Determine Rate Law from Reaction Mechanisms, Fast then Slow Step: Part I
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- Опубліковано 8 вер 2024
- How to determine the rate law (equation) using reaction mechanisms. This example has the fast step precede the slow step.
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So just to clarify, we set the forward rate and reverse rate equal to each other and we solve for the intermediate.
Then, what we solve for the intermediate, we will plug that into the overall reaction rate.
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Hi, this is a really helpful video! I just would like to ask, shouldn't it be K1[NO2Cl]=K-1[NO2][Cl]+K2[NO2Cl][Cl], since the second slow step is also consuming Cl.
Hi Mr. Rabbit, there is only multiplication and division in rate laws, no addition or subtraction. Thank you for your comment!
@@OldSchoolChemistry Hi, thank you so much for your reply. I am still a little confused over the subject, since an article published in Chem Libretexts does indeed involve addition. If you have time, could you please look at it? The name of the article is here: 4.12: Steady-State Approximation.
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For step 1, why did you say to write the overall equation when we dont use it?
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Because it shows us what is not included in the final reaction - the intermediates and/or catalysts. Those cannot be in the final rate law. Thanks!
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Why did we solve for Cl? I thought we did not include the intermediates into our overall rate law? Thank you!
Hi Ayleen, we solved for Cl so we could then SUBSTITUTE for Cl and remove it from the overall rate law. I hope that helps. Thanks!
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Hello. What happened to that k constant at the bottom at 6:20 you didn't include it. Did it just disappear or ?(:
The k1 and k-1 just all become one k in the final rate law. They are simply each a number that combine to make one final number at the end. I should have written that k-1. Honestly, it does not matter as long as you write k on the final rate law. I hope that helps. Thanks Beatriz!
Fantastic. Thank you
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How we can plot Langmuir Adsorption Isotherms and what data is necessary for that??
Hi Shafi, I am sorry I am not familiar with Langmuir Adsorption Isotherms! Good luck:)
Can u elaborate on why slow step is rate determining step ? Do we neglect the time consumed by fast step?
Hi UfC, pretend 4 students are simultaneously doing 4 different tasks. Everyone is released when each task for all 4 people is finished. Pretend the third person takes the longest to complete their task. Everyone gets to leave when person 3 is finished, the other 3 people sat and waited because they finished their tasks sooner. That is not a perfect analogy but at least gives you an idea of the rate determining step. You are correct, the time for the fast step is negligible in comparison to the slow step. I hope that helps ! Please subscribe if you haven’t already.😊
@@OldSchoolChemistry Thank you mam.
😊👍
Hi madam.May i know how many elementary steps in the reaction? Is the reversible reaction considered as 1 elementary step? Since are there k1 and k-1 , i confuse if the reversible reaction has 1 or 2 elementary steps .anyway, thank you for the explanation in the videoo ,it helps me a alot!
Hi Izy, it is considered one elementary step that is reversible and reaches equilibrium. I hope that helps! Thanks:)
Hi, so I am a student taking his MCAT soon. That has nothing to do with my question, I just wanted to announce it because... I am in literal hellfire. Anyways, I was wondering how the rate law is the law for the fast reaction mechanism and not for the slow step or even the overall step? I hope you understand what I am asking. Essentially, why find the rate law of the fast step only, or am I missing something? Thank you in advance.
Hi Just Your Average Cup of Joe: Great job preparing for the MCAT! It will be worth the pain:) You only find the fast step rate laws (both forward and reverse) if there are 2 conditions met. 1. The slow step comes after the fast step. And 2. The slow step includes an intermediate. If both conditions are met then you must solve for the intermediate by setting the forward and reverse fast step rate laws equal to each other. You then substitute that solution into the intermediate species in the slow step rate law. I hope that helps!
What if you have two slow steps?
HI potato man, you will only have one slow step, the slowest step. Everything else will be at least slightly faster. The slow step determines the overall rate of the reaction because the reaction cannot go faster than the slow step. I hope that helps!
I can't help but notice that professor looks like Courtney Cox from friends
Hi Harikesh, that is a kind comment. Thank you. I hope your chemistry class goes well:)
Just perfect!want you here in India💜.where you from mam?
Hi Payne, thank you for your comment! I would love to visit India. I am from the United States of America.💗 God bless you in your home!
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@@payne7792 You can leave a comment in UA-cam. Thanks!
@@OldSchoolChemistry okay thank you mam!
Hi Payne, I’m glad this was helpful. I’m in the USA. Grinds to you in beautiful India!
if the rate determining step is the second reaction, will the first fast reaction always be in equilibrium like this?
Hi Tess! It is nice to hear from you again. Yes! The fast step preceding the slow step is always in equilibrium. Thanks😊
@@OldSchoolChemistry ok noted! thank you again for all your help and content that got me through the semester. i can't wait to start watching your ochem videos in a few weeks to prep for my class this fall!!!
I’m so proud of your hard work!!
How do you know if its 1st order, 2nd order, etc, if it is not given?
Hi Lily, If given reaction mechanisms you count the total coefficients on the reactants for the reaction order in the forward direction. If you only have one reactant with a coefficient of 1 then the reaction order is first order for the forward direction. I hope that helps! Thanks:)
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I know - there is a lot to digest in this. Go over it slowly, practice, and make it make sense in your mind. Good luck!
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