## Linearization of a Second Order Reaction

$\frac{d[R]}{dt}=-k[R]^{2}; \frac{1}{[R]}=kt + \frac{1}{[R]_{0}}; t_{\frac{1}{2}}=\frac{1}{k[R]_{0}}$

Kyle Sheu 1C
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### Linearization of a Second Order Reaction

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?

Anh Nguyen 2A
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### Re: Linearization of a Second Order Reaction

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.

Kyle Sheu 1C
Posts: 87
Joined: Fri Jun 23, 2017 11:39 am

### Re: Linearization of a Second Order Reaction

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?

Bansi Amin 1D
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### Re: Linearization of a Second Order Reaction

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.