## Slope of second order reaction

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

Carlos Gonzales 1H
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### Slope of second order reaction

Why is the slope of a second order reaction positive compared to first and zero order reactions?

Annalise Eder 2L
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Joined: Sat Jul 22, 2017 3:01 am

### Re: Slope of second order reaction

The variable for a second order reaction that create a linear graph are 1/[A] over time while for a first order reaction it is ln[A] over time and for a zero order reaction it is [A] over time. The 1/[A] creates a positive slope because as the concentration of A decreases 1/[A] increases.

Haocheng Zhang 2A
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Joined: Fri Sep 29, 2017 7:04 am

### Re: Slope of second order reaction

For the second order reaction, k is proportional to 1/[A]. As the time of reaction increases, the concentration [A] decreases and 1/[A] increases, so the slope is positive.

Christy Zhao 1H
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### Re: Slope of second order reaction

The integrated rate laws of the zero and first-order have a -kt, while the second-order has a +kt. Since the integrated rate laws correspond to y=mx+c (m=k), the slope depends on k.