## Slope = -k

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

Olivia Young 1A
Posts: 60
Joined: Fri Sep 28, 2018 12:24 am

### Slope = -k

Is there a specific reason both zero order reactions and first order reaction graphs have a slope of -k, while the slope of a second order reaction is k?

Melody P 2B
Posts: 35
Joined: Fri Sep 29, 2017 7:06 am

### Re: Slope = -k

Differential-and-Integrated-Rate-Laws1.pdf

Chloe Qiao 4C
Posts: 65
Joined: Fri Sep 28, 2018 12:27 am

### Re: Slope = -k

I believe it is because of the integrated rate laws derived (specific sign because of the math).

805087225
Posts: 30
Joined: Thu Jun 07, 2018 3:00 am

### Re: Slope = -k

This only means that integrating gives different values, and k would be negative or positive, depending on the equation.