15.39a - "0.37[A]o"


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Chris Freking 2G
Posts: 60
Joined: Fri Sep 28, 2018 12:29 am

15.39a - "0.37[A]o"

Postby Chris Freking 2G » Fri Mar 15, 2019 5:51 pm

15.39 Determine the time required for each of the following second-order reactions to take place: (a) 2 A -> B + C, for the concentration of A to decrease from 0.10 mol/L to 0.080 mol/L, given that k = 0.015 L-1 mol min for the rate law expressed in terms of the loss of A.

In the 6th edition solutions manual, [A]t is calculated as 0.055molA/L, but then calculated to 0.37[A]o.

Why is this information relevant if we only need and already have [A]t, [A]o, and k to solve for t in the second order integrated rate equation?

Madelyn Cearlock
Posts: 72
Joined: Fri Sep 28, 2018 12:19 am

Re: 15.39a - "0.37[A]o"

Postby Madelyn Cearlock » Fri Mar 15, 2019 7:22 pm

i am also confused by this.. the concentrations used to find that answer of part a is from the information given in part b

Madelyn Cearlock
Posts: 72
Joined: Fri Sep 28, 2018 12:19 am

Re: 15.39a - "0.37[A]o"

Postby Madelyn Cearlock » Fri Mar 15, 2019 7:30 pm

i figured it out. That information pertains to the second part of the problem. it finds the concentration of A to put into the rate law equation and solve for t. i dont think it has anything to do with part a.


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