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## Chemistry library

### Unit 15: Lesson 2

Enthalpy- Calorimetry and enthalpy introduction
- Enthalpy
- Heat of formation
- Hess's law and reaction enthalpy change
- Worked example: Using Hess's law to calculate enthalpy of reaction
- Bond enthalpy and enthalpy of reaction
- Bond enthalpies
- 2015 AP Chemistry free response 7

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# Hess's law and reaction enthalpy change

Using Hess's Law and standard heats of formation to determine the enthalpy change for reactions. Created by Sal Khan.

## Want to join the conversation?

- In the video Sal uses a table to find the energy. Why do some of these formations absorb energy and some give them off?(9 votes)
- Some substances form more easily than others. For instance, it's not hard to form oxygen gas - gaseous O2 is in its natural state, so its heat of formation is zero. On the other hand, it takes a lot of energy to oxygen into a liquid, so its heat of formation is going to be very positive. Therefore, some of these reactions absorb a lot of energy because they're unlikely to occur under standard conditions, and other reactions give off a lot of energy out of a kind of "eagerness" to break their atomic bonds.(27 votes)

- how do you do a problem like this on a quiz or something when you are not given the heat of formation?(4 votes)
- you don't. or you just look it up in a table if youre allowed to use one(17 votes)

- At13:24, Sal says "this side of the reaction has less heat than this side of the reaction." Heat is not a state variable, you can't have heat (Sal spent so much time riffing on this in the Internal Energy lectures). My question is, did I misunderstand that, or should Sal be saying "this side of the reaction has less ENERGY than this side of the reaction" (or something else)?(8 votes)
- Heat q is not a state variable, but enthalpy H is.

Enthalpy used to be called "heat content"; that's why it still has the symbol H.

For the purposes of Hess's Law, he is treating ΔH**as if**it were a reactant or product in the reaction, purely for convenience.(1 vote)

- If the reaction yields a negative energy then it is exothermic, if it is posative it is endothermic? What one is it heating up? The way im thinking of it is, it gains energy, so it must heat up. But it loses energy so the surrounding is heating up, therefor its heating up? I don't know if that makes sence but I cant seem to understand this...(4 votes)
- Exothermic means releasing heat. Everything around it, including the reactants themselves heat up. Endothermic means absorbing heat. Everything around it, including reactants, cools down. In all cases energy is being conserved, but it is changing form from thermal energy to energy bound in chemical bonds.(4 votes)

- How come the hess cycle was not complete as in full triangular diagram with arrows from or to the reactants and the products.(5 votes)
- Well, it could, considering equilibrium. But it appears Sal wanted to focus on the reactants.

Hope this helps! :D(2 votes)

- why is the heat of formation of any element equals zero ??(2 votes)
- Heat of formation is basically the amount of energy either released or taken to form a different compound. Elements are originals.

Let me make it easier to understand. You see, in the questions in the video, most of the things are compounds (different molecules). Compounds are made up of different elements. The elements require 0 heat of formation to form themselves, but forming compounds either gives off or takes energy. Bonds like ionic bonds and covalent bonds only form with the use or release of energy even if it is very tiny. Since elements are the basic building blocks and their is nothing before them to create them, you cannot describe their heat of formation. I hope I have answered this correctly. Thank you!(3 votes)

- How do we determine the route taken when calculating the enthalpy change?(2 votes)
- The route is irrelevant. For example, you can go from A to B to C to D but the "net" (the overall path) is from A to D.

In Chemistry, what you often do is take a compound and break it down to individual atoms first because you know the enthalpies for bond formations or otherwise enthalpies of individual reactions. Then you take these atoms and form the products made. You effectively do: Reactants -> Atoms -> Products, which is the equivalent of Reactants -> Products(3 votes)

- I am assuming that the reason for not including the negative sign in the final -2219.85 kJ/mol rxn is because the negative sing is simply expressing that this reaction is an exothermic reaction, verses a positive endothermic reaction. Am I thinking the correctly?(2 votes)
- At12:45? Yes you are thinking about this correct. The negative sign implys that the equation is exothermic and 2219kJ is "lost" from the equation, meaning the energy goes to another place in this case heat and light.(2 votes)

- Is it possible to calculate heat of formation?(2 votes)
- Most of the enthalpy of formation are calculated from enthalpy of reaction.

(Note the difference between enthalpy of Reaction and formation)

Enthalpy of Reaction= Enthalpy of Formation of product - Enthalpy of formation of reactant -eq1.

U'll can find Enthalpy Of Reaction by running that experiment in a calorimeter. and plug that value in the eq-1.

Now suppose someone already previously in different reaction had found enthalpy of formation for product. Now plug that value in the equation. Only unknown is now your reactant.(2 votes)

- When Sal says that no change in enthalpy is needed for a pure, elemental state, does this include polyatomic ions as well? I noticed how O2, although a molecule, had no change in enthalpy as it was in its purest elemental state. Does this also include polyatomics? For example, if I have one mole of sulfate (SO3), would I need to break that down into its constituent atoms, one sulfur and three oxygens, and find the heat of formation for sulfate? Or should I just leave the sulfate as it is and not notate a change in enthalpy for formation?(3 votes)
- O2 is the standard for the oxygen thus he uses zero. Standard enthaply of formation is amount of enthalpy(heat) required to produce molecules from its most stable elementary form at 1 bar pressure. Forming oxygen gas ( 02) from its standard form which is also (02) will be zero as they are the same thing. At 1 bar O2 gas is most stable. It's not 03(ozone)

Thus to form O3 he will require some enthalpy.

3(O2)---> 2(O3)(1 vote)

## Video transcript

Now that we know a little bit
about the formation and enthalpy change, and what
enthalpy is, we can talk a little bit about Hess's Law. And what this tells us is that
the energy change of a process is independent of how we get
from one state to another. And really, that's a by-product
of the fact that energy is a state variable. Whether we're talking about
enthalpy or internal energy, they're state variables. And we've talked multiple times
that it's independent of how many steps it takes to
get there, or what path you happen to take. But how is that useful to us
when we're dealing with everyday reactions? So let me just make up some
reaction where I have A plus B yields, oh, I don't know,
let's just say this yields C plus D. And I wanted to figure out
what was the change in enthalpy of this reaction? Or essentially, how much heat
is absorbed or released by this reaction. I don't know what it is. I haven't measured it. And all I have are the
heats of formation. So all I know is, how do you
go-- so I know the heat of formation of A-- so let me call that the heat of formation. Remember, H isn't for heat. Even though we kept calling
it heat of formation, it's actually the change
in enthalpy. And it's the standard
change in enthalpy. But the change in enthalpy
we know as heat. So it's heat, change in enthalpy
of formation was the same thing as heat
of formation. This little naught sign
tells us it's a standard heat the formation. We can look up that in a table,
and let's say that that's some number. And then we have our heat of
formation of B-- delta heat of formation, let me
call it, of B. This Is heat of formation
of A, and it's a standard heat of formation. And we could look up in a table
that heat of formation of C, which is change
in enthalpy. And then the heat of
formation for D. So all of these things we can
look up in a table, right? And we'll do that in a second. Now, what has Hess's law tells
us is that the change in energy, the change in-- and
enthalpy is what we're measuring here-- the change in
enthalpy here is independent of what we're doing. 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. Let's go back to our constituent
products, so kind of the elemental
form of these. So you know, if this was like
carbon dioxide, you'd be going back to the carbon and
the oxygen molecules. So you'd go back to the
elemental form. And how much energy, or what's
the change in enthalpy, as you go back to the elemental form? The heat of formation is what
you get from the element of form to A, or the elemental
form to B. So to get A and B back to the
elemental form is going to be the minus of those. You're going to take
the reaction in the other direction. So this change is going to take
minus delta-- the heat of, I guess, of forming A, or it
could be the minus the heat of deconstructing A, you
can almost view it. And it would also be minus
the same thing for B. And then, this is just
the elemental form. And now we can go from
the elemental form back to the products. Because we have the
same atoms here. They're just rearranging
themselves into two different sets of molecules. So now we can go back from the
elemental form and go up here. And we know what those are. We know how much energy it takes
to go from the elemental form to C and D. That's their heats
of formation. So Hess's Law tells us that
delta H of this reaction, the change in enthalpy of this
reaction, is essentially going to be the sum of what it takes
to decompose these guys, which is the minus heat of formations
of these guys, plus what it takes to reform
these guys over here. So we can just write it as delta
H of formation for C plus delta H of formation
for D. So the heat of formations for
these guys minus these guys. This is what it took you to
get to the elemental form. So minus delta heat of formation
of A, minus delta heat of formation of B. And then you'll have the
heat of the reaction. And if it's negative, we would
have released energy. And if this number is positive,
then that means that there's more energy here than on
this side, so we would have to absorb energy for this
reaction to happen, and it would be endothermic. So this is all abstract and
everything, and I've told you about Hess's Law. Let's actually apply it
to some problems. So let's say I have this
reaction right here, where I start with ammonia. And it's ammonia gas. And I'm going to react that with
molecular oxygen to yield some nitrogen monoxide, 4 moles
of it, and some water. So what's the heat of this
reaction right here? So what we do, is we just look
up the heats of formation of each of these. So let's just look them up. Let's start with the ammonia. What's the heat of formation
of ammonia? And it's always given in
kilojoules per mole, so they'll say to form one
mole of ammonia. So to form 1 mole of ammonia--
let's look up here. This is all cut and paste
from Wikipedia. And am I starting in the gaseous
or the aqueous state? Well, I think I just-- see, I'm starting the gaseous state. I've added that G there. So ammonia in the gaseous state
has a heat of formation of minus 45.9 per joule. So what is that going--
so minus 45.9 kilojoules per mole. That's just for one mole of ammonia, the heat of formation. It's in kilojoules. I'll just look them
all up right now. Now what's the heat of
formation of oxygen? And I'm not going to look it up
right now, because oxygen is in its elemental form. So if you see something in the
form that it just always takes, before you do anything
to it, its heat of formation is 0. So if you see O2, its heat
of formation is 0. If you see hydrogen, if you
see H2, its heat of formation is 0. If you see carbon by itself,
heat of formation is 0. Carbon in the solid state, heat
of formation is 0, at standard temperature
and pressure. Now what about nitrogen
monoxide? Let's look that up. I have it right here. Nitrogen monoxide. Heat of formation. It's positive, 90.29. And finally, what's the heat
of formation of water? Well, let me see. Liquid water. Minus 285.83. Now you might tempted
to say, OK. Hess's Law says that if if we
want the delta H for this reaction, we just take this plus
this, and subtract that. And you'd be almost right, but
you'd get the problem wrong. Because these are the heat
of formation per mole. But we notice in this reaction,
we have 4 moles of this, plus 5 moles of this,
yields 4 moles of this plus 6 moles of that. So we have to multiply this
times the number of moles. 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. I don't even worry about
multiplying 0 times 5, because it's just going to be 0. So now we can apply Hess's Law
to figure out the delta H of this reaction. So the delta H of this reaction
is going to be equal to, 4 times the heat of
formation of nitrogen monoxide-- so 4 times 90.29,
plus 6 times the heat of formation of water. So plus, I'll switch colors,
6 times minus 285.83. And just as a side note, given
that the heat of formation of nitrogen monoxide is positive,
that means that you have to add heat to a system to get this
to its elemental form. So it has more energy than
its elemental form. So it won't just happen
by itself. And water, on the other hand,
it releases energy when you form it from its
elemental form. So in some ways, it's
more stable. But anyway let me-- So these are the heats of
formations of the products. And then we want to subtract out
the heats of formation of the reactants in our reaction. So here it's 4 times 45.9--
Let me make sure. It's a minus 45.9. Right? That ammonia had a minus
45.9 heat of formation. So what did we end up with? Let me get the calculator out. So I have-- let me make sure
I put it over here. I have to be able to read it. Well, I'll just do it off the
screen, because my screen is getting filled up. So I have-- let me
just do it here. 4 times 90.29 plus 6 times
285.83 negative is equal to-- so so far, we're
at minus 1,353. Does that sound about right? That looks about right. And now we want to subtract from
that 4 times minus 45.9. So we want to subtract-- so
minus 4 times 45.9 negative is equal to minus 1,170. So our delta h of this reaction
is equal to minus 1,170 kilojoules for
this reaction. And all we did is, we took the
heat of formation of the products, multiply it times
the number of moles, and subtracted out the
heat of formation of the actual reactants. There you go. Let's do one more of these. Let's say I had some propane. I had some propane, and I'm
going to combust it. I'm going to oxidize the propane
to yield some carbon dioxide in water. Well, it's the same drill. What's the heat of formation
of propane? Look it up here. It is amazing how exhaustive
these lists really are. Propane is down here in
its liquid state. Heat of formation minus 104.7. So let me write that down. Minus 104.7. Heat of formation of oxygen
in its elemental state. That's how you always
find oxygen. So it's just 0. Heat of formation of carbon
dioxide-- Let's see. Carbon dioxide, and as
a gas, minus 393.5. And water. We already figured that out. It's minus 285.83. So how much heat is formed when
we combust one mole of propane right here? So let's see. We have to figure out the heat
of the products, the heat of formation of the products-- so
it's going to be 3 times this. Because we formed
3 moles of this. For every mole, we release
this much energy. And then plus 4 times
this, and then subtract out 1 times this. So what do we get? We get 3 times 393.5
and that's a negative, is equal to that. Plus 4 times 285.83 negative
is equal to minus 2,300 kilojoules, roughly. And then we have to subtract
out 1 times this. Or we could just add 104.7. So let me just do that. So plus 104.7 is equal
to minus 2,200. So here my heat of this
reaction, is equal to minus 2,219 kilojoules as we
go in this direction. For every mole of propane that
I combust, I will actually produce this much energy
on the other side. Because this right here has
roughly 2,200 less kilojoules than this side right there. So I could actually rewrite this
reaction where I write all that, and I could
have added-- actually, let me do it. I could rewrite this reaction
is C3H8 propane, plus 5 oxygens, yields 3 carbon
dioxides plus 4 waters plus 2,219 kilojoules. That's actually what's released
by this reaction. It's exothermic. This side of the reaction has
less heat than this side, and that-- it didn't
just disappear. It got released. And this is where
it got released. Now sometimes you'll see a
question where they say, hey. Fair enough. You figured out the heat
of this reaction. How much heat is going to be
released if I were to hand you, I don't know, let's
say I were to hand you 33 grams of propane? Well, then you just start
thinking, oh, well, how many moles of propane this is. Because if I combust one
mole of propane, I get this much heat. So how many moles of propane
is 33 grams? Well, how much does
1 mole weigh? The 1 mole of carbon weighs 12
grams. 1 mole of hydrogen weighs 1 gram. So 1 mole of propane is going to
be 3 times 12-- so times 3, because we have 3 carbons there
and 8 hydrogens, so times 8-- so it's going to
be equal to 36 plus 88. So it's going to be 44. So this is going to be 44
grams per mole, right? This is, let me write
that down. 44 grams per mole. Now, if I give you 33
grams, how many moles am I giving you? Well, 33 grams times, I guess we
could say, 1 over 44 moles per gram-- I don't have to write
the whole gram there. And then the grams cancel. I'm giving you 33 over
44 of a mole, or I'm giving you 0.75 moles. So if one mole produces this
much energy, 3/4 of a mole is going to produce 3/4 of this. So we just multiply
that times 0.75. And you get 1,664. So times 0.75 is
equal to 1,664. So if I were to give you 1 mole
of propane, and I were to combust it with enough oxygen,
I'll produce 2,200 kilojoules that's released from
the system. So this side of system has
less energy left over. But if I were to only give you
33 grams, which is 3/4 of a mole, then you're going to
release roughly 1,600 kilojoules. Anyway. Hopefully you found
that helpful.