Hessy's Law

[00:00:00]Everyone, welcome to my YouTube channel.

[00:00:02]So, here today we are going to be looking at Hess's law. We are going to be solving all these problems, which will equip you for an exam.

[00:00:10]So now, let's define what Hess's law states. So, Hess's law just simply states that the total enthalpy change for a chemical reaction is the same regardless of the roots taken, provided that the initial and the final conditions are the same.

[00:00:26]So, let me explain the definition first.

[00:00:29]So, let me just rephrase it. So, we are saying that the total enthalpy change for a chemical reaction is the same regardless of the roots taken provided that the initial and final conditions are the same. So, what you can just think of it is uh something like this. So, just think of Hess's law as uh you are at point A, you want to go at point D. So, if someone goes like that, goes like that, they will reach D.

[00:01:01]It is the same way even if someone goes like this.

[00:01:06]We are just interested in the initial and the final position. So, this as a result is what we call Hess's law. So, the we are saying that the total enthalpy change for a chemical reaction is the same regardless of the roots taken provided that the initial and the final conditions are the same. So, you can just think of it like this. So, if you are asked Hess's law, you can just draw A and your B, then draw two paths like that, then come up with your definition using those two paths.

[00:01:45]So, that is Hess's law. So, as long as you add them, regardless of how you add them, but if you reach the same conclusion, that simply means to say you are going to have the same enthalpy change. So, how do we do some calculations under Hess's law? Let's begin with the simplest one.

[00:02:02]So, let's say we have our A plus our B plus our C is the change in enthalpy is equal to -50. We have our C plus D is equal to E. The change in enthalpy is equal to 30. So, they can give you two equations and ask you calculate the enthalpy change for A plus B plus D. So, how do you do this? So, if C is here and C is here, that means to say you can be able to cancel your C like that. After you cancel your C, since they are on the opposite direction, then you add your A plus your B plus your D that is remaining plus your D that is remaining should be able to give you your E like that. So, the change in enthalpy, as long as you have done that, you are just going to add them. So, it will be your negative uh negative 50 negative 50 minus the 30. So, you're just adding them normally. So, you have added these two. So, this plus this will give you negative 80. So, this is Hess's law. So, kilojoules.

[00:03:07]So, basically, this is Hess's law. So, the that is how we do that. So, that means to say under Hess's law, if you have let's say your A plus your B is equal to uh to produce let's say C. If C is uh let's say negative 30 kilojoules.

[00:03:28]30 kilojoules. That means if you have to change, you're beginning now with your C to produce A plus your B, this negative changes to a positive 30 kilojoules like that. So, this is how we do Hess's law calculations. Let's now begin answering questions that are examinable.

[00:03:49]One thing that I also forgot to mention is if you have your A plus your B should be equal to C if you have a 50 like that kJ. If you multiply two to these letters where you have your 2A plus your 2B should be equal to 2C.

[00:04:07]Even this one must be multiplied by two which will give you 100 kJ. Likewise, if you multiply by half so that the equation becomes half of A plus the half of B should be able to produce half of C, then even this one it will be half times 50 will give you a 25 kJ. So that is how we do with these calculations.

[00:04:35]So let's now answer this one. So the question says that use the reaction enthalpies to determine the enthalpy change for the following. So we have our P4 a gas plus the 10 Cl2 a gas also to produce four moles of uh phosphorus pentachloride.

[00:05:01]So how do we answer this one? So this is what we want to arrive at. We want to arrive at this answer. So since this is what we are want to arrive at we are going to be making these two equations be similar to the answer we expect. So let's write these equations.

[00:05:20]So we have PCl5 to produce PCl3 plus Cl2. The change in enthalpy is equal to 157 kJ per mole.

[00:05:34]Then we also have the other equation is P4 P4 plus 6 Cl2 to produce to produce 4 PCl 3 the change in enthalpy is equal to -1207 kJ per mole like that.

[00:05:58]So now we want this this this these two equations to form this. So you look at your first one. So let's call this one as our equation one as our equation two.

[00:06:08]So if you look, PCL5 is on this other side. That means to say you have to reflect the equation. So we are going to say flip equation one.

[00:06:22]So you are flipping this equation one.

[00:06:25]Look, is there any number on PCL5? There is four, so and you are going to say and multiply by four.

[00:06:36]by four.

[00:06:39]Or I can just say to save time multiply four. So we are going to say this equation will Uh so we are going to begin with this.

[00:06:48]So it will be PCL3 plus CL2 to produce our PCL5 like that. Then this enthalpy is negative is 157, so it will be a negative the change in enthalpy.

[00:07:08]Change in enthalpy.

[00:07:12]Since you have reflected the equation, it will be -157 like that. So it will be 157 kJ like that.

[00:07:24]kJ per mole. So then we are going to get the other one, which is our We are going to look at this. So we have P4.

[00:07:36]We have P4.

[00:07:39]So that means if we have P4 and the required just P4 without a number, that means we just add. So, we are going to just get this equation P 4 plus it is 6 Cl2 to produce 4 PCl3 4 PCl3 the change in enthalpy is equal to negative 1,000 and 7 kJ/mol.

[00:08:16]So, this gives us a value that will be equivalent to So, in this regard, we are going to cancel whatever is common. So, as you can see as you can see So, as we said, we have to multiply by 4 to this equation.

[00:08:35]So, multiplying by 4, we are going to write this Let Let me keep it.

[00:08:40]So, multiplying by 4 means Let me cut it. So, multiplying by 4, we are going to have 4 PCl3 plus we have our 4 Cl2 should be able to produce 4 PCl5.

[00:08:59]Then, even the the enthalpy change should be multiplied by 4. So, as multiplying by 4 means we are going to have 157 times uh 4. It's negative for that matter. So, we are going to have negative uh 628 kJ/mol like that.

[00:09:21]Then, when we do this, what we are going to get is that what we are we are going to get the second equation. Since this second equation has got P4 of which what they want is P4 there. Then, this side they they have this which is also found on the other side. So, we are going to get that equation and put it like that. So, when we do that we are going to compare. If there is anything common on different sides, you cancel it if the coefficients are the same. So, as we can see this and these are the same and they have the same number like that.

[00:09:59]So, we are going to cancel them. So, we are going to do this. When we cancel we are remaining with our P4 plus 4 plus 6. They're on the same side, so it will be 10 Cl2 should be able to produce our 4 PCl5.

[00:10:20]Then we have the change in enthalpy will be equal to So, the outcome will be equal to -1,835 kJ as our answer like that.

[00:10:32]So, this is our answer for this problem.

[00:10:35]Let's move on to our next one.

[00:10:37]So, looking at this other one, how do you go about this one? So, here you need to be very careful. So, you identify what do they want? They want NO or and NO2 at the end and this equation has that.

[00:10:52]But one thing is it has got O3 and the first one has got O3. Now, you know to say you cannot flip this equation because you want NO to be the first thing. So, that means to say you will have to flip the second equation.

[00:11:08]So, you're going to say or the first one. So, this is the first one. So, the first equation is we have our 2 O3 to produce our 3 O2 then we also have our O2 to produce our 2 O like that.

[00:11:27]Then the last one is saying that we have our NO plus our O3 to produce our NO2 NO2 plus our O2.

[00:11:42]So, what we are going to do is this is our equation one, equation two, equation three like that. So, we are going to say that flip equation one. So, we are going to flip equation one. So, flipping the equation one means So, the change in enthalpy for this one is equal to we have a negative 427.

[00:12:04]This one we have the change in enthalpy is equal to we have uh 495.

[00:12:10]And the last one, the change in enthalpy is equivalent to let's reduce this the size.

[00:12:17]So, the change in enthalpy for this one is equivalent to negative 199.

[00:12:27]If I'm not mistaken, negative 199 kilojoules.

[00:12:31]So, what we are going to do is flip equation one so that we have our three O2 will be able to become we have our two O3 like that.

[00:12:44]And this one will become positive. So, the change in enthalpy for this is a positive 427.

[00:12:51]Instead of a negative, it will become positive 427 because we have flipped the equation.

[00:12:58]So, this is what we have.

[00:13:01]Now, we are going to now compare with what you really want. So, as we can see here, this equation has got O3, but it doesn't have a two.

[00:13:13]Now, we can choose to multiply by half on both equations, but uh since three, if we multiply by half, it will be three over two, we are going to avoid doing that. So, any any way that you do it, it's still okay as long as you arrive at the same answer. So, what we are going to do is write the three equations now. We are going to have our O2.

[00:13:38]O2, it's saying that O2 should be able to become 2 O like that.

[00:13:47]So, the change in enthalpy for this one is equivalent to 495.

[00:13:55]Now, look at the final answer. O must be this side. So, we are going to say on top, we are going to flip this equation.

[00:14:04]So, flip flip equation flip equation two so that we have our 2 O should be able to become our O2. The change in enthalpy will now become negative since we have flipped the equation like that.

[00:14:24]Then, what we are going to do next is the third equation. Third equation has got O3.

[00:14:31]And this one has O3 this side, but it has got a two. So, we are going to say multiply two to equation three.

[00:14:48]So, to equation three. So, the equation three, which is our NO Let me just lift this equation.

[00:14:58]So, this is the equation in which we are going to multiply the two.

[00:15:05]So, multiplying the two, it will be equal to So, we are going to multiply by a two like that.

[00:15:20]So, this gives us 2 NO + 2 O3 to produce 2 NO2 + 2 O2.

[00:15:31]So, this change in enthalpy will be equal to 2 * -199.

[00:15:36]This will be equal to It's going to give us a value that will be equal to -398 kJ like that.

[00:15:46]So, this is what we have. So, after doing that, we can now add them. So, don't worry about here looking at the coefficient. It's one. You can always divide at the end.

[00:15:58]So, now we have uh we're going to say adding the three equations.

[00:16:06]So, adding the three equations, we're going to get all these three equations and add them. So, the first equation that we flipped is this with its with its enthalpy change.

[00:16:20]So, we are going to do this.

[00:16:22]Put it there.

[00:16:25]So, this is what you do. You You also write the other equation, which is this one.

[00:16:34]So, this is what we have. Then, the last equation, which is uh this one.

[00:16:41]We're going to take this equation and put it there.

[00:16:49]So, after we do this, after we do this, we can now add them.

[00:16:54]So, cancel whatever is common.

[00:16:57]So, we have uh things that are common on different sides. We have uh this and this.

[00:17:04]So, that thing needs to be canceled.

[00:17:06]What else can be identified?

[00:17:09]We have So, as you can see, O2 + 2 O2 is 3 O2, which is here.

[00:17:16]So, we can say that this is common because O2 plus 2 O2 is 3 O2. So, we are going to have those. Then, just cancel.

[00:17:26]We have this. So, 1 + 2 is 3. That's why we cancel that. Then, this and this.

[00:17:33]After doing that, just add them. So, we are going to have We are remaining with our 2 O, but we can begin with this. So, we have our 2 NO plus our 2 O like that. So, this is going to give us what we are remaining with is this side it's just our 2 NO2.

[00:17:54]Then, the change in enthalpies, you just add them. So, you are going to press your 427 minus your 495.

[00:18:03]Then, you have minus your 398 like that. So, this is going to give us the change in enthalpy will be equal to -466 kJ like that. Then, remember it is similar to what they want in the equation, but not quite. So, we need just one to be on every coefficient.

[00:18:27]That means to say we multiply by half.

[00:18:31]So, getting this equation and multiplying by half gives us gives us uh a value.

[00:18:41]So, we say multiply half.

[00:18:48]So, we are going to have our half multiplies everything. Half by this will give you just NO. Half by this, it will give you O. Then, half by this two, it will give you NO2. The change in enthalpy will be equal to Even half by 466, it is going to give us a value of -233 kJ like that. So, this is uh how we answer that problem.

[00:19:21]Let's move on to our next problem.

[00:19:24]Looking at this other one, we have used the reaction and the enthalpy changes to predict the standard enthalpy change for this one.

[00:19:33]So, how do we do this? So, we want to find this equation using this equation. So, we're going to use this equation. So, we have our half N2 a gas plus our O2 a gas as well to produce our NO2. Then, the enthalpy change is equal to -33.2 kJ.

[00:19:57]So, we have the kilojoules.

[00:20:04]Then, this other one, we have our H2 a gas plus half of O2 to produce H2O, which is uh the change in enthalpy is equal to -241.8 kJ.

[00:20:19]We have our N2 a gas plus our 2H2 to produce N2H4.

[00:20:31]The change in enthalpy is equal to 47 kJ.

[00:20:36]So, the equation where we want to get Let me copy it and put it here.

[00:20:43]So, using these equations, we want to get to here.

[00:20:47]So, identify the first equation. You have half N2.

[00:20:52]N2 NO2 is anything showing here? We have NO2 here, and NO2 is this side. So, what we're going to do is Let's call this one as our equation one, this one as our equation two, this one as our equation three. So, as you can see this N doesn't have a two and here it has a two. So, you and it is on the other side. So, we are going to say flip flip equation one and multiply two.

[00:21:28]So, that means getting this equation.

[00:21:31]So, I'll just be getting the equations to be faster. So, getting this equation, we are going to say this equation by we multiply our two.

[00:21:47]So, multiplying a two we have the two like that. So, two by half is just uh So, before that, let's flip this equation. So, flipping this equation means it will be NO2 to produce half of H two plus our O2.

[00:22:09]Then, the change in enthalpy will be equal to -30. Instead of negative, it will be positive 33.2 kJ like that.

[00:22:18]So, after doing that, we multiply by the two. So, multiplying by two, it will be 2NO2 should be able to produce two by half is just this and this will cancel. So, it will be N2 plus our 2O2 like that. So, the change in enthalpy will change to a value such as 66. 4 kJ like that.

[00:22:44]So, 66.4 kJ. So, we have one equation that is similar to the answer that we expect.

[00:22:54]So, with this equation, what are we going to do?

[00:22:58]We go to the next equation, which is this one. So, this one has got water. Do we have water here? So, yes, we have water. And water, according to this, it should have a two in front, and it should be on the other side, as you can see it has a two and on the other side.

[00:23:15]So, we are going to say flip equation two and multiply by two. So, we are going to say flip equation two and multiply multiply by two. So, this gives us taking the second equation we're going to have our value.

[00:23:45]>> [snorts] >> So, this value will be equal to we have this. So, we have to flip it, so it will be our H O our H2O should be able to become H2 plus half of O2, then the change in enthalpy will now become positive 241.8 kJ. So, we are now going to multiply by two. So, it will be 2H2O should be able to produce 2H2, then we have plus half by two by half, it will just be O2, then the change in enthalpy will be multiplied by two. So, this value when we multiply it will be so, [snorts] our 241.8 times our two will be equal to 483.6 kJ.

[00:24:51]So, 486 uh 483, sorry, 0.6 kJ.

[00:24:58]Joules.

[00:25:01]Then after that, we look at our final equation. So, this is the other equation.

[00:25:15]So, now after we do that, we look at our final equation, which is uh this one. So, this equation it has N2.

[00:25:27]H2.

[00:25:29]We don't have any of these, but we have this. So, as you can see, it's on the same position as where it is in the ori- in this other equation, and it has got the same number. So, you just lift it.

[00:25:40]So, lifting it means uh you're going to put it there. So, it will just be the same equation maintained.

[00:25:49]Where is it? It's here. So, this equation So, you can just add a comment to say maintaining equation two.

[00:26:01]Or maintain equation three.

[00:26:06]Maintain equation three.

[00:26:10]We're going to have something like this. Then finally, add the three equations.

[00:26:19]Adding equations.

[00:26:24]So, we're going to get all these equations, the three of them.

[00:26:28]We're going to have our value, which is going to be here.

[00:26:39]And the other equation, which will be here.

[00:26:51]And the final equation, which is this one.

[00:27:05]So, with this done, we are going to say >> [sighs] >> let's [snorts] cancel whatever is common. So, what is common?

[00:27:15]So, we have our Let me be using the same color for what is common. Let's use blue.

[00:27:24]Okay, so [clears throat] where do we have things that are common?

[00:27:35]water not common We have hydrogen and hydrogen here.

[00:27:45]They are common. What else is common?

[00:27:50]So, basically, that's all the things that are common.

[00:27:58]Okay, so looks like these are the only things that are common.

[00:28:05]I hope there's no mistake.

[00:28:07]So, from that, let's just add now. Let's see what is the answer it expects N H 2.

[00:28:14]Okay, so adding now means this we have our Even N 2 is common.

[00:28:27]So, we can now cancel them. So, this and this goes.

[00:28:31]What else goes?

[00:28:33]This one, they are on the opposite side, so you can cancel them. So, then we are just remaining with our two N O N O 2 plus the water there on this side. That those are the only things remaining. Then on the other side, you have your 2 + 1. So, this is both oxygen. 2 + 1 is 3. So, we are going to say should produce 3 O2 plus our H2 N2H4, like this. Then the change in enthalpy will be the addition, which will be equal to So, if we add this, we are getting a value that is equivalent to This will give us a value that will be equivalent to 5 97.

[00:29:33]So, the change in enthalpy should be equal to 597.6 kJ.

[00:29:41]So, we have kilojoules.

[00:29:47]So, that is our answer for that one.

[00:29:49]Let's look at our last two problems.

[00:29:53]So, this one is one of the most challenging in this video. So, we have the enthalpy change of the reaction at 27° C are these.

[00:30:05]Now, from the above, calculate the heat change. They are not asking for the enthalpy change, heat change. But first, let's calculate the the enthalpy change and then use this formula. The change in enthalpy is equal to we are going to have the the change in the heat, which is U, the internal energy, plus the Sorry, minus.

[00:30:31]Unless it's the other side, that's when you subtract. So, minus the that gas formula. So, where R are you, you have to use the 3 8.314.

[00:30:42]So, first find the change in enthalpy, then make the change in heat to be the subject of the formula by making this cross the other side. But, let's first find the change in enthalpy.

[00:30:53]Which we will use which we are finding using the same usual means. So, the change in enthalpy will be equal to we have our uh N NA plus our half Cl 2 to be able to produce this.

[00:31:16]Then, we have our change in enthalpy of formation is equal to According to this, we have -411 kJ/mol.

[00:31:30]The next one, it is our hydrogen plus our sulfur should plus our Sorry, not hydrogen plus sulfur plus 2 O2 can make a mistake.

[00:31:44]2 O2 to be able to produce this.

[00:31:53]The enthalpy of formation is equal to 811 kJ kJ per mol.

[00:32:10]Then, the other equation, which is our equation three which is our C. So, we have our 2 NA plus our sulfur I'm just copying, so let me just copy.

[00:32:29]So, this is the required equation. So, as we can see, the first thing is that identify. So, this is your equation A.

[00:32:36]So, this is your A, your B, it's just that space. So, we have our A there.

[00:32:42]We have our A there. We have our B, then we have our C, then our D like that. So, they want this equation. So, sodium should be this side. Look at your equation one. So, your equation one, sodium is the other side and it lacks a two. So, you're going to say flip flip A and multiply multiply two. So, flip A and multiply two. So, you're going to get this equation. Flip it, so it will be sodium chloride like that to become to become our sodium plus the half of chlorine like that.

[00:33:33]Then, multiply your two.

[00:33:35]First, let's write the change in enthalpy.

[00:33:42]So, the change in enthalpy by flipping this equation means it will change from 411. Was it negative or positive? It was negative. So, from -411 it will be positive 411 kJ per mole like that. So, after we do that the next thing that we are going to do is we multiply by two.

[00:34:12]So, as we have said, we are going to multiply by the two which will give us a value of two by sodium chloride, it will be two sodium chloride to produce two by sodium, it will be this plus two by half, it would be that.

[00:34:37]Then the change in enthalpy will be equal to So, if you multiply your 411 by two, it will be 822 kJ per mole.

[00:34:51]Then the next one, it will be equal to So, this is the equation that we are going to get for this problem.

[00:35:02]Then let's look at our B. Are we going to do anything with our B? So, our B is the H2 plus S plus And then let's look at the sulfuric acid. Is it showing in the equation? So, sulfuric acid is showing and it only has one.

[00:35:17]There's one sulfuric acid and here there's one. So, the only thing we're going to do is we're going to flip the equation.

[00:35:25]So, just flipping equation B, we're going to have without multiplying anything. So, flip flip B.

[00:35:36]So, flipping B means we are going to have Where's our B?

[00:35:46]This equation.

[00:35:48]So, flipping it, it will be sulfuric acid to become hydrogen, sulfur, and oxygen. So, we have sulfuric acid. So, that is as flipping that equation should be able to become our hydrogen gas plus our sulfur plus our oxygen, which is our O2.

[00:36:09]Like this.

[00:36:11]So, now the equation is balanced just the way it was.

[00:36:15]Like that. We have just flipped the equation. The change in enthalpy was uh negative 811, but it will now be positive 811. So we are going to say our positive the change in enthalpy is equal to positive 811.

[00:36:34]So it will be 811 kJ per mole.

[00:36:43]So with this we have now our C. So if we look at our C our C has got sodium sulfate. Do we have sodium sulfate in the equation? So we have sulfate on this side.

[00:36:58]Like that. So this equation it means to say we're going to maintain our C because we already have the sodium sulfate this side where it's supposed to be. So we can say maintain C.

[00:37:14]So maintaining C is us just getting this equation and putting it where it should be.

[00:37:23]So this equation will be We have this equation maintaining it here.

[00:37:35]Finally we have our D.

[00:37:38]So our D is equivalent to this.

[00:37:45]We have sodium chloride or hydrochloric acid. It is lacking it to this side but it is on the other side.

[00:37:52]So we just maintain it but multiply by two. So we are going to say multiply two two D.

[00:38:05]So multiplying two two D means getting the same equation which is this one.

[00:38:18]Now just multiply the two.

[00:38:22]So it will be two by half, it will just be H2 plus Cl2 should be able to produce two HCl.

[00:38:33]Then we have the change in enthalpy should be equal to -92 * 2. So, 92 * 2 to be -184 kJ like that.

[00:38:50]So, after we do that, we are going to get our value, which will be equivalent to uh this. So, these equations are now are now what we want. Then, adding the four equations, that is what we are going to say adding the four equations, [snorts] we are going to have So, what we are going to do is that we are going to cancel now whatever common on different sides. So, what do we have?

[00:39:26]So, this is our sodium.

[00:39:29]So, sodium has got the same number, so we cancel it.

[00:39:34]What else can we cancel?

[00:39:36]So, looking at the sulfur is the same here.

[00:39:41]Even this one was there was a two here, if you recall.

[00:39:48]Two.

[00:39:49]So, we have two O2.

[00:39:54]So, we cancel.

[00:39:57]Then we also look at chlorine is common, we cancel.

[00:40:03]Hydrogen is common, we cancel.

[00:40:06]After canceling everything that is common, we can now be able to add the two reactions. So, we have two moles of sodium chloride plus the only remaining is the our sulfuric acid. This will be able to become the only remaining thing is our sodium sodium sulfate plus our HCl. Two moles of HCl for that matter.

[00:40:41]So, we have our two moles of HCl.

[00:40:47]>> [snorts] >> Two moles of HCl. So, then the change in enthalpy will be equal to So, remember we have everything the same things.

[00:41:00]Now, we just have to add our change in enthalpy.

[00:41:05]So, this was uh This is supposed to be a three there.

[00:41:09]So, we have Adding so, we have our 822 plus 811 minus 1,000 82.

[00:41:22]We have 251. 251 minus our 184. So, we have our 67. So, it will be the change in enthalpy will be equal to 67. But, they want the change in heat.

[00:41:36]So, remember what we said. So, we have the change in heat.

[00:41:41]We are going to have the change in enthalpy should be equal to the change in the internal energy or heat which plus the n the number of moles times the the number of moles of a gas the change in the number of moles of a gas times the error times the temperature at that period. So, that means to say if this goes the other side it will be change in enthalpy minus the n the change in the number of moles of the gas times the error times our T.

[00:42:14]So, the gas gas exchange from So, the the change of the gas the change in NRT will be equal to the change in the number of moles of the gas from the equation is such that if you look at this equation, the only gases that we have is we have two moles of this gas.

[00:42:36]But, here we don't have any gas. So, we have our two minus zero. So, the change in number of moles of the gas is equal to two from the product. So, we can say product minus the reactant, which will be two minus zero. So, that's how come we have two.

[00:42:54]Then, our R for this type of a problem our R for this type of a problem you shall use the 8.31 4. Then, the temperature it should be in Kelvins.

[00:43:13]So, the temperature in Kelvins should be equal to temperature in degree Celsius plus 273. Just remember remember that >> [snorts] >> K is higher in the alphabet, so you needs to add. So, we have the temperature in Kelvins is equal to we have we have the temperature that we are provided with from the equation is 27 degree Celsius. So, we add our 27 degrees 27 degrees plus our 273. This gives us a value of 27 plus 273. This gives us a 300 Joules Kelvins. So, with that we can now calculate our change in the number of moles of the gas R times T, which will be equal to we have the change in the number of moles is two times the R 8.1.

[00:44:11]8.314.

[00:44:13]Then we have times our value of 300 Kelvins. This is going to give us two times our 8.314.

[00:44:25]Then we multiply this by our 300. This gives us a value of four our four nine eight 8.4 Joules.

[00:44:40]Now remember for us to use as you can see this is kilojoules. So we need to convert this value into a kilojoules. So to convert this into kilojoules kilo is 10 10 to the power three. So you just move three steps backwards 1 2 3.

[00:44:56]So the n g times r t will be equal to 4.9 um it will be 4.99 kilo If you want you can say times 10 to the power three. But where 10 to the power three is we are going to say kilojoules.

[00:45:18]>> [clears throat] >> So we have the kilojoules like that. Therefore finally the change in heat will be equal to the change in enthalpy minus the the change in the number of moles of the gas times r t which will be our 67 kilojoules minus our 4.99 kilojoules.

[00:45:41]This gives us our 67 minus 4.99.

[00:45:47]This gives us a value of 62.01 kilojoules.

[00:45:53]So that is the change in the heat according to the question that was required. If they ask for the change in the change in enthalpy we could have ended uh the changing enthalpy there, but we go further because of that. Let's look at our last problem, which is in form of a statement.

[00:46:15]So, we have this one.

[00:46:30]So, we are told the heat of combustion for benzene and acetylene -3,900 and -642 J respectively. Find the heats of the reaction per mole for the reaction.

[00:46:49]So, for this reaction they want us to reach >> [snorts] >> uh this reaction using the information they have given us. So, we are just going to say benzene.

[00:47:05]Sorry, respectively. So, you should not benzene. Benzene is this one, C6H6.

[00:47:11]So, we are going to say this reaction. So, combustion meaning bent in oxygen. So, we are going to say C6 H6 when it is bent in oxygen.

[00:47:25]You are going to produce the two byproducts are carbon dioxide plus water only.

[00:47:32]So, now this is the reaction that you're going to write. Then, you're going to now balance the reaction. So, looking at this, we have how many moles of We have six carbons, so we are going to put a six here.

[00:47:49]So, make sure that you don't affect this one.

[00:47:51]Don't affect the C6H6.

[00:47:55]Just balance these other things. So, we have our six there.

[00:47:59]Then, which else can we balance? So, we have hydrogens. We have six this side, so we should have a three here. So, that three plus times two is six. Oh, yeah, six, which is here. Then, count the number of oxygens. Oxygen here is six times two is 12 plus three, that's 15.

[00:48:19]So, even here we need 15 oxygens, so we can put a 15, but 15 times two is 30.

[00:48:26]So, you put 15 over two, which is 7.5.

[00:48:29]15 over two times two is this and this cancels.

[00:48:33]We're going to remain with our 15 like that. So, after doing that, now you can write your change in enthalpy according to the information you're given. The change in enthalpy for this reaction is equivalent to negative uh 3,000 900.

[00:48:54]So, the change in enthalpy is found to be negative 3,000 900 like that.

[00:49:02]So, if this is the case, remember, it should be this other side.

[00:49:07]So, we are going to flip.

[00:49:11]So, let's call this one as our our A.

[00:49:16]Flip A.

[00:49:20]So, that we have our six CO2 plus our plus our three H2O to produce our C6H6 plus our 15 over two our O2.

[00:49:39]Then, we have our change in enthalpy will change to be positive like that.

[00:49:45]Then, after after we have done that, we are going now to deal with this one.

[00:49:51]So, this one also as a tiling, which is the our C2H2 will burn in oxygen. So, C2H2 will burn in oxygen to produce to produce carbon dioxide plus water only.

[00:50:13]So, what we have is what we have is this then let's balance. So, we need two carbons.

[00:50:28]Two hydrogens already there.

[00:50:31]Oxygens already balanced.

[00:50:34]So, we have three oxygens.

[00:50:36]We have three uh okay, we don't have three oxygens here.

[00:50:41]This side we have two, this side we have five.

[00:50:45]So, two by two is four plus one, that's five. So, we're going to put our five over two so that we have five over two times two. This will cancel to leave a five there.

[00:50:57]So, that is what we have. So, the change in enthalpy as long as you don't affect this the one that is being combusted will be the exact value they have given you, which is equal to We have um just a bit just reducing this a bit. So, the change in enthalpy for this one is equivalent to negative six 42.

[00:51:26]So, the change in enthalpy is equal to negative six 42 kilojoules like that.

[00:51:35]So, it's uh joules, not kilojoules, but joules, not kilojoules.

[00:51:42]So, we have the joules there.

[00:51:47]We have the jaws here.

[00:51:52]What do they want?

[00:51:55]So, calculate the heat heat reaction following this.

[00:52:05]The heat of reaction following this.

[00:52:08]So, we are going to have these two equations. So, adding the two equations, we are going to have So, this equation and this other equation.

[00:52:51]Now, when we look at this equation, we need two of them according to what we have, or we need three of them according to what we have. We are We need three here. So, we say multiplying by three, this equation multiplying by three, it will be equal to We have something like this.

[00:53:16]multiplied by three.

[00:53:22]So, when we multiply by three, it will be equal to 3 C2H2 [snorts] plus 3 by 5 it will be equal to 15 over 2 O2, which will be equal to 3 * 2 is 6 CO2 plus our 3 times H2O is 3H2O.

[00:53:45]Then we have the change in enthalpy should be equal to So, multiplying here, it will give us a value that will be equivalent to that will be equivalent to we have our three times our 642, negative for that matter. It is giving us a negative 1,926.

[00:54:14]So, this is what we are supposed to write here.

[00:54:18]So, when we are adding these two equations, we have this equation.

[00:54:53]So, we have what is common here.

[00:54:57]We cancel.

[00:55:00]We have six carbon dioxide. So, this and this cancels. What else is common?

[00:55:06]Oxygen, we have 15. We have 15 here.

[00:55:10]So, those are the two things that are common even even water is common. So, we are just remaining with our 3C2H2, which should be equal to We are going to have our C6H6.

[00:55:29]Then the change in enthalpy will be equal to We have our 3,900 minus 1,926.

[00:55:39]This is going to give us 1,974 Joule.

[00:55:46]So, that is how we deal with such type of a problem. Thank you very much for watching, but also just a reminder, if the change in enthalpy gives you a positive value, 1,000, let's say like this, then that means to say it is endothermic because endothermic involves heat or rather the the change in enthalpy is positive. If it gives you -184 as your final answer, as the change in enthalpy, it means to say this is exothermic because heat is being lost to the surrounding. So, this is uh that for this video. See you in the next tutorial.