This would be the amount of energy that's essentially released. This is our change in enthalpy. So if this happens, we'll get our carbon dioxide. Now we also have-- and so we would release this much energy and we'd have this product to deal with-- but we also now need our water.
And this reaction right here gives us our water, the combustion of hydrogen. So we have-- and I haven't done hydrogen yet, so let me do hydrogen in a new color.
That's not a new color, so let me do blue. Will give us H2O, will give us some liquid water. Now, before I just write this number down, let's think about whether we have everything we need.
To make this reaction occur, because this gets us to our final product, this gets us to the gaseous methane, we need a mole. Or we can even say a molecule of carbon dioxide, and this reaction gives us exactly one molecule of carbon dioxide. So that's a check. And we need two molecules of water. Now, this reaction only gives us one molecule of water. So let's multiply both sides of the equation to get two molecules of water.
So this is a 2, we multiply this by 2, so this essentially just disappears. And then you put a 2 over here.
So I just multiplied this second equation by 2. So I just multiplied-- this is becomes a 1, this becomes a 2. And if you're doing twice as much of it, because we multiplied by 2, the delta H now, the change enthalpy of the reaction, is now going to be twice this. Let's get the calculator out. It's now going to be negative Because we just multiplied the whole reaction times 2.
So negative So it's negative Now, let's see if the combination, if the sum of these reactions, actually is this reaction up here. And to do that-- actually, let me just copy and paste this top one here because that's kind of the order that we're going to go in. You don't have to, but it just makes it hopefully a little bit easier to understand. So let me just copy and paste this. Actually, I could cut and paste it. Cut and then let me paste it down here.
That first one. And let's see now what's going to happen. To see whether the some of these reactions really does end up being this top reaction right here, let's see if we can cancel out reactants and products. Let's see what would happen. So this produces carbon dioxide, but then this mole, or this molecule of carbon dioxide, is then used up in this last reaction. So this produces it, this uses it. So those cancel out. Let me do it in the same color so it's in the screen.
This reaction produces it, this reaction uses it. Now, this reaction right here produces the two molecules of water. And now this reaction down here-- I want to do that same color-- these two molecules of water. Now, this reaction down here uses those two molecules of water.
Now, this reaction right here, it requires one molecule of molecular oxygen. This one requires another molecule of molecular oxygen. So these two combined are two molecules of molecular oxygen. You will notice that I haven't bothered to include the oxygen that the various things are burning in. The amount of oxygen isn't critical because you just use an excess anyway, and including it really confuses the diagram. Why have I drawn a box around the carbon dioxide and water at the bottom of the cycle?
I tend to do this if I can't get all the arrows to point to exactly the right things. In this case, there is no obvious way of getting the arrow from the benzene to point at both the carbon dioxide and the water.
Drawing the box isn't essential - I just find that it helps me to see what is going on more easily. Notice that you may have to multiply the figures you are using. For example, standard enthalpy changes of combustion start with 1 mole of the substance you are burning.
In this case, the equations need you to burn 6 moles of carbon, and 3 moles of hydrogen molecules. Forgetting to do this is probably the most common mistake you are likely to make. How were the two routes chosen? Remember that you have to go with the flow of the arrows. Choose your starting point as the corner that only has arrows leaving from it. Choose your end point as the corner which only has arrows arriving.
So why is this answer different? The main problem here is that I have taken values of the enthalpies of combustion of hydrogen and carbon to 3 significant figures commonly done in calculations at this level. That introduces small errors if you are just taking each figure once.
However, here you are multiplying the error in the carbon value by 6, and the error in the hydrogen value by 3. If you are interested, you could rework the calculation using a value of So why didn't I use more accurate values in the first place? Because I wanted to illustrate this problem! Answers you get to questions like this are often a bit out. The reason usually lies either in rounding errors as in this case , or the fact that the data may have come from a different source or sources.
Trying to get consistent data can be a bit of a nightmare. In this case, we are going to calculate the enthalpy change for the reaction between ethene and hydrogen chloride gases to make chloroethane gas from the standard enthalpy of formation values in the table.
If you have never come across this reaction before, it makes no difference. Note: I'm not too happy about the value for chloroethane! The data sources I normally use give a wide range of values. This uncertainty doesn't affect how you do the calculation in any way, but the answer may not be exactly right - don't quote it as if it was right. Select basic ads. Create a personalised ads profile.
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Develop and improve products. List of Partners vendors. Share Flipboard Email. Todd Helmenstine. Todd Helmenstine is a science writer and illustrator who has taught physics and math at the college level. He holds bachelor's degrees in both physics and mathematics. Updated December 03, Featured Video. Cite this Article Format. Helmenstine, Todd. Enthalpy Definition in Chemistry and Physics. Balanced Equation Definition and Examples.
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