Activation Energy Diagrams

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Old School Chemistry

Old School Chemistry

Күн бұрын

Пікірлер: 12
@MariachristinalouisaAnnejoymae
@MariachristinalouisaAnnejoymae 6 ай бұрын
The best explanation I found on YT thank you😊❤🙏👍
@OldSchoolChemistry
@OldSchoolChemistry 6 ай бұрын
Hi Krueger, I am so glad. Thank you for your comment:)
@hoanhtrinh4863
@hoanhtrinh4863 3 жыл бұрын
Thank you for the knowledge you've imparted
@OldSchoolChemistry
@OldSchoolChemistry 3 жыл бұрын
cr7, My pleasure. I hope your class goes well!!
@AbarJames
@AbarJames 10 ай бұрын
Thank you for helping understanding 🙏God bless you I really appreciate that
@OldSchoolChemistry
@OldSchoolChemistry 10 ай бұрын
Hi Abar, Think you for the blessing! I am proud of your hard work. God bless you in your efforts:)
@asisababun7543
@asisababun7543 3 жыл бұрын
Thanks for the video great apron explanation
@OldSchoolChemistry
@OldSchoolChemistry 3 жыл бұрын
Hi Asis, I am so glad you liked the apron example. I think the contextualizations helps students. Please subscribe if you have not already. Thanks!
@roxanneal-nimri8534
@roxanneal-nimri8534 3 жыл бұрын
You are sooooo helpful! Thank you
@OldSchoolChemistry
@OldSchoolChemistry 3 жыл бұрын
Hi Roxanne, I’m so glad. Thank you for your comment!!😊
@roxanneal-nimri8534
@roxanneal-nimri8534 3 жыл бұрын
@@OldSchoolChemistry of course! I have a question though. If you have a 3 step mechanism and you have 2 fast ones, do you solve for the intermediate in both of them and then plug that in to the slow step rate law?
@OldSchoolChemistry
@OldSchoolChemistry 3 жыл бұрын
@@roxanneal-nimri8534 Hi Roxanne, you only use the fast step that occurs before the slow step. Only fast steps that precede the slow step will go into equilibrium. There is usually only one fast step before the slow step. If there are multiple fast steps before the slow step then you are correct. When finding the overall rate law you can ignore the fast steps that happen after the slow step. I hope that helps!
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