You’re famous in our university in Saudi Arabia 🇸🇦 Your lectures are understandable and so easy❤️👌 thanks for making Bioc easy for us.

My friend right now I'm a student and I'm broke af, but the moment I graduate ad I get a job I'm fucking going to hit that 200$ donation button, thank you for the immense help you are giving us, *ak lectures is the best!!* a big hug from Italy :)

@himariyabelayneh1937

3 жыл бұрын

I hope you came through

Saved my day again, by the way thank you so much literally, all the students in my university watch your lectures and I got all my friends to watch your lectures too , keep up the good work ❤️

Your lectures are literally on par with my biochemistry lectures haha, your lectures are a lot easier to understand. Thank you!

@AKLECTURES

9 жыл бұрын

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

@shauncheese

9 жыл бұрын

of course :), University of Waterloo Ontario

@AKLECTURES

9 жыл бұрын

Sean Chih very cool! thanks! :)

@miguel2001p

2 жыл бұрын

@@AKLECTURES I go to university of california san diego and i also find your videos very helpful and needed even after watching course lectures

Your tutorials are amazing, tnx god for ppl like you !!

waw waw waw ... thank you so much you bro deserve more than 400K subscribes ,, you deserve to talk on the red carpet

I really love this channel. it makes things so easy to learn...I like the lecture methodology...

what a lovely and clear your language. thank you very much

You're the real MVP!!

You just make so much sense. Thank you!

Thank you, this was perfectly explained.

if i was taught like this from begining i would be a very very intelligent student. but sorry to say that our teachers have a lot knowledge but they have not the potential to teach like this. thank you sir . love from india.

Wow …😱😱what a great lecture thank you❤️

you are amazing. thanks so much.

best so far n till now

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.

Great work :)

Amazing! Thank you

Thankyou thankyou thankyou to universe for making u sir

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

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

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

great job !!!!!

Excellent 🔥

Thanks very much 🔥

really helpful.. thanku ..

Thank you so much

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

2 жыл бұрын

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

U are best bro My new.sir keep it up

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

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

7 жыл бұрын

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

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

8 жыл бұрын

+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

8 жыл бұрын

Yeah!thankyou...

Thanku sir

thank you so much! :)

thanks you rock !!!

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

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

2 жыл бұрын

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

Love from india thank you sir

GREAT VIDEO

@AKLECTURES

9 жыл бұрын

Lior Marko Thanks Lior :)

thank u thank u thank u

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

Good

Why no sound???

cant hear sound:(

Brain goals

No sound 🙁

No sound

I love you

5-stars

👍👍👍👍

Sound

😔No sound

@AKLECTURES

7 жыл бұрын

Turn your volume on? ;-)

@carolinemangare3071

7 жыл бұрын

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

7 жыл бұрын

your explanation is wonderful 😊

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

@criticalmindset8471

6 жыл бұрын

Ekram M no me too