## 2nd order

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

Raquel Ruiz 1K
Posts: 30
Joined: Tue Nov 14, 2017 3:02 am

### 2nd order

Can someone explain why a second order graph has a positive slope? Does this mean that the concentration is increasing?

Ashley Kim
Posts: 62
Joined: Fri Sep 28, 2018 12:19 am
Been upvoted: 1 time

### Re: 2nd order

The second order graph has nothing to do with the concentration, but rather the second derivative.

Maya_Peterson1C
Posts: 62
Joined: Fri Sep 28, 2018 12:28 am

### Re: 2nd order

The y-axis is the inverse of the concentration of A. Thus, the slope shows that as time increases, the inverse of [A] increases (not [A] itself).

Priscilla Okaiteye
Posts: 61
Joined: Tue Nov 21, 2017 3:02 am

### Half-life qraphs

Do we also have to know or study the graphs for half-life?

Nicolle Fernandez 1E
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Joined: Fri Sep 28, 2018 12:25 am
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### Re: 2nd order

Since the graph is 1/A v time and when you integrate that you'll get 1/[A] = kt + 1/[A]o, and when you plot 1/[A] against t you'll get a positive slope because k is positive