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Topic: second order rate constants  (Read 7728 times)

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kellycox

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second order rate constants
« on: February 13, 2006, 10:52:01 PM »
I have struggled with a problem for over 1 1/2 days. Here's the question.
   "The reaction A->B+C. is 2nd order in A. When [A]0=0.100M, the reaction is 20% complete in 40.0 minutes. Calculate the value of the rate constant(in L/min *mol)."
    The formula that I have been working with is Half-life
                                  t1/2 = 1/k[A]0
    I am solving for "k", but I keep getting an answer of 1.00 x 10-1. I don't think that I am using the right formula for the information that I have. Is there another formula that I should be using to solve this question? Thanks for any direction/help that you can give me.

Offline pantone159

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Re:second order rate constants
« Reply #1 on: February 14, 2006, 01:32:31 AM »
Half-life is for first order reactions, not second.
There would be a different formula for second order.

Offline Donaldson Tan

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Re:second order rate constants
« Reply #2 on: February 14, 2006, 06:45:48 AM »
first order reactions exhibit constant half-life.

second order reactions exhibit half-life which is a function of the concentration. The derrivation is shown below:

let A be the concentration of component A.
-dA/dt = k.A2 => -dA/A2 = k.dt
integrate from Ao to 0.5Ao ( -dA/A2 )  = integrate from 0 to t1/2 ( k.dt )
=> 1/Ao = k.t1/2 => t1/2 = 1/k.Ao

Kelly's equation is correct. But there is no point using half-life to calculate k. That is because the reaction is 20% complete and not 50% complete.

Since we are given the time taken for reaction to complete 20%, then
concentration of A at 20% complete = (1 - 0.2)Ao = 0.8 Ao
t0.2 = 40min = 40x60 s = 2400s
let A be the concentration of component A.
-dA/dt = k.A2 => -dA/A2 = k.dt
integrate from Ao to 0.8Ao ( -dA/A2 )  = integrate from 0 to t0.2 ( k.dt )
=> 1/0.8Ao - 1/Ao = k.t0.2 => (1/Ao)(0.8-1-1) = k.t0.2

The rate constant k is this given as k = 1/Ao.t0.2)(0.8-1-1)
« Last Edit: February 14, 2006, 06:52:03 AM by geodome »
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