Actually Khan Academy is sorta overated imo

Glad you found this helpful Krishna and I'm more than happy to have them available here! Happy Studying!

@@ChadsPrep Plz dont stop uploading. You are growing very fast believe me.

@@microcomputers5609 Slow and steady!

@@ChadsPrep I am an Indian and i should be honest with you. We have an entrance exams ( JEE Advanced ) which allows us into the IIT's and i am sure you know about it. Most of the teachers here in India are incomponent of teaching its level. And i am sure professor, you can teach upto that level. If you raised level till JEE advanced, believe me you will recieve huge crowd from India and you will be helping alot of students. Also this level will be automatically covered. Its my suggestion 🥰😍🤩

Wow, thanks! Glad you found it helpful!

Better late than never!

Glad you're finding them helpful Cody! And I have no idea what I'd do if I wasn't teaching!

You're welcome - Glad you found it helpful.

Glad it was helpful, Jamyson!

Glad you found us, Dashing Planet - you're welcome!

Thanks for watching, Abdulrahman.

Glad you found us!

That's great to hear, Taran kaur!

Thanks - and you, too!

Great!

You're welcome, eunoia, glad it helped!

Glad that connected with you Christian. Some people connect with the plain math of the equation, but this is my preferred way of looking at to also! 🙂

Yes!

You're welcome.

Thanks!

This is copy of a video on youtube. Not thankyou to the man who upload the video ,thankyou to the teacher who gives lecture.

Yeah I think so that's why the teacher is amazing but there are so less comments n subscribers

Yaaaaa...

@@romanreigns542 Hey there, Roman - I am the teacher/lecturer, but I also uploaded the video. Glad you find the video helpful.

You are welcome so much!

Glad to hear that, amine - thanks for your comment.

Thanks M! Glad you're finding my videos helpful!

Wow - glad you think so - Happy Studying!

Thank you.

@@ChadsPrep 😊

:)

Thanks, Micah!

Thanks, Kora - glad the channel is helping!

Excellent - glad to hear it!

Thanks Vasu! Happy Studying :)

😀 😀 😀

I don't know what else to say besides THANK YOU Bon!

You're welcome Danny! Glad you found this lesson helpful!

Thanks!

Hope you do well, J.Z.

@@ChadsPrep I am back to testify this video was very helpful and guess what? I got it all right! Thanks very much.

Congrats Jonathan and you're welcome!

can you clarify what you mean by a mixed species system?

Think about the combustion of glucose: C6H12O6(s) + 6O2(g) --> 6CO2(g) + 6H2O(g) The delta H is negative and delta S is positive which means it is spontaneous at all temperatures. Now consider the reverse reaction: 6CO2(g) + 6H2O(g) --> C6H12O6(s) + 6O2(g) This has a postive delta H and negative delta S and is not spontaneous at any temperature. But this doesn't mean it won't ever happen it just means it won't happen on its own. It will have to be accompanied by a huge input of energy from an external source. This is the reaction plants do during photosynthesis, and that big energy requirement is met by photons from the sun! Hope this helps!

Glad the channel is helping you - Happy Studying!

You're right in saying that when every reaction reaches equilibrium that delta G will be zero. But the conclusion you come to seems to be then why do we care? And the idea is that if you have a system which is not at equilibrium but is spontaneous (negative delta G), then the absolute value of delta G under those conditions will tell you the maximum amount of work that can be performed by the reaction as it goes to equilibrium. This is very valuable information for say getting an idea of the maximum amount of work that could be performed by a combustion reaction (like in a combustion engine) as it goes to equilibrium. Ultimately, the more negative a delta G value the more energy that will be released as that reaction goes to equilibrium. Hope this helps!

@@ChadsPrep So how we can count total gibbs E of some reaction if every reaction achieve equilibrium at the end, that means we can count for some reaction more gibbs E depends at which point our rection is, but when our reaction achieves equilibrium we can't count the gibbs E because it becomes 0

So the further a reaction is from equilibrium the greater the magnitude of delta G and the more work that can be done in getting to equilibrium. But once equilibrium is reached the reaction has reached its minimum free energy and no more work can be done by the reaction.

@@ChadsPrep Thanks

Thanks Ari and that is definitely true! I'm working or refilming the entire playlist this school year. I just finished the 1st semester topics (up through chapter 13) and will be working on the 2nd semester topics in the Spring! Happy Studying!

Thanks!

Fun with CHEMISTRY. 😁 😁 😁

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Not to mention: less criminal activity and more teaching!

Thanks Dunya! Helping is all that matters!