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

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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.

Yu HuangPost authorthank you soooooooo much, i've been sit in my chem class since school starts and now i understand everything just by this 15 min video..

Daniel ColyerPost authorThank you for all of your help. When I'm rich I will donate many treasures to you.

MJ5347Post authorThis was so helpful, THANK YOU SO MUCH! you have saved me from hours of confusion.

Tanya PilaPost authorthat was so helpful! thank you so much khanacademy! π

David Imanuel WPost authorstudied 2 week @school. 15 min with khan academy

TheKhulKidPost authorYou prove why all teachers should be fired and why we should simply learn from amazing teachers at home like you.

Meldon FernandesPost authorcan u pls tell me how to calculate the enthalpy of combustion of glucose .

thank u.

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

Denver BrandtsPost authorwikipedia. khanacademy approved

GyriPost authorI LOVE YOU, SAL

Candycanes02Post authorI understood almost nothing till I watched this video. I wont fail my test tomorrow!

Very Grateful!!! π

CEBPost authorGod 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.

thomas grantPost authorFor propane, you had it written as a gas, but used the heat of formation for it's liquid state

This is a mistake?

stefanperanovicPost authorDon't worry dude he is right, Wikipedia just made a mistake that is the enthalpy change of formation for propane in the gaseous state π Just go to Wikipedia and write Propane (Data Page) π

choogiesaurPost authoramen brotha. in your situation right now

dimi4threePost authorwow i love you

James BarnettPost authorMy brain adores you.

donothackPost authorThis video helped me a ton, thanks

Marcus OngPost authorkhanacademy, why is the state symbol for H2O (liquid) instead of gas?

Your awesomePost authorTHANK YOU THANK YOU THANK YOU

Dandy LionPost authorh2o is water… water is a liquid?

Yaa Ofori-MarfohPost authorKhan you used the heat of formation of propane in its liquid state…but your equation has propane in a gaseous state….

crmnmaPost authoryour voice is strangely soothing

Paddy NeshPost author2 hours is pretty extensive for a chemistry class. Mine is 50 min…

fchoudhury13Post authorWow, try 3 hours ._.

GiftPost authorI 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!

SmileyAnniiPost authorMy god, your videos always save me!!! Thank you so much

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

Brenna RansomPost authorYOU ARE A GOD!!!!!! THANK YOU SOOOOOO MUCH!!!! YOU ARE GOING TO SAVE MY HONORS CHEMISTRY GRADE π

Allen SpacerPost authorYOUR ARE MY HERO!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

josh cruddosPost authorThank you so much !!!

blackblur1Post authorAwesome, saved my ass for honors chem tomorrow

AlbanianMusic LovePost authorYEEESSS!!! Now it makes sense! Thank You π π

stefanjenouskiPost authorWhen 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.

Yasir SolejaPost authorhi

Shibaz HussainPost authorYOU ARE A HERO

binghua quiPost authorgreat job

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

00PapyrusPost author<3

Cathy SJ9211WPost authorYou forgot the minus sign at 11:08, this has caused me great confusion. =(

Anna NPost authorWow this video was great! Thanks so much

William TurnerPost authorHoly shit, now it's clear to me. Thank you!

OdiakoptisPost authorwait is hess`s correct

Victoria JaroshPost authorFantastic

oneuponzeroPost author+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?

Shopana BPost authorThanks so much for thisππ½

FrancisPost authorthis really helped, thanks!

Karolina ZgodziΕskaPost authorthank you!

Ken The GPost authorOMG. If i had found this like 2 hours ago, so much time would have been saved. Thanks so much, this was very helpful.

Some Random FellowPost author10:44 thats gaseous propane, not liquid

Bobs BurgerPost authorwhy the ringing sound

Chuck HarrellPost authorGod bless your soul Sal

Kara OPost authorThis 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!

AndrΓ© PrielaPost authorThe 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

BethanPost authorSaved my chemistry as level! Thanks:)

Chaser ZXPost authorthanks really helpful. as the other person mentioned, if only I had found it 2 hours ago.

Youssef M. Al-AasarPost authorthe propane is in a gaseous state, not liquidous.

Mitch KPost authorHey 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?

EmkayPost authorat 12.20 why is 104.7 added and not subtracted to the products?

Sierra JarmonPost authorWow 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!

Josiah GPost authorHess' law *

Mindless GamingPost authorSo much easier than drawing the stupid diagrams. Thanks

Real GuyPost authorwhy did you keep oxygen zero. doesn't oxygen have bond energy.

A MPost authorno one else can explain it properlt, thank you for simplifying and not using unnecessary arrows all over tha place

Jordyn AshlockPost authorjust got an A- on my AP Chem final. Couldn't have done it without you

JasminePost authorWould you get the same answer doing it this way as drawing the arrow diagrams?

Christian PopovskiPost authorthe propane should be liquid

The Gayest Person on YouTubePost authorThank a tha you for this my good

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TulopPost authorwhere do your numbers come from again?

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

PawlosPost authorwhy 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?

Diamond MindPost authorHess' law.

Ehsun Malik MalikPost author7 years laters this video still a saver

GibraltarPost authori love you <3 ……..like bromance…..

Sleven KelevraPost authorThere is a mistake in calculations, no? I got 1537.42 in the end

Jaya LishwarPost authoru r the best in the world

Aarnav ChauhanPost authorGreat

Rutuja KalePost authorekdam bore

Ulvi BektashiPost authorHi 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

Norbert LimPost authorexcuse me but why for O2 is zero

Karolina SPost authorI'm still confused π

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

Madeline BadePost authorI still don't get it

TheOutdoorAnglerPost authorAre the heats of formation in the reference tables?

Aqif bhatPost authorfull clarity .explained very well.thanks

Just a Single Drone VideoPost authorπ thanks

128 StrangePost authorAnyone 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 .

jaux4Post authorI love you. All I can say.

Van NgoPost authorYou are a legend mate. Thankyou for explaining topics so clearly

Rukudzo MunyezaPost authorThanks man. Youβre gifted

Katestrophe SPost authorWhy was the 285.83 a negative when you multiplied it in your calculator?

BNZPost authorThank you, very helpful

susan hPost authorthis video remains great after 9 years! #teamkhanacademy

Waka WakaPost authorI need help whit this one:

Fe2O3(s) + 3CO(g) -> + 2Fe(s) +3CO2(g)

Gabrielle SpritePost authorkJ/mol!

I shouted that out every time he said just kJ. Other than that, this is on point.

Kenneth OnumakuPost author"Hope you found that helpful"…are you kidding me, just helpful? you are a hero indeed!

DexterPost authorThis guys deserves a knighthood or something π

Nunya BusinessPost authorWelp, I might fail my chem test.

jai parkashPost authorthanks it helped me a lot

TimmyQ RevolutionPost authorI 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!!!π thanks so much