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Topic: Second order rate law equation  (Read 4291 times)

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Offline slwJAMS

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Second order rate law equation
« on: April 07, 2008, 09:18:16 PM »
i am doing a kenetic factor homework problem and i was givin the k, the t, and the initial concentration. i don't understand why my book changed the equation around the way they did.

origional 2nd order rate law equation:

 1  _  1 = kt
[A]   [A]0

to solve for the ending concentration, my book swiched it around like this:

[A]= [A]0
     1+[A]0kt


why is the initial rate on both the top and the bottom? i'm puzzled.

Offline Rabn

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Re: Second order rate law equation
« Reply #1 on: April 07, 2008, 09:42:35 PM »
solve the original equation for [A]..what happens? It isn't magic, it's algebra.

Offline slwJAMS

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Re: Second order rate law equation
« Reply #2 on: April 07, 2008, 10:04:24 PM »
yeah i know. but the way i would rearrange it isnt that equation.

i would usually use the origional equation and solve it algebraically and i wasnt getting the same answer. if i were to rearrange it, this is how i would have done it:

[A]= 1/(kt + 1/[A]0)

wait. now im getting the right answer. and i type it into my calculator step by step and not all at once. but what i'm still seeing is that initial concentration ([A]0) not on the top and bottom like the equation that my book gave me.

Offline Rabn

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Re: Second order rate law equation
« Reply #3 on: April 07, 2008, 10:17:31 PM »
The first step in solving for [A] is to add (1/[A]0) to the right side.  To get the equation in your book, give the term on the right side a common denominator and invert both sides.  That gives you the equation in the book.

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