How to write a formation reaction video
Well, I think I just-- see, I'm starting the gaseous state. Because you've released heat, you can say that the methane is in a lower energy state, or it has a lower potential energy, than these guys did.
Write an equation for the formation of co2(g) from its elements in their standard states.
So here I have to multiply this times 4, 4 here, and I have to multiply it times 4 here, and I have to multiply it times 6 here. So let's just look them up. I have to be able to read it. This little naught sign tells us it's a standard heat the formation. And what all of this does is, it gives us a framework so that if we know how much heat it takes to form or not form certain products, then we can kind of predict how much heat will either be released, or how much heat will be absorbed by different reactions. This tells us that this state has more potential than this state. If you have a bunch of nitrogen, it's going to be N2. So you'd have to say, plus joules. And I just showed you that that's the exact same thing as the change in enthalpy. Now, if you want to go the other way-- let's say we wanted to go from methane and go back, you'd have to add heat into the reaction.
And then, I don't know, maybe I shake it up or something to make them react, and then I get a bunch of methane gas. And I copied all of this stuff-- actually, let me get the actual tables from Wikipedia down here.
And then you'll have the heat of the reaction.
That's how you always find oxygen. It's positive, So it has more energy than its elemental form. And so here I'm going to touch on another notion.
Enthalpy of formation video
And then, I don't know, maybe I shake it up or something to make them react, and then I get a bunch of methane gas. So instead of saying this reaction, we could say hey, let's go from this reaction, and go back-- let me do it in a different color. So what do we get? I'll do it in gray. It gets released. So what did we end up with? I'll do it in the yellow of that container. So how many moles of propane is 33 grams? So how can we think about this? So this has to be lower than this by 74 kilojoules. And what's interesting here is, is not so much what the absolute value of this enthalpy is here, or what the absolute value of this enthalpy here is. There you go. If we assume constant pressure, we saw that the change in enthalpy becomes the heat added to the system at that constant pressure. It's a minus And I wanted to figure out what was the change in enthalpy of this reaction?
And that's because you always use the elemental form as your reference point. And it's the standard change in enthalpy. Because we formed 3 moles of this.
But we know that this is a valid state variable.
based on 104 review