Hess’s law and reaction enthalpy change | Chemistry | Khan Academy


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.




Comments
  1. You prove why all teachers should be fired and why we should simply learn from amazing teachers at home like you.

  2. yeah…they have the same effect as caffeine based diet pills. You could, like, lose weight from not thinking about eating. *dramatic conclusion music*

  3. I understood almost nothing till I watched this video. I wont fail my test tomorrow!
    Very Grateful!!! ๐Ÿ˜€

  4. God bless you, seriously. I had this concept explained to me by so many people but I did not get it until I actually stumbled upon this video.

    I wish I had known about this channel at the beginning of the semester.

  5. For propane, you had it written as a gas, but used the heat of formation for it's liquid state
    This is a mistake?

  6. Don't worry dude he is right, Wikipedia just made a mistake that is the enthalpy change of formation for propane in the gaseous state ๐Ÿ™‚ Just go to Wikipedia and write Propane (Data Page) ๐Ÿ™‚

  7. Khan you used the heat of formation of propane in its liquid state…but your equation has propane in a gaseous state….

  8. I have been trying to understand for almost 4 years now. I actually finally get it and it feels amazing. I don't think you know how much you are changing lives!

  9. thanks it helps me understand more. Can you talk about different types of enthalpies such as atomisation, bond, enthalpy of solution, lattice enthalpy.

  10. When you give us .75 mol of propane, don't you just multiply that by propanes given Delta Hr and then subtract that from your products? in the video you divided the whole thing by .75 mol.

  11. in the second equation why did you plus the products and the reactants standard heat formation instead of subtracting it like the equation stated.

  12. +William Turner –I walk into a casino with an exothermic amount of cash. I hope to walk out endothermically with a different amount equal to the same amount when I walked in plus some extra $$ inside it. Right or wrong?

  13. This video series saved my chemistry grade this unit. A two-hour lecture and meeting with the teacher after school didn't stick, but somehow this made everything click. Thank you!

  14. The given in the reactant side is -104.7, you forgot to put the negative sign I think so the answer must be different. I'm kinda confused but thanks anyway, you help college student from dying lol

  15. Hey I think the sum of the enthalpy of formation on the product side would be -36.5kJ/mol making the total change in enthalpy for the reaction 68.2kJ/mol right?

  16. Wow calculating enthalpy really isn't that hard like I thought it was. I guess the way my book explained it and my professor explained it just didn't make any sense to me. Thank you so much, this was extremely helpful!

  17. For the exam board AQA anyone, is it essential to know how to draw the diagrams? They make no sense to me.

  18. why do you use the standard change of formation when more than 1 mole of compound is formed? standard change of formation is the enthalpy change when one mole of compound is formed from its element. can anyone explain?

  19. Hi Mr Salman Khan, first off I love your work and what you do. You are an amazing person. Now not to get all grammar Nazi on you or whatever but; Hess'. you are welcome

  20. in an exothermic reaction if change of enthalpy is smaller than activation energy does the temperature of ambient decreases? thx in advandance

  21. Anyone can tell me why the unit of the enthalpy change of reaction is not KJmol^(-1) but KJ instead ? I get a little bit confuse .

  22. I missed a lecture on this topic and understanding it mainly on the textbook was kind of complex, but after watching this simplified powerful video with examples, I am now adept at this topic!!!๐Ÿ˜ thanks so much

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