Second Order Kinetics with Two different Reactant I Chemical Kinetics I Physical Chemistry

[@vinayakmathad6257]

thanks and awesome explanation,integration of partial fraction was very diffcult ,you made it very easy

[@GuruprakashAcademy]

You may like my video on alkene: ruclips.net/video/MWOy19e4kxE/видео.html

[@FAEM-ji5rv]

The only problem is that by replacing the variables A and B with x you must assume that both a and B have decreased by the same amount which would work only for equations where the reactants A and B have the same coefficient.

[@GuruprakashAcademy]

Thanks for watching

[@anitarathod5530]

4. Find the differential equation for bimolecular reaction: A + B -> P sir iska solution mil skta h

[@tanishqshastri9961]

Sir maybe you forgot to explain from where did ln(a/b) came into the reaction. It came from the "c" which we will get after integration that is "kt+c" and from here the value of "c" is calculated by putting t=0 and x=0 i.e the value of "c" in the "kt+c" corresponds to the initial concentration when time was equal to 0.

[@GuruprakashAcademy]

You may like this video on alkene: ruclips.net/video/MWOy19e4kxE/видео.html

[@markcavendish7148]

Best lecture on whole of you tube. Please continue posting new videos.

[@GuruprakashAcademy]

Thanks for watching

[@raghavendrahpt5820]

Excellent sir

[@GuruprakashAcademy]

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[@ddoa]

Hello, before I watched your video I consulted other sources (eg, chemwiki.ucdavis.edu/Physical_Chemistry/Kinetics/Reaction_Rates/Second-Order_Reactions) wherein the integration of 1/(a-x)(b-x) = (1/b-a)(ln(1/a-x)-ln(1/b-x)). I have also consulted a table of integrals and it agrees with what is shown in the books. Can you explain to me why you get (1/a-b)(ln(1/a-x)-ln(1/b-x))? Thank you!

[@krzysztofhaagan6109]

In this video integration of 1/(a-x)(b-x) = [1/(a-b)](ln(a-x)-ln(b-x)) not "(1/a-b)(ln(1/a-x)-ln(1/b-x))". And [1/(a-b)](ln(a-x)-ln(b-x)) = [1/(b-a)]{ln[1/(a-x)]-ln[1/(b-x)]} so derivation is OK

[@jagannathprasadnayak9304]

Good explanation

[@mahbubulhasanshihab3051]

please,give some problems with solutions related with it

[@GuruprakashAcademy]

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[@gustavopetro1194]

Thaaaaaaaaaaaaaaanks sir you are my hero

[@ujjwalmandal8069]

Superb explanation 😊

[@GuruprakashAcademy]

Please to share with your friends as well

[@dineshkoppula359]

How we identify that term as partial fractioms,how u know tell me.

[@GuruprakashAcademy]

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[@sumitkumarsahoo7601]

great=prakash academy

[@sritamamaiti7203]

But what if the order with respect to A and B are 3/2and1/2 or anything else (that sums into 2), instead of 1 and 1??

[@GuruprakashAcademy]

steps can be repeated.

[@manjushadeshmukh5634]

thank you for this video .... explanation is best

[@rafaelscur2687]

good explanation, dude

[@vinithk1217]

Really awesome.Easy & clear cut explaination.Continue it sir.

[@GuruprakashAcademy]

Thanks.

[@blackpen1524]

Great lecture,sir.....😍😍😍

[@GuruprakashAcademy]

If you find it useful please don't forget to share with others, so that they can also get benefit of it

[@shankhadwiptalukdar2928]

sir how can we write the relationship of k and initial concentration of the reactants

[@GuruprakashAcademy]

use rate, k and concentration.

[@hruaiteirenthlei3753]

Good lecture👍

[@thorsen273]

Thanks for the video, it was really helpful!

[@GuruprakashAcademy]

Thanks

[@_MUTIARAFERBRI]

it seems your explanation is very good, but unfortunately i can't speak English😭

[@GuruprakashAcademy]

Thanks

[@shankhadwiptalukdar2928]

sir what about b>a

[@GuruprakashAcademy]

In that case a will be b and b will be a. That is one will be greater than other.

[@shankhadwiptalukdar2928]

@@GuruprakashAcademy thanks sir and if a = b, in which step we need to change a with b

[@abdurrehman9568]

nice

[@sdmhussain4579]

Thank you..

[@GuruprakashAcademy]

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[@light_dv4767]

Thank you!

[@GuruprakashAcademy]

You're welcome!

[@oedrees]

I love your voice

[@zkhan3480]

omar edrees 😄😄😄