have eight hydrogen on left side in the right side. And for the number of oxygen atoms, we have four, and we have four and eight. So we know that this right now is not correct. Um, and so we can change this so that it is. Yeah. So right now, our equation is balanced because we have again two carbon atoms on the left and right side. We have eight hydrogen atoms on the left and right side, and we have eight Oxygen's on both sides. So this is the fully balanced equation. And again, this is a combustion reaction because you have oxygen reacting with a few to create CEO tube and water as your products on DSO to how to make sure this concept is clear. So let's say you have something like sugar, which has a chemical formula of C 12, h 22 0 11. Sorry, this is not gas. This is a solid reacting with oxygen gas to create some amount of C 02 as well as some amount of water. So again, using the same method as before, you kind of want to go back and forth between figuring out well, what ratios work by working with one atom at a time. Eso first, let's start off with the number of carbons so we can see on the left side we have 12 carbons, and on the right side we have just one. So right now we can add another coefficient in front of CO two to make sure that the number of carbons are the same. So now, looking at the number of hydrogen atoms, we have 22 on the left side and two on the right side. And so we can add a coefficient of 11 to make sure that the number of hydrogen, our balance on the left and right side, and now to account for the number of oxygen's, we can see that we have 11 plus two here and we can see that on the side. We have 11 oxygen's as well as 12 times. Um, the number of oxygen's in carbon dioxide, and we can see that this formula is actually not too bad to Dio because we have the 11 from the sugar, so we can add a number of coefficients, Um, based on the number, um, in front of co two. And in this case, it is what? So here we have a balanced equation where one sugar molecule can react with 12 molecules of oxygen to create 12 molecules of carbon dioxide and 11 molecules of water. And again, this is a combustion reaction because you have oxygen reacting with a fuel to create CO two and water.