7B.13 Numerator

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Emil Velasco 1H
Posts: 96
Joined: Wed Nov 21, 2018 12:19 am

7B.13 Numerator

Postby Emil Velasco 1H » Tue Mar 10, 2020 8:05 pm

For this problem, when it asks for one sixteenth or one fourth, we change the numerator (e.g. to 16/[A]o).
What's the math behind this?

Alicia Lin 2F
Posts: 83
Joined: Wed Sep 18, 2019 12:17 am

Re: 7B.13 Numerator

Postby Alicia Lin 2F » Tue Mar 10, 2020 8:12 pm

This is because, for example, part a is asking for 1/16 of the original concentration. In the solution manual, they are setting [A]=(1/16)[A]0. Plugging this value into the 2nd order linear equation, you get 1/([A]0/16)=kt+(1/[A]0). The left side of this equation written in simpler terms is 16/[A]0. Use this same logic for the rest of the problem too.

Sue Bin Park 2I
Posts: 52
Joined: Mon Jun 17, 2019 7:24 am

Re: 7B.13 Numerator

Postby Sue Bin Park 2I » Tue Mar 10, 2020 11:14 pm

could we not just set [A] = 1/16 and [A]initial = 1?
EDIT: oh nvm we are given [A]initial. my bad

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