Kinetics Question (spent over 7 hours on this)

[bengalkitties]

Homework Statement

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 5.2×10−2 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?

Homework Equations

1/A = kt + 1 / A initial

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=9.8×10−2,9.8×10−2. Please tell me what I am doing wrong. Thanks!

 

[chrisk]

k=\frac{rate}{[AB]}

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:

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

The units should be

\frac{1}{Ms}

If 1/[AB] vs. time is linear then k is not linear. Is this a first order reaction?

 

[bengalkitties]

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]

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.