Question that I was looking up on Google

[bengalkitties]

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!

 

[rrogers]

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))

 

[quicknote]

It appears that you have made a mistake in your calculations. The correct equation to use in this situation is 1/[AB] = kt + 1/[AB] initial. This equation is derived from the integrated rate law for a first-order reaction, which is the type of reaction that is being monitored in this experiment. By rearranging this equation, we can solve for [AB] at any given time, which can then be used to determine the concentrations of A and B.

Using the given information, we can set up the equation as follows:

1/[AB] = (5.2×10−2 M/s)(80 s) + 1/0.210 M

Solving for [AB], we get 1/[AB] = 1.04 + 4.76 = 5.8 M

Since [AB] = [A] + , we can divide this concentration by 2 to get the individual concentrations of A and B. Therefore, after 80 s, the concentrations of A and B are both 2.9 M.

I'm not sure where the answer of 9.8×10−2 for the concentrations of A and B came from. It is possible that there was a mistake in the given information or in the calculation process. I would recommend double-checking your calculations and the given information to ensure accuracy. If you are still unsure, it may be helpful to consult with a colleague or mentor for further assistance.