Thermodynamic versus Kinetic Control

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  • เผยแพร่เมื่อ 14 ม.ค. 2025

ความคิดเห็น • 40

  • @sarahholland5980
    @sarahholland5980 4 ปีที่แล้ว +4

    this made so much more sense than my professor or textbook. thank you!!!!

  • @jkpp.
    @jkpp. 10 ปีที่แล้ว +3

    Thank you so much. It's rare when I get that, "FUCK, I GET IT NOW" feeling. The clear explanation is much appreciated.

  • @vasanthakumarg4538
    @vasanthakumarg4538 4 ปีที่แล้ว +5

    Crisp and clear explanation of concepts, well done and thank you

  • @joenyc
    @joenyc 13 ปีที่แล้ว +5

    you rock. thanks for the AMAZING explanation. i was clueless about the difference. now it all makes so much sense!

  • @sciencenerd8879
    @sciencenerd8879 8 ปีที่แล้ว +8

    Life makes sense again. Thank you!!!

  • @chiman3778
    @chiman3778 5 ปีที่แล้ว

    I just wanted to thank you for putting this together

  • @razasyed575
    @razasyed575 2 ปีที่แล้ว

    my main man micheal evans coming to the rescue thank you!

  • @oguzcangokce6959
    @oguzcangokce6959 2 ปีที่แล้ว +1

    Thanks for the brief and clear explanation!

    • @mevansthechemist
      @mevansthechemist  2 ปีที่แล้ว

      You’re welcome and thanks for watching!

  • @danielgladish2502
    @danielgladish2502 4 ปีที่แล้ว +2

    Great explanation. What I don't understand is how the factor reaction time works? Like why does the kinetic product eventually convert into the thermodynamic product? No new energy is added to overcome the larger activation energy barrier of the thermodynamic product.
    Thanks!

    • @mevansthechemist
      @mevansthechemist  4 ปีที่แล้ว +1

      Great question! You're absolutely right that no new energy is added, but keep in mind that the distribution of molecules in the reaction changes over time. It's all based on relative activation energies. Early in the reaction we have mostly the starting material (SM), which converts mostly into the kinetic product (KP) due to the lower activation energy leading to it. A little bit of thermodynamic product (TP) forms. Now, if we focus on the activation energies of the REVERSE reactions, we see that the KP reverts back to SM faster than TP. Thus, KP will eventually be funneled to TP due to its greater rate of reversion to the starting material.

  • @sanadaltarawneh3550
    @sanadaltarawneh3550 ปีที่แล้ว

    Good stuff

  • @Hands4Surgery
    @Hands4Surgery 12 ปีที่แล้ว

    Is your activation energy correct? (0:20 - 0:25)
    I think activation energy is the whole value from the energy level of the starting material, all the way up to the peak of the transition state. Your second (right hand side) activation energy seems correct proceeding from left to right.

    • @mevansthechemist
      @mevansthechemist  7 ปีที่แล้ว +3

      Good question-for anyone still confused by this, note that the graph is showing two different reactions proceeding outward from the center. Thus, two separate activation energies are shown.

  • @jameelsahadath217
    @jameelsahadath217 9 ปีที่แล้ว

    how is kinetic control used to affect the outcome of a chemical reaction where several reaction paths are available

  • @Elevatone
    @Elevatone 11 ปีที่แล้ว

    What organic chemistry software is that? Link please?

  • @redarmuhammad1953
    @redarmuhammad1953 10 ปีที่แล้ว

    very nice explanation.. Thanks

  • @saritar8819
    @saritar8819 11 ปีที่แล้ว

    Thank you for simplifying it!

  • @panoskb9405
    @panoskb9405 7 ปีที่แล้ว +1

    If you have a reaction that A+B->C, and it is only reversible then the only conclusion that you can make is that is kinetically controlled right?

    • @mevansthechemist
      @mevansthechemist  7 ปีที่แล้ว +1

      If the reaction as a whole is reversible, it's thermodynamically controlled, not kinetically controlled. "Reversible" in essence means that the forward and reverse reactions are proceeding at equal rates; i.e., that the reaction system is in equilibrium (caveat: this is the organic chemist's "back of the envelope" definition. The definition in physical chemistry is more rigorous.)

  • @adosar7261
    @adosar7261 4 ปีที่แล้ว

    If we change the temperature doesn't delta G of both reactions change? So how do we know that after changing the temperature p1 isn't more thermodynamicaly favorable than p2?

  • @parkjane116
    @parkjane116 6 ปีที่แล้ว

    This helped a lot. Thank you so so much.

  • @RBaleog
    @RBaleog 13 ปีที่แล้ว

    Good explanation, thanks!

  • @aoifewest
    @aoifewest 10 ปีที่แล้ว

    Great presentation. Thank you very much?

  • @ruthmasih9047
    @ruthmasih9047 2 ปีที่แล้ว

    amazing! thanks so much.

  • @camichael74
    @camichael74 11 ปีที่แล้ว

    Thank you! great video.

  • @zaya421
    @zaya421 3 ปีที่แล้ว

    I still can't get it.. 😢

  • @skyyward111
    @skyyward111 9 ปีที่แล้ว

    Thanks a lot! Very helpful!

  • @TheBitili
    @TheBitili 14 ปีที่แล้ว

    Great,,,

  • @davidbohadana8274
    @davidbohadana8274 10 ปีที่แล้ว

    Grate job!!!

  • @FatimaSayes-en5bc
    @FatimaSayes-en5bc 7 หลายเดือนก่อน

    Finally i understood.

  • @Aryan_sg
    @Aryan_sg ปีที่แล้ว

    Thanks bro 😉

  • @mmarycalvin6395
    @mmarycalvin6395 ปีที่แล้ว

    Thank you.

  • @anoopkumarsonar1001
    @anoopkumarsonar1001 8 ปีที่แล้ว

    Thanks Bro !

  • @jingyiang9786
    @jingyiang9786 5 ปีที่แล้ว +2

    CN2116

  • @LunaDanu1
    @LunaDanu1 8 ปีที่แล้ว

    thank you

  • @Ace0077
    @Ace0077 11 ปีที่แล้ว

    thanks, helped

  • @arqumabdullah2462
    @arqumabdullah2462 3 ปีที่แล้ว

    whose here from AAMC C/P SB #20