Writing Rate Laws of Reaction Mechanisms Using The Rate Determining Step - Chemical Kinetics
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- Опубликовано: 21 июл 2024
- This chemistry video tutorial provides a basic introduction into reaction mechanisms within a chemical kinetics setting. It explains how to write the rate law expression for a reaction mechanism. A reaction mechanism consist of a series of elementary steps or elementary reactions whose rate law can be written from its molecularity - that is from the coefficients of the balanced reaction. The rate of a reaction mechanism is completely dependent on the slow step or the rate-determining step. This video explains how to substitute an intermediate when writing rate law expressions. It contains plenty of examples and practice problems.
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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
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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.
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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'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?
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in the overall reaction, why didnt you raise the rate overall for NO2 to the power 2? maybe i missed out on something
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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
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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? :(
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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
Thank u soooo much man
Can a fast first step ever be non-equilibrium?
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12:54 for example they gave the rate and they gave the overall equation, how do we then break them down into slow reaction and fast reaction
They have to tell you which one is the fast step and which one is the slow step
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.
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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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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?
how do u know which step is slow step and which step is fast step if neither is provided
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For the overall reaction the NO2 has a coefficient of 2. when wrinting the rate law isnt it supposed to be to the power 2 making it 3rd order?
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18:24 Is there someone who has solved the problem number two? Would anyone please walk me through what must be done in that problem? First reaction is reversible, so how does it affect our calculations? Is the law of mass action any helpful for this somehow? Plus am I to put down the reactions separately: one the forward and the next to be backward?
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Overall reaction has a coefficient of 2 on hydrogen peroxide,are we not supposed to say concentration of hydrogen peroxide raised to power 2 since there is two on the overall equation?
my confusion also
13:17 can someone explain why we did not write "2" for power of NO2 at the last rate formula?
It is based on the slow reaction, not the overall reaction. The slow reaction only has one NO2.
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So why is the rate order directly matched with the stoichiometry? This has to be determined experimentally and cannot just be assigned like this right?
And why is iodide not in the total rate law? Surely it speeds up the reaction.
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"ter" in "termolecular" is the root of "tertiary" or "third in order or level"
Is the first step always the slow step?