This made everything so easy. Thank you so much! There’s another part which I wished to see, second order with two reactants instead of one. That part is very difficult right now, and I wished to see it in this video.
No, sir. It's correct. Finishing the integration you get to the equation of second order. Or maybe I just did not understand what part you said that it was wrong
I think what you're focusing on are the negatives, but because everything becomes negative, you simply multiply both sides by negative one, and you get the integrated rate law. His one mistake was not finishing the 2nd order example to completion to show how that happens.
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I suck at memorizing math. I like to prove what I’m doing. Which can Be a pain when doing general chemistry so thank you so much for this video
Glad the channel is helping you - Happy Studying!
nice to see proof search process for formulas in Gen-Chem, a very rare thing to see :) thanks for the clearance
You're welcome and Thank You.
This made everything so easy. Thank you so much!
There’s another part which I wished to see, second order with two reactants instead of one. That part is very difficult right now, and I wished to see it in this video.
Glad you found it helpful. I'll keep that in mind for a future video.
Yay I’m in calculus and so chem I can do this now I don’t have to remember the integrated rate laws
Excellent!
You're my academic guardian angel 😇
😇😇😇
4:58 1/[A] integrates into ln[A], but why did you ignore the dA doesn’t that integrate into A ?
Had this derivation in our chem textbook
Great explanation!
Thank you!
You are saving my chem grade rn.thank you!!
Glad the channel is helping - Happy Studying!
keep it up sir
You bet!
Rate law for second order =-1/2× dA/dt and rate=k[A]^2
You are a great man!
Thank you - Happy Studying!
Fight the good fight of faith bro 😉☺️
Indeed!
bless you sir
Thank you.
ty!
yw!
Thank youu
You're welcome!
Derivation for second order is wrong....check the rate law
No, sir. It's correct. Finishing the integration you get to the equation of second order. Or maybe I just did not understand what part you said that it was wrong
I think what you're focusing on are the negatives, but because everything becomes negative, you simply multiply both sides by negative one, and you get the integrated rate law. His one mistake was not finishing the 2nd order example to completion to show how that happens.