Transcript of the video

Before we start talking about the Hess’s law which is used to calculate the enthalpy of a chemical reaction, we are first going to learn about something that’s called State Functions. These are very important quantities and one example of the state functions can be the altitude. So, let’s see why the altitude is considered a state function and how they can generally be described. For example, in this case, if we have two hikers which are going to this point A and they’re taking two different routes; the hiker one is going on this route, going directly and fast to the summit of the mountain and then going down to this point A, while on the other hand, the second hiker is taking a longer path, he is going through the forest maybe taking some rest so just taking longer time and taking a longer route but eventually getting to this point A. What is important here is that once they’re here, they are on the same altitude. So, let’s say this is the altitude and attitude does not depend on the route that the hikers have taken but it only depends on the initial and final states. Initially they are at this altitude and then now at this altitude A. So, if we want to generalize the State functions, these are quantities that only depend on the initial and final state of the system. They are not dependent on how the initial and final states have been achieved.

Other examples of state functions are for example of the pressure and the volume. Why is volume a state function? Because it doesn’t matter if initially this bottle is empty and I fill half of it or if it was full and I’ve taken out half of it. I’m still going to get half a bottle of water, so volume is a state function. Same for the pressure. Internal energy is also a state function and, perhaps more importantly for our purpose here, the enthalpy is also a state function and we can use the same principle of the dependence of only final and initial states to calculate the enthalpy of chemical reaction.

The way we can do it is by looking at this example. If the hiker first is going to the summit of the mountain, all the way to the top (let’s say this is the altitude 1) then he’s going to down again to this point and then he’s climbing to this point (let’s say B) and we don’t know the altitude of point B but we know the altitude of the summit of the mountain, we know this altitude, we know this height and also this one. So, we can first add this and then we can subtract this height and then we can add decide and we’ll get the altitude of B. So, let’s say this is X and this is Y and this is Z. If I add X and Y and Z together, I’ll get to the point B. The only thing that we will have to pay attention to is the signs. First, I am going to the point A, which means that the height X will be plus here. Then I’m going to down so Y is negative and then Z is plus again. So, if I do all of this together then I’ll get the altitude of the point B.

Now, let’s see how this is related to chemical reactions. For example, if we have a reaction of A and B and the products are C and D and we need to find enthalpy of the reaction (which is here, the Δ*H*) and we don’t have a direct way to do it which means that we cannot do it experimentally and we cannot use the enthalpies of formation, we can still go back to the Hess’s law and determine the enthalpy of the reaction considering that we have given data for some other reactions that are starting from A and B or A and B are the products. Let’s see how it works. For example if we put the energy level here (A and B, so this will be a different reaction going to product X) and we know the enthalpy of this reaction and also, on the other hand, we know that X can, in a different reaction, give the product Y and we also know that the product Y can undergo a certain type of reaction and get to Z and we know that Z can give another product, W and W can undergo a reaction to finally give these product C and D which is our target, which is what we wanted to determine – the enthalpy of this reaction.

So, like for the mountains, if we know the X and we know the Y and Z, the heights or the altitude of these, and we add everything together, we get the B. We do exactly the same here. If I add the enthalpy of the reaction, for example, this is 5 kilojoules and this is let’s say 8 and then we have this going down, let’s say this is 3 and we can put a sign here, so it’s actually -3 , and we know that here we have 4 going up and then here going down and we can say that the height is minus 7. If I add all of these together, the Δ*H* of the reaction will be the sum of the enthalpies of all of these reactions. So, we have, for the reaction one, -5 and then for the second one, I have eight and then I have here minus 3 – an exothermic reaction and I can add 4 here and I’m going down, -7 again. This will give me the total process, the enthalpy of the reaction, which seems to be 7 kJ. So, if we generalize, we can say that the enthalpy of the reaction can be calculated from enthalpies of other reactions as long as we can find a way how to add everything together, all the other reactions, to get to the target products we have here.

As an example, suppose I need to find the enthalpy of this reaction. How the nitrogen dioxide can be formed from nitrogen and oxygen knowing the enthalpies of these two reactions? So, when you are solving a Hess’ law problem, what you want to do is take a look at what is given. These are the reactions that are given and you have the data and then you are looking at your target reaction which is this and you trying to find a way how to add these reactions together to make the target reaction. When we are saying add the reactions, you may need to reverse the reactions in order just to match the molecules because remember, the hiker can go up, can go down, and then up again. All that matters is that finally, you want to get to this point A which is the nitrogen dioxide in this case. So, if I’m looking at my reaction, the nitrogen dioxide is the product and then I am going to look for it in the other two reactions. It is not here but it’s here and here it is a reactant but I want it to be a product which means that what I need to do is to reverse this reaction. So when I reverse it, the products and reactants will be switched. I’m going to move the nitrogen monoxide and oxygen to this side and I’m going to write down that two NO reacting with oxygen will produce two nitrogen dioxides, so when I reverse it, I need to change the sign here because if the hiker is going up – it is plus, if he’s going down – it’s minus.

So, I’m going to write down that the Δ*H *of this reaction is equal to -112 kJ. So, generally, these are the two rules that you need to remember.

- If you are going to reverse the reaction, you also need to change the sign of the enthalpy.
- If you are going to multiply the reaction by any factor, you need to multiply the enthalpy of the reaction by the same factor.

Now lets’ check what we have. So, if I add these two reactions together (for a moment we kind of forget about this because we don’t need it anymore, so we have two reactions- reaction one and reaction two) this will be my product and what I want in the target I want the nitrogen and the oxygen be the reactants, which are reactants here. So let’s try to add these two reactions together. If I add reaction one to reaction two, what I will have is N_{2} + O_{2,} the reactants, and I’m taking the reactants from this side + 2N_{2} + O_{2} they’re going to give me the products and products are here. So I have nitrogen monoxide and I have nitrogen dioxide The enthalpy of this reaction is going to be the sum of the Δ*H*_{1} plus Δ*H*_{2 }, of course, we are going to consider the signs as well. Before doing that, let’s check and see if there are actually some extra molecules on both sides of the equation. Nitrogen monoxide and nitrogen monoxide, so you can leave this out and here you have two oxygens and one nitrogen so if I just rewrite it slightly differently, I can say that nitrogen reacting with two oxygens is producing nitrogen dioxide and the enthalpy of this reaction is the sum of the two so if I add these two together it is going to be 180 minus 112 kilojoules which is going to be equal to 68 kilojoules and this is the enthalpy of my target reaction.

So, that’s the idea, anytime you have a problem on Hess’s law, you are going to manipulate this equation, you may need to reverse them, you may need to multiply by any number just to match target equation, match the product and match the reactants and then you also add the enthalpies.

#### Practice

Calculate the enthalpy for the oxidation of CO to CO_{2} using the enthalpy of reaction for the combustion of C to CO (Δ*H* = -221.0 kJ) and the enthalpy for the combustion of C to CO_{2} (Δ*H* = -393.5 kJ).

2CO(g) + O_{2}(*g*) → 2CO_{2}(g) **Δ H = ?**

C(s) + O_{2}(*g*) → CO_{2}(g) Δ*H* = -393.5 kJ

2C(s) + O_{2}(*g*) → 2CO(g) Δ*H* = -221.0 kJ

-566 kJ

Calculate the enthalpy for the combustion reaction of sulfur to sulfur trioxide using the enthalpies of the two reactions shown below:

2S(*s*) + 3O_{2}(*g*) → 2SO_{3}(*g*) ** Δ H = ?**

S(*s*) + O_{2}(*g*) → SO_{2}(*g*) Δ*H* = -297 kJ

2SO_{3}(*g*) → 2SO_{2}(*g*) + O_{2}(*g*) Δ*H* = 198 kJ

-792 kJ