## Straight-Line Plot for Second-Order Reactions

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

Annalyn Diaz 1J
Posts: 61
Joined: Fri Sep 28, 2018 12:15 am

### Straight-Line Plot for Second-Order Reactions

Why is the straight-line plot for second-order reactions have a positive slope rather than a negative one like zero-order and first-order reactions?

Michael Novelo 4G
Posts: 64
Joined: Fri Sep 28, 2018 12:28 am

### Re: Straight-Line Plot for Second-Order Reactions

because the integrated 2nd order reaction equation is 1/[A] = kt + 1/[A] where kt is the slope using the equations y = mx + b if mx is a positive then the slope is increasing. The first and zero order have a -kt value in the integrated equation so its slope is negative and is decreasing with time.