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Heat of Formation - Concept

Teacher/Instructor Kendal Orenstein
Kendal Orenstein

Rutger's University
M.Ed., Columbia Teachers College

Kendal founded an academic coaching company in Washington D.C. and teaches in local area schools. In her spare time she loves to explore new places.

Heat of formation is the change in enthalpy when one mole of a compound is formed at 25 degrees Celsius and 1 atm from its elements. Heat of formation has been measured in labs and recorded so that we can look it up when we need it.

Alright so when you're dealing with different types of thermo chemical equation you're going to be dealing with different types of delta h's or enthalpy changes. Well we're going to talk about the heat of formation or the formation of a substance. So we're going to hold that delta h with a subscript of f saying we're forming something. That little 0 sign says we're forming something in its standard state I'll talk about that in just a second. Okay so let's just read about what this means, this delta hf is a change in enthalpy that accompanies the formation of 1 mol of that compound or whatever substance we're dealing with 1 mol that's it, in their standard state from it's elements in their standard state. So basically what we're talking about is we're forming something, we form it fro it's elements and in their standard state means whatever state of matter they are in at 25 degrees Celsius. So that's in little 0 thing says at the top, that means everything I'm promising you, I'm telling you that everything that I'm dealing with is in its standard state. So like for example water at room temperature is going to be a liquid H2O in room temperature is a liquid so I want to make sure that the water was in liquid form.

Okay so when we're dealing with elements because we're going from their elements to the substance free elements in their standard states are assigned delta hf of 0 kilo joules so because we do not form them, we don't form elements, elements are just here on earth we are not going to give them a delta h value they aren't formed from anything. But if we, from those 2 things that they, from elements if we get a substance and it requires energy the delta h is going to be a positive number it's going to be higher than this. If the delta is from its element it releases energy it's going to have a negative delta h or beneath it, so releasing energy it's exothermic. So this is going to be our like arbitrary standard or line where we decide if it's going to be exo or we decide what the delta h is going to be. So let's go over here and talk about what I'm dealing with.

Alright so we have 3 different, I put up here 3 different heats of formation so if we're dealing with for example hydrogen fluoride to form it from it's elements hydrogen and fluorine it's going to release 273 kilo joules of heat. So notice also that I've a fraction coefficients, in heats of formation reaction that's totally okay because we want to make sure the most important thing is that we have 1 product or 1 mol of the product. So notice nothing in the product side has coefficient in it. So that allows us to have fractional coefficients and don't forget the elements in their standard state so hydrogen gas is a gas with 25 degrees Celsius, fluorine is a gas with 25 degrees Celsius and they're also diatomic meaning that they have, they can not stand alone they must have 2 of them together and making molcules.

And that's totally fine because that's actually true and then we have the formation of sulfur hexafluoride and that comes from their standard elements and it releases 1,220 kilo joules of heat and then we have the formation of hydrogen sulfide coming from its 2 elements and it actually, it releases 21 kilo joules of heat. So why do we need to know this, like what kind of information can we gain from these types of things? Well we can actually have, I like that we have a reaction and we don't know how much energy is going to be released or gained from this reaction and let's say that we haven't done this and we're kind of worried over how much energy is going to be released or you'll worry you don't have enough energy for this reaction to occur so we need to figure out the delta h before we actually do this in practice. So how do we do that? Well we can employ Hess's law, Hess's law don't forget is when you add up and manipulate the different types of reactions and then you can add up the delta h's to figure them out. So let's actually employ Hess's law in this to figure out what the delta h of this reaction is. So over here we have hydrogen sulfide as a reactant okay so here is my hydrogen sulfide it's a product so I'm going to flip this to make it the reactant. So I'm going to have H2S gas yields sulfur solid plus H2 gas and because I flipped it that means I'm going to have to make this an endothermic reaction and it's 21 kilo joules so my delta h is going to be 21 a positive 21 kilo joules. This is going to be since I flipped it this sign is going to be flipped as well.

Okay so that helps me with that, let's go over here to do HF if we write, I can find HF here it's a product so I don't have to flip it but this does has a 2 coefficients so I'm going to multiply this whole thing by 2 to make this have a 2 coefficient as well. So I'm going to say okay one half times 2 is 1 so it's H2 and that's a gas plus one half times 2 is 1 fluorine 2 is a gas and that's going to yield 2HF and that's going to mean, I'm going to multiply this by 2 because this is telling me how much it is for 1 to be formed but I need to know how much 2 mols is going to be so my HF is going to be multiplied by 2 so negative 273 times 2 is going to give me negative 546 okay. Then let's go to SF6 the product here's a product 2 so I can just use the same equation and go down. So I have sulfur solid plus 3 fluorine gases yields sulfur hexafluoride and since I'm not manipulating this equation in any way I can just say my delta h is the same negative 1220 kilo joules okay.

Then everything we're going to add them up should equal this equation in blue. So I want to make sure I get rid of my intermediate so this case it's my hydrogen over here and my hydrogen up here can we cross out because one is a product one is a reactant. I have my, what else can I cross out? My sulfur is a product here, my sulfur is a product here that can be crossed out and I think that's pretty much it. So I'm going to write my final reaction H2S gas plus 3+1 is 4 fluorine gases yields 2HF and that should be a gas should be a gas there plus SF6 and that's a gas. And then I can add this up 21-546-1220 kilo joules is going to give me a total of 1745, negative 1745 kilo joules. So it's releasing 1745 kilo joules when this reaction occurs according to this manipulation in Hess's law. Okay and we can do this every time and this will work so my delta h is this and this will work but there's an actual easier way to do this and I'll prove that as well. So if we go over here our easy way, our delta h over our reaction is equal to the sum of and that symbol mean sum of, the delta h of our product minus the sum of the delta h of the reactant.

Okay so let's prove that and make sure that's true, okay the delta h is the product so I go over here my products are 2HF and SF6. So I have 2HF so this, I'm going to have to multiply this by 2 which is 546 so negative 546 kilo joules plus I have SF6 and that's negative 1220 kilo joules and I'm going to add all these together minus my delta h in the reactants and my reactants is, one of them is an element so I know that 0 I don't have to include and then H2S is here so it's negative 21 kilo joules, so if I do the negative 546 plus the negative 1220 minus negative 21 kilo joules if we do this actually all out it should be negative 1745 kilo joules which is exactly what we got over here. So this might be a short cut instead of having to use Hess's law, this [IB] be a short cut and you'll be able to use the heat of formation when dealing with the heat of the reaction an over all reaction.