Derivations of 0th, 1st & 2nd order integrated rate law

Derivations of 0th, 1st, 2nd order integrated rate law

[@abhijeet_ghosh]

Sir, this was extremely helpful. My chemistry class is made up of a majority of students who dont know calculus, and therefore the teacher just gave us the formulas and a brief explanation. This helped me understand 🙏

[@MrGrodskiChemistry]

Glad it helped!

[@tomasb7645]

This us one SWEET calculus-based explanation at deriving the integrated rate laws! THANK YOU!!!

[@MrGrodskiChemistry]

You're very welcome!

[@jeridahketty8280]

This is now the best have been looking for, thank you.

[@MrGrodskiChemistry]

You're welcome!

[@AstrogeezTube]

Incredible

[@MrGrodskiChemistry]

thanks!

[@AstrogeezTube]

@@MrGrodskiChemistry i will subscribe for this video

[@nedeisenberg]

Thanks for the thorough and descriptive video - really helped.

[@travisyoung6055]

Amazingly comprehensive for the maths layman such as me

[@peterquiroz8899]

Thank you sooo much. This video was very useful for a paper of mine!

[@LochanKutz]

hello sir, your sign for the integral is wrong. the video was good either way.

[@MrGrodskiChemistry]

You are right! Great edit! For some reason I always wrote them that way.

[@LochanKutz]

@@MrGrodskiChemistry 😁👍

[@anonimoalasad9581]

In case anyone missed it as i did, during the proof of the 1st order law the integer of d[a]/[a] is equal to ln[a] due to the fact that if we derived ln[a] we would have to apply the power rule and take the derivative of the ln (1/a) times the function embbeded inside the function, [a].

[@daniellaobikunle1440]

Very precise, thank you.

[@MrGrodskiChemistry]

You're very welcome

[@debisadebela9263]

thank you verry much i appreciate you guys u are a genius man!

[@sakshishukla41]

Hi, thanks for the video. However, I am still confused why [A] is constant (on the right side of the equation) in the differentiation for the first order equation?

[@copiryte9535]

-d[a]/dt=k[a] in order to integrate both sides we need to get the same variables on both sides of the equation first lets get both [a] on one side and dt on the other side like so: d[a]/dt=-k[a] [a]*d[a]/dt = -k divide by [a] [a]*d[a] = -k*dt multiply by dt now plug in your integrals and you can solve!

[@dernayt2728]

Video is really helpfull, idk why but the textbook I use just wrote the equations without derivation. Btw, is it legal to integrate both sides with different variables in 14.33

[@doer6002]

Thank you so much.

[@funnymalia8581]

the only problem you have is that you uploaded this really long before people were interested in this :) If u had posted this sometime in the past year or so, you would surely have at least 50k views :) great work man!! you've done a really fine job! I sincerely hope more people find this gem.

[@SS-pn7ss]

Thankyou so much

[@MrGrodskiChemistry]

You're welcome!

[@jarencascino7862]

paying thousands to go to college just so I can look up a free video that does a better job of teaching it

[@alexanderstephenson4120]

thank you so much :)

[@MrGrodskiChemistry]

Thanks! You're welcome!

[@achiriu1987]

Thank you

[@MrGrodskiChemistry]

You're welcome!

[@x69WINNING69x]

i love it, now i dont have to remember these!