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Kinetics Question (spent over 7 hours on this)

by bengalkitties
Tags: hours, kinetics, spent
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bengalkitties
#1
Jan31-09, 09:46 AM
P: 8
1. The problem statement, all variables and given/known data
The following reaction was monitored as a function of time:
AB---> A + B
A plot of 1/[AB] versus time yields a straight line with slope 5.210−2 M \s.

If the initial concentration of AB} is 0.210 M, and the reaction mixture initially contains no products, what are the concentrations of {A} and {B} after 80 s?


2. Relevant equations

1/A = kt + 1 / A initial

3. The attempt at a solution

1/A = (5.2x10^-2)(80 s) + 1/0.21 M
*When I did this equation it was wrong. The correct answer was that the concentration of both A and B=9.810−2,9.810−2. Please tell me what I am doing wrong. Thanks!
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chrisk
#2
Jan31-09, 10:17 AM
P: 288
[tex]k=\frac{rate}{[AB]}[/tex]

and [AB] at t=0 can be used because k is a constant ratio. The rate is the slope of the line. However, the problem appears to have an error:

[tex]\frac{1/[AB]}{t}=5.2x10^{-2}M/s[/tex]

The units should be

[tex]\frac{1}{Ms}[/tex]

If 1/[AB] vs. time is linear then k is not linear. Is this a first order reaction?
bengalkitties
#3
Jan31-09, 10:40 AM
P: 8
Thanks for your response. I think the rate is second order because given was: Rate=k,[AB]^2

I just plugged in those numbers for first order and I am still getting the answer wrong. This question is frustrating me to the max. Any help would be hugely appreciated.

chrisk
#4
Jan31-09, 10:58 AM
P: 288
Kinetics Question (spent over 7 hours on this)

Since it's a second order reaction, your equation for 1/[AB] is correct. Use conservation of mass. Your answer gives the concentration of AB after 80 seconds so this means what was converted is

.21 - [AB] = 9.8 x 10-2 M = concentration of products.


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