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The discussion revolves around a chemical reaction where AB decomposes into A and B, monitored over time. The reaction follows a second-order kinetics, as indicated by the linear relationship between 1/[AB] and time, with a slope of 5.2×10−2 M/s. Given an initial concentration of AB at 0.210 M, the goal is to determine the concentrations of A and B after 80 seconds. The correct approach involves using the integrated rate law for a second-order reaction, which is 1/[AB] = kt + 1/[AB0]. The user initially misapplied the formula, leading to incorrect calculations. The correct concentrations of both A and B after 80 seconds are found to be 9.8×10−2 M each, indicating that the user needed to correctly apply the rate law to derive the final concentrations from the initial conditions and the reaction's kinetics.
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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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