Integrated Rate Law: Second Order Reaction

A second order reaction has integrated rate law
1/{A] = -kt + 1/[A]0
And so a graph with 1/[A] on the y-axis and time on the x-axis will give a line with a slope of -k and a y-intercept of 1/[A]0
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[@emlmm88]

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[@matt96533]

Very helpful and loved the evil laugh at 3.17

[@ee4life623]

Cool video. so it's negative at the beginning because it disappears, if it was the opposite for some reason it would be positive rate of change, correct?

[@wfk2757]

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[@chemistNATE]

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[@hadeerkareem8320]

thanks..but how i can using conversion in this subject??

[@umamahamir42]

But it is for 1 reactant involved in 2nd order ..when 2 reactant involved ?

[@phhnd97]

Thank you, it was really helpful 🙏🏻

[@desdemonajoseph2515]

Why is rate not equal to (-dA ÷dT)×(1÷2)?

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[@m.aassuhidis3840]

great video guide, thanks

[@user-pv8ok7tj2z]

Sir , just a clarification where does negative sign come from after integration of the equation :-1/[A] .

[@amitanshupattanayak9065]

sir, why didn't you take two products in consideration while finding the equation of the second order rate law?

[@ppmuscimol01]

shhh ...no one else below noticed that! ... just for you: in the end there is a '2' missing next to the 'K'

[@sdadeveloper5223]

The result would be similar

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Canadian?

[@craxmanxidig9919]

my teacher thank you
but after integration where is ln because the first order after you reach integration you change ln
I wish you to answer this question my teacher?!!!!

[@ayushchaudhary2988]

Perfect explanation

[@ratankanade7395]

thanks it was really very helpfull

[@acasale]

This is an example for A+A = P..... what about A+B = P?

[@truebeliever174]

This is very higher level equation. Don't worry about that for now.

[@RoughAccount-x9w]

What happened to the [A] left aside

[@twiggytimes5778]

does it depend on the initial concentration? for example if the reaction started out like 2A --> product
then rate becomes -1/2 dA/dt

[@chemistNATE]

The coefficient does not matter for Rate law. So the answer is NO

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It was a very good video.

[@nurdayana2427]

sir please help me to solve for 1.5 order

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do it for different initial concentration

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[@craxmanxidig9919]

my teacher thank you
but after integration where is ln because the first order after you reach integration you change ln
I wish you to answer this question my teacher?!!!!

[@eniola7722]

thanks

[@craxmanxidig9919]

my teacher thank you
but after integration where is ln because the first order after you reach integration you change ln
I wish you to answer this question my teacher?!!!!