Chemical Kinetics - Free Formula Sheet: bit.ly/3zYmDjy Direct Link to The Full 53-Minute Video: bit.ly/3vPPUad Chapter 12 - Video Lessons: www.video-tutor.net/chemical-kinetics.html
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13:15 there are two NO2's so why is it not third order. when you had 2A + B -> C you did [A]^2 [B], so why dont you also put it to the power of 2 in that one. Thanks
Question: at time 13:20 Since there is one molecule of O3 and 2 molecules of NO2, Shouldn't k= [O3] [NO2]^2? Then the reaction would be first order with respect to O3, second order with respect to NO2 and third order overall?
No, the rule that you are thinking of, about the molecularity and how the coefficients equal the reactant order, only applies to "elementary reactions". It does not apply to this overall reaction because it is a multistep reaction. He says this at 9:05 . The reaction order(s) of an overall reaction that is multistep depends on the rate law of the determining step.
@@keanupie8399 It's been awhile for me, but I think they will usually either give you the rates of each step, or they will simply tell you one of them is the slow step depending on the question.
Professor Organic Chemistry Tutor, thank you for showing/explaining How to write Rate Law expression for a reaction Mechanism in AP/General Chemistry. Writing the Rate Law reaction for mechanism is not a difficult process, however finding the catalyst and intermediate in the reactions is confusing. I will rewatch and review this material from start to finish. This is an error free video/lecture on UA-cam TV with the Organic Chemistry Tutor.
It would be, except that he adds that the first step is the "slow" step, and the second step is the "fast" step. Since the slow step always determines the speed of the reaction (think "you can't go faster than the slowest person in line"), then you only use what's in the slow step, which, in this case, is just one H2O2. If it's still confusing, go back and check the previous example that he showed, and compare its answer to his first example, which didn't use the "slow" or "fast" rates.
For a multi-step reaction, the slow step (rate limiting/rate determining step) determines the rate of the whole chemical reaction even when your overall reaction has different molar coefficient/s. In other words, stick to the slow step at all times when writing the rate law of the reaction.
I've watched many of your videos, which were free on UA-cam ever since. But now, you have moved your full content to Patreon. I understand that, your time is worth a lotta bucks. But, poor students who were benefited by your videos will no longer be able to gain that benefit anymore. And that's sad.
@@Garret_bruh_homey Yes he has a lot of free stuff. But still, most of his good content is now paid. You won't understand unless you benefited from his great content and now don't. And most of the content the UA-cam has is nowhere close to this man's content in quality. I will reiterate that poor students are missing a great opportunity, that they had before.
I'm just a bit confused. So the coefficients don't matter when determining what type of overall reaction it is? How would you be able to distinguish bimolecular reaction from a unimolecular reaction?
idk if this will help but overall rate law can only be determined experimentally. So like looking at the graphs or tables. so I think concluding the order of each reactant for overall rate law was a mistake. Elementary reaction on the other hand, you can calculate by looking at molecularity. One cannot determine the order of each reactants in overall balanced equation simply by looking at coefficients. Does that help?
what happened to that other fast/slow step video where u had an example of a slow step being in the middle of the reaction? im trying to find that vid and i cant find it did you delete it? in that vid you showed us why the overall order is different from whats in the slow/fast steps, it was rlly helpful, on this one u gloss over it like in 13:18.. why is NO2 not raised to the 2nd power? :(
NO2 isn't raised to the second power because it depends on the slow step of the reaction. Usually at this level they give you all the information about the system in the question . Rate determining step = first slow step
The slow step in elementary reaction is considered to be the rate determining step. It is always the slow step for the basis of overall rxn. The reason why we only have [NO2] in the overall.
10:20 Sir Why u took only 1st reaction as slow reaction and 2nd one as fast reaction...Please reply sir I am in confusion please And sir I like your teaching sir....Thankyou for this lecture sir....I understood everything except that one thing please sir answer my question sir
Hey he said the first reaction they give you is usually the slow step unless they specify otherwise. At A- level dont bother learning the actually systems they will tell you a you need to know in the question
Should the orders be experimentally determined and not by the coefficients? This is what I’ve learned in rate law. Are we just assuming that the coefficients are the order or reaction? But may show a different order in an experiment? Thanks
Please, why is the overall rate of reaction for the second question, first order? H2O2 has 2 Infront of it in the overall reaction. Please explain further, I don't get it.
Chemical Kinetics - Free Formula Sheet: bit.ly/3zYmDjy
Direct Link to The Full 53-Minute Video: bit.ly/3vPPUad
Chapter 12 - Video Lessons: www.video-tutor.net/chemical-kinetics.html
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For everyone wondering on why it’s not 2h2o2 for the last one it’s because overall reaction is solely determined on slow process
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Isn’t having I in rate equation a violation to one of the conditions? And then replace it using rate formed = rate reversed?
If it is soley determined on slow process, why is there no I-?
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Since the rate law is determined by the coefficient of the reactants…
Then why is 2H2O2 written as “k[H2O2]^1” instead of “k[H2O2]^2”
That’s what I was thinking
Turns out it’s because we only write what the slow step is. Since the slow step only has H2O2 we only write that since we follow the slow step only
Now does mean also the overall reaction also depends on the slow reaction or what??😢
@@RichardMwaba-g4e exactly that's what he said in the video "The rate will always depend on the slow step"
Its because the rate law is affected by the stoichiometric coefficients only for the elementary reactions, not for the overall reaction
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13:15 there are two NO2's so why is it not third order. when you had 2A + B -> C you did [A]^2 [B], so why dont you also put it to the power of 2 in that one. Thanks
Lol tell me why this was the perfect timing. I was looking for your explanation on this for so long.
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Question: at time 13:20
Since there is one molecule of O3 and 2 molecules of NO2, Shouldn't k= [O3] [NO2]^2?
Then the reaction would be first order with respect to O3, second order with respect to NO2 and third order overall?
So you only look at the rate determining step when writing rate law. The rate determining step will al always be the slow step.
No, the rule that you are thinking of, about the molecularity and how the coefficients equal the reactant order, only applies to "elementary reactions". It does not apply to this overall reaction because it is a multistep reaction. He says this at 9:05 . The reaction order(s) of an overall reaction that is multistep depends on the rate law of the determining step.
@@kingpyrrhusofepirus6686 how do we know which is the slow step?
@@keanupie8399 It's been awhile for me, but I think they will usually either give you the rates of each step, or they will simply tell you one of them is the slow step depending on the question.
@@kingpyrrhusofepirus6686 thank you!
Professor Organic Chemistry Tutor, thank you for showing/explaining How to write Rate Law expression for a reaction Mechanism in AP/General Chemistry. Writing the Rate Law reaction for mechanism is not a difficult process, however finding the catalyst and intermediate in the reactions is confusing. I will rewatch and review this material from start to finish. This is an error free video/lecture on UA-cam TV with the Organic Chemistry Tutor.
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Question for the last example wouldn’t the rate constant be= k[H2O2]^2?
That's what I would've thought to
It would be, except that he adds that the first step is the "slow" step, and the second step is the "fast" step. Since the slow step always determines the speed of the reaction (think "you can't go faster than the slowest person in line"), then you only use what's in the slow step, which, in this case, is just one H2O2. If it's still confusing, go back and check the previous example that he showed, and compare its answer to his first example, which didn't use the "slow" or "fast" rates.
For a multi-step reaction, the slow step (rate limiting/rate determining step) determines the rate of the whole chemical reaction even when your overall reaction has different molar coefficient/s. In other words, stick to the slow step at all times when writing the rate law of the reaction.
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I've watched many of your videos, which were free on UA-cam ever since. But now, you have moved your full content to Patreon. I understand that, your time is worth a lotta bucks. But, poor students who were benefited by your videos will no longer be able to gain that benefit anymore. And that's sad.
Correct
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He still has plenty of free videos on here. And the internet is abundant with free resources. This guilting and moralizing isn’t appropriate.
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I'm just a bit confused. So the coefficients don't matter when determining what type of overall reaction it is? How would you be able to distinguish bimolecular reaction from a unimolecular reaction?
idk if this will help but overall rate law can only be determined experimentally. So like looking at the graphs or tables. so I think concluding the order of each reactant for overall rate law was a mistake. Elementary reaction on the other hand, you can calculate by looking at molecularity. One cannot determine the order of each reactants in overall balanced equation simply by looking at coefficients. Does that help?
@@na-kumlee6317 Thank you!
I’m soooo confused on why he did [h2o2] [i^-] instead of [h2o2] [no2] for 17:27
also confused on why the IO and I canceled out instead of h2o2
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Why we can't write the intermediate in rate law even its rate-determining approximations?
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In 13:20 wouldn't it be Fast Rate = K [ NO3 ] [ NO2 ] rather than Rate = K [ O3 ] [ NO2 ]?
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what happened to that other fast/slow step video where u had an example of a slow step being in the middle of the reaction? im trying to find that vid and i cant find it did you delete it? in that vid you showed us why the overall order is different from whats in the slow/fast steps, it was rlly helpful, on this one u gloss over it like in 13:18.. why is NO2 not raised to the 2nd power? :(
He's forcing you to pay for the longer videos now on Patreon.
NO2 isn't raised to the second power because it depends on the slow step of the reaction. Usually at this level they give you all the information about the system in the question . Rate determining step = first slow step
shouldn't [NO2] in the overall rate be [NO2]^2 ? since it has 2 NO2 in the equation?
The slow step in elementary reaction is considered to be the rate determining step. It is always the slow step for the basis of overall rxn. The reason why we only have [NO2] in the overall.
@@teacherdanz3794 ohh right right
@@teacherdanz3794 is the first step always the slow step?
@@kjay5587 yes
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How can you identify the the slow and fast reactions?
It will be given, you are not required to identify
anyone know why the H2O2 is in the first order for overall reaction and not 2 ? confuse a bit
It doesn't have a number in front
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18:03 this contradicts Khan Academy. They included catalysts in the overall rate law
khan academy is correct. He probably made a mistake
@@shimshonbalakhani5446 so would the overall rate law be rate=k[H202][I^-]
same here it contradicts my book
Very helpful. thanks!
in the overall reaction, why didnt you raise the rate overall for NO2 to the power 2? maybe i missed out on something
You always save me!
Wait at the end....why the rate equation doesn't become "rate=k [H2O2]^2" but only to the power of 1??
Yes I have also noticed, because the coefecient of the overall reaction is #2
I think it’s simply because we’re to consider the slow reaction, and not the overall equation.
@@quyumkehinde2683 on the contrary we are only supposed to see the slow reaction . I feel you meant slow but typed fast .
@@aadygoel4837 Exactly! Thank you.
@@quyumkehinde2683 but that is the overall reaction?
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10:20 Sir Why u took only 1st reaction as slow reaction and 2nd one as fast reaction...Please reply sir I am in confusion please
And sir I like your teaching sir....Thankyou for this lecture sir....I understood everything except that one thing please sir answer my question sir
Hey he said the first reaction they give you is usually the slow step unless they specify otherwise. At A- level dont bother learning the actually systems they will tell you a you need to know in the question
@@kxkxsxi6305 Oh.....okay Thank you Now its clear for me
at 12:47 why is NO2 not raised to the power of 2 in the rate law?
Perfect timing !!!]
Thank you soo much
Hey, I think your rate law at 13:19 is off. Aren't there 2 NO2's?
No, he said that the reaction is dependent on the slow step, therefore since the slow step had a order of 1, the overall reaction has to be 1.
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Euclidean Algorithm please!!
Should the orders be experimentally determined and not by the coefficients? This is what I’ve learned in rate law. Are we just assuming that the coefficients are the order or reaction? But may show a different order in an experiment?
Thanks
I believe in elementary reactions, the coefficients correspond to the order, although they are normally determined experimentally
Thank u soooo much man
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Please, why is the overall rate of reaction for the second question, first order?
H2O2 has 2 Infront of it in the overall reaction.
Please explain further, I don't get it.