THERMOCHEMISTRY: HESS LAW.

Added: 2026-05-12

[00:00:00]Hi guys. So, in today's video we are going to be talking about Hess's law.

[00:00:06]We're going to talk about Hess's law under our thermochemistry.

[00:00:12]So, I have some written examples. To download the question, just check the description below this video. You download the question from there.

[00:00:21]Mhm.

[00:00:22]They gave us this question.

[00:00:29]They gave us this equation.

[00:00:30]2 H2O reacts to giving 2 H2 plus O2 with the enthalpy of -572.

[00:00:51]Enthalpy of -572.

[00:00:55]And O2 giving 2 O2 plus O2 And they asked us to find the enthalpy of this equation. H2 plus O2 giving H2O.

[00:01:29]We're going to find the enthalpy.

[00:01:33]So, according to Hess's law, if you are given a question like this, you are going to reform these two equations in a way that is going to resemble this one.

[00:01:43]We We modify these two equations in a way that are going to resemble this one.

[00:01:47]Then we add the enthalpy together.

[00:01:50]So, to do this, it's easy.

[00:01:53]You just look at this equation. This The we are looking at.

[00:01:56]This is the one we are following.

[00:01:58]So, we have H2O here.

[00:02:01]We have H2O here. So, we just look at the equation that has H2.

[00:02:07]And that's equation one.

[00:02:08]Equation one has H2.

[00:02:11]So, and H2 is in the reactant side here, but in the product side here. But, we want this H2 to be in the reactant side.

[00:02:18]So, what do we do? We reverse the equation.

[00:02:21]And you can see here that H2 is one molecule. Oh, here is two molecules. So, we divide through by two.

[00:02:28]So, we reverse this equation one and divide through by two.

[00:02:32]So, after reversing, we get equation one, we get H2 plus O2 half O2 because we divided by two giving H H2.

[00:02:49]Then, the enthalpy Since we divided everything Since we divided everything by two, the enthalpy two we divide it by two.

[00:02:58]Since we divided all this side of the equation by two, the enthalpy two divided by two.

[00:03:06]So, that will be 572 divided by two. 572 divided by two.

[00:03:11]That's 286.

[00:03:13]Well, mind you, we reversed the equation. So, because we reversed the equation, the sign of the enthalpy will change. So, that will be minus 286.

[00:03:22]286 kJ.

[00:03:24]And 286 kJ. Well, I said it I said it in my last video that when you reverse the equation, the sign of the enthalpy will change. And if you multiply by any factor or you divide by any factor, you also have to divide the enthalpy by that factor. So, here because we reversed the equation, the sign changed to minus. And because we divided all through by two, the sign changed to the We also divided the enthalpy by two. And why do you divide by two? Because what you are looking for is H2, not 2 so we have to divide this one by 2 and if you divide this one by 2 you have to divide everything else by 2.

[00:03:58]So that's -26.

[00:04:01]So here we have 82 + O.

[00:04:05]82 + O.

[00:04:07]If you also look at this equation let's consider equation 2. O is here.

[00:04:13]O O is here O is not here. O is here but O is in the product side. we need it to be in the reactant side. So what do you do? You also reverse the equation. And this one is 2 O not O. So we have to divide all 2 by 2. So when you reverse the equation equation 2 we get O giving half O2.

[00:04:34]And the change in H we also have to divide the change in H the enthalpy we also have to divide by 2 because we're dividing all 2 by 2. And because we reversed it the sign will be changed.

[00:04:45]So that will be minus that's 495 divided by 2.

[00:04:53]That will be minus 247 minus 247.5.

[00:05:01]Minus 247.5 kJ as you can see.

[00:05:05]So this is not the final answer. The final answer for us to get everything we now need to add the equations together.

[00:05:12]So that that one will show whether you are correct or not. So here you add the equations together.

[00:05:16]Everything must cancel in the way that they cancel until you get this equation back. Like these two things these two equations these modified equations must cancel in a way that you get these two back. So let me show you what I'm talking about.

[00:05:31]Yeah. Look at here we have half O3 here.

[00:05:34]We also have half O2 here.

[00:05:36]So that's this one will cancel this one.

[00:05:38]Because this one is in the reactant side and this one is in the product side.

[00:05:41]So when this has canceled this the overall reaction will be can you see? The overall reaction will be 82 + O.

[00:05:48]write everything on the reactant side 82 + O giving H2. We can cancel with this one. So, this is giving H2.

[00:05:57]You can see that our our final equation resembles the one that they said we should get.

[00:06:04]So, if your final equation is is exactly like this one, then you are correct. So, what next? We just add What next? We just add the two the two enthalpies together. To get the enthalpy, just add the same enthalpy.

[00:06:18]That will be minus 286 minus 247.5.

[00:06:26]And that will give you minus minus 5 93.5 kJ.

[00:06:36]So, that is your solution.

[00:06:39]Yeah, so let's take a look at this question.

[00:06:41]They said N2 plus 3H2 gives two molecules of ammonia. When the change in H, the enthalpy is negative 92 kJ. And H2 gives hydrogen plus half oxygen. When the enthalpy is positive 286 kJ. Then they said we should find the enthalpy of this reaction.

[00:07:08]So, the first thing we need to do is to modify these two equations in a way that it should look like your overall equation. Because these ones that we are driving at, so our our modification was the it must resemble this one. It must be exactly like this one after canceling everything out. Then we add the enthalpy.

[00:07:28]So, the first thing to do here is let's take a look at ammonia.

[00:07:32]Ammonia is here. It's in the reactant side of this our standard equation. Let's call it standard equation.

[00:07:38]But in the equation here in equation one and two our ammonia is in the in the product side.

[00:07:46]And since it is only since it is only in equation one, what we do is that we reverse the equation. We reverse the equation of equation one. We reverse equation one to have the ammonia in the reactant side.

[00:07:59]Because ammonia is in the reactant side here, but here it is in the product side. So, we have to reverse this equation so ammonia can be in the reactant side. And here, we have 4 NH3, but here it is 2 NH3. So, we have to times everything by two so you can have 4 NH3. So, we have to times equation by two, equation one by two, and reverse it.

[00:08:17]So, what we have is when we reverse it, we have Let's say 2 NH3 2 NH3 giving N2 N2 plus 3 H2.

[00:08:33]But that is not all. We need to times times two. So, when you times everything by two, here you get 4 NH3.

[00:08:39]You get 4 2 * 2 is 4.

[00:08:42]2 * 1 times two times two and 2 * 3 is 6.

[00:08:46]That's 6 H2.

[00:08:49]That's 6 H2.

[00:08:51]Don't forget, since we reversed it, in the enthalpy, the sign will change to positive. And since we times by two, that will now be 92 * 2.

[00:09:02]So, that will be positive positive 92 * 2. That is 184.

[00:09:10]That is positive 184 kJ.

[00:09:15]So, we are done with ammonia.

[00:09:18]What we do next is let's look at let's look at oxygen.

[00:09:23]Yes, let's look at oxygen. Oxygen is also in the reactant side, but in equation two, oxygen is in the oxygen is in the product side.

[00:09:32]Now, let me give you one hint. When you see hydrogen, like even when you see an element that is present in this equation, don't use that element because it's going to give you a lot of problem.

[00:09:42]Just use the one that is present in one equation and it's not present in the other equation, so it will make it easier. So, oxygen is present here, but it's not present here.

[00:09:49]But, in the standard equation, oxygen is in the reactant side, while here it's in the product side. So, what you have to do is you reverse it.

[00:09:56]But, now you have 3 O2, you have three molecules of oxygen here, but here it's half half O2. So, what are we going to multiply by half to give us three? That would be six because six times half is three. So, you have to multiply all these equations by six and reverse it. So, when we reverse it and multiply it, what you are going to get is here will be 3 O2.

[00:10:18]3 O2 + 6 H2. Then we are multiplying by six.

[00:10:23]Then it will give give us six here. We have six H2O.

[00:10:30]And don't forget, since we reversed it, the enthalpy will change. Now, it will be minus minus that's negative negative then 266 * 6. Since we multiplied everything by six, as 266 * 6 is 1716.

[00:10:46]Um -1716 kJ. Now, let's add our equations together. So, when we add our equations together, we find out that hydrogen here six hydrogen is here, six hydrogen is here. It's going to cancel out because it's in the reactant side and in the product side here, so let's cancel it out. As it's canceled out then there's nothing else to cancel, so let's add together. That's 4 NH3 NH3 + 3 O2 to give you 2 N2 2 N2 + 6 H2.

[00:11:22]You can see that our final our final equation it resembles this one that they they gave us in the question. So, if it's exactly the same, then you're on the right path. So, what's next?

[00:11:33]You have to change in H.

[00:11:35]Then the changing H just add this together.

[00:11:40]When it's 4 when it's 4 plus minus 1716 that was negative 15 1532 negative 1532. So, that's the answer.

[00:11:52]But, let me give you since for your test or exam is going to be OBJ. So, this one is because I'm explaining that's why I'm writing this full step.

[00:12:01]For your test or exam, you don't need to write all these full things. What you just need to do is you don't even need to be adding all these equations. What you just need to do is if you see that this one is having six and I want to get this one, just reverse the equation, multiply it.

[00:12:15]Is this this value they are going to ask in the exam. They'll say find enthalpy.

[00:12:19]So, it's this value they'll find. So, you don't need to be doing all these equation stuff. Just know how to manipulate only the value and get your answer. So, when you reverse it, just know that the value the it will change in sign. When you multiply, you multiply this is also the enthalpy. So, that's how to get your answer fast when you are dealing with OBJ because your test is OBJ. You cannot be writing all these long long equations.

[00:12:42]Yeah, so this question is from the slide. It's directly from the slide. It says to find the the enthalpy for this thermochemical equation. And they give us these three equations. So, I want to show you the fastest way to solve this instead of just writing equations, so many top and wasting your time. And when time is running in the test or exam.

[00:12:59]So, the first thing we need to do is look at this equation. Look at the the elements in this equation. Look for them here.

[00:13:08]And just manipulate. So, let's start.

[00:13:09]So, we have carbon here. Carbon is in the reactant side. In all these three equations, in this equation A, we have carbon. But, carbon is two here. Carbon is one here. So, what do we do? We multiply this one times two. But, since they are in the reactant reactants, we don't need to reverse. So, that is changing H.

[00:13:26]Changing H is equal to Since you multiply this one by two, that's -393.5 * 2.

[00:13:34]That's -787.

[00:13:36]-787.

[00:13:38]Then we've done for carbon. For hydrogen, hydrogen is in the hydrogen is in the upper side. Yeah, hydrogen is also in the upper side and one one mole. So, we leave it like that.

[00:13:48]That's changing H.

[00:13:51]I see in the reactant and reactant is going to give us a gain. So, changing is just writing the latter. -385.8.

[00:13:58]Write the last one.

[00:14:00]For C2H2, for C2H2 we have C2 But here, C2H2 is in the product side. Yeah, C2H2 is in the reactant side. So, what do you do? We reverse this equation. And here is one mole, here is two moles. So, after reversing this equation, we now divide by two.

[00:14:16]When you reverse this equation, you get plus +2598.8. But we're dividing by two, that's 2598.8 / 2.

[00:14:25]That's +129 1299.4.

[00:14:32]So, after that, just add all your all your enthalpy. That's -787 -285.8 +1299.4.

[00:14:43]That will give you 226.

[00:14:46]226.6.

[00:14:48]So, that's how to do it. That's how to do it in the fastest way. Since we know that and if you divide here or you multiply here or you reverse it, you're still going to do it here. So, instead of doing the equation, just do it directly to the enthalpy and just add it. That's how it works in the exam. They're not going to ask for the final equation.

[00:15:04]They'll just ask you for the enthalpy.

[00:15:06]Because the final equation is still this way.

[00:15:08]So, they'll just ask you for the enthalpy straight.

[00:15:11]I'm going to stop here.