## Derivations

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

Ardo 2K
Posts: 31
Joined: Fri Sep 29, 2017 7:05 am

### Derivations

When deriving the integrated law for a second order reaction why isn't the kt negative, and also does a positive kt value mean that the slope of the line of 1 over the concentration of reactant plotted against time will be positive as well?

Zane Mills 1E
Posts: 80
Joined: Fri Sep 25, 2015 3:00 am
Been upvoted: 1 time

### Re: Derivations

Because k must be positive (positive slope in the graph), answering your second question (yes).

Michaela Capps 1l
Posts: 50
Joined: Thu Jul 27, 2017 3:00 am

### Re: Derivations

For the first question, in the integrated rate law, kt is positive, so that is why we use pos slope.

Michaela Capps 1l
Posts: 50
Joined: Thu Jul 27, 2017 3:00 am

### Re: Derivations

It is postitive because y=mx+b so the slope is positive.

Christina Cen 2J
Posts: 53
Joined: Sat Jul 22, 2017 3:01 am

### Re: Derivations

the graph of 1/[reactant] over time has a positive slope which is equal to k for 2nd order reactions