Writing Rate Laws of Reaction Mechanisms Using The Rate Determining Step - Chemical Kinetics

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[@masterno8576]

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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[@ElisaFalsafi]

Isn’t having I in rate equation a violation to one of the conditions? And then replace it using rate formed = rate reversed?

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If it is soley determined on slow process, why is there no I-?

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[@pelumiemmanuel7965]

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”

[@Kamal17824]

That’s what I was thinking

[@Kamal17824]

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

[@RichardMwaba-g4e]

Now does mean also the overall reaction also depends on the slow reaction or what??😢

[@TheNobleQuran11462]

@@RichardMwaba-g4e exactly that's what he said in the video "The rate will always depend on the slow step"

[@JuayryahRU]

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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[@sheetalkonda7280]

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?

[@kingpyrrhusofepirus6686]

So you only look at the rate determining step when writing rate law. The rate determining step will al always be the slow step.

[@morathicasas7520]

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]

@@kingpyrrhusofepirus6686 how do we know which is the slow step?

[@kingpyrrhusofepirus6686]

@@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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[@6lubia]

Question for the last example wouldn’t the rate constant be= k[H2O2]^2?

[@julianescobar1672]

That's what I would've thought to

[@prestonrossi7597]

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.

[@gerryaraujo7852]

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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[@ericmora6287]

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?

[@na-kumlee6317]

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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[@flyorwalk1743]

Wait at the end....why the rate equation doesn't become "rate=k [H2O2]^2" but only to the power of 1??

[@taongakasanda9487]

Yes I have also noticed, because the coefecient of the overall reaction is #2

[@quyumkehinde2683]

I think it’s simply because we’re to consider the slow reaction, and not the overall equation.

[@aadygoel4837]

@@quyumkehinde2683 on the contrary we are only supposed to see the slow reaction . I feel you meant slow but typed fast .

[@quyumkehinde2683]

@@aadygoel4837 Exactly! Thank you.

[@wajdy2620]

@@quyumkehinde2683 but that is the overall reaction?

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[@christiatannous2877]

shouldn't [NO2] in the overall rate be [NO2]^2 ? since it has 2 NO2 in the equation?

[@teacherdanz3794]

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.

[@christiatannous2877]

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[@kjay5587]

@@teacherdanz3794 is the first step always the slow step?

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[@Justzoy-j2r]

Why we can't write the intermediate in rate law even its rate-determining approximations?

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[@animefan7628]

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

[@AD-wg8ik]

18:03 this contradicts Khan Academy. They included catalysts in the overall rate law

[@shimshonbalakhani5446]

khan academy is correct. He probably made a mistake

[@jailen.e]

@@shimshonbalakhani5446 so would the overall rate law be rate=k[H202][I^-]

[@mohanned4279]

same here it contradicts my book

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[@01FrozenFuze]

In 13:20 wouldn't it be Fast Rate = K [ NO3 ] [ NO2 ] rather than Rate = K [ O3 ] [ NO2 ]?

[@quintustheophilus9550]

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[@diablopm.1]

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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[@slackclover5]

Bruh we just started rate laws how u know?

[@carpediem2789]

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

[@elijah4145]

I believe in elementary reactions, the coefficients correspond to the order, although they are normally determined experimentally

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[@Ray-jv3go]

Rate law is based entirely on the slowest elementary step. Not on the overall reaction. Catalyst do affect the rate of reaction so it is part of the rate law. The last problem was not done right

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[@ofosuaanyarko6153]

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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[@dumebidonald]

Why are the stoichiometric number not considered in rate expression? E.g 2H2 + O2 = 2H2O Rate for H20 = [H20] = ½[H2]²[O2]

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[@rohann6199]

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

[@kxkxsxi6305]

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

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[@Joe0x7F]

I think the "overall" rate law equation is wrong. I think it is supposed to be the exact same as the rate law equation of the slowest elementary step, which includes the catalyst. Please recheck for us, please.

[@kaleabyohannes7909]

Tnx very very much

[@wandilesihle9220]

why is the overall rate law of the second question not raised to the power 2 because the overall reaction has 2 molecules of H2O2 as a reactant?

[@jeffreymoses7131]

I thought so too

[@jeffreymoses7131]

I was just about to comment that

[@juliegermaine2031]

Apparently you only use the coefficients for the elementary reactions not the overall reaction

[@aiman_shakir]

Thank you so much🌹🌹

[@debrizzy5638]

Is the first step always the slow step?

[@alfred4177]

Thank you soo much

[@kevincorrigan1754]

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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[@kxkxsxi6305]

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

[@eriknelson2559]

Can a fast first step ever be non-equilibrium?

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[@lovemorenkomo732]

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?

[@wandilesihle9220]

my confusion also

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