Gibbs free energy example | Thermodynamics | Chemistry

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Gibbs free energy example | Thermodynamics | Chemistry | Khan Academy

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15 жыл бұрын

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Determining if a reaction is spontaneous by calculating the change in Gibbs free energy. Also calculates the change in entropy using table of standard entropies. Created by Sal Khan.
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Пікірлер: 56

@luchis117 11 жыл бұрын

I hope they restructure 'higher education' soon. It needs to be updated and interactive and tailored to the individual so EVERYONE can learn these things. I'm sick of professors explaining things as if we already know the material. Thanks for taking us a step in the right direction Sal.

@jmmtcidc 12 жыл бұрын

I don't know why, but your seemingly random changes in pen colour actually make this a lot clearer. Thanks for your videos!

@hierapresmidi 12 жыл бұрын

You've just managed to condense my couple of hours of IB revision into 10 minutes... amazing.. THANKYOU!!

@yuvanm359 3 жыл бұрын

Watching this on 2021, extremely helpful

@rickandsteve1 12 жыл бұрын

This just made my AP Chemistry notes so much more understandable. A LOT more helpful than my teacher. Thanks!

@rhcorr 14 жыл бұрын

Hey! At first I was having a very hard time on exactly comprehending what the so-called Gibbs "Free Energy" really was. It actually is no "real" energy, but an arbitrary definition *based* on state functions. Maybe this can help others with similar uncertainties: (double-ues)•chem1•com/acad/webtext/thermeq/TE4•html (replace with proper formatting symbols) And maybe include that on your next Free Energy videos? And, of course, very nice work there!

@kendo512 9 жыл бұрын

The last few sentences about temperature relative to entropy change really hit it home for me. If you're still having trouble understanding this intuitively, listen to the last minute or two a few times.

@MuziMzwandileMavuhHovah 12 жыл бұрын

really helpful especially to 1st year students even postgraduates because we all human

@kylehenryricarse7397 9 жыл бұрын

kJ = is kiloJoules which is a unit of enthalpy.. for entropy, you have Joule/Kelvin.. so, why divide -242.2 by 1000 when it is not J/kilogram for the entropy.. kilogram and Kelvin are pretty much not of the same spectrum for you to convert one with respect to the other. confused.

@kylehenryricarse7397 9 жыл бұрын

+Khan Academy

@DieTreppenwitz 8 жыл бұрын

+kyle henry enthalpy is kilojoule per mole. i.e. 1 kJ/mol is 1000 joule per mole. entropy is joule/kelvin which means 1 J/K.mol is a change of 1 joule per degree change per mole. to account for that 1000 joule difference we divide by 1000. no kilograms here. kilo merely means a factor of 1000.

@SmashBrosBrawl 8 жыл бұрын

+Arjun Mohan I calculate -242.7J as Delta H, how did Khan get -890kJ?

@Redmau 13 жыл бұрын

great video!!! thanks!

@medic8038 3 жыл бұрын

Thank you very very much That was exactly what I want

@danzavier03 14 жыл бұрын

Brilliant, you are the savior of my chemistry module.

@MichelJosephCardin Жыл бұрын

Would ice be able to compress a gas enough to melt it and when released would it be cold enough to freeze it back up until the same hardness is achieved?

@sciencenerd7639 2 жыл бұрын

very helpful review, thanks

@virupannamedikinal Жыл бұрын

Thank you so much.

@tedpianoman 12 жыл бұрын

You have possibly saved my life. Thank you

@tavito92 13 жыл бұрын

Excellent video man, thanks! You're helping alot of people out!

@Bernadettefull 13 жыл бұрын

Thank you so much!!!! I love how you are so clear and quick! Absolutely marvelous!

@RakeshSharmarajj 7 жыл бұрын

dear me i would like to ask how Di oxygen has standard hear of formation 205 Kj permole it must be zero

@joelyeohjierong553 7 жыл бұрын

Thats not heat of formation, its standard molar entropy

@echo0204 12 жыл бұрын

Sal, how can we calculate Gibbs free energy of the reaction which is not at the standard pressure and temperature condition? and when how can we create this condition? is the delta H and delta S should be changed based on the new Temp and pressure??

@Zachtml 13 жыл бұрын

Wow you helped me so much thanks.

@Aeynx 15 жыл бұрын

You are a savior.

@sciencenerd7639 2 жыл бұрын

wouldn''t the water be vapor (gas) and not liquid though?

@ad2181 15 жыл бұрын

Sal, How do you remove the energy without increasing the temperature of the reaction?

@lulita1234 11 жыл бұрын

will they have the heats formation numbers in an exam? where do u find them?

@vladklimovich5213 7 жыл бұрын

I made it. I used inplix scripts for that. It was pretty easy to make it

@MadM0nte 8 жыл бұрын

I'm sorry but I really don't get how you got -890.3 from those numbers. How do you get kj from joules that aren't even in the thousands.

@huntermcquary8711 8 жыл бұрын

+MadM0nte The -890.3 referred to the Enthalpy (delta H) while the numbers using jules/k(mol) are referring to the entropy of each of the systems (delta S). They are two different things.

@97GoldDust 9 жыл бұрын

how did you find delta H in the beginning of the video? When I do the math I get the same value as (delta) S. To find (delta)H did you use Hess's law? The part where you have 3 equations and you add up the heat for the 3 equations and that sort of stuff?

@qwertydnfbryeu 9 жыл бұрын

Hess's law. dH = H(products) - H(reagents) H for simple substance equal 0 CH4 + 2O2 -> CO2 + 2H2O dH = H(CO2) + 2H(H2O) - H(CH4)

@Bridscape 9 жыл бұрын

PLEASE ANSWER; ok so for the delta S and delta H values, are you just grabbing them from a table? is there a way to calculate them without using a table? please explain!

@kylehenryricarse7397 9 жыл бұрын

+q.Edits same is true with all of them. summation of products - reactants. but take into consideration their heats of formation.. those data is tabulated and are given, just subs. them with the molar coeff. and you are good to go.

@thegeffc 12 жыл бұрын

great video thanks

@alexasmithy 12 жыл бұрын

brilliant video

@arschoi3 9 жыл бұрын

THANK YOU SO MUCH

@filipbahunek2504 7 жыл бұрын

This makes very little sense to me. This would mean that methane gas would spontaneously combust with the oxygen in our atmosphere, however, it doesn't, does it?Also, wouldn't the water produced by the reaction be in a gas phase rather than a liquid phase at the temperatures produced by the reaction itself?

@michaeldepauw7446 7 жыл бұрын

Just because a reaction is spontaneous does not mean that it occurs quickly. At room temperature without a match or spark of some kind, you could be waiting for years before an appreciable amount of products would form. This is because the activation energy is very high. As such, a catalyst or initiator is generally needed to speed things up. For example, because the reaction is EXTREMELY exothermic, if even a small spark which is a highly concentrated tiny locus of energy causes a group of these molecules to gain sufficient energy to begin reacting, the HUGE amount of energy they release provides the energy for a large number of additional molecules to react, triggering a chain reaction which explodes the entire sample. As far as your second question, you are absolutely correct that the water formed would be in the gas phase when the sample is burned. However, if the reaction were to occur via the aid of a catalyst as opposed to a spark or flame, liquid water may condense on the catalyst instead. I know your comment was months ago, but hopefully this provides some closure. :o)