Enthalpy change Calculations

[00:00:00]Welcome to my YouTube channel. So, here today we are going to be looking at the change in enthalpy calculations. So, we are going to solve these two uh problems which mostly don't miss in an exam.

[00:00:14]So, now the change in enthalpy is given by the formula the change in enthalpy is equal to the summation of the of the reactants reactants minus the summation of the products. So, this is the change. So, you can just think of it as uh for you to know the change in the energy, you must begin with the reactants. How much reactants did you begin with minus the products. So, this is uh how we are going to be solving it.

[00:00:49]Let's begin with this question. So, this question um our A says that we have acetal dehyde which is ethanol has ethanol with an A has a fruit aroma and is naturally contained in food such in foods such as such as fruits and fructose.

[00:01:17]Ethanol has the structural the structure shown.

[00:01:25]>> [clears throat] >> Gaseous bends by the equation.

[00:01:28]So, it bends by this equation. So, the structure is this.

[00:01:32]So, they have given us two diagrams, but you are supposed to know how you can come up with these two diagrams. So, the equations that they have given us are these. So, we have our CH3 CHOCH O plus two and a half which is uh 2.5 or two to produce what we have uh, our our water plus our two H2O. Then, we also we are also told that we have all these other equations. This is just the same equation, but written in a different form in which the bonds are showing. So, plus two and a half, then we have oxygen and oxygen with a double bond. This should be able to produce, according to what we have, it should be able to produce our our oxygen.

[00:02:29]It should be able to produce two oxygen to oxygen, then plus two we have our oxygen double bonded with carbon double bonded like that. So, this is, uh, the bonds that we have.

[00:02:44]Then, the best way to answer this is to express all these into structural formulas.

[00:02:52]So, what we have is what we have is this structure, according to what we are given, it's this, but you can, uh, still be able to draw it even if they haven't given you.

[00:03:04]So, we have the the bonds are like this.

[00:03:10]So, carbon is bonded like that. We have the methyl, then we have the carbon double bond oxygen. So, we have the double bond here, oxygen, then we have the hydrogen like that. So, this is the ethanol. Then, we have plus the two and a half.

[00:03:29]Then, we have inside the brackets, we have our O double bonded like that.

[00:03:35]Then, let's, uh, let's reduce the size of this. So, reducing the size of this, we are going to get something like, uh, so this should be equal to two, then the oxygen oxygen there should be a double bond.

[00:03:53]So, this is supposed to be water, not like this. So, it's supposed to be the the equation is supposed to be like not like this, but So, where have we written it? It's supposed to be like this.

[00:04:10]So, since it's water, it will be our our oxygen, then we have H and we have our H like that.

[00:04:21]Let me just write properly. So, we have our H bonded with oxygen, bonded with our hydrogen like that. Then we have plus that. So, that we have our our bonds like that. Now, here you needs to be very careful, but before that, let's uh just draw the way they are, like that. Then we have plus two. Then we have our oxygen bonded with with with our carbon bonded with our our another oxygen. So, with this done, we're going to have we're going to have something like this.

[00:04:57]So, now they give the standards of enthalpy. So, the standards of enthalpy for this question is we have the carbon bonded with hydrogen, we have our carbon bonded with carbon, we have our carbon bond double bonded with our oxygen, then we have oxygen bonded with oxygen double bond.

[00:05:20]Okay, so the values first.

[00:05:23]Since we we are writing even on that side, so we have this one we have 413.

[00:05:27]Carbon to carbon single bond we have 347.

[00:05:31]Then we have the carbon bonded with oxygen a double bond we have 736.

[00:05:39]Then, according to this, if we have oxygen double bonded with oxygen, we have a 498.

[00:05:46]So, we have a 198.

[00:05:48]Then, when we have our oxygen bonded with hydrogen, we have a 464.

[00:05:55]Then, last but not the least, carbon bonded with our oxygen, this should be equal to a 736 like that.

[00:06:03]So, this is uh This is uh what we are going to have.

[00:06:09]Therefore, if this is the case, we're going now to look at our bonds.

[00:06:17]So, the first question says, "Use the mean bond energy data to calculate the enthalpy change which occurs when the when the bonds in the reactants shown above the equation are broken." So, we can say that the summation of bonds of reactants So, the reactants that we have are We are going to look at this. So, look at the number of Let's begin with hydrogen since the one that is common here is the hydrogen and the the what's this?

[00:06:57]Carbon with hydrogen, so let's begin with that. So, if we do that, we're going to have the first bond, carbon to hydrogen, number one. Bonds two.

[00:07:08]Bonds number three.

[00:07:10]Bonds number number four, that is hydrogen to carbon.

[00:07:14]So, we are going This is how you represent your work to avoid uh making a mistake. So, you're going to say four four [snorts] times the carbon to the hydrogen like this. Then, we're going to say plus. What other bonds can you see? So, the other bonds that we can see are uh carbon to carbon.

[00:07:38]So, we have the carbon to carbon there.

[00:07:41]So, this carbon to carbon is one bond.

[00:07:43]So, we are going to say plus the carbon to carbon like that. Then we have plus then another bond that we can be able to see carbon double bonded with oxygen here.

[00:07:57]So, we are going to have plus our carbon double bonded with our oxygen. That is they want us to find for the reactants like that. So, the reactants we have carbon [snorts] double bonded with oxygen there. Then last but not the least, we have the oxygen bonded with oxygen, which is here. Then times the 2.5. So, we are going to say we are mo- we are going to move this and have plus the 2.5, which is 2 and 1/2 is the same as 2.5 times the oxygen oxygen double bonded with oxygen like that. So, that means to say we are going to get a a value that would be equivalent to we have four. Then take the carbon to hydrogen bond. Carbon to hydrogen bond according to this is equal to 413.

[00:08:53]So, we are going to get our 413 413 like that.

[00:08:58]kilojoules So, these are in kilojoules per mole.

[00:09:03]So, we have this. Then we are going to say plus what is our carbon to carbon?

[00:09:09]So, the carbon to carbon bonds is um equal to according to this carbon to carbon is our our value is 374.

[00:09:19]So, we are going to add our 374 370 347 or 374? 347. Then we have plus our carbon double bonded with oxygen. This produces what we have what we are going to have is uh carbon double bonded with oxygen double bonded with oxygen is a our four 90. Okay. Carbon bonded with oxygen gives us carbon bonded with oxygen. Where is it?

[00:09:55]Carbon bonded with oxygen double bond is 736. So, we are going to put our 736 like that.

[00:10:04]Then, we have plus our 2.5.

[00:10:08]So, this will be our 2.5.

[00:10:13]2.5 by oxygen double bonded with oxygen.

[00:10:18]This is going to give us a value of Remember, using this oxygen double bonded with oxygen is 498. So, we are going to say by our 498 like that. So, adding these First, let's multiply. So, we have our 4 * our 413. This gives us a value of 1,652 plus our 347 plus our 736 736 plus our our 2.5 * 4 498. This gives us 1,245.

[00:10:58]So, adding, we are going to get a 1,652 plus our 347 plus our 736 plus our 1,245.

[00:11:13]So, this is going to give us the summation of uh of the bonds of reactants.

[00:11:22]Reactants should be equal to 3980 kJ. So, according to this, you obtain your five marks like that.

[00:11:35]So, after that, let's move on to the next one.

[00:11:38]So, we have calculate the enthalpy change which occurs when all the bonds in the production in the above equation are for So, for this one, we have calculate the change in the enthalpy that occurs when all the bonds in the production in the above equation are formed.

[00:11:56]So, to to look at we are looking at the product. So, here you need to be very careful. If it is double bonded, that means to say we have oxygen bonded with this. That's one bond. The up the bond is um oxygen bonding with this other one. So, here there is a two.

[00:12:17]So, same applies here. We have one and number two. So, that means to say you multiply another two. So, what this means is that for the bonds of the products, so looking at this, let me just lift this.

[00:12:33]So, for the products, we are going to say So, the summation of bonds of products bonds of products will be equal to We are going to have this two that is outside times another two, then the O H bonds. Remember, there are two O H bonds. That's how come I have split them. There is There is this bond O bonding with this hydrogen and O bonding with this hydrogen. So, together they are two bonds of that type. Even here, we have one bond, then the second bond is this one. So, we are going to say that will be our bonds there. Then we are going to add plus our two that is outside. Then there are two type two bonds of the same type. So, we have our two, then we have our O double bond with C like that. So, with this we can say two 2 * 2 is 4, so we can say 4 4 then multiplied by what is our OH bond? So, the bond energy for OH single bond is 464, so we can say 464 like that kilojoules. Then we have and then we have plus our 2 by 2 is 4 then times our oxygen double bonded with carbon gives us a value of our 464.

[00:14:05]So, we can say 464 like that.

[00:14:09]So, 464 like that. So, with this done Okay, so is this the same?

[00:14:18]Oxygen with hydrogen and oxygen with carbon Oxygen with carbon double bonded is 736 not not what I wrote 736.

[00:14:33]So, here it will be 736.

[00:14:37]736 like that.

[00:14:40]So, 736 kilojoules.

[00:14:44]So, this gives us what we are going to have as 4 * 464. This gives us 1,856 plus we have our 736 by 4. This gives us a value that is equivalent to our 2,944.

[00:15:05]So, when we add these we are going to get 2,000.

[00:15:10]So, we have 1,856 then we have plus two nine four four So, this is going to give us a value that is to 4,800 kJ. So, this is the enthalpy of the products, like that.

[00:15:31]Then, finally, the final question for this uh question is saying that hence hence So, we have hence calculate the enthalpy change for the complete combustion of of ethanol as shown in the diagram. So, the enthalpy change of the reaction will be equal to the enthalpy change of of the reactants.

[00:16:06]So, you must subtract the reactants minus the enthalpy change of the products.

[00:16:13]products. So, if you wants to be remembering this, remember RTP. So, just this will help you. It will be R minus T. So, RTP is room temperature and pressure. So, just doing this. Then, uh for T, just say minus so that you know which ones comes first. So, we are going to have our our change in the reactants was found to be 3,980 kJ minus the products, which is 4,800 kJ.

[00:16:49]800 kJ, like that. So, this gives us a value of 3,980 minus 4,800.

[00:16:58]This gives us a value of negative 820 kJ, like that. So, the enthalpy of the reaction should be equal to this amount of the kJ.

[00:17:11]So, with that done, with that done, we have wrapped up this question. Let's look at our final question so that we are conversant with these types of questions. They carry a lot of marks and just you doing this will help you. So we have this one. Calculate the value of the of the enthalpy change of combustion of ethanol given the following. So enthalpy combustion of ethanol. So combustion meaning to say you burn it in oxygen. So you're going to have this is we have the six.

[00:17:53]So you have uh two carbons. So we can say C2.

[00:17:58]Then we have how many hydrogens? 1 2 3 4. Five those that do not have OH. So we have H5. Then we have the OH there. Then it burns in oxygen. So when alcohols burn in oxygen, they'll produce carbon dioxide plus water only like that. So with this done, we are now going to balance the equation by putting our values such as uh So carbon, there are two carbons. Make sure that this one is intact. It is just one more. Then here we have since there are two carbons here, we are going to put two carbons. Since there are five six hydrogens, so if you add this five plus this one, it will be equal to we have uh six. [snorts] So we have to put a three here so that three by two is uh three by two gives us a value that is equivalent to six hydrogens. So now to count the oxygens, two by two by two is four plus the three, it will be seven.

[00:19:07]Here we have two. So if we add a three here, it will be three by two is six plus one that's seven.

[00:19:14]So, we have balanced the equation in that regard. So, with that done, what are we going to do? So, what we are going to do is that we are going to have we are going to have our value our value which will be equal to So, before that, let's write this in terms of the compound like that. So, we have carbon carbon then we have hydrogen hydrogen hydrogen hydrogen then we have our OH, we have our H, we have our H, we have our H, we have our H, we have our H then we have plus our three the double bond of oxygen here. So, oxygen since oxygen there are two of them, they form a double bond.

[00:20:01]So, oxygen likes to form two bonds like that. So, they will form the double bond like that. Then we have this should be able to produce we have carbon carbon and oxygen will form the double bond like that.

[00:20:18]So, this is what we have plus the three then water also we have the hydro oxygen there hydrogen hydrogen like that. So, this is basically how it's going to be for water like that. So, if this done, we are now going to to do So, they want us to find the cal calculate the standard enthalpy of combustion meaning to say they want us to find the change in enthalpy of the reaction. So, we will find the bonds then the the reactants then the bonds. So, we can say that the summation of the summation of bonds of reactants reactants will be equal to the change in enthalpy of the reactants will be equal to Okay, so we have the change in the enthalpy of the reactants should be equal to So, look at the number of bonds. So, we have one two, that's the hydrogens, three bonded to to to carbon four five, like that.

[00:21:40]So, we have five of them, so we can say our five by our carbon bonded to carbon bonded to hydrogen, like that.

[00:21:54]Then we have plus our three, like that.

[00:21:58]Then our oxygen bonded with another oxygen. Oh, sorry.

[00:22:04]We are not done with all the bonds of this. So, the other will be carbon with carbon. So, carbon with another carbon is uh So, we only have one bonds there, so we are going to say plus the carbon to carbon plus the final bond is carbon and So, even here, be very very careful.

[00:22:29]So, even here you're going to have something like this. So, oxygen bonded like that. So, remember, they haven't given us ox- oxygen um carbon with OH, but each one is having a bond, so you separate it like that. So, now we are on carbon with uh with oxygen. So, the change in enthalpy for that Our value for this one is that we have carbon bonded with oxygen is 360. So, we can get the 360 and write it. Oh, before that, we have our carbon single bonds like that plus then last but not the least is carbon oxygen bonded with hydrogen.

[00:23:14]Which will be something like this. So, oxygen bonded with carbon.

[00:23:21]Oxygen No, not not that, but our oxygen bonded with hydrogen there.

[00:23:31]We have uh hydrogen like that, oxygen bonded with hydrogen with a single bond like that. So, we are just basically getting all the bonds. Then the bonds we can now plug in the numbers. This prevents you from making any mistake plus the three, then we have the oxygen double bonded with another oxygen like that. So, this gives us we have our five.

[00:24:00]Let me even put it here. We have our five.

[00:24:03]Then the carbon with hydrogen bond, carbon with hydrogen bond is equal to We have the carbon hydrogen bond is 412, so we're going to say 412 kJ. Then we have plus our carbon to carbon bond. Carbon to carbon bond is 348 kJ like that. So, we have 348 kJ, then this We have plus carbon with oxygen. Carbon with oxygen is uh So, carbon with our oxygen is uh This is the one that we have written, 360. Okay, we haven't written it, but 360. Carbon to carbon 348 was it? Yes.

[00:24:56]So, we have done that. Then, we are going to write our plus our oxygen with hydrogen. This one is equal to oxygen with hydrogen.

[00:25:09]We have our 463.

[00:25:11]So, we we are going to write our 460 463 like that. Plus, then the last one will be three. Oxygen double bonded with oxygen is uh equivalent to oxygen bonded with oxygen is equivalent to 496.

[00:25:30]So, we have our 496.

[00:25:35]496.

[00:25:37]So, we are going to write our 496 kJ like that. Then, just add them. So, we have our five five by 412.

[00:25:48]Then, we have uh 412 plus So, five by 412 is our 2060 kJ plus our 348 kJ plus our 360 plus our 463 plus our three times 496 gives us 1,488.

[00:26:16]Then, this when we add them, they are going to give us So, 2060 plus 348 plus 360 plus our 463 plus our 1,488 we are going to get a value that will be equivalent to So, the enthalpy is the enthalpy of the reactants reactants should be equal to you add this, it's 4,100 and um like that. If you do this, it will be very very much easy for you not to make a mistake and you would get a correct answers. Then, finally, we are going to say the summation of the the bonds of products.

[00:27:20]So, we can say the enthalpy of products should be equal to we are going to say our products as we can see from this. We have Again, we have this is having one bond there.

[00:27:37]Then, the second bond there. So, we have one bond. Then, the second bond is here.

[00:27:43]So, it will carry a two. So, it will be this two times the number of bonds there. So, we have we have our Okay, let's first write our two times our two, then the carbon is bonded like that. There are two bonds of this type.

[00:28:00]Then, plus Even water itself, it is it is having one the first bond, then the other bond is this hydrogen with oxygen hydrogen with oxygen like that. So, it will also carry a two. So, we have a two outside already.

[00:28:17]We have a two a three, sorry. So, we have a three outside. So, we can say three, then we are going to open. We have two types of two bonds of the type of the hydrogen the the oxygen bonding with hydrogen.

[00:28:35]For that type of a bond, then we are going to have our two times two is four.

[00:28:41]So, we have four, then times our carbon with our oxygen double bond is equivalent to carbon with oxygen, so you have to look. So, your carbon and oxygen double bond is 803.

[00:28:57]So, we are going to write our 803.

[00:29:00]So, 803 like that.

[00:29:03]Then, we have uh plus our 3 * 2 gives us a value of 6. So, we have our oxygen with our hydrogen.

[00:29:16]Oxygen with our hydrogen, it will be 463.

[00:29:21]So, we have our 463.

[00:29:24]463 kJ.

[00:29:27]Then, we are going to add. So, our 4 * our 803, this gives us a value of 3,212 plus our 6 * uh 463 gives us a value of 2,700 and 78 like that kJ.

[00:29:49]So, we have something like that. So, adding these our 3,212 plus 2,700 and 78, we are going to have 5,000 and 5,990 kJ like that. So, this is the change in enthalpy of the products. Then, finally, to find the change in enthalpy of the reaction, which is simplified as RXN, should be equal to the summation of the enthalpy of the reactants reactants minus the summation of the enthalpy of products.

[00:30:41]So, this gives us a value that will be equivalent to we have the summation of the reactants, which will be So, the reactants, if you recall, it was 4,719.

[00:30:55]So, we have 4,719 kJ minus the reactants, which is 599 0 kJ. So, this gives us a value that will be equivalent to 4,719 minus our 5990.

[00:31:15]This gives us our -1,271 kJ like that. So, this is the enthalpy of the reaction like that.

[00:31:25]So, this is our answer for that one.

[00:31:29]Then, for the B part, we are told to use using your answer in C1 above, calculate how much energy is in is evolve involved or released or gained when 12 g of ethanol is burned in excess oxygen.

[00:31:52]So, you look at the equation, make sure that it's balanced, and look which one do they want.

[00:32:01]So, looking at this, using the answer from C1 above, calculate how much energy is involved in is involved, released, or gained when 12 g of ethanol is burned in excess oxygen.

[00:32:15]So, you're going to get your equation.

[00:32:17]So, this is the your equation. So, this equation is equation you're going to use and do this.

[00:32:26]So, simply put this equation is balanced by one, and you have found that your change in enthalpy for this equation is equivalent to you have the change in enthalpy change in enthalpy is found to be equal to -1071 kJ. So, now what we are going to say is that so simple analogy which I used in the previous video. If you have four let's say if you if they are selling if you have a two quarter, let's say you can buy four pieces of bread what if you have one quarter? So, you will be able to say you know that it you will be able to buy two. So, it will be X you don't know then cross multiply two times X it will be 2x is equal to 1 * 4 it will be 4.

[00:33:19]Then divide by two divide by two. So, our X value will be equal to two like that.

[00:33:26]So, our X value will be equal to two like that.

[00:33:30]So, with this done with this done this is the principle we are going to use. So, the one that they are telling us is uh is uh for combustion of the one that they mentioned is the one you use the above calculate how much is involved when 12 g of ethanol. So, the ethanol is burned. So, we are going to say find the mass of this ethanol in this reaction.

[00:33:59]So, it will be C2H5OH.

[00:34:03]So, it's going to be ox carbon or hydrogen is 1.01 as if you're just finding the molar mass grams. But, you will just treat it as your mass. Then oxygen is 16 g. Then here we have five times we have five of them.

[00:34:20]So, five times hydrogen has got the molar mass of 1.01.

[00:34:26]Then we have our two times carbon is 12.01.

[00:34:31]So, we are going to have 1.01.

[00:34:34]We have our 16. We have our 5.05 when you multiply this. Then, we have our 24.02 like that. So, when you add these, you are going to get a value that is equivalent to So, 1.01 plus our 16 plus our 5.05.

[00:34:54]Then, we have uh plus our 24.02.

[00:34:59]So, this gives us a value that will be equivalent to 46.08 um grams.

[00:35:07]Then, just do your math.

[00:35:10]So, this was the whole thing that is able to produce this. So, you're going to say 46.08 g is the one that is able to produce our three it is the one that is able to produce our -1,271 kJ like that.

[00:35:31]Then, how about if you have um how much of it?

[00:35:36]How about if you have How about if you have the 12 g that they are telling you?

[00:35:44]You're going to say, "How about the 12 g? How much energy will it produce?"

[00:35:49]You're going to say X. Cross multiply.

[00:35:52]So, it will be 46.0 46.08 g X should be equal to 12 g * our 1,271 kJ like that. Then, divide the 46.08 g over the 46.08 g. So, this gives you a value of X is equal to We have our 12.

[00:36:20]our 12 * our 1,271 like that.

[00:36:26]Which gives us a negative 15,252 over our 46.08 g. So, our X value should be equal to we have our divided by our 46.08.

[00:36:44]This gives us a value of our 300 and 30.99 kJ like that. So, this is how we deal with those questions. Thank you very much for watching. Hope you would like, subscribe, and share for more. See you in the next tutorial.