We can use either reaction speed or reaction extent to control product distribution. Which aspect is actually controlling depends on reaction conditions.
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!
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.
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.)
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?
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.
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.
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Crisp and clear explanation of concepts, well done and thank you
Thank you so much. It's rare when I get that, "FUCK, I GET IT NOW" feeling. The clear explanation is much appreciated.
this made so much more sense than my professor or textbook. thank you!!!!
Life makes sense again. Thank you!!!
you rock. thanks for the AMAZING explanation. i was clueless about the difference. now it all makes so much sense!
my main man micheal evans coming to the rescue thank you!
I just wanted to thank you for putting this together
Thanks for the brief and clear explanation!
You’re welcome and thanks for watching!
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!
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.
amazing! thanks so much.
Good stuff
very nice explanation.. Thanks
Finally i understood.
Great presentation. Thank you very much?
Thank you for simplifying it!
Thank you! great video.
Good explanation, thanks!
This helped a lot. Thank you so so much.
Thanks a lot! Very helpful!
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?
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.)
Grate job!!!
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?
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.
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.
how is kinetic control used to affect the outcome of a chemical reaction where several reaction paths are available
Thank you.
Thanks bro 😉
Great,,,
I still can't get it.. 😢
Thanks Bro !
What organic chemistry software is that? Link please?
thanks, helped
thank you
CN2116
whose here from AAMC C/P SB #20