Enzymes' Effect on Activation Energy and Free Energy

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Sean Chih lol thats awesome and thanks! spread the word to your classmates. what school are you in?

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of course :), University of Waterloo Ontario

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Sean Chih very cool! thanks! :)

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[@nobruco]

Hi, thanks for this lecture about Thermodynamics of chemical reactions. I have a small concern with the upper right part of your board; It seems that you forgot the standard gibbs free energy (DelatG°). In fact "DeltaG = DeltaG° + RTLnKeq", and when reaching the equilibrium, DeltaG equals zero means that DeltaG° = -RTlnKeq (but does not mean that k1=k2). Otherwise all equilibrium constants would be equal to 1. In other words reaching the equilibrium allows one to quantify Keq, the equilibrium constant (which is also : Keq=k1/k2) and this value not necessarily equals to one. I would rather say that reaching the equilibrium means that k1*[substrate]=k2*[product] (forward velocity equals reverse velocity) The case you are mentioning, (i.e. when k1=k2) is just a particular case where reactants and products have the same level of energy and they are both at the same concentration at equilibrium (50% of reactants + 50% of products at equilibrium). Or that "k1=k2" means DeltaG = DeltaG° and DeltaG is independent of the temperature at wich the reaction takes place. But I agree with the overall message that at equilibrium the kinetic of the reaction has stopped and DeltaG equals zero.

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Thankyou thankyou thankyou to universe for making u sir

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Wow …😱😱what a great lecture thank you❤️

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best so far n till now

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The best ever 🙏🙏🙏🙏🙏🙏u are such a blessing to us

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Thank you, this was perfectly explained.

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Thanks very much 🔥

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Excellent 🔥

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Thank you so much

[@NutriHubbyDtMaram]

U are best bro My new.sir keep it up

[@alvinjesus100]

Thanks for explaning....You do a great job....i had a question regarding activation energy of surface chemistry... amongst physisorption and chemisorption,which of the two has greater activation energy?

[@rejathbenny1964]

+Akshat Bajpai It is evidently Chemisorption , simply because it involves a chemical reaction and needs activation energy . On the other hand , there is NO reaction taking place between the adsorbant and adsorbate during physisorption (but physical adsorption of gases involve condesation ). Am I clear ?

[@alvinjesus100]

Yeah!thankyou...

[@freeflyer9482]

GREAT VIDEO

[@AKLECTURES]

Lior Marko Thanks Lior :)

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what's the difference between endergonic,exergonic and endothermic , exothermic..thank u by the way,ur videos are so helpful

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Good

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Great work :)

[@1marthag]

you are amazing. thanks so much.

[@nondumisondhlovu9181]

Please help me understand this. I was able to follow until the 11th minute where you say that the enzyme lowers the activation energy hence allows the reaction to reach equilibrium faster (this I understand) but now what confuses me is that you said that lowering of the activation energy results in stabilising the transition state. Now, won't stabilising the transition state mean that it will take longer for the equilibrium to be reached because the components of the transition state are no longer unstable - isn't it that instability is associated with readiness to react and stability takes a longer time to reach?

[@kwindat.i.3560]

I think is late to answer this question but in principle you asked a very good question. What I understand about catalyst or rather enzymes is that they lower the barrier which is the hill and this allow more molecules to overcome it to form product at a faster rate. The term stabilization has to do with lowering of activation energy but in principle the molecules on the top of the hill are still not thermodynamically stable and eventually will react to form product with lower Gibbs free energy and now the driving force is the reduction of Gibbs free energy. Note, that catalyst only remove or reduce the barrier but don't determine which product is to be formed. That has to be determined by thermodynamics. So enzymes or catalysts only affect kinetics but not thermodynamics. So to answer your question, it will not stay at the hill for longer time due to its instability, it will further decompose to be at the lowest energy state i.e. thermodynamic stable state

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thank u thank u thank u

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You are choo cute AK..i love your lectures😍

[@idilfarah2798]

Really great video!! I finally understand this! Thaaaanks :) Best regards, Idil from Denmark

[@rajkumarraghupathy8235]

really helpful.. thanku ..

[@taehoonlee4287]

great job !!!!!

[@anooprao6380]

At 8:32, if you don't know what the transition state looks like, how do you make transition state analogs? Thanks again for the video :)

[@okpolihenry983]

u can use an analog of the substrate to trap the reaction at its transition state , that's how we study the structure of the transition

[@trishasales714]

Hello, I know this isn't directly related to the video but you have talked about this before in another video so I'd like to ask- does chymotrypsin also cleave at the c terminus of methionine? Or does it only cleave after large aromatic residues? Also, I read somewhere that similar to typsin, chymotrypsin only cleaves at the c terminus of large hydrophobic residues ** UNLESS THE next amino acid is proline- is this true?

[@dphunk87]

In my class we learned that it cleaves c terminus of large hydrophobic residues, methionine included as well as the aromatic ones.

[@pragatipandey7489]

Thanku sir

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5-stars

[@alejandrosalazar8332]

thanks you rock !!!

[@palakvarma7720]

Is activation energy the difference between the energy of transition state and energy of reactants ?

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I love you

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thank you so much! :)

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[@JonathanR1994]

Brain goals

[@abdulbasitghauri8934]

Can anyone tell me that what is K1 and K2 ??

[@criticalmindset8471]

Why no sound???

[@sierraandrewcavazos8371]

cant hear sound:(

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No sound 🙁

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No sound

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Sound

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Why doesn’t this video have sound or is it just me🤨

[@criticalmindset8471]

Ekram M no me too

[@carolinemangare3071]

😔No sound

[@AKLECTURES]

Turn your volume on? ;-)

[@carolinemangare3071]

AK LECTURES I did but still couldn't hear anything. Though I read your notes on the board. You are amazing. Thank you for your lectures!

[@chemisthmh198]

your explanation is wonderful 😊