Question that I was looking up on Google

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The discussion centers on the kinetics of the reaction AB → A + B, where the rate law is determined by the equation 1/[AB] = kt + 1/[AB0]. The slope of the plot of 1/[AB] versus time is given as 5.2×10−2 M−1s−1. After 80 seconds, the concentrations of both products A and B are calculated to be 9.8×10−2 M. The user initially misapplied the formula, leading to incorrect results.

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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?


Here's what I did:
1/A = kt + 1 / A initial

*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!
 
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How about 1/AB=kt+1/AB0? As your given graph.
so AB=1/(kt+(1/AB0))
in terms of moles I think
A=B=1/(kt+(1/AB0))
 

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