Kinetics: Initial Rates and Integrated Rate Laws
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- Опубліковано 5 сер 2024
- Who likes math! Oh, you don't? Maybe skip this one on kinetics. Unless you have to answer this stuff for class. Then yeah, watch this.
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Hey everyone, UA-cam is deleting annotations, and there is an error at 5:41 that I had fixed with an annotation so unfortunately I must simply list the correction here! The values at the bottom right should read: 0.04 = k(0.01)(0.250) and k = 16/(M^2 * s), when making the video I forgot to square [NO] so please make that correction in your minds, sorry about that!
Professor Dave Explains thank you!
Sir do u mean 'top right'?
Towkir, play the video at the given time, not pause it. You'll see it on bottom right
@@Reforitor Thank you.
You're welcome bro
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If your exam was in 30 mins you wouldnt exactly have the time to type a comment I would imagine
@@ffire3795 😂
5:31
"the reaction order HAPPENS to match the stoichiometric ... but this won't always be the case"
right there, professor dave has done what no fucking textbook in the fucking world has ever done. it has pointed out something that a student might mistake, by thinking the way a student might.
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This is exactly why I was getting my answers wrong because my teacher NOR the textbook explained that the reaction order was not in relation to the stoichiometric coefficients!!
i have a question, so the overall order is the the third order?
@@mohamadhanan5706 same doubt bro
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I still don't get it
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Explained it better in ten minutes than my uni lecturers did in a week - thank you XD
2:30
When the concentration is doubled and the rate then also doubles, the reaction is first order with respect to that reactant. When the concentration is doubled and the rate quadruples as a result, the reaction is second order with respect to that reactant. If changes in the concentration doesn't affect the rate, the reaction is zero order with respect to that reactant.
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This is amazing. I didn't have any idea about Initial Rates and Integrated Rate Laws before watching the video. You are a miraculous savior. Thank you sir.
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Hello professor,
For the reaction A-->B and 2A-->2B, do we get 2 different rate constants (irrespective of the reaction order)?
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Hey, how is the reaction order for [ClO2] 1st order?
The concentration was tripled which then quadrupled the rate? I didnt see this explained. Can someone explain?
[ClO2] was quadrupled! .01*4=.04
When clo2 was quadrupled the rate also quadrupled ,therefore the order is 1
If the rate is multiplied by 16 then it's 2nd order .....
I've got my exam in a couple of minutes and this just helped me grasp the topic better..
Should NO be squared when finding its rate since its 2nd order? k(,100)(.250) or k(.100)^2 (.250)?
Excellent explanation
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Thankyou for your work.
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when determining the reaction order by changing the initial concentration of the reactant, how short the time interval has to be in order to represent the true reaction rate of the reactant? thank you in advance
Thank you very much professor Dave , could you please answer how is the rate law {rate= k * [A]’ [B]” [C]”’ }derived ?
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