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### Linearization of a Second Order Reaction

Posted: **Sun Mar 04, 2018 3:58 pm**

by **Kyle Sheu 1C**

For a reactant A that is 2nd order, when we graph 1/[A] vs. time, which mathematical property justifies the fact that taking the inverse of the concentration yields a straight line?

### Re: Linearization of a Second Order Reaction

Posted: **Sun Mar 04, 2018 4:04 pm**

by **Anh Nguyen 2A**

The integrated rate law of second order reaction is 1/[A]t = 1/[A]o + kt which is in the form of a linear equation y= ax + b (b=1/[A]o and a=k) so graphing 1/[A] vs time would yield a straight line with the slope k and y-intercept=1/[A]o.

### Re: Linearization of a Second Order Reaction

Posted: **Sun Mar 04, 2018 4:17 pm**

by **Kyle Sheu 1C**

Thank you, this makes sense! However, if we were to graph [A]_{t} = kt + [A]_{0}, we would not have a straight line.

Perhaps a better way to phrase my question is: why do we take the inverse of [A] for a 2nd order reactant in order to linearize the graph?

### Re: Linearization of a Second Order Reaction

Posted: **Sun Mar 04, 2018 4:32 pm**

by **Bansi Amin 1D**

The inverse is taken because we equate the rate to k[A]^2, and when we bring the terms over to one side, we are left with the inverse of [A].

(-1/a)*(d[A]/dt)=k[A]^2

at a=1:

(d[A]/[A]^2)=-k*dt

Integration leads to the inverse. We take the inverse because that's what it mathematically turns out to be.